## Big Ideas Math Book Geometry Answer Key Chapter 7 Quadrilaterals and Other Polygons

The concept of Learning the concepts of Quadrilaterals and Other Polygons says that if both pairs of opposite sides of a quadrilateral are parallel, then it is a parallelogram. Check out the list of topics before you start solving the problems. We have presented the list of topics as per the latest syllabus. Thus click on the links provided below and solve the problems. Check whether the solutions are correct or not with the help of Big Ideas Math Book Geometry Answer Key Chapter 7 Quadrilaterals and Other Polygons.

### Quadrilaterals and Other Polygons Maintaining Mathematical Proficiency

Solve the equation by interpreting the expression in parentheses as a single quantity.

Question 1.
4(7 – x) = 16
3

Explanation:
Given,
4 (7 – x) = 16
Transformation of 4 from L.H.S to R.H.S
7 – x = $$\frac{16}{4}$$
7 – x = 4
7 – 4 = x
x = 3
So, the value of x is 3.

Question 2.
7(1 – x) + 2 = – 19
4

Explanation:
Given,
7 (1 – x) + 2 = -19
Transformation of 4 from L.H.S to R.H.S
7 (1 – x) = -19 – 2
7 (1 – x) = -21
1 – x = –$$\frac{21}{7}$$
1 – x = -3
1 + 3 = x
x = 4
So, the value of x is 4.

Question 3.
3(x – 5) + 8(x – 5) = 22
7

Explanation:
Given,
3 (x – 5) + 8 (x – 5) = 22
Take (x – 5) common, so the equation is as follows,
(x – 5) (3 + 8) = 22
(x – 5) 11 = 22
Transformation of 11 from L.H.S to R.H.S
x – 5 = $$\frac{22}{11}$$
x – 5 = 2
Transformation of -5 from L.H.S to R.H.S
x = 2 + 5
x = 7
So, the value of x is 7.

Determine which lines are parallel and which are perpendicular.

Question 4.

line c and d are perpendicular lines.

Explanation:

From the given figure,
The coordinates of line a are: (-2, 2), (4, -2)
The coordinates of line b are: (-3, -2), (0, -4)
The coordinates of line ‘c’ are: (-3, 0), (3, -3)
The coordinates of line ‘d’ are: (1, 0), (3, 4)
Compare the given coordinates with (x1, y1), (x2, y2)
We know that,
The slope of the line (m) = $$\frac{y2 – y1}{x2 – x1}$$
The slope of line a = $$\frac{-2 – 2}{4 + 2 }$$
= $$\frac{-4}{6}$$
= –$$\frac{2}{3}$$
The slope of line b = $$\frac{-4 + 2}{0 + 3}$$
= $$\frac{-6}{3}$$
= -2
The slope of line c = $$\frac{-3 – 0}{3 + 3}$$
= $$\frac{-3}{6}$$
= –$$\frac{1}{2}$$
The slope of line d = $$\frac{4 – 0}{3 – 1}$$
= $$\frac{4}{2}$$
= 2
So, line c and d are perpendicular lines.

Question 5.

line a and b are parallel lines.
line c and d are parallel lines.
line b and c and line a and c are perpendicular lines.

Explanation:

From the given figure,
The coordinates of line a are: (3, 1), (0, -3)
The coordinates of line b are: (0, 1), (-3, -3)
The coordinates of line ‘c’ are: (2, 1), (-2, 4)
The coordinates of line ‘d’ are: (4, -4), (-4, 2)
Compare the given coordinates with (x1, y1), (x2, y2)
We know that,
The slope of the line (m) = $$\frac{y2 – y1}{x2 – x1}$$
The slope of line a = $$\frac{-3 – 1}{0 – 3}$$
= $$\frac{-4}{-3}$$
= $$\frac{4}{3}$$
The slope of line b = $$\frac{-3 – 1}{-3 – 0}$$
= $$\frac{-4}{-3}$$
= $$\frac{4}{3}$$
The slope of line c = $$\frac{2 + 4}{-4 – 4}$$
= $$\frac{6}{-8}$$
= –$$\frac{3}{4}$$
The slope of line d = $$\frac{4 + 2}{-4 – 4}$$
= $$\frac{6}{-8}$$
= –$$\frac{3}{4}$$
So, line a and b; line c and d are parallel lines.
line b and c; line a and c are perpendicular lines.

Question 6.

line b and c are parallel lines.
line b and d; line c and d are perpendicular lines.

Explanation:

From the given figure,
The coordinates of line a are: (4, -4), (-2, -2)
The coordinates of line b are: (-3, -2), (-2, 2)
The coordinates of line ‘c’ are: (3, 1), (2, -3)
The coordinates of line ‘d’ are: (0, 3), (-4, 4)
Compare the given coordinates with (x1, y1), (x2, y2)
We know that,
The slope of the line (m) = $$\frac{y2 – y1}{x2 – x1}$$
The slope of line a = $$\frac{-2 + 4}{-2 – 4}$$
= $$\frac{2}{-6}$$
= –$$\frac{1}{3}$$
The slope of line b = $$\frac{2 + 2}{-2 + 3}$$
= $$\frac{4}{1}$$
= 4
The slope of line c = $$\frac{-3 – 1}{2 – 3}$$
= $$\frac{-4}{-1}$$
= 4
The slope of line d = $$\frac{4 – 3}{-4 – 0}$$
= $$\frac{1}{-4}$$
= –$$\frac{1}{4}$$
So, line b and c are parallel lines.
line b and d; line c and d are perpendicular lines.

Question 7.
ABSTRACT REASONING
Explain why interpreting an expression as a single quantity does not contradict the order of operations.
In the order of operations, “Parenthesis” occupies the top position according to the BODMAS rule
So, the interpreting of an expression as a single quantity or as different quantities don’t change the result.
Hence, the interpreting of an expression as a single quantity does not contradict the order of operations.

### Quadrilaterals and Other Polygons Mathematical Practices

Monitoring Progress

Use the Venn diagram below to decide whether each statement is true or false. Explain your reasoning.

Question 1.
Some trapezoids are kites.
False.

Explanation:
The given statement is that,
Some trapezoids are kites.

From the given Venn diagram, there is no relation between trapezoids and kites.
So, that the given statement is false.

Question 2.
No kites are parallelograms.
True.

Explanation:
The given statement is,
No kites are parallelograms.

From the given Venn diagram, there is no relation between kites and parallelograms.
Hence, the given statement is true.

Question 3.
All parallelograms are rectangles.
False.

Explanation:
The given statement is:
All parallelograms are rectangles.

From the given Venn diagram, we observe that rectangles are a part of parallelograms.
But not all parallelograms are rectangles because parallelograms contain rhombuses, squares, and rectangles.
So, that the given statement is false.

Question 4.
True.

Explanation:
The given statement is:
From the given Venn diagram,

We observe that squares are a small part of quadrilaterals and quadrilaterals contain other than squares like,
parallelogram, rhombuses, rectangles trapezoids and kites.
So, the given statement is true.

Question 5.
Example 1 lists three true statements based on the Venn diagram above. Write six more true statements based on the Venn diagram.
The below ae the three true statements based on the Venn diagram given,.
Some squares are part of rectangles.
No kites are parallelograms.
The given Venn diagram is:

From the above Venn diagram,
The six more true statements based on the Venn diagram are:
a. Some parallelograms are rhombuses.
b. Some parallelograms are squares.
c. Some parallelograms are rectangles.

Question 6.
A cyclic quadrilateral is a quadrilateral that can be circumscribed by a circle so that the circle touches each vertex. Redraw the Venn diagram so that it includes cyclic quadrilaterals.

The Venn diagram that includes cyclic quadrilaterals are:

Explanation:
Given that a cyclic quadrilateral is a quadrilateral that can be circumscribed by a circle so that the circle touches each vertex.
So, from the above Venn diagram the cyclic quadrilaterals have,
trapezoids, parallelograms, rectangles, squares, rhombuses, kites and also cyclic quadrilaterals.

### 7.1 Angles of Polygons

Exploration 1

The Sum of the Angle Measures of a Polygon

Work with a partner. Use dynamic geometry software.

a. Draw a quadrilateral and a pentagon. Find the sum of the measures of the interior angles of each polygon.
Sample

The sum of the interior angles of quadrilateral is 360°
The sum of the interior angles of pentagon is 360°

Explanation:
The representation of the quadrilateral and the pentagon are:

From the above figures,
The angle measures of the quadrilateral are 90°, 90°, 90°, 90°, and 90°
So, the sum of the angle measures of a quadrilateral = 90° + 90° + 90° + 90° = 360°
From the above figures,
The angle measures of a pentagon are: 108°, 108°, 108°, 108°, and 108°
So, the sum of the angle measures of a pentagon = 108° + 108° + 108° + 108° + 108° = 540°

b. Draw other polygons and find the sums of the measures of their interior angles. Record your results in the table below.

Explanation:
The sum of the angle measures of a polygon = 180° (n – 2)
Where, n is the number of sides.
The completed result of the sums of the internal measures of their internal angles are:
For second term,
180° (n – 2)
= 180° (4 – 2)
= 180° x 2
= 360°

For third term,
180° (n – 2)
= 180° (5 – 2)
= 180° x 3
= 540°

For fourth term,
180° (n – 2)
= 180° (6 – 2)
= 180° x 4
= 720°

For fifth term,
180° (n – 2)
= 180° (7 – 2)
= 180° x 5
= 900°

For sixth term,
180° (n – 2)
= 180° (8 – 2)
= 180° x 6
= 1080°

For seventh term,
180° (n – 2)
= 180° (9 – 2)
= 180° x 7
= 1260°

c. Plot the data from your table in a coordinate plane.

Explanation:
The table from part (b) is:

The representation of the data in the table in the coordinate plane is shown above:
Each point can be identified by an ordered pair of numbers;
a number on the x-axis called an x-coordinate, and a number on the y-axis called a y-coordinate.
Ordered pairs are written in parentheses (x-coordinate, y-coordinate).
The origin is located at (0,0).

d. Write a function that fits the data. Explain what the function represents.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to reason inductively about data.
y = 180 (x – 2)

Explanation:
From part (c),
When we observe the coordinate plane,
The function that fits the data is:
y = 180 (x – 2)
Where x is the number of sides and,
y is the sum of the measures of the internal angles.

Exploration 2

The measure of one Angle in a Regular Polygon

Work with a partner.

a. Use the function you found in Exploration 1 to write a new function that gives the measure of one interior angle in a regular polygon with n sides.
$$\frac{180° (n – 2)}{n}$$

Explanation:
From Exploration 1,
From part (d),
The function that fits the sum of the angle measures of the internal angles of n sides is:
y = 180° (n – 2) ——-(1)
To find the one interior angle in a regular polygon with n sides,
Divide equation (1) by n
So, the measure of an interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$

b. Use the function in part (a) to find the measure of one interior angle of a regular pentagon. Use dynamic geometry software are to check your result by constructing a regular pentagon and finding the measure of one of its interior angles.
108°

Explanation:
From part (a),
We know that the measure of an interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$
n = 5 (pentagon has 5 sides)
The measure of an interior angle of a regular pentagon = $$\frac{180° (5 – 2)}{5}$$
= $$\frac{180° (3)}{5}$$
= 36° × 3
= 108°
Hence, the measure of an interior angle of a regular pentagon is 108°

c. Copy your table from Exploration 1 and add a row for the measure of one interior angle in a regular polygon with n sides. Complete the table. Use dynamic geometry software to check your results.
The completed table along with the column of “Measure of one interior angle in a regular polygon” is:

Explanation:
The given table from Exploration 1

Then add a row for the measure of one interior angle in a regular polygon with n sides as shown above in answer.
Measure of interior angle = sum of the angles ÷ number of sides.
180 ÷ 3 = 60
360 ÷ 4 = 90
540 ÷ 5 = 108
720 ÷ 6 = 120
900 ÷ 7 = 128.5
1080 ÷ 8 = 135
1260 ÷ 9 = 140

Question 3.
What is the sum of the measures of the interior angles of a polygon?
180° (n – 2)

Explanation:
We know that, the sum of the measures of the interior angles of a polygon = 180° (n – 2)
Where “n” is the number of sides.

Question 4.
Find the measure of one interior angle in a regular dodecagon (a polygon with 12 sides).
150°

Explanation:
We know that, the measure of one interior angle in a regular polygon = $$\frac{180° (n – 2)}{n}$$
Where “n” is the number of sides
So, a regular dodecagon has 12 sides.
The measure of one interior angle in a regular dodecagon = $$\frac{180° (12 – 2)}{12}$$
= $$\frac{180° (10)}{12}$$
= 15 × 10
= 150°
Hence, the measure of one interior angle in a regular dodecagon is 150°

### Lesson 7.1 Angles of Polygons

Monitoring Progress

Question 1.
The Coin shown is in the shape of an 11-gon. Find the sum of the measures of the interior angles.

1620°

Explanation:
It is given that the coin shown is in the shape of an 11-gon
So, the number of sides in a coin (n) = 11
We know that, the sum of the measures of the interior angles = 180° (n – 2)
Where “n” is the number of sides.
= 180° (11 – 2)
= 180° (9)
= 1620°
Hence, the sum of the measures of the interior angles in a coin is: 1620°

Question 2.
The sum of the measures of the interior angles of a convex polygon is 1440°. Classify the polygon by the number of sides.
Decagon

Explanation:
It is given that the sum of the measures of the interior angles of a convex polygon is 1440°
We know that, the sum of the measures of the interior angles of a polygon = 180° (n – 2)
Where “n” is the number of sides.
1440° = 180° (n – 2)
n – 2 = $$\frac{1440}{180}$$
n – 2 = 8
n = 8 + 2
n = 10
Hence, the polygon with 10 sides is called “Decagon”

Question 3.
The measures of the interior angles of a quadrilateral are x°, 3x°. 5x°. and 7x° Find the measures of all the interior angles.
22.5°, 67.5°, 112.5°, and 157.5°

Explanation:
Given that the measures of the interior angles of a quadrilateral are x°, 3x°, 5x°, and 7x°
We know that, the sum of the measures of the interior angles of a quadrilateral is: 360°
So, x° + 3x° + 5x° + 7x° = 360°
16x° = 360°
x° = $$\frac{360}{16}$$
x° = 22.5°
The measures of all the interior angles of a quadrilateral are:
x° = 22.5°
3x° = 3 (22.5)° = 67.5°
5x° = 5 (22.5)° = 112.5°
7x° = 7 (22.5)° = 157.5°
Hence, the measures of the internal angles of a quadrilateral are:
22.5°, 67.5°, 112.5°, and 157.5°

Question 4.
Find m∠S and m∠T in the diagram.

∠S = ∠T = 103°

Explanation:
From the given figure below,

We can observe that the number of the sides are 5.
We know that,
The sum of the measures of the interior angles of the pentagon = 540°
Let, ∠S = ∠T = x°
93° + 156° + 85° + x° + x° = 540°
2x° + 334° = 540°
2x° = 540° – 334°
2x° = 206°
x° = $$\frac{206}{2}$$
x° = 103°
So, ∠S = ∠T = 103°

Question 5.
Sketch a pentagon that is equilateral but not equiangular.

Explanation:
We know that,
The “Equilateral” means all the sides are congruent.
The “Equiangular” means all the angles are congruent.
A pentagon that is equilateral but not equiangular is shown above.

Question 6.
A convex hexagon has exterior angles with measures 34°, 49°, 58°, 67°, and 75°. What is the measure of an exterior angle at the sixth vertex?
77°

Explanation:
Given that a convex hexagon has exterior angles with measures 34°, 49°, 58°, 67°, and 75°
We know that, hexagon has 6 sides.
The sum of the measures of the exterior angles of any polygon is: 360°
Let the exterior angle measure at the sixth vertex of a convex hexagon be: x°
34° + 49° + 58° + 67° + 75° + x° = 360°
283° + x° = 360°
x° = 360° – 283°
x° = 77°
So, the measure of the exterior angle at the sixth vertex of the convex hexagon is: 77°

Question 7.
An interior angle and an adjacent exterior angle of a polygon form a linear pair. How can you use this fact as another method to find the measure of each exterior angle in Example 6?
90°

Explanation:
Given that an interior angle and an adjacent exterior angle of a polygon form a linear pair i.e,
it forms a supplementary angle.
Interior angle measure + Adjacent exterior angle measure = 180°
So by using the above property,
The sum of the angle measure of the exterior angle of any polygon = 180° × 4 = 360°
The measure of each exterior angle = $$\frac{360}{4}$$ = 90°

### Exercise 7.1 Angles of Polygons

Question 1.
VOCABULARY
Why do vertices connected by a diagonal of a polygon have to be nonconsecutive?

Explanation:
A segment connecting consecutive vertices of a polygon is not diagonal,
because it is actually a side of a polygon.
We define diagonal as a segment that connects two non-consecutive vertices in a polygon.
Hence, vertices connected by a diagonal of a polygon have to be nonconsecutive.

Question 2.
WHICH ONE DOESNT BELONG?
Which sum does not belong with the other three? Explain your reasoning.

 The sum of the measures of the interior  angles of a quadrilateral The sum of the measures of the exterior angles of a quadrilateral The sum of the measures of the interior  angles of a pentagon The sum of the measures of the exterior angles of a pentagon

The statement (c) does not belong.

Explanation:
From the above given statements,
The sum of the angle measures of the interior angles of a polygon = 180° (n – 2)
Where “n” is the number of sides.
The sum of the angle measures of the exterior angles of any polygon is: 360°
We know that,
The number of sides of a quadrilateral is 4.
The number of sides of a pentagon is 5.
The sum of the angle measures of the interior angles of a quadrilateral = 180° ( 4 – 2)
= 180° (2)
= 360°
The sum of the angle measures of the interior angles of a pentagon = 180° ( 5 – 2)
= 180° (3)
= 540°
Hence, the statement (c) does not belong with the other three.

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, find the sum of the measures of the interior angles of the indicated convex polygon.

Question 3.
nonagon
1260°

Explanation:
The sum of the angle measures of the interior angles of a polygon = 180° (n – 2)
Where “n” is the number of sides.
The polygon nonagon has 9 sides.

Question 4.
14-gon
2160°

Explanation:
The given convex polygon is: 14-gon
The number of sides of 14-gon is: 14
We know that,
The sum of the angle measures of the interior angles of a polygon = 180° (n – 2)
Where “n” is the number of sides.
sum of a polygon = 180° (14 – 2)
= 180° (12)
= 2160°

Question 5.
16-gon
2520°

Explanation:
The sum of the angle measures of the interior angles of a polygon = 180° (n – 2)
Where “n” is the number of sides.
The polygon 16-gon has 16 sides.

Question 6.
20-gon
3240°

Explanation:
The given convex polygon is: 20-gon.
The number of sides of 20-gon is: 20.
We know that, the sum of the angle measures of the interior angles of a polygon = 180° (n – 2)
Where “n” is the number of sides.
The sum of the angle measures of the interior angles of 20-gon = 180° (20 – 2)
= 180° (18)
= 3240°

In Exercises 7-10, the sum of the measures of the interior angles of a convex polygon is given. Classify the polygon by the number of sides.

Question 7.
720°
Hexagon is a polygon with 6 sides.

Explanation:
The sum of the angle measures of the interior angles of a polygon = 180° (n – 2)
Where “n” is the number of sides.

Question 8.
1080°
Octagon is a polygon with 8 sides.

Explanation:
The sum of the angle measures of the interior angles of a polygon = 180° (n – 2)
Where “n” is the number of sides.
1080° = 180° (n – 2)
n – 2 = $$\frac{1080}{180}$$
n – 2 = 6
n = 6 + 2
n = 8
Hence, the given polygon is Octagon with 8 sides.

Question 9.
2520°
16-gon is a polygon with 16 sides.

Explanation:
The sum of the angle measures of the interior angles of a polygon = 180° (n – 2)
Where “n” is the number of sides.

Question 10.
3240°
20-gon is a polygon or Icosagon with 20 sides.

Explanation:
The sum of the angle measures of the interior angles of a polygon = 180° (n – 2)
Where “n” is the number of sides.
3240° = 180° (n – 2)
n – 2 = $$\frac{3240}{180}$$
n – 2 = 18
n = 18 + 2
n = 20
Hence, the given polygon is 20-gon or Icosagon with 20 sides.

In Exercises 11-14, find the value of x.

Question 11.

x = 64°

Explanation:
Given that,

Question 12.

x = 66°

Explanation:

From the above given figure, the polygon has 4 sides.
The given angle measures of a polygon with 4 sides are:
103°, 133°, 58°, and x°
The sum of the angle measures of the interior angles of a polygon with 4 sides = 180° (4 – 2)
= 180° (2)
= 360°
So, 103° + 133° + 58° + x° = 360°
294° + x° = 360°
x° = 360° – 294°
x° = 66°
Hence, the value of x is 66°

Question 13.

x = 89°

Explanation:
Given that,

Question 14.

x  = 99°

Explanation:

From the given figure, the polygon has 4 sides.
The given angle measures of a polygon with 4 sides are:
101°, 68°, 92°, and x°
The sum of the angle measures of the interior angles of a polygon with 4 sides = 180° (4 – 2)
= 180° (2)
= 360°
So, 101° + 68° + 92° + x° = 360°
261° + x° = 360°
x° = 360° – 261°
x° = 99°
Hence, the value of x is 99°

In Exercises 15-18, find the value of x.

Question 15.

x = 70°

Explanation:

Question 16.

x = 117°

Explanation:

From the given figure, the polygon has 5 sides.
The given angle measures of a polygon with 5 sides are:
140°, 138°, 59°, x°, and 86°
The sum of the angle measures of the interior angles of a polygon with 5 sides = 180° (5 – 2)
= 180° (3)
= 540°
So, 140° + 138° + 59° + x° + 86° = 540°
423° + x° = 540°
x° = 540° – 423°
x° = 117°
Hence, x is 117°

Question 17.

x = 150°

Explanation:

Question 18.

x = 88.6°

Explanation:

From the given figure, the polygon has 8 sides.
The given angle measures of a polygon with 8 sides are:
143°, 2x°, 152°, 116°, 125°, 140°, 139°, and x°
The sum of the angle measures of the interior angles of a polygon with 8 sides = 180° (8 – 2)
= 180° (6)
= 1080°
So, 143° + 2x° + 152° + 116° + 125° + 140° + 139° + x° = 1080°
815° + 3x° = 1080°
3x° = 1080° – 815°
3x° = 265°
x° = $$\frac{265}{3}$$
x° = 88.6°
Hence, x is: 88.6°

In Exercises 19 – 22, find the measures of ∠X and ∠Y.

Question 19.

m∠X = m∠Y = 92°

Explanation:

Question 20.

∠X = ∠Y = 142°

Explanation:
The given figure is:

We can observe that, the above polygon has 5 sides.
We know that, if the angles are not mentioned in a polygon, then consider that angles as equal angles.
So, the given angle measures of a polygon with 5 sides are:
47°, 119°, 90°, x°, and x°
The sum of the angle measures of the interior angles of a polygon with 5 sides = 180° (5 – 2)
= 180° (3)
= 540°
So, 47° + 119° + 90° + x° + x° = 540°
256° + 2x° = 540°
2x° = 540° – 256°
2x° = 284°
x° = $$\frac{284}{2}$$
x° = 142°
Hence, ∠X = ∠Y = 142°

Question 21.

m∠X = m∠Y = 100.5°

Explanation:

Question 22.

∠X = ∠Y = 140°

Explanation:
The given figure is:

We can observe that, the polygon has 6 sides.
We know that, if the angles are not mentioned in a polygon, the consider that angles as equal angles.
The given angle measures of a polygon with 6 sides are:
110°, 149°, 91°, 100°,  x°, and x°
The sum of the angle measures of the interior angles of a polygon with 6 sides = 180° (6 – 2)
= 180° (4)
= 720°
So, 110° + 149° + 91° + 100 +  x° + x° = 720°
440° + 2x° = 720°
2x° = 720° – 440°
2x° = 280°
x° = $$\frac{280}{2}$$
x° = 140°
Hence, ∠X = ∠Y = 140°

In Exercises 23-26, find the value of x.

Question 23.

X = 111°

Explanation:
The sum of the angle measures = 360°

Question 24.

x = 53°

Explanation:
The given figure is:

We can observe that, the above polygon has 7 sides
The given angle measures of a polygon with 7 sides are:
50°, 48°, 59°, x°, x°, 58°, and 39°
The sum of the angle measures of the exterior angles of any polygon is: 360°
So, 50° + 48° + 59° +  x° + x° + 58° + 39° = 360°
254° + 2x° = 360°
2x° = 360° – 254°
2x° = 106°
x° = $$\frac{106}{2}$$
x° = 53°
Hence, the value of x is: 53°

Question 25.

x = 32°

Explanation:
We can observe that, the above polygon has 5 sides
The given angle measures of a polygon with 5 sides are:
71°, 85°, 44°, 3x°, 2x°
The sum of the angle measures of the exterior angles of any polygon is: 360°

Question 26.

x° = 66°

Explanation:
The given figure is:

We observe that the above polygon has 5 sides.
The given angle measures of a polygon with 5 sides are:
45°, 40°, x°, 77°, and 2x°
The sum of the angle measures of the exterior angles of any polygon is: 360°
So, 45° + 40° +  x° + 77° + 2x° = 360°
162° + 3x° = 360°
3x° = 360° – 162°
3x° = 198°
x° = $$\frac{198}{3}$$
x° = 66°
Hence, the value of x is 66°

In Exercises 27-30, find the measure of each interior angle and each exterior angle of the indicated regular polygon.

Question 27.
pentagon
x = 72°

Explanation:

Question 28.
18-gon
The measure of each interior angle of 18-gon is: 160°
The measure of each exterior angle of 18-gon is: 20°

Explanation:
The given polygon is: 18-gon.
The number of sides of 18-gon is: 18
We know that,
The measure of each interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$
The measure of each exterior angle of a polygon = $$\frac{360°}{n}$$
The measure of each interior angle of 18-gon = $$\frac{180° (18 – 2)}{18}$$
= $$\frac{180° (16)}{18}$$
= 160°
The measure of each exterior angle of 18-gon = $$\frac{360°}{18}$$
= 20°
Hence, The measure of each interior angle of 18-gon is: 160°and exterior angle of 18-gon is: 20°

Question 29.
45-gon
The measure of each interior angle of 45-gon is: 172°
The measure of each exterior angle of 45-gon is: 8°

Explanation:

Question 30.
90-gon
The measure of each interior angle of 90-gon is: 176°
The measure of each exterior angle of 90-gon is: 4°

Explanation:
The given polygon is: 90-gon
So, the number of sides of 90-gon is: 90.
We know that,
The measure of each interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$
The measure of each exterior angle of a polygon = $$\frac{360°}{n}$$
The measure of each interior angle of 90-gon = $$\frac{180° (90 – 2)}{90}$$
= $$\frac{180° (88)}{90}$$
= 176°
The measure of each exterior angle of 90-gon = $$\frac{360°}{90}$$
= 4°
Hence, the measure of each interior angle of 90-gon is: 176° and exterior angle of 90-gon is: 4°

ERROR ANALYSIS
In Exercises 31 and 32, describe and correct the error in finding the measure of one exterior angle of a regular pentagon.

Question 31.

Explanation:
The given polygon is pentagon.
The number of sides of pentagon is 18.
We know that,
The measure of each interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$
Where “n” is the number of sides.
$$\frac{180° (n – 2)}{n}$$
= $$\frac{180° (5 – 2)}{5}$$
= $$\frac{180° (3)}{5}$$
= $$\frac{540°}{5}$$
= 108°
So, the interior angle is 108°
The measure of each exterior angle of a polygon = $$\frac{360°}{n}$$
= $$\frac{160°}{5}$$
= 72°
So, the measure of each exterior angle 72°
= 20°
Hence,  it is proved that the measure of exterior angle shown in the question is wrong.

Question 32.

Yes, the measure of each angle is 36°

Explanation:
It is given that there are 10 exterior angles, two at each vertex.
So, the number of sides based on 10 exterior angles are 10.
We know that,
The measure of each exterior angle of any polygon = $$\frac{360°}{n}$$
The measure of each exterior angle of a polygon with 10 sides = $$\frac{360°}{10}$$
= 36°
Hence, the measure of each exterior angle in a polygon of 10 sides is: 36°

Question 33.
MODELING WITH MATHEMATICS
The base of a jewelry box is shaped like a regular hexagon. What is the measure of each interior angle of the jewelry box base?
120°

Explanation:
Given, The base of a jewelry box is shaped like a regular hexagon.
So, the shape of the box has 6 sides.

Question 34.
MODELING WITH MATHEMATICS
The floor of the gazebo shown is shaped like a regular decagon. Find the measure of each interior angle of the regular decagon. Then find the measure of each exterior angle.

Explanation:
Given, The floor of the gazebo shown is shaped like a regular decagon.
So, the number of sides of a regular decagon is: 10
We know that,
The measure of each interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$
The measure of each interior angle of a regular decagon = $$\frac{180° (10 – 2)}{10}$$
= $$\frac{180° (8)}{10}$$
= 144°
The measure of each exterior angle of a regular decagon = $$\frac{360°}{10}$$
= 36°
Hence, the measure of each interior angle is 144° and exterior angle of a regular decagon is 36°

Question 35.
WRITING A FORMULA
Write a formula to find the number of sides n in a regular polygon given that the measure of one interior angle is x°.
n = $$\frac{360°}{180° – x°}$$

Explanation:

Question 36.
WRITING A FORMULA
Write a formula to find the number of sides n in a regular polygon given that the measure of one exterior angle is x°.
n = $$\frac{360°}{x°}$$

Explanation:
Given that the measure of one exterior angle is x°
We know that,
The measure of each exterior angle of a polygon = $$\frac{360°}{n}$$
Where “n” is the number of sides.
x° = $$\frac{360°}{n}$$
n = $$\frac{360°}{x°}$$
The formula to find the number of sides n in a regular polygon given that the measure of one exterior angle is,
n = $$\frac{360°}{x°}$$

REASONING
In Exercises 37-40, find the number of sides for the regular polygon described.

Question 37.
Each interior angle has a measure of 156°.
15°

Explanation:
Given, Each interior angle has a measure of 156°.

Question 38.
Each interior angle has a measure of 165°.
24°

Explanation:
Given that each interior angle has a measure of 165°
We know that,
The measure of each interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$
165° = $$\frac{180° (n – 2)}{n}$$
165n = 180 (n – 2)
165n = 180n – 360
180n – 165n = 360
15n = 360
n = $$\frac{360}{15}$$
n = 24
Hence, the number of sides with each interior angle 165° is 24°.

Question 39.
Each exterior angle has a measure of 9°.
40°

Explanation:
Given, Each exterior angle has a measure of 9°.
We know that,
The measure of each exterior angle of a polygon = $$\frac{360°}{n}$$

Question 40.
Each exterior angle has a measure of 6°.
60°

Explanation:
Given that each exterior angle has a measure of 6°
We know that,
The measure of each exterior angle of a polygon = $$\frac{360°}{n}$$
6° = $$\frac{360°}{n}$$
6°n = 360°
n = $$\frac{360}{6}$$
n = 60°
Hence, the number of sides with each exterior angle 6° is 60°

Question 41.
DRAWING CONCLUSIONS
Which of the following angle measures are possible interior angle measures of a regular polygon? Explain your reasoning. Select all that apply.
(A) 162°
(B) 171°
(C) 75°
(D) 40°
Option A & B

Explanation:

Question 42.
PROVING A THEOREM
The Polygon Interior Angles Theorem (Theorem 7.1) states that the sum of the measures of the interior angles of a convex n-gon is (n – 2) • 180°. Write a paragraph proof of this theorem for the case when n = 5.

540°

Explanation:
Polygon Interior Angles Theorem:
Given statement:
The sum of the measures of the interior angles of a convex n-gon is: 180° (n – 2)
Proof of the Polygon Interior Angles Theorem:

ABCDE is an “n” sided polygon.
Take any point O inside the polygon.
Join OA, OB, OC.
For “n” sided polygon, the polygon forms “n” triangles.
We know that the sum of the angles of a triangle is equal to 180 degrees.
The sum of the angles of n triangles = n × 180°
From the above statement, we can say that,
Sum of interior angles + Sum of the angles at O = 2n × 90° ——(1)
But, the sum of the angles at O = 360°
Substitute the above value in (1), we get
Sum of interior angles + 360°= 2n × 90°
So, the sum of the interior angles = (2n × 90°) – 360°
Take 90 as common,
Then the sum of the interior angles = (2n – 4) × 90°
Therefore, sum of “n” interior angles is (2n – 4) × 90°
When n= 5,
The sum of interior angles = ([2 × 5] – 4) × 90°
= (10 – 4) × 90°
= 6 × 90°
= 540°

Question 43.
PROVING A COROLLARY
Write a paragraph proof of the Corollary to the Polygon Interior Angles Theorem (Corollary 7. 1).
The Corollary states that the sum of the interior angles of any quadrilateral is equal to 360°

Explanation:

Question 44.
MAKING AN ARGUMENT
Your friend claims that to find the interior angle measures of a regular polygon. you do not have to use the Polygon Interior Angles Theorem (Theorem 7. 1). You instead can use the Polygon Exterior Angles Theorem (Theorem 7.2) and then the Linear Pair Postulate (Postulate 2.8). Is your friend correct? Explain your reasoning.

Explanation:
Given that your friend claims that to find the interior angle measures of a regular polygon.
you do not have to use the Polygon Interior Angles Theorem.
You instead can use the Polygon Exterior Angles Theorem and then the Linear Pair Postulate.
We know that in a polygon,
The sum of the angle measures of exterior angles + The sum of the angle measures of interior angles = 180°
The sum of the angle measures of interior angles = 180° – (The sum of the angle measures of exterior angles)
According to Linear Pair Postulate,
The sum of the exterior angle and interior angle measures is 180°
The angle measures of the exterior angles can be found out by using the “Polygon Exterior Angles Theorem”
Hence, the claim of your friend is correct.

Question 45.
MATHEMATICAL CONNECTIONS
In an equilateral hexagon. four of the exterior angles each have a measure of x°. The other two exterior angles each have a measure of twice the sum of x and 48. Find the measure of each exterior angle.
21°, 21°, 21°, 21°, 138° and 138°

Explanation:
Given, an equilateral hexagon.

Question 46.
THOUGHT-PROVOKING
For a concave polygon, is it true that at least one of the interior angle measures must be greater than 180°? If not, give an example. If so, explain your reasoning.
Yes, it is true that at least one of the interior angle measures must be greater than 180°

Explanation:
We know that, for a concave polygon,
The angle measure of at least one interior angle should be greater than 180°
The word “Concave” implies that at least 1 interior angle is folding in.
So, this “Folding in” should be greater than 180° to produce the required shape.
Hence, it is true that at least one of the interior angle measures must be greater than 180°

Question 47.
WRITING EXPRESSIONS
Write an expression to find the sum of the measures of the interior angles for a concave polygon. Explain your reasoning.

The sum of the interior angles of a concave polygon is (n – 2) x 180,
where n is the number of sides.

Explanation:

Question 48.
ANALYZING RELATIONSHIPS
Polygon ABCDEFGH is a regular octagon. Suppose sides $$\overline{A B}$$ and $$\overline{C D}$$ are extended to meet at a point P. Find m∠BPC. Explain your reasoning. Include a diagram with your answer.
∠BPC = 90°

Explanation:
Given that polygon ABCDEFGH is a regular polygon and $$\overline{A B}$$ and $$\overline{C D}$$ are extended to meet at a point P.
So, the representation of the regular octagon is,

We know that,
The angle measure of each exterior angle = $$\frac{360°}{n}$$
Where “n” is the number of sides.
The number of sides of the Octagon is: 8
The angle measure of each exterior angle = $$\frac{360°}{8}$$
= 45°
From the given figure,
We can observe that ΔBPC is an Isosceles triangle
∠B = ∠C = 45°, ∠P = x°
We know that,
the sum of the angle measures of a given triangle is: 180°
45° + x° + 45° = 180°
x° + 90° = 180°
x° = 180° – 90°
x° = 90°
So, ∠BPC = 90°

Question 49.
MULTIPLE REPRESENTATIONS
The formula for the measure of each interior angle in a regular polygon can be written in function notation.

a. Write a junction h(n). where n is the number of sides in a regular polygon and h(n) is the measure of any interior angle in the regular polygon.
b. Use the function to find h(9).
c. Use the function to find n when h(n) = 150°.
d. Plot the points for n = 3, 4, 5, 6, 7, and 8. What happens to the value of h(n) as n gets larger?

Question 50.
HOW DO YOU SEE IT?
Is the hexagon a regular hexagon? Explain your reasoning.

Question 51.
PROVING A THEOREM
Write a paragraph proof of the Polygon Exterior Angles Theorem (Theorem 7.2). (Hint: In a convex n-gon. the sum of the measures of an interior angle and an adjacent exterior angle at any vertex is 180°.)

Question 52.
ABSTRACT REASONING
You are given a convex polygon. You are asked to draw a new polygon by increasing the sum of the interior angle measures by 540°. How many more sides does our new polygon have? Explain your reasoning.
3 more sides.

Explanation:
Given, a convex polygon and asked to draw a new polygon by increasing the sum of the interior angle measures by 540°
We know that, the sum of the angle measures of the interior angles in a polygon = 180° (n – 2)
Let the number of sides of a new polygon be x.
180° (x – 2) + 540° = 180° (n – 2)
180x – 360° + 540° = 180n – 360°
180x – 180n = -540°
180 (x – n) = -540°
x – n = $$\frac{540}{180}$$
x – n = -3
n – x = 3
n = x + 3
So, we have to add 3 more sides to the original convex polygon.

Maintaining Mathematical Proficiency

Find the value of x.

Question 53.

101°

Explanation:
Sum of the angles = 180°

Question 54.

113°

Explanation:
The given figure  has two corresponding angles as 113° and x°
According to the “Corresponding Angles Theorem”, the corresponding angles are congruent i.e., equal
x° = 113°
So, the value of x is 113°

Question 55.

16°

Explanation:
Sum of the angles = 180°

Question 56.

9.6°

Explanation:
From the given figure,
We observe that, (3x + 10)° and (6x – 19)° are the corresponding angles.
According to the “Corresponding angles Theorem”, the corresponding angles are congruent i.e., equal
(3x + 10)° = (6x – 19)°
6x – 3x = 19° + 10°
3x° = 29°
x° = $$\frac{29}{3}$$
x° = 9.6°
So, the value of x is 9.6°

### 7.2 Properties of Parallelograms

Exploration 1

Discovering Properties of Parallelograms

Work with a partner: Use dynamic geometry software.

a. Construct any parallelogram and label it ABCD. Explain your process.
Sample

The representation of parallelogram ABCD is:

Explanation:
Draw a straight line on one side of the first line and tangent to both arcs.
We know that the two parallel lines form two sides of the parallelogram.
Draw another straight line through both parallel lines.
This will be the third side of the parallelogram.
Finally join the other side of the polygon to form a parallelogram.

b. Find the angle measures of the parallelogram. What do you observe?
The representation of parallelogram ABCD with the angles are,

Explanation:
The opposite interior angles of a parallelogram are equal.
∠A = 105°, ∠B = 75°, ∠D = 105°, and ∠C = 75°
The angles on the same side of the transversal are supplementary, that means they add up to 180 degrees.
Hence, the sum of the interior angles of a parallelogram is 360 degrees.

c. Find the side lengths of the parallelogram. What do you observe?
The representation of the parallelogram ABCD along with the side lengths are,

Explanation:
From the parallelogram ABCD,
The opposite side lengths of a parallelogram are equal.
So, AB = CD = 5cm, and AC = BD = 2.8cm

d. Repeat parts (a)-(c) for several other parallelograms. Use your results to write conjectures about the angle measures and side lengths of a parallelogram.
The representation of parallelogram ABCD along with its angles and the side lengths are,

Explanation:
From the parallelogram ABCD we observe that,
The opposite sides AB and CD; AC and BD are congruent i.e., equal.
The opposite angles A and C; B and D are congruent i.e., equal.

Exploration 2

Discovering a Property of Parallelograms

Work with a partner: Use dynamic geometry software.

a. Construct any parallelogram and label it ABCD.
The representation of the parallelogram ABCD is:

Explanation:
Draw a straight line on one side of the first line and tangent to both arcs.
We know that the two parallel lines form two sides of the parallelogram.
Draw another straight line through both parallel lines.
This will be the third side of the parallelogram.
Finally join the other side of the polygon to form a parallelogram.

b. Draw the two diagonals of the parallelogram. Label the point of intersection E.

The representation of the parallelogram ABCD along with its diagonals are,

Explanation:
From the above parallelogram ABCD,
We observe that, the diagonals of parallelogram ABCD are: AC and BD
The intersection point of AC and BD is E.

c. Find the segment lengths AE, BE, CE, and DE. What do you observe?
The representation of the parallelogram ABCD along the segment lengths are,

Explanation:
From the above parallelogram ABCD,
We observe that, the opposite side lengths of a parallelogram are equal.
AE = DE = 1.9 cm
CE = BE = 2.7 cm

d. Repeat parts (a)-(c) for several other parallelograms. Use your results to write a conjecture about the diagonals of a parallelogram.
The representation of parallelogram ABCD along with its side lengths are,

Explanation:
From the parallelogram ABCD we observe that,
The opposite sides AB and CD; AC and BD are congruent i.e., equal.
The opposite angles A and C; B and D are congruent i.e., equal.

MAKING SENSE OF PROBLEMS
To be proficient in math, you need to analyze givens, constraints, relationships, and goals.
The representation of the parallelogram ABCD along with the length of the diagonals are,

Explanation:
from the above given parallelogram ABCD,
The diagonals bisect each other.
So, the lengths of the diagonals are:
A = 1.9; D = 1.9; B = 2.7; C = 2.7
AD = 3.8 cm and BC = 5.4 cm

Question 3.
What are the properties of parallelograms?

The properties of a parallelograms are:
a. The opposite sides are parallel.
b. The opposite sides are congruent.
c. The opposite angles are congruent.
d. Consecutive angles are supplementary.
e. The diagonals bisect each other.

### Lesson 7.2 Properties of Parallelograms

Monitoring progress

Question 1.
Find FG and m∠G.

FG = 8 and ∠G = 60°

Explanation:
From the given figure,
EH = 8 and ∠E = 60°
We know that in a parallelogram,
The opposite sides and angles are congruent.
FG = HE and GH = FE
∠G = ∠E and ∠H = ∠F
Hence, FG = 8 and ∠G = 60°

Question 2.
Find the values of x and y.

The values of x and y are: 25° and 15.

Explanation:
From the above given parallelogram,
JK = 18 and LM = y + 3
∠J = 2x° and ∠L = 50°
We know that in a parallelogram,
The opposite sides and angles are congruent.
JK = LM and ∠J = ∠L
y + 3 =18
y = 18 – 3
y = 15
2x° = 50°
x° = $$\frac{50}{2}$$
x° = 25°
Hence, the values of x and y are: 25° and 15.

Question 3.
WHAT IF?
In Example 2, find in m∠BCD when m∠ADC is twice the measure of ∠BCD.
∠BCD = 70°

Explanation:
In Example 2, it is given that ABCD is a parallelogram.
In the parallelogram ABCD, ∠ADC = 110°
We know that, the sum of the angle measure of the consecutive angles are supplementary.
In Example 2, We observe that,
∠BCD = 180° – 110°
= 70°
Hence, ∠BCD = 70°

Question 4.
Using the figure and the given statement in Example 3, prove that ∠C and ∠F are supplementary angles.

Question 5.
Find the coordinates of the intersection of the diagonals of STUV with vertices S(- 2, 3), T(1, 5), U(6, 3), and V(3, 1).
(2, 3)

Explanation:
The given coordinates of the parallelogram STUV are:
S (-2, 3), T (1, 5), U (6, 3), and V (3, 1)
Compare the given points with (x1, y1), (x2, y2)
We know that, the opposite vertices form a diagonal.
In the parallelogram STUV, SU and TV are the diagonals.
We know that, the intersection of the diagonals means the midpoint of the vertices of the diagonals,
because diagonals bisect each other.
The midpoint of SU = ($$\frac{x1 + x2}{2}$$, $$\frac{y1 + y2}{2}$$)
= ($$\frac{6 – 2}{2}$$, $$\frac{3 + 3}{2}$$)
= ($$\frac{4}{2}$$, $$\frac{6}{2}$$)
= (2, 3)
Hence, the coordinates of the intersection of the diagonals of parallelogram STUV is: (2, 3)

Question 6.
Three vertices of ABCD are A(2, 4), B(5, 2), and C(3, – 1). Find the coordinates of vertex D.
(10, 3)

Explanation:
The given vertices of parallelogram ABCD are:
A (2, 4), B (5, 2), and C (3, -1)
Let the fourth vertex of the parallelogram ABCD be: (x, y)

We know that, the diagonals of a parallelogram bisect each other i.e., the angle between the diagonals is 90°
So, the diagonals are the perpendicular lines.
Hence, AC and BD are the diagonals.
Slope of AC = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{-1 – 4}{3 – 2}$$
= $$\frac{-5}{1}$$
= -5
Slope of BD = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{y – 2}{x – 5}$$
AC and BD are the perpendicular lines.
The product of the slopes of the perpendicular lines is equal to -1
(Slope of AC) × (Slope of BD) = -1
-5 × $$\frac{y – 2}{x – 5}$$ = -1
$$\frac{y – 2}{x – 5}$$ = $$\frac{1}{5}$$
Equate the numerator and denominator of both expreesions
We get,
y – 2 = 1                    x – 5 = 5
y = 1 + 2                   x = 5 + 5
y = 3                          x = 10
So, the coordinates of the vertex D are: (10, 3)

Exercise 7.2 Properties of Parallelograms

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Why is a parallelogram always a quadrilateral, but a quadrilateral is only sometimes a parallelogram?

Question 2.
WRITING
You are given one angle measure of a parallelogram. Explain how you can find the other angle measures of the parallelogram.
The opposite angles of a parallelogram are equal.

Explanation:

It is given that parallelogram has one angle measure of a polygon.
Let the one angle measure of the given parallelogram be ∠A
We know that in a parallelogram,
The opposite angles are congruent i.e., equal
The sum of the consecutive angles are 180°
So, ∠A = ∠C and ∠B = ∠D
Hence, We can measure the other angles of the parallelogram by using the property of “The opposite angles are congruent”.

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, find the value of each variable in the parallelogram.

Question 3.

x = 9; y = 15

Explanation:
We know that the opposite sides of a parallelogram are equal.

Question 4.

n = 12; m = 5

Explanation:
From the above given figure is:
We know that, the opposite sides of a parallelogram are equal.
AB = CD and AD = BC
n = 12          m + 1 = 6
n = 12          m = 6 – 1
n = 12          m = 5
Hence, the values of m and n are 12 and 5.

Question 5.

z = 128; d = 126

Explanation:
From the above given figure is:
We know that, the opposite sides and angles of a parallelogram are equal.

Question 6.

g° = 61° and h = 9

Explanation:
Explanation:
From the above given figure is:
We know that, the opposite sides and angles of a parallelogram are equal.
AB = CD and AD = BC
According to the parallelogram Opposite Angles Theorem,
∠A = ∠C and ∠B = ∠D
Hence, from the figure,
(g + 4)° = 65°             16 – h = 7
g° = 65° – 4°                h = 16 – 7
g° = 61°                       h = 9
Hence, the values of g and h are 61° and 9.

In Exercises 7 and 8. find the measure of the indicated angle in the parallelogram.

Question 7.
Find m∠B.
m∠B = 129°

Explanation:
The sum of the angles = 180°
m∠A = 51°; m∠B = x

Question 8.
Find m ∠ N.
∠N = 85°

Explanation:
From the above given figure,
We know that, the sum of the consecutive angle measures is equal to180°
∠M + ∠N = 180°
95° + ∠N = 180°
∠N = 180° – 95°
∠N = 85°

In Exercises 9-16. find the indicated measure in LMNQ. Explain your reasoning.

Question 9.
LM

Question 10.
LP
LP = 3.5

Explanation:
From the above given figure,
LN = 7
We know that, in a parallelogram the diagonals bisect each other.
So, the length of each diagonal will be divided into half of the value of the diagonal length.
LN can be divided into LP and PN.
LP = $$\frac{LN}{2}$$
LP = $$\frac{7}{2}$$
LP = 3.5

Question 11.
LQ

Question 12.
MQ
MQ = 8.2

Explanation:
The given figure is:

Hence,
From the given figure,
The diagonal of a parallelogram is MQ.
MQ = 8.2

Question 13.
m∠LMN
m∠LMN = 80°

Explanation:

Question 14.
m∠NQL
∠NQL = 80°

Explanation:
The given figure is:

According to the parallelogram Consecutive angles Theorem,
The sum of the consecutive angle measures is: 180°
∠L + ∠M = 180°
100° + ∠M = 180°
∠M = 180° – 100°
∠M = 80°
We know that, the opposite angles are congruent i.e., equal.
So, ∠M = ∠Q
∠Q = 80°
Hence, the value of ∠NQL is 80°

Question 15.
m∠MNQ
m∠MNQ = 100°

Explanation:
The given figure is:

Question 16.
m∠LMQ
∠LMQ = 80°

Explanation:
The given figure is:

We know that, the sum of the consecutive angle measure is 180°
∠L + ∠M = 180°
100° + ∠M = 180°
∠M = 180° – 100°
∠M = 80°
Hence, the value of ∠LMQ is 80°

In Exercises 17-20. find the value of each variable in the parallelogram.

Question 17.

n = 110°; m = 35°

Explanation:
Sum of angles of parallelogram is 180°

Question 18.

b = 90°

Explanation:
The given figure is:

We know that, the opposite angles of parallelogram are equal.
∠A = ∠C and ∠B = ∠D
∠A + ∠B = 180° and ∠C + ∠D = 180° and ∠A + ∠D = 180° and ∠B + ∠C = 180°
(b – 10)° + (b + 10)° = 180°
2b° = 180°
b° = $$\frac{180}{2}$$
b° = 90°

Question 19.

k = 7; m = 8

Explanation:
From the given figure,
we know that the diagonals of a parallelogram are equal.

Question 20.

u = 6 and v = 18

Explanation:
From the given figure,

The diagonals bisect each other as shown above.
$$\frac{v}{3}$$ = 6                       2u + 2 = 5u – 10
v = 6 (3)                                                     5u – 2u = 10 + 2
v = 18                                                        3u = 18
v = 18                                                         u = $$\frac{18}{3}$$
v = 18                                                          u = 6
Hence, the values of u and v are 6 and 18.

ERROR ANALYSIS
In Exercises 21 and 22, describe and correct the error in using properties of parallelograms.

Question 21.

m∠v = 130°

Explanation:

Question 22.

$$\overline{G F}$$ = $$\overline{F J}$$

Explanation:
According to the properties of the parallelogram,
GJ and HK are the diagonals of the parallelogram.
F is the perpendicular bisector of the diagonals.
GF = FJ and KF = FH
So, quadrilateral GHJK is a parallelogram.
$$\overline{G F}$$ = $$\overline{F J}$$

PROOF
In Exercises 23 and 24, write a two-column proof.

Question 23.
Given ABCD and CEFD are parallelograms.
Prove $$\overline{A B} \cong \overline{F E}$$

Question 24.
Given ABCD, EBGF, and HJKD are parallelograms.
Prove ∠2 ≅∠3

Given that,
ABCD, EBGF, and HJKD are parallelograms.
According to transitive property of he transitive property of equality x = y and y = z, then x = z.
Where x, y and z belongs to the same category elements.
Hence, ∠2 ≅ ∠3 is proved.

In Exercises 25 and 26, find the coordinates of the intersection of the diagonals of the parallelogram with the given vertices.

Question 25.
W(- 2, 5), X(2, 5), Y(4, 0), Z(0, 0)
(1, 2.5)

Explanation:

Question 26.
Q(- 1, 3), R(5, 2), S(1, – 2), T(- 5, – 1)
(0, $$\frac{1}{2}$$)

Explanation:
The given coordinates of the parallelogram are:
Q (-1, 3), R (5, 2), S (1, -2), and T (-5, -1)
Compare the given coordinates with (x1, y1), and (x2, y2)
We know that, the diagonals of a parallelogram bisect each other.
The diagonals of the parallelogram are: QS and RT
The midpoint of QS = ($$\frac{x1 + x2}{2}$$, $$\frac{y1 + y2}{2}$$)
= ($$\frac{1 – 1}{2}$$, $$\frac{3 – 2}{2}$$)
= (0, $$\frac{1}{2}$$)
The midpoint of RT = ($$\frac{x1 + x2}{2}$$, $$\frac{y1 + y2}{2}$$)
= ($$\frac{5 – 5}{2}$$, $$\frac{2 – 1}{2}$$)
= (0, $$\frac{1}{2}$$)
So, the coordinates of the intersection of the diagonals of the given parallelogram are (0, $$\frac{1}{2}$$)

In Exercises 27-30, three vertices of DEFG are given. Find the coordinates of the remaining vertex.

Question 27.
D(0, 2), E(- 1, 5), G(4, 0)
F(3, 3)

Explanation:

Question 28.
D(- 2, – 4), F(0, 7), G(1, 0)
Fourth vertex is (12, -2)

Explanation:
The given vertices of parallelogram are:
D (-2, -4), F (0, 7), and G (1, 0)
Let the fourth vertex of the parallelogram be: (x, y)
In a parallelogram, the diagonals bisect each other i.e., the angle between the diagonals is 90°
In the given parallelogram,
DF and EG are the diagonals.
Slope of DF = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{7 + 4}{0 + 2}$$
= $$\frac{11}{2}$$
Slope of EG = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{y – 0}{x – 1}$$
= $$\frac{y}{x – 1}$$
We know that, DF and EG are the perpendicular lines.
The product of the slopes of the perpendicular lines is equal to -1
(Slope of DF) × (Slope of EG) = -1
$$\frac{11}{2}$$ × $$\frac{y}{x – 1}$$ = -1
$$\frac{y}{x – 1}$$ = –$$\frac{2}{11}$$
Equate the numerator and denominator of both expressions.
We get,
y = -2                        x – 1 = 11
y = -2                        x = 11 + 1
y = -2                        x = 12
Hence, the coordinates of the fourth vertex are: (12, -2)

Question 29.
D(- 4, – 2), E(- 3, 1), F(3, 3)
G(2, 0)

Explanation:
Given coordinates are:
D(- 4, – 2), E(- 3, 1), F(3, 3)

Question 30.
E (1, 4), f(5, 6), G(8, 0)
fourth vertex are: (8, 13)

Explanation:
The given vertices of a parallelogram are:
E (1, 4), F (5, 6), and G (8, 0)
Let the fourth vertex of the parallelogram be: (x, y)
We know that in a parallelogram,
The diagonals bisect each other i.e., the angle between the diagonals is 90°
In the given parallelogram, FH and EG are the diagonals.
Slope of FH = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{y – 6}{x – 5}$$
Slope of EG = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{0 – 4}{8 – 1}$$
= $$\frac{-4}{7}$$
= –$$\frac{4}{7}$$
We know that, FH and EG are the perpendicular lines.
So, the product of the slopes of the perpendicular lines is equal to -1
(Slope of FH) × (Slope of EG) = -1
–$$\frac{4}{7}$$ × $$\frac{y – 6}{x – 5}$$ = -1
$$\frac{y – 6}{x – 4}$$ = $$\frac{7}{4}$$
Equate the numerator and denominator of both expressions.
We get,
y – 6 = 7                    x – 4 = 4
y = 7 + 6                   x = 4 + 4
y = 13                        x = 8
Hence, the coordinates of the fourth vertex are: (8, 13)

MATHEMATICAL CONNECTIONS
In Exercises 31 and 32. find the measure of each angle.

Question 31.
The measure of one interior angle of a parallelogram is 0.25 times the measure of another angle.
The angles are 36° and 144°

Explanation:
Given that,
The measure of one interior angle of a parallelogram is 0.25 times the measure of another angle.

Question 32.
The measure of one interior angle of a parallelogram is 50 degrees more than 4 times the measure of another angle.
The angle measures are: 26° and 154°

Explanation:
Given that, the measure of one interior angle of a parallelogram is 50 degrees more than 4 times the measure of another angle.
So, the measure of one interior angle is: x°
The measure of another interior angle is: 50° + 4x°
We know that, the opposite angles of the parallelogram are equal.
The sum of the angles of the parallelogram is: 360°
x° + 4x + 50° + x° + 4x + 50° = 360°
10x° + 100° = 360°
10x° = 360° – 100°
10x° = 260°
x° = $$\frac{260}{10}$$
x° = 26°
The angle measures of the parallelogram are:
4x° + 50° = 4 (26°) + 50°
= 104° + 50°
= 154°
Hence, the angle measures are: 26° and 154°

Question 33.
MAKING AN ARGUMENT
m∠B = 124°, m∠A = 56°, and m∠C = 124°.

Question 34.
ATTENDING TO PRECISION
∠J and ∠K are Consecutive angles in a parallelogram. m∠J = (3t + 7)°. and m∠K = (5t – 11)°. Find the measure of each angle.
The measure of each angle is: 76° and 104°

Explanation:
Given that ∠J and ∠K are the consecutive angles in a parallelogram.
So, ∠J + ∠K = 180°
It is given that
∠J = (3t + 7)° and ∠K = (5t – 11)°
(3t + 7)° + (5t – 11)° = 180°
8t° – 4 = 180°
8t° = 180° + 4°
8t° = 184°
t° = $$\frac{184}{8}$$
t° = 23°
So, ∠J = (3t + 7)°
= 3 (23)° + 7
= 69° + 7°
= 76°
∠K = (5t – 11)°
= 5 (23)° – 11
= 115° – 11°
= 104°
Hence, the measure of each angle is: 76° and 104°

Question 35.
CONSTRUCTION
Construct any parallelogram and label it ABCD. Draw diagonals $$\overline{A C}$$ and $$\overline{B D}$$. Explain how to use paper folding to verify the Parallelogram Diagonals Theorem (Theorem 7.6) for ABCD.

Question 36.
MODELING WITH MATHEMATICS
The feathers on an arrow from two congruent parallelograms. The parallelograms are reflections of each other over the line that contains their shared side. Show that m ∠ 2 = 2m ∠ 1.

Question 37.
PROVING A THEOREM
Use the diagram to write a two-column proof of the Parallelogram Opposite Angles Theorem (Theorem 7.4).

Given ABCD is a parallelogram.
Prove ∠A ≅ ∠C, ∠B ≅ ∠D

Question 38.
PROVING A THEOREM
Use the diagram to write a two-column proof of the Parallelogram Consecutive Angles Theorem (Theorem 7.5).

Given PQRS is a parallelogram.
Prove x° + y° = 180°
Given:
PQRS is a parallelogram
Prove:
x° + y° = 180°

Question 39.
PROBLEM-SOLVING
The sides of MNPQ are represented by the expressions below. Sketch MNPQ and find its perimeter.
MQ = – 2x + 37 QP = y + 14
NP= x – 5 MN = 4y + 5
52 units.

Explanation:
Given that,
The sides of MNPQ are represented by the expressions below
MQ = – 2x + 37 QP = y + 14
NP= x – 5 MN = 4y + 5

Question 40.
PROBLEM SOLVING
In LMNP, the ratio of LM to MN is 4 : 3. Find LM when the perimeter of LMNP is 28.
LM = 8

Explanation:
Given that, in the parallelogram LMNP,
The ratio of LM to MN is: 4 : 3
The perimeter of the parallelogram LMNP is: 28
Let the length of LM be 4x
Let the length of MN be 3x
We know that, the opposite sides of the parallelogram are equal and the perimeter is the sum of all the sides.
4x + 3x + 4x + 3x = 28
8x + 6x = 28
14x = 28
x = $$\frac{28}{14}$$
x = 2
The length of LM = 4 x
= 4 (2)
= 8

Question 41.
ABSTRACT REASONING
Can you prove that two parallelograms are congruent by proving that all their corresponding sides are congruent? Explain your reasoning.

Question 42.
HOW DO YOU SEE IT?
The mirror shown is attached to the wall by an arm that can extend away from the wall. In the figure. points P, Q, R, and S are the vertices of a parallelogram. This parallelogram is one of several that change shape as the mirror is extended.

a. What happens to m∠P as m∠Q increases? Explain.
When ∠Q increases, ∠P has to decrease.

Explanation:
From the given above figure, we can observe that,
∠P and ∠Q are the consecutive angles.
So, ∠P + ∠Q = 180°
To make the sum 180°,
if one angle measure increases, then the other angle measure has to decrease.

b. What happens to QS as m∠Q decreases? Explain.
As ∠Q decreases, the length of QS may also decrease or may also increase.

Explanation:
From the above given figure,
QS is a diagonal of the parallelogram.
Q and S are the opposite angles.
We know that, the opposite angles are equal in a parallelogram.
So, ∠Q decreases, the length of QS may also decrease or may also increase.

c. What happens to the overall distance between the mirror and the wall when m∠Q decreases? Explain.
Increases.

Explanation:
From the above given figure,
We can observe that, the angle between Q and the wall increases.
So, the overall distance between the mirror and the wall increase,
as the opposite angles in the parallelogram are equal.

Question 43.
MATHEMATICAL CONNECTIONS
In STUV m∠TSU = 32°, m∠USV = (x2)°, m∠TUV = 12x°, and ∠TUV is an acute angle. Find m∠USV.

m∠USV = 16°

Explanation:
Given that,
In STUV m∠TSU = 32°, m∠USV = (x2)°, m∠TUV = 12x°,
∠TUV is an acute angle.

Question 44.
THOUGHT-PROVOKING
Is it possible that any triangle can be partitioned into four congruent triangles that can be rearranged to form a parallelogram? Explain your reasoning.
Yes, it is possible that any triangle can be partitioned into four congruent triangles that can be rearranged to form a parallelogram.

Explanation:
We know that,
In any quadrilateral the diagonals bisect each other and the angles may or may not be 90° in the diagonals.
So, after the bisecting with the diagonals in a quadrilateral.
We observe that the quadrilateral is divided into four triangles.
Hence, it is possible that any triangle can be partitioned into four congruent triangles that can be rearranged to form a parallelogram.

Question 45.
CRITICAL THINKING
Points W(1. 2), X(3, 6), and Y(6, 4) are three vertices of a parallelogram. How many parallelograms can be created using these three vertices? Find the coordinates of each point that could be the fourth vertex.
The fourth vertex is (8, 8)

Explanation:
Given, Points W(1. 2), X(3, 6), and Y(6, 4) are three vertices of a parallelogram.

Question 46.
PROOF
In the diagram. $$\overline{E K}$$ bisects ∠FEH, and $$\overline{F J}$$ bisects ∠EFG. Prove that $$\overline{E K}$$ ⊥ $$\overline{F J}$$. (Hint: Write equations using the angle measures of the triangles and quadrilaterals formed.)

Question 47.
PROOF
Prove the congruent Parts of Parallel Lines Corollary: If three or more parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal.

Given
Prove $$\overline{H K}$$ ≅ $$\overline{K M}$$
(Hint: Draw $$\overline{K P}$$ and $$\overline{M Q}$$ such that quadrilatcral GPKJ and quadrilateral JQML are parallelorams.)

Maintaining Mathematical Proficiency

Determine whether lines l and m are parallel. Explain your reasoning.

Question 48.

l and m are parallel lines.

Explanation:
From the above given figure,
We observe that the given angles are the corresponding angles.
According to the Corresponding Angles Theorem, l and m are parallel lines.

Question 49.

Explanation:
From the above given figure,
We observe that the given angles are the corresponding angles i.e., an interior angle and an exterior angle.
According to the Corresponding Angles Theorem, l and m are parallel lines.

Question 50.

l is not parallel to m.

Explanation:
From the given figure,
We observe that the given angles are the consecutive interior angles.
We know that, the sum of the angle measures of the consecutive interior angles is 180°
But, from the given figure,
The sum of the angle measures is not 180°
Hence, l is not parallel to m.

### 7.3 Proving That a Quadrilateral is a Parallelogram

Exploration 1

Proving That a Quadrilateral is a Parallelogram

Work with a partner: Use dynamic geometry software.

a. Construct a quadrilateral ABCD whose opposite sides are congruent.

Explanation:
Given, to construct a quadrilateral ABCD whose opposite sides are congruent.
We know that the opposite sides of a quadrilateral are parallel and equal.

Yes, quadrilateral a parallelogram when the opposite sides are congruent and each angle measure is not equal to 90° and the diagonals bisected each other.

Explanation:
The diagonals bisected each other in the given polygon are equal to each other as shown below.
The representation of the quadrilateral ABCD as a parallelogram is shown below,

c. Repeat parts (a) and (b) for several other quadrilaterals. Then write a conjecture based on your results.
If the opposite sides of a quadrilateral are equal, then the opposite angles of a quadrilateral are equal.

Explanation:

If the opposite sides of a quadrilateral are equal, then the opposite angles of a quadrilateral are equal.
AB = CD and AC = BD
∠C = ∠B and ∠A = ∠D

d. Write the converse of your conjecture. Is the Converse true? Explain.

REASONING ABSTRACTLY
To be proficient in math, you need to know and flexibly use different properties of objects.
If the opposite sides of a quadrilateral are equal, then the opposite angles of a quadrilateral are equal.

Explanation:
From part (c),
If the opposite sides of a quadrilateral are equal, then the opposite angles of a quadrilateral are equal
The converse of your conjecture is:
If the opposite angles of a quadrilateral are equal, then the opposite sides of a quadrilateral are equal
From the below figure,

We observe that ∠B and ∠D are 90°,
Hence, the converse of the conjecture from part (c) is true

Exploration 2

Proving That a Quadrilateral Is a Parallelogram

Work with a partner: Use dynamic geometry software.

a. Construct any quadrilateral ABCD whose opposite angles are congruent.

Explanation:
We know that the opposite angles of a quadrilateral are congruent.
So, ∠B = ∠D

Yes, quadrilateral ABCD is a parallelogram.

Explanation:
From the given figure below,

We know that, the opposite sides are equal and the opposite angles are equal.
Hence, we conclude that the given quadrilateral ABCD is a parallelogram.

c. Repeat parts (a) and (b) for several other quadrilaterals. Then write a conjecture based on your results.
From parts (a) and (b),
The opposite angles of a quadrilateral are equal, then the opposite sides of a quadrilateral are equal.

d. Write the converse of your conjecture. Is the converse true? Explain.
From part (c),
If the opposite angles of a quadrilateral are equal, then the opposite sides of a quadrilateral are equal.
Hence,
The converse of the conjecture of the quadrilaterals are,
If the opposite sides of a quadrilateral are equal, then the opposite angles of a quadrilateral are equal.

Question 3.
How can you prove that a quadrilateral is a parallelogram?
The number of ways to prove that a quadrilateral is a parallelogram are:
a. The opposite sides of a parallelograms are congruent.
b. The opposite angles of a parallelograms are congruent.
c. The opposite sides of a parallelograms are parallel.
d. The consecutive angles of a parallelograms are supplementary.
e. An angle is supplementary to both its consecutive angles.

Question 4.
The quadrilateral at the left is the “Parallelogram”.

Explanation:
From the given figure, we know that the opposite angles are equal.
From the conjecture of the quadrilateral,
If the opposite angles of the quadrilateral are equal, then the opposite sides of the quadrilateral are equal.
We know that,
if the opposite angles are equal and the angles are not 90°, then the quadrilateral is called the “parallelogram”
Hence, the quadrilateral ate the left is the “Parallelogram”.

### Lesson 7.3 Proving that a Quadrilateral is a Parallelogram

Monitoring Progress

Question 1.
In quadrilateral WXYZ, m∠W = 42°, m∠X = 138°, and m∠Y = 42°. Find m∠Z. Is WXYZ a parallelogram’? Explain your reasoning.
Yes, WXYZ a parallelogram.

Explanation:
∠W = 42°, ∠X = 138°, and ∠Y = 42°
Let ∠Z = x°
We know that, the sum of the angles of a quadrilateral is 360°
So, ∠W + ∠X + ∠Y + ∠Z = 360°
42° + 138° + 42° + x° = 360°
84° + 138° + x° = 360°
x° = 360° – 222°
x° = 138°
So, ∠Z = 138°
We know that,
If the opposite angles of a quadrilateral are equal and the angle is not equal to 90°,
then that quadrilateral is called the “Parallelogram”.
Hence, WXYZ is a parallelogram.

Question 2.
For what values of x and y is quadrilateral ABCD a parallelogram? Explain your reasoning.

The quadrilateral ABCD is a parallelogram

Explanation:
The given figure is:

We know that, for a quadrilateral to be a parallelogram the opposite angles are equal.
4y° = (y + 87)°                                    2x° = (3x – 32)°
4y° – y° = 87°                                      3x° – 2x° = 32°
3y° = 87°                                              x° = 32°
y° = 87 / 3                                            x° = 32°
y° = 29°                                                x° = 32°
Hence, the quadrilateral ABCD is a parallelogram.

State the theorem you can use to show that the quadrilateral is a parallelogram.

Question 3.

Yes, the quadrilateral is a parallelogram.

Explanation:
The given figure is:

We observe that the opposite sides are equal and parallel to each other.
According to the “Opposite sides parallel and congruent Theorem”,
So, the given quadrilateral is a parallelogram.

Question 4.

Yes, the quadrilateral is a parallelogram.

Explanation:
The given figure is:

We observe that the opposite sides are congruent and parallel to each other.
According to the “Opposite sides parallel and congruent Theorem”,
So, the given quadrilateral is a parallelogram.

Question 5.

Yes, the quadrilateral is a parallelogram.

Explanation:
The given figure is:

We observe that the opposite angles are equal and parallel to each otherr.
According to the “Opposite angles parallel and congruent Theorem”,
So, the given quadrilateral is a parallelogram.

Question 6.
For what value of x is quadrilateral MNPQ a parallelogram? Explain your reasoning.

Yes, the quadrilateral is a parallelogram.

Explanation:
The given figure is:

From the “Parallelogram Diagonals Converse Theorem”,
MP and NQ bisect each other.
So, MP = NQ
It is given that
MP = 10 – 3x
NQ = 2x
So, 2x = 10 – 3x
2x + 3x = 10
5x = 10
x = $$\frac{10}{5}$$
x = 2
Hence, x = 2, the quadrilateral MNPQ is a parallelogram.

Question 7.
Show that quadrilateral JKLM is a parallelogram.

The quadrilateral JKLM is a parallelogram according to the “Opposite sides parallel and congruent Theorem”.

Explanation:
From the above given coordinate plane,
The coordinates of the quadrilateral JKLM are:
J (-5, 3), K (-3, -1), L (2, -3), and M (2, -3)
For the quadrilateral JKLM to be a parallelogram,
The opposite sides of the quadrilateral JKLM must be equal.
So, JL = KM
We know that the distance between the 2 points = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
JL = $$\sqrt{(5 + 2)² + (3 + 3)²}$$
= $$\sqrt{(7)² + (6)²}$$
= $$\sqrt{49 + 36}$$
= $$\sqrt{85}$$
= 9.21
KM = $$\sqrt{(5 + 2)² + (3 + 3)²}$$
= $$\sqrt{(7)² + (6)²}$$
= $$\sqrt{49 + 36}$$
= $$\sqrt{85}$$
= 9.21
Hence, the quadrilateral JKLM is a parallelogram according to the “Opposite sides parallel and congruent Theorem”.

Question 8.
Refer to the Concept Summary. Explain two other methods you can use to show that quadrilateral ABCD in Example 5 is a parallelogram.
Concept Summary
Ways to Prove a Quadrilateral is a Parallelogram

The other ways to prove a quadrilateral parallelogram are:
By proving that the sum of the consecutive angles is supplementary.
By proving that an angle is supplementary to both the consecutive angles.

### Exercise 7.3 Proving that a Quadrilateral is a Parallelogram

Vocabulary and Core Concept Check

Question 1.
WRITING

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.

 Construct a quadrilateral with opposite sides congruent. Construct a quadrilateral with one pair of parallel sides. Construct a quadrilateral with opposite angles congruent, Construct a quadrilateral with one pair of opposite sides congruent and parallel.

Statements a., b., and d are needed for constructing a parallelogram,
whereas statement c. is unnecessary.

Explanation:
The given statements are as follows,
a. Construct a quadrilateral with opposite sides congruent.
b. Construct a quadrilateral with opposite angles congruent.
c. Construct a quadrilateral with one pair of parallel sides.
d. Construct a quadrilateral with one pair of opposite sides congruent and parallel.
For constructing a parallelogram we need,
2 pairs of opposite sides congruent and parallel lines.
2 pairs of opposite angles congruent and parallel lines.
So, statements a., b., and d are needed for constructing a parallelogram.

Monitoring Progress and Modeling with Mathematics

In Exercises 3-8, state which theorem you can use to show that the quadrilateral is a parallelogram.

Question 3.

Explanation:
According to the parallelogram opposite angles converse theorem,
the opposite sides of a parallelogram are congruent.
If each of the diagonals of a quadrilateral divides the quadrilateral into two congruent triangles,
then the quadrilateral is a parallelogram.
If the opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram.

Question 4.

Yes, the given quadrilateral is a parallelogram.

Explanation:
From the given figure is:

We observe that the opposite sides are congruent and parallel to each other.
According to the “Parallelogram Opposite sides Congruent and Parallel Theorem”,
We can conclude that the given quadrilateral is a parallelogram.

Question 5.

Explanation:
According to the parallelogram diagonals converse theorem the diagonals of a parallelogram bisect each other. Given ABCD, let the diagonals AC and BD intersect at E,
we must prove that AE ∼ = CE and BE ∼ = DE.
So, if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

Question 6.

Yes, the given quadrilateral is a parallelogram.

Explanation:
From the given figure is:

We observe that the opposite sides are congruent and parallel to each other.
According to the “Parallelogram Opposite sides Congruent and Parallel Theorem”,
we conclude that the given quadrilateral is a parallelogram.

Question 7.

Explanation:
From the above given figure,
if both pairs of opposite sides of a quadrilateral are congruent,
then the figure is a parallelogram.

Question 8.

Yes, the given quadrilateral is a parallelogram.

Explanation:
From the given figure below,

We observe that the diagonals of the quadrilateral bisect each other,
according to the “Parallelogram Diagonals Converse Theorem”.
Hence, the given quadrilateral is a parallelogram.

In Exercises 9-12, find the values of x and y that make the quadrilateral a parallelogram.

Question 9.

Explanation:
From the above given figure,
we observe that the opposite angles of a quadrilateral are congruent.
According to the parallelogram opposite angles converse theorem the figure is said to be a parallelogram.

Question 10.

Yes, the given quadrilateral is a parallelogram.
The values of x = 16 and y = 9

Explanation:
From the given figure below,

we observe that the lengths of the opposite sides of the given quadrilateral are equal.
So, the given quadrilateral is a parallelogram.
Hence, the values of x and y are:
x = 16 and y = 9

Question 11.

Yes, the given quadrilateral is a parallelogram.
The values of x = 3 and y = 4

Explanation:

Question 12.

Yes, the given quadrilateral is a parallelogram.
The values of x = 25° and y = 15°

Explanation:
From the figure given below, quadrilateral is a parallelogram.

According to the “Opposite angles Theorem”,
The opposite angles of the parallelogram are equal
(4x + 13)° = (5x – 12)°
4x° – 5x° = -12° – 13°
x° = -25°
x° = 25°
(3x – 8)° = (4y + 7)°
3x – 8 – 7 = 4y°
4y° = 3 (25°)  15
4y° = 75 – 15
4y° = 60°
y° = $$\frac{60}{4}$$
y° = 15°
So, the values of x and y are:
x = 25° and y = 15°

In Exercises 13-16, find the value of x that makes the quadrilateral a parallelogram.

Question 13.

Yes, the given quadrilateral is a parallelogram.
The value of x = 8

Explanation:
From the above figure,
We observe that the diagonals of the quadrilateral bisect each other,

Question 14.

Yes, the given quadrilateral is a parallelogram.
The value of x = 4

Explanation:
The given figure quadrilateral is a parallelogram.

According to the “Parallelogram Opposite angles parallel and congruent Theorem”,
the lengths of the opposite sides of the parallelogram are equal.
2x + 3 = x + 7
2x – x = 7 – 3
x = 4

Question 15.

Yes, the given quadrilateral is a parallelogram.
The value of x = 7

Explanation:
The given figure quadrilateral is a parallelogram.

Question 16.

Yes, the given quadrilateral is a parallelogram.
The value of x = $$\frac{2}{3}$$

Explanation:
The given figure quadrilateral is a parallelogram.

According to the “Parallelogram Diagonals Converse Theorem”,
the diagonals bisect each other.
So, 6x = 3x + 2
6x – 3x = 2
3x = 2
x = $$\frac{2}{3}$$

In Exercises 17-20, graph the quadrilateral with the given vertices in a coordinate plane. Then show that the quadrilateral is a parallelogram.

Question 17.
A(0, 1), B(4, 4), C(12, 4), D(8, 1)

Question 18.
E(- 3, 0), F(- 3, 4), G(3, – 1), H(3, – 5)
According to the “Parallelogram opposite sides parallel and congruent Theorem”,
the quadrilateral with the given vertices is a parallelogram.

Explanation:
The given vertices of the quadrilateral are,
E (-3, 0), F (-3, 4), G (3, -1), and H (3, -5)
The representation of the vertices of a quadrilateral in the coordinate plane are,

We know that the length of the opposite sides of a parallelogram are congruent and parallel to each otherr.
The distance between 2 points = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
EF = $$\sqrt{(4 + 0)² + (3 – 3)²}$$
= $$\sqrt{(4)² + (0)²}$$
= $$\sqrt{16 + 0}$$
= 4
GH = $$\sqrt{(5 – 1)² + (3 – 3)²}$$
= $$\sqrt{(4)² + (0)²}$$
= 4
Hence, the quadrilateral with the given vertices is a parallelogram.

Question 19.
J(- 2, 3), K(- 5, 7), L(3, 6), M(6, 2)

Question 20.
N(- 5, 0), P(0, 4), Q(3, 0), R(- 2, – 4)
According to the “Parallelograms Opposite sides parallel and congruent Theorem”,
the quadrilateral with the given vertices is a parallelogram.

Explanation:
The given vertices of the quadrilateral are:
N (-5, 0), P (0, 4), Q (3, 0), R (-2, -4)
The representation of the vertices of the quadrilateral in the coordinate plane are;

The distance between 2 points = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
NP = $$\sqrt{(4 – 0)² + (0 + 5)²}$$
= $$\sqrt{(4)² + (5)²}$$
= $$\sqrt{16 + 25}$$
= 6.40

QR = $$\sqrt{(4 + 0)² + (3 + 2)²}$$
= $$\sqrt{(4)² + (5)²}$$
= $$\sqrt{16 + 25}$$
= 6.40

ERROR ANALYSIS
In Exercises 21 and 22, describe and correct the error in identifying a parallelogram.

Question 21.

Question 22.

Yes, The quadrilateral JKLM is said to be a parallelogram.

Explanation:
From the above given quadrilateral JKLM,
The lengths of the 2 sides are given.
We observe that the given 2 lengths are congruent and are parallel to each other.
So, the quadrilateral JKLM is said to be a parallelogram based on the “Parallelogram Opposite sides parallel and congruent Theorem”.

Question 23.
MATHEMATICAL CONNECTIONS

x = 5

Explanation:

Question 24.
MAKING AN ARGUMENT
Your friend says you can show that quadrilateral WXYZ is a parallelogram by using the Consecutive Interior Angles Converse (Theorem 3.8) and the Opposite Sides Parallel and Congruent Theorem (Theorem 7.9). Is your friend correct? Explain your reasoning.

Explanation:
The given figure is:

We observe that the given angles are the consecutive angles and the sum of the angles are supplementary.
The quadrilateral WXYZ to be a parallelogram, any one of the below condition has to be satisfied:
a. The opposite sides are congruent
b. The sum of the consecutive interior angles are supplementary
So, he is correct according to Converse and the Opposite Sides Parallel and Congruent Theorem.

ANALYZING RELATIONSHIPS
In Exercises 25-27, write the indicated theorems as a biconditional statement.

Question 25.
Parallelogram Opposite Sides Theorem (Theorem 7.3) and Parallelogram Opposite Sides Converse (Theorem 7.7)

Explanation:
If both pairs of opposite sides of a quadrilateral are congruent, then the figure is a parallelogram.
If each of the diagonals of a quadrilateral divides the quadrilateral into two congruent triangles,
then the quadrilateral is a parallelogram.

Question 26.
Parallelogram Opposite Angles Theorem (Theorem 7.4) and Parallelogram Opposite Angles Converse (Theorem 7.8)
A quadrilateral is a parallelogram, if both pairs of opposite angles are congruent.

Explanation:
The given Theorems are:
Parallelograms Opposite angles Theorem and Parallelogram Opposite Angles Converse Theorem.
The two theorems represents a biconditional statement.
As quadrilateral is a parallelogram, because both pairs of opposite angles are congruent.

Question 27.
Parallelogram Diagonals Theorem (Theorem 7.6) and
Parallelogram Diagonals Converse (Theorem 7.10)

Explanation:
According to the Parallelogram Diagonals Theorem and Parallelogram Diagonals Converse Theorem,
each diagonal bisects the parallelogram into two congruent triangles.
So, the Sum of the square of all the sides of a parallelogram is equal to the sum of the square of its diagonals.

Question 28.
CONSTRUCTION
Describe a method that uses the Opposite Sides Parallel and Congruent Theorem (Theorem 7.9) to construct a parallelogram. Then construct a parallelogram using your method.
First show that a pair of sides are congruent and parallel.
Then apply the Opposite Sides Parallel and Congruent Theorem.
Construct a parallelogram, given two sides and an angle.
Draw a parallelogram.
Label the congruent angles.

Explanation:
The steps to construct a parallelogram using the Opposite Sides parallel and Congruent Theorem are:
a. Draw a segment of length x cm and name the segment as AB.
b. Take one endpoint as B and draw another segment of length y cm and name the segment as BC.
c. By using the Opposite sides parallel and congruent Theorem,
The lengths of AB and CD must be equal.
So, AB = CD = x cm
d. By using the Opposite sides parallel and congruent Theorem,
The lengths of BC and DA must be equal.
So, BC = DA = y cm
Hence, the representation of the parallelogram by using the above steps is shown above.

Question 29.
REASONING
Follow the steps below to construct a parallelogram. Explain why this method works. State a theorem to support your answer.
Step 1: Use a ruler to draw two segments that intersect at their midpoints.

Step 2: Connect the endpoints of the segments to form a parallelogram.

Question 30.
MAKING AN ARGUMENT
Your brother says to show that quadrilateral QRST is a parallelogram. you must show that $$\overline{Q R}$$ || $$\overline{T S}$$ and $$\overline{Q T}$$ || $$\overline{R S}$$. Your sister says that you must show that $$\overline{Q R} \cong \overline{T S}$$ and $$\overline{Q T} \cong \overline{R S}$$. Who is correct? Explain your reasoning.

Both are correct.

Explanation:
Given that the quadrilateral QRST is a parallelogram.
According to the Opposite Sides Parallel and Congruent Theorem,
QR ≅TS and QT ≅ RS
QR || TS and QT || RS
You have to show that
QR || TS and QT || RS
You have to show that
QR ≅TS and QT ≅ RS
Hence, the quadrilateral QRST is a parallelogram.

REASONING
In Exercises 31 and 32, our classmate incorrectly claims that the marked information can be used to show that the figure is a parallelogram. Draw a quadrilateral with the same marked properties that are clearly not a parallelogram.

Question 31.

Question 32.

A new quadrilateral with the same marked properties that are clearly not a parallelogram is drawn below,

Question 33.
MODELING WITH MATHEMATICS
You shoot a pool ball, and it rolls back to where it started, as shown in the diagram. The ball bounces off each wall at the same angle at which it hits the wall.

a. The ball hits the first wall at an angle of 63°. So m∠AEF = m∠BEH = 63°. What is m∠AFE? Explain your reasoning.
b. Explain why m∠FGD = 63°.
c. What is m∠GHC? m∠EHB?

Question 34.
MODELING WITH MATHEMATICS
In the diagram of the parking lot shown, m∠JKL = 60°, JK = LM = 21 feet, and KL = JM = 9 feet.

a. Explain how to show that parking space JKLM is a parallelogram.
According to the Opposite sides congruent and parallel Theorem,
the opposite sides of a given figure are congruent and parallel to each other.

Explanation:
Given that,
JK = LM = 21 feet
KL = JM = 9 feet
From the parking lot JKLM,
We observe that the shape of the parking lot is a quadrilateral.
JK and LM are the opposite sides.
JM and KL are the opposite sides.
Given that,
JK = LM and JM = KL
If the opposite sides are congruent and parallel according to the Opposite sides congruent and parallel Theorem,
the parking space JKLM is a parallelogram.

b. Find m∠JML, m∠KJM, and m∠KLM.
∠K = ∠M = 60°
∠J = ∠L = 120°

Explanation:
Given that,
∠JKL = 60°
We know that the parking space JKLM is a parallelogram.
So, the opposite sides and angles of a parallelogram are equal.
∠K = ∠M and ∠J = ∠L
∠K = ∠M = 60°
The sum of the consecutive interior angles is 180°
∠K + ∠L = 180°
∠L = 180° – ∠K
∠L = 180° – 60°
∠L = 120°
So, ∠J = ∠L = 120°
Hence, from the above derivation,
∠K = ∠M = 60°
∠J = ∠L = 120°

c. $$\overline{L M}$$||$$\overline{N O}$$ and $$\overline{N O}$$ || $$\overline{P Q}$$ which theorem could you use to show that $$\overline{J K}$$ || $$\overline{P Q}$$?
According to the Opposite sides parallel and Congruent Theorem,
JK || PQ

Explanation:
Given that,
LM || NO and NO || PQ
So, from the given data JKPQ is a parallelogram.
According to the Opposite sides parallel and Congruent Theorem,
the opposite sides of a parallelogram are equal to each other.
So, JK || PQ

REASONING
In Exercises 35-37. describe how to prove that ABCD is a parallelogram.

Question 35.

Question 36.

Yes, the given quadrilateral ABCD is a parallelogram.

Explanation:
From the above given figure,
We observe that the opposite angles are congruent.
So, ∠B = ∠D
According to the Opposite Angles Converse Theorem,
the given quadrilateral ABCD is a parallelogram.

Question 37.

Question 38.
REASONING
Quadrilateral JKLM is a parallelogram. Describe how to prove that ∆MGJ ≅ ∆KHL.

Question 39.
PROVING A THEOREM
Prove the Parallelogram Opposite Angles Converse (Theorem 7.8). (Hint: Let x° represent m∠A and m∠C. Let y° represent m∠B and m∠D. Write and simplify an equation involving x and y)
Gien ∠A ≅ ∠C, ∠B ≅∠D
Prove ABCD is a parallelogram.

Question 40.
PROVING A THEOREM
Use the diagram of PQRS with the auxiliary line segment drawn to prove the Opposite Sides Parallel and Congruent Theorem (Theorem 7.9).
Given $$\overline{Q R}$$ || $$\overline{P S}$$. $$\overline{Q R} \cong \overline{P S}$$
Prove PQRS is a parallelogram.

Question 41.
PROVING A THEOREM
Prove the Parallelogram Diagonals Converse (Theorem 7.10).
Given Diagonals $$\overline{J L}$$ and $$\overline{K M}$$ bisect each other.
Prove JKLM is a parallelogram.

Question 42.
PROOF
Write the proof.
Given DEBF is a parallelogram.
AE = CF
Prove ABCD is a parallelogram.

Question 43.
REASONING
Three interior angle measures of a quadrilateral are 67°, 67°, and 113°, Is this enough information to conclude that the quadrilateral is a parallelogram? Explain your reasoning.

Question 44.
HOW DO YOU SEE IT?
A music stand can be folded up, as shown. In the diagrams. AEFD and EBCF are parallelograms. Which labeled segments remain parallel as the stand is folded?

AE || FD and AD || EF
EB || CF and EF || BC

Explanation:
The music stand when folded and not folded is given:

So, AEFD and EBCF are the parallelograms.
The segments that remain parallel as the music stand folded are:
In the parallelogram AEFD,
AE || FD and AD || EF
In the parallelogram EBCF,
EB || CF and EF || BC

Question 45.
CRITICAL THINKING
In the diagram, ABCD is a parallelogram, BF = DE = 12, and CF = 8. Find AE. Explain your reasoning.

Question 46.
THOUGHT-PROVOKING
Create a regular hexagon using congruent parallelograms.

Explanation:
A hexagon is a closed two-dimensional polygon with six sides.
Hexagon has 6 vertices and 6 angles also.
Hexa means six and gonia means angles.
From the above figure ABCF and CDEF are the consecutive parallelograms.

Question 47.
WRITING
The Parallelogram Consecutive Angles Theorem (Theorem 7.5) says that if a quadrilateral is a parallelogram, then its consecutive angles are supplementary. Write the converse of this theorem. Then write a plan for proving the converse. Include a diagram.

Question 48.
PROOF
Write the proof.
Given ABCD is a parallelogram.
∠A is a right angle.
Prove ∠B, ∠C, and ∠D are right angles.

Question 49.
ABSTRACT REASONING
The midpoints of the sides of a quadrilateral have been joined to turn what looks like a parallelogram. Show that a quadrilateral formed by connecting the midpoints of the sides of any quadrilateral is always a parallelogram. (Hint: Draw a diagram. Include a diagonal of the larger quadrilateral. Show how two sides of the smaller quadrilateral relate to the diagonal.)

Question 50.
CRITICAL THINKING
Show that if ABCD is a parallelogram with its diagonals intersecting at E, then you can connect the midpoints F, G, H, and J of $$\overline{A E}$$, $$\overline{B E}$$, $$\overline{C E}$$, and $$\overline{D E}$$, respcetively, to form another parallelogram, FGHJ.

Maintaining Mathematical proficiency

Question 51.

Question 52.

Yes, the given quadrilateral is a rectangle.

Explanation:
The given figure is:

We observe that the opposite sides are congruent and parallel and the angles are 90°.
The above figure has opposite sides that are equal and parallel.
It has four angles, equal to 90 degrees.
Hence, a quadrilateral that has congruent and parallel opposite sides and an angle 90° is called a “Rectangle”.
Therefore the given quadrilateral is a rectangle.

Question 53.

Question 54.

Yes, the given quadrilateral is a rhombus.

Explanation:
The given figure is:

We observe that all the sides are congruent but the angle measures are not 90°
We know that,
In a rhombus, opposite sides are parallel and the opposite angles are equal.
All the sides of a rhombus are equal in length, and the diagonals bisect each other at right angles.
But the given quadrilateral has 4 congruent sides but not an angle equal to 90°.
Hence, the given quadrilateral is a rhombus.

### 7.1 – 7.3 Quiz

Find the value of x.

Question 1.

x° = 80°

Explanation:
The given figure is:

We know that, the sum of the angle measures of a polygon = 180° (n – 2)
Where “n” is the number of sides.
The sum of the angle measures of a polygon = 180° (4 – 2)
= 180° (2)
= 360°
So, 115° + 95° + 70° + x° = 360°
280° + x° = 360°
x° = 360° – 280°
x° = 80°

Question 2.

x° = 135°

Explanation:
The given figure is:

We know that the sum of the angle measures of a polygon = 180° (n – 2)
Where “n” is the number of sides.
The sum of the angle measures of a polygon = 180° (5 – 2)
= 180° (3)
= 540°
So, 150° + 120° + 60° + x° + 75° = 540°
405° + x° = 540°
x° = 540° – 405°
x° = 135°

Question 3.

x° = 97°

Explanation:
The given figure is:

We know that the sum of the exterior angles of any polygon is: 360°
So, 60° + 30° + 72° + 46° + 55° + x° = 360°
263° + x° = 360°
x° = 360° – 263°
x° = 97°

Find the measure of each interior angle and each exterior angle of the indicated regular polygon.

Question 4.
decagon
The measure of each interior angle of Decagon is: 144°
The measure of each exterior angle of Decagon is:36°

Explanation:
The given polygon is: Decagon
The number of sides of Decagon is 10.
The measure of each interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$
The measure of each exterior angle of a polygon = $$\frac{360°}{n}$$
Where “n” is the number of sides.
The measure of each interior angle of Decagon = $$\frac{180° (10 – 2)}{10}$$
= $$\frac{180° (8)}{10}$$
= 144°
The measure of each exterior angle of a Decagon = $$\frac{360°}{10}$$
= 36°

Question 5.
15-gon
The measure of each interior angle of 15-gon is: 156°
The measure of each exterior angle of 15-gon is: 24°

Explanation:
The given polygon is: 15-gon
The number of sides of 15-gon is 15.
The measure of each interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$
The measure of each exterior angle of a polygon = $$\frac{360°}{n}$$
Where “n” is the number of sides.
The measure of each interior angle of 15-gon = $$\frac{180° (15 – 2)}{15}$$
= $$\frac{180° (13)}{15}$$
= 156°
The measure of each exterior angle of a 15-gon = $$\frac{360°}{15}$$
= 24°

Question 6.
24-gon
The measure of each interior angle of 24-gon is: 165°
The measure of each exterior angle of 24-gon is: 15°

Explanation:
The given polygon is: 24-gon
The number of sides of 24-gon is 24.
The measure of each interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$
The measure of each exterior angle of a polygon = $$\frac{360°}{n}$$
Where “n” is the number of sides.
The measure of each interior angle of 24-gon = $$\frac{180° (24 – 2)}{24}$$
= $$\frac{180° (22)}{24}$$
= 165°
The measure of each exterior angle of a 24-gon = $$\frac{360°}{24}$$
= 15°

Question 7.
60-gon
The measure of each interior angle of 60-gon is: 174°
The measure of each exterior angle of 60-gon is: 6°

Explanation:
The given polygon is: 60-gon
The number of sides of 60-gon is 60.
The measure of each interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$
The measure of each exterior angle of a polygon = $$\frac{360°}{n}$$
Where “n” is the number of sides.
The measure of each interior angle of 60-gon = $$\frac{180° (60 – 2)}{60}$$
= $$\frac{180° (58)}{60}$$
= 174°
The measure of each exterior angle of a 60-gon = $$\frac{360°}{60}$$
= 6°

Find the indicated measure in ABCD. Explain your reasoning.

Question 8.
CD
The length of the CD is 16.

Explanation:
The given parallelogram is:

We know that the lengths of the opposite sides of the parallelogram are equal.
In parallelogram ABCD,
AB = CD
AB = 16
So, CD = 16

Question 9.
The length of the AD is 7.

Explanation:
The given parallelogram is:

We know that the lengths of the opposite sides of the parallelogram are equal.
In parallelogram ABCD,
BC = 7

Question 10.
AE
The length of the AE is 7.

Explanation:
The given parallelogram is:

We know that the diagonals of the parallelogram bisect each other.
In parallelogram ABCD,
AE = EC
EC = 7
So, AE = 7

Question 11.
BD
The length of the BD is 20.4.

Explanation:
The given parallelogram is:

We know that the diagonals of the parallelogram bisect each other.
In parallelogram ABCD,
AC ⊥ BD
ED = 10.2
BD = BE + ED
BE = ED
BD = 10.2 + 10.2
BD = 20.4

Question 12.
m∠BCD
m∠BCD  = 120°

Explanation:
The given parallelogram is:

We know that the opposite angles of the parallelogram are equal.
In parallelogram ABCD,
∠A = ∠C
In the given figure,
∠A = 120°
So, ∠C = 120°
Hence, the value of m∠BCD is 120°

Question 13.
m∠ABC
m∠ABC = 60°

Explanation:
The given parallelogram is:

We know that the sum of the consecutive angles of the parallelogram are supplementary.
In parallelogram ABCD,
∠A + ∠B = 180°
In the given figure,
∠A = 120°
∠B = 180° – 120°
= 60°
Hence, the value of m∠ABC is 60°

Question 14.

Explanation:
The given parallelogram is:

We know that the opposite angles of the parallelogram are equal.
In parallelogram ABCD,
∠B = ∠D
From Exercise 13,
∠B = 60°
So, ∠D = 60°
Hence, the value of m∠ADC is 60°

Question 15.
m∠DBC
m∠DBC is 60°

Explanation:
The given parallelogram is:

We know that the opposite angles of the parallelogram are equal.
In parallelogram ABCD,
∠B = ∠D
In the given figure,
∠D = 60°
So, ∠B = 60°
Hence, the value of m∠DBC is 60°

State which theorem you can use to show that the quadrilateral is a parallelogram.

Question 16.

Parallelograms Opposite sides congruent and parallel Theorem.

Explanation:

We observe that the length of the opposite sides are equal and parallel to each other.
So, according to the Parallelograms Opposite sides congruent and parallel Theorem,
the given quadrilateral is a parallelogram.

Question 17.

Diagonals Congruent Converse Theorem.

Explanation:

We observe that the diagonals bisect each other.
So, according to the Diagonals Congruent Converse Theorem,
the given quadrilateral is a parallelogram.

Question 18.

Parallelograms Opposite Angles Theorem.

Explanation:

We observe that, the opposite angles of the given quadrilateral are congruent and the angle measures are not 90°.
So, according to the Parallelograms Opposite Angles Theorem,
the given quadrilateral is a parallelogram.

Graph the quadrilateral with the given vertices in a coordinate plane. Then show that the quadrilateral is a parallelogram.

Question 19.
Q(- 5, – 2) R(3, – 2), S(1, – 6), T(- 7, – 6)
The given vertices of a quadrilateral are not parallelogram as shown below.

Explanation:
Given vertices of a quadrilateral are;
Q (-5, -2), R (3, -2), S (1, -6), and T (-7, -6)
From the given vertices of a quadrilateral, QS and RT are the opposite vertices.
We know that, the given quadrilateral to be a parallelogram.
The lengths of the opposite sides must be equal and parallel.
The slope between 2 points = $$\frac{y2 – y1}{x2 – x1}$$
The distance between 2 points = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
QS = $$\sqrt{(6 – 2)² + (1 + 5)²}$$
= $$\sqrt{(4)² + (6)²}$$
= $$\sqrt{16 + 36}$$
= 7.2
RT = $$\sqrt{(6 – 2)² + (3 + 7)²}$$
= $$\sqrt{(4)² + (10)²}$$
= $$\sqrt{16 + 100}$$
= 10.77
Therefore,  we can conclude that the given vertices of a quadrilateral are not parallelogram since the lengths of the opposite sides are not equal.

Question 20.
W(- 3, 7), X(3, 3), Y(1, – 3), Z(- 5, 1)
The given vertices of a quadrilateral are not parallelogram as shown below,

Explanation:
Given vertices of a quadrilateral are:
W (-3, 7), X (3, 3), Y (1, -3), Z (-5, 1)
From the given vertices of a quadrilateral, WY and XZ are the opposite vertices.
If the given quadrilateral to be a parallelogram,
The lengths of the opposite sides must be equal and parallel.
The slope between 2 points = $$\frac{y2 – y1}{x2 – x1}$$
The distance between 2 points = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
WY = $$\sqrt{(1 + 3)² + (7 + 3)²}$$
= $$\sqrt{(4)² + (10)²}$$
= $$\sqrt{16 + 100}$$
= 10.77
XZ = $$\sqrt{(3 – 1)² + (3 + 5)²}$$
= $$\sqrt{(2)² + (8)²}$$
= $$\sqrt{4 + 64}$$
= 8.24
Therefore, the given vertices of a quadrilateral are not parallelogram since the lengths of the opposite
sides are not equal.

Question 21.
A stop sign is a regular polygon. (Section 7.1)

a. Classify the stop sign by its number of sides.
Yes, stop sign is a regular polygon.
Looks like Hexagon.

Explanation:
The given stop sign is:

We observe that the number of sides are 6.
So, the polygon which is six sided is known as Hexagon.

b. Find the measure of each interior angle and each exterior angle of the stop sign.
The measure of each interior angle of a hexagon = 120°
The measure of each exterior angle of a hexagon = 60°

Explanation:
The measure of each interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$
The measure of each exterior angle of any polygon = $$\frac{360°}{n}$$
From part (a),
The stop sign is in the shape of a regular hexagon with 6 sides.
The measure of each interior angle of a hexagon = $$\frac{180° (6 – 2)}{6}$$
= 120°
The measure of each exterior angle of a hexagon = $$\frac{360°}{6}$$
= 60°

Question 22.
In the diagram of the staircase shown, JKLM is a parallelogram, $$\overline{Q T}$$ || $$\overline{R S}$$, QT = RS = 9 feet, QR = 3 feet, and m∠QRS = 123°.

a. List all congruent sides and angles in JKLM. Explain your reasoning.
JK || LM and JM || KL
∠J = ∠L and ∠K = ∠M

Explanation:
We know that,
In a parallelogram the opposite sides and the angles are congruent.
In the parallelogram JKLM, the congruent sides are:
JK || LM and JM || KL
The congruent angles are:
∠J = ∠L and ∠K = ∠M

b. Which theorem could you use to show that QRST is a parallelogram?
Parallelograms Opposite sides congruent and parallel Theorem.

Explanation:
Given that,
QT = RS = 9 feet; QR = 3 feet.
According to the “Parallelograms Opposite sides congruent and parallel Theorem”,
If both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral is a parallelogram.
Hence, QRST is a parallelogram.

c. Find ST, m∠QTS, m∠TQR, and m∠TSR. Explain your reasoning.
ST = 3 feet
m∠QTS = 123°
m∠TQR = 57°
m∠TSR = 57°

Explanation:
Given,
QT = RS = 9 feet; QR = 3 feet; m∠QRS = 123°
From part (b), we know that QRST is a parallelogram.
QR = ST = 3 feet
By using the opposite angles Theorem,
∠T = ∠R and ∠Q = ∠S
It is given that, ∠R = 123°
So, ∠T = 123°
From the parallelogram QRST,
∠Q + ∠R = 180°
∠Q = 180° – 123°
∠Q = 57°
∠S = 57°
Therefore, ST = 3 feet
m∠QTS = 123°
m∠TQR = 57°
m∠TSR = 57°

### 7.4 Properties of Special Parallelograms

Exploration 1

Work with a partner: Use dynamic geometry software.

Sample

a. Draw a circle with center A.

Explanation:
The representation of a circle with center A is shown above.
We know that the center of a circle is the point equidistant from the points on the edge.

b. Draw two diameters of the circle. Label the endpoints B, C, D, and E.

Explanation:
The representation of the two diameters of the circle is shown above,
Diameter is twice the radius of a circle.
The two end points are BC and DE.

Explanation:
The representation of quadrilateral BDCE is shown above,
The quadrilateral which is circumscribed in a circle is called a cyclic quadrilateral.
It means that all the four vertices of quadrilateral lie in the circumference of the circle.
Hence, a quadrilateral is a plane figure that has four sides or edges and four corners or vertices.
The angles are present at the four vertices or corners of the quadrilateral.

d. Is BDCE a parallelogram? rectangle? rhombus? square? Explain your reasoning.
BDCE is a rhombus.

Explanation:
We observe that the opposite sides and the opposite angles are equal to each other.
So, the given quadrilateral BDCE is a rhombus.

e. Repeat parts (a)-(d) for several other circles. Write a conjecture based on your results.
The quadrilaterals formed will be either a rhombus, a parallelogram, or a  square.

Explanation:
From parts (a) – (d) of the circles shown above,
the quadrilaterals formed will be either a rhombus, a parallelogram, or a  square.

Exploration 2

Work with a partner: Use dynamic geometry software.

Sample

a. Construct two segments that are perpendicular bisectors of each other. Label the endpoints A, B, D, and E. Label the intersection C.

Explanation:
The representation of a line segment and its perpendicular bisector is are shown above,
Firstly, draw a line segment with suitable length.
Then, take a compass and draw arcs above and below the line segment.
Repeat the same step with end.

Explanation:
The above quadrilateral is a plane figure with four sides or edges and four corners or vertices.
The angles are present at the four vertices or corners of the quadrilateral.

c. Is AEBD a parallelogram? rectangle? rhombus? square? Explain your reasoning.
AEBD is a rhombus.

Explanation:
The representation of the quadrilateral AEBD is shown above:
The quadrilateral AEBD is a rhombus because all the sides are equal and the angles bisect each other at 90°.

d. Repeat parts (a)-(c) for several other segments. Write a conjecture based on your results.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to make conjectures and build a logical progression of statements to explore the truth of your conjectures.
From parts (a) – (c),
the quadrilaterals formed from the perpendicular bisectors are only “Rhombus”.

Question 3.
What are the properties of the diagonals of rectangles, rhombuses, and squares?
The properties of diagonals of rectangles, rhombuses, and squares are follows as;
The two diagonals of the rhombus are perpendicular to each other.
In rectangles the diagonals are congruent and bisect to each other.
The diagonals in squares bisect each other and the angle is 90°.

Question 4.
Is RSTU a parallelogram? rectangle? rhombus? square? Explain your reasoning.

RSTU is a rhombus.

Explanation:
The given figure is:

We observe that the diagonals bisect each other at 90 degrees and are congruent to each other.
Hence, the given quadrilateral is a rhombus.

Question 5.
What type of quadrilateral has congruent diagonals that bisect each other?
Rectangle.

Explanation:
We know that the quadrilateral has congruent diagonals that bisect each other.
Hence, the quadrilateral is a “Rectangle”.

### Lesson 7.4 Properties of Special Parallelograms

Monitoring Progress

Question 1.
For any square JKLM, is it always or sometimes true that $$\overline{J K}$$ ⊥ $$\overline{K L}$$? Explain your reasoning.
For any square JKLM, is always true.

Explanation:
We know that the diagonals of a square bisect each other at 90°.
So, all the 4 angles of a square are equal i.e., 90°
The angles between any 2 adjacent sides in a square is also 90° i.e., the two adjacent sides are perpendicular to each other.
Hence, for any square JKLM, it is always true that,
$$\overline{j K}$$ ⊥ $$\overline{K L}$$

Question 2.
For any rectangle EFGH, is it always or sometimes true that $$\overline{F G} \cong \overline{G H}$$? Explain your reasoning.
For any rectangle EFGH, is always false.

Explanation:
We know that the opposite sides of a rectangle are congruent to each other.
In the rectangle EFGH,
EF and GH are the parallel sides.
FG and EH are the parallel sides.
EF ≅GH
FG ≅ EH
Hence, for a rectangle EFGH, is always false that $$\overline{F G} \cong \overline{G H}$$.

Question 3.
A quadrilateral has four congruent sides and four congruent angles. Sketch the quadrilateral and classily it.

Explanation:
A quadrilateral that has 4 congruent sides and 4 congruent angles is called a “Square”
So, ABCD is a square as shown above.

Question 4.
In Example 3, what are m∠ADC and m∠BCD?

Explanation:

From the given figure,
We know that, the sum of the angle measures of a triangle is 180°
From ΔBCD,
∠B + ∠C + ∠D = 180°
By using the Vertical Angles Theorem,
∠D = ∠4
By using the Corresponding Angles Theorem,
∠4 = ∠C
∠4 = ∠D = 61°
∠C = 61°
So, ∠ADC = ∠BCD = 61°

Question 5.
Find the measures of the numbered angles in rhombus DEFG.

Question 6.
Suppose you measure only the diagonals of the window opening in Example 4 and they have the same measure. Can you conclude that the opening is a rectangle? Explain.
Yes, the opening window is a rectangle.

Explanation:
The representation of the window opening as mentioned in Example 4 is:

From the given figure,
We observe that the opposite sides of the opening window are equal.
It is given that the diagonals of the opening window have the same measure.
According to the Rectangle Diagonals measure Theorem,
A parallelogram is said to be a rectangle only when the length of the diagonals are congruent.
Hence, the opening window is a rectangle.

Question 7.
WHAT IF?
In Example 5. QS = 4x – 15 and RT = 3x + 8. Find the lengths of the diagonals of QRST.
QS = RT = 77

Explanation:
The given lengths of the diagonals of the rectangle QRST are,
QS = 4x – 15 and RT = 3x + 8
According to the Rectangle Diagonal Measure Theorem,
The diagonals of a rectangle are congruent.
In a rectangle QRST,
QS = RT
So, 4x – 15 = 3x + 8
4x – 3x = 15 + 8
x = 23
QS = 4x – 15
= 4 (23) – 15
= 92 – 15
= 77
RT = 3x + 8
= 3 (23) + 8
= 69 + 8
= 77
Hence, the lengths of the diagonals of the rectangle QRST are,
QS = RT = 77

Question 8.
Decide whether PQRS with vertices P(- 5, 2), Q(0, 4), R(2, – 1), and S(- 3, – 3) is a rectangle, a rhombus, or a square. Give all names that apply.
The parallelogram ABCD is a square since the diagonals are congruent and all the sides are congruent.

Explanation:
The given vertices of the parallelogram PQRS is,
P (-5, 2), Q (0, 4), R (2, -1), and S (-3, -3)
Compare the given points with the co-ordinates (x1, y1), and (x2, y2)
So, the representation of the given vertices of the parallelogram PQRS in the coordinate plane is:

We know that,
The distance between the 2 points = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
PR = $$\sqrt{(2 + 5)² + (2 + 1)²}$$
= $$\sqrt{(7)² + (3)²}$$
= $$\sqrt{49 + 9}$$
= 7.61
QS = $$\sqrt{(0 + 3)² + (4 + 3)²}$$
= $$\sqrt{(7)² + (3)²}$$
= $$\sqrt{49 + 9}$$
= 7.61
So, the diagonals of the parallelogram PQRS are congruent.
So, the parallelogram PQRS must be either a rectangle or a square.
PQ = $$\sqrt{(0 + 5)² + (4 – 2)²}$$
= $$\sqrt{(5)² + (2)²}$$
= $$\sqrt{25 + 4}$$
= 5.38
QR = $$\sqrt{(4 + 1)² + (2 – 0)²}$$
= $$\sqrt{(5)² + (2)²}$$
= $$\sqrt{25 + 4}$$
= 5.38
RS = $$\sqrt{(2 + 3)² + (3 – 1)²}$$
= $$\sqrt{(5)² + (2)²}$$
= $$\sqrt{25 + 4}$$
= 5.38
SP = $$\sqrt{(3 + 2)² + (5 – 3)²}$$
= $$\sqrt{(5)² + (2)²}$$
= $$\sqrt{25 + 4}$$
= 5.38

Exercise 7.4 Properties of Special Parallelograms

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What is another name for an equilateral rectangle?

Question 2.
WRITING
What should you look for in a parallelogram to know if the parallelogram is also a rhombus?
To know if the parallelogram is also a rhombus,
a. Check whether the diagonals are not congruent.
b. Check whether the angle measures will not be equal to 90°.

Monitoring Progress and Modeling with Mathematics

In Exercises 3-8, for any rhombus JKLM, decide whether the statement is always or sometimes true. Draw a diagram and explain your reasoning.

Question 3.
∠L ≅∠M

Question 4.
∠K ≅∠M
∠K is always congruent with ∠M.

Explanation:
We know that, the opposite angles of a rhombus are always congruent.

So, from the given figure ∠K is always congruent with ∠M.

Question 5.
$$\overline{J M} \cong \overline{K L}$$

Question 6.
$$\overline{J K} \cong \overline{K L}$$
$$\overline{J K} \cong \overline{K L}$$ is sometimes true when the rhombus is a square.

Explanation:
We know that, In a rhombus the opposite sides are congruent.

So, JK and KL are the adjacent sides.
Hence, $$\overline{J K} \cong \overline{K L}$$ is sometimes true when the rhombus is a square.

Question 7.
$$\overline{J L} \cong \overline{K M}$$

Question 8.
∠JKM ≅ ∠LKM
∠JKM ≅ ∠LKM is always true.

Explanation:
We know that, the diagonals of a rhombus bisect each other at the right angles i.e., 90°.

Hence, ∠JKM ≅ ∠LKM is always true.

Question 9.

Question 10.

The given figure is in the form of a rectangle.

Explanation:
The given figure is:

We observe that the opposite sides are congruent and all the angles are the right angles in the given figure.
A rectangle has opposite sides that are congruent and all the angles are 90°.
Hence, the given figure is in the form of a rectangle.

Question 11.

Question 12.

The given figure is rhombus.

Explanation:

From the given figure,
We observe that, the adjacent angle measures are equal to 180°.
The opposite sides and angles are congruent.
The diagonals are not congruent.
Hence, the given figure is a rhombus.

In Exercises 13-16. find the measures of the numbered angles in rhombus DEFG.

Question 13.

Question 14.

Question 15.

Question 16.

In Exercises 17-22, for any rectangle WXYZ, decide whether the statement is always or sometimes true. Draw a diagram and explain your reasoning.

Question 17.
∠W ≅ ∠X

Question 18.
$$\overline{W X} \cong \overline{Y Z}$$
$$\overline{W X} \cong \overline{Y Z}$$ is always true

Explanation:
A rectangle has the congruent opposite sides
From the rectangle WXYZ,
The opposite sides are WX, YZ, WZ, and XY as shown in the figure.

Hence, $$\overline{W X} \cong \overline{Y Z}$$ is always true.

Question 19.
$$\overline{W X} \cong \overline{X Y}$$

Question 20.
$$\overline{W Y} \cong \overline{X Z}$$
$$\overline{W Y} \cong \overline{X Z}$$ is always true.

Explanation:
We know that, the length of the diagonals is the same in a rectangle.
In the rectangle WXYZ, the diagonals are:
WY and XZ as shown in the figure.

Hence, $$\overline{W Y} \cong \overline{X Z}$$ is always true.

Question 21.
$$\overline{W Y}$$ ⊥ $$\overline{X Z}$$

Question 22.
∠WXZ ≅∠YXZ
∠WXZ ≅∠YXZ is always true

Explanation:
We know that, the diagonals of a rectangle bisect each other at the right angle i.e., 90°
In the rectangle WXYZ,
∠WXZ ≅∠YXZ bisect each other as shown in the figure.

In Exercises 23 and 24, determine whether the quadrilateral is a rectangle.

Question 23.

Question 24.

The given quadrilateral is not a rectangle.

Explanation:
The given figure is:

We can observe that, the length of the opposite sides are congruent and the angle is right angle.
But we don’t know anything about the other three angles.
Hence, the given quadrilateral is not a rectangle.

In Exercises 25-28, find the lengths of the diagonals of rectangle WXYZ.

Question 25.
WY = 6x – 7
XZ = 3x + 2

Question 26.
WY = 14x + 10
XZ = 11x + 22
The length of the diagonals WY = XZ = 66

Explanation:
We know that, the length of the diagonals are congruent in a rectangle.
WY = XZ are the diagonals.
14x + 10 = 11x + 22
14x – 11x = 22 – 10
3x = 12
x = $$\frac{12}{3}$$
x = 4
WY = 14 (4) + 10
= 56 + 10
= 66
XZ = 11 (4) + 22
= 44 + 22
= 66
Hence, the length of the diagonals are:
WY = XZ = 66

Question 27.
WY = 24x – 8
XZ = – 18x + 13
WY = XZ = 4

Explanation:

Question 28.
WY = 16x – 2
XZ = 36x – 6
WY = XZ = 1

Explanation:
We know that, the length of the diagonals are congruent in a rectangle.
WY = XZ are the diagonals.
16x – 2 = 36x – 6
16x – 36x = -6 + 2
-20x = -4
20x = 4
x = $$\frac{4}{20}$$
x = $$\frac{1}{5}$$
WY = 16 ($$\frac{1}{5}$$) – 2
= 3.2 – 2
= 1
XZ = 36 ($$\frac{1}{5}$$) – 6
= 7 – 6
= 1
Hence, the length of the diagonals are,
WY = XZ = 1

In Exercises 29-34, name each quadrilateral – parallelogram, rectangle, rhombus, or square – for which the statement is always true.

Question 29.
It is equiangular.

Question 30.
It is equiangular and equilateral.
The given quadrilateral that is both equiangular and equilateral is a “Square”.

Question 31.
The diagonals are perpendicular.

Question 32.
The opposite sides are congruent.
Parallelogram, Rectangle, Square, and Rhombus are the quadrilaterals,
where the opposite sides are congruent to each other.

Question 33.
The diagonals bisect each other.

Question 34.
The diagonals bisect opposite angles.
Square, and Rhombus are the quadrilaterals that the diagonals bisect opposite angles.

Question 35.
ERROR ANALYSIS
Quadrilateral PQRS is a rectangle. Describe and correct the error in finding the value of x.

x° = 143°

Explanation:

Question 36.
ERROR ANALYSIS
Quadrilateral PQRS is a rhombus. Describe and correct the error in finding the value of x.

x° = 143°

Explanation:
The sum of the adjacent angles of a rhombus is: 180°
∠Q + ∠R = 180°
37° + x° = 180°
x° = 180° – 37°
x° = 143°

In Exercises 37 – 42, the diagonals of rhombus ABCD intersect at E. Ghen that m∠BAC = 53°, DE = 8, and EC = 6, find the indicated measure.

Question 37.
m∠DAC
m∠DAC = 53°

Explanation:
In a rhombus, diagonals bisect each other at right angles.

Question 38.
m∠AED
m∠AED = 53°

Explanation:
According to the Rhombus Opposite Angles Theorem,
the diagonals bisect each other at right angles.
∠A = ∠E
∠E = 53°
Hence, the value of m∠AED is 53°

Question 39.

Explanation:

Question 40.
DB
DB = 16

Explanation:
We know that, the diagonals of a rhombus are equal
DB = 2 (DE)
It is given that, DE = 8
DB = 2 (8)
DB = 16
Hence, the length of DE is 16.

Question 41.
AE
AE = 6

Explanation:
The diagonals of a parallelogram bisect each other.
Given ABCD, let the diagonals AC and BD intersect at E.
If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

Question 42.
AC
AC =12

Explanation:
We know that, the diagonals of a rhombus are equal.
AC = 2 (AE)
It is given that, AE = 6
AC = 2 x 6
AC = 12
Hence, the length of AC is 12.

In Exercises 43-48. the diagonals of rectangle QRST intersect at P. Given that n∠PTS = 34° and QS = 10, find the indicated measure.

Question 43.
m∠QTR
m∠QTR = 56°

Explanation:
We know that, the diagonals of a rectangle bisect each other at the right angle i.e., 90°

Question 44.
m∠QRT
m∠QRT = 56°

Explanation:
We know that, the opposite angles of a rectangle are equal.
From Exercise 43,
We can observe that, ∠QTR = 56°
According to the Rectangle opposite angles Theorem,
∠QRT = 56°
Hence, ∠QRT = 56°

Question 45.
m∠SRT
m∠SRT = 56°

Explanation:
The sum of the three angles in a triangle is 180°

Question 46.
QP
QP = 5

Explanation:
We know that, the diagonals of a rectangle are congruent and bisect each other
Given that, QS = 10
According to the Diagonals Congruent Theorem,
QP = PS are the diagonals.
QP = $$\frac{10}{2}$$
QP = 5
Hence, the length of QP is 5

Question 47.
RT
RT = 10

Explanation:
The diagonals of a parallelogram bisect each other.
If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

Question 48.
RP
RP = 5

Explanation:
We know that, the diagonals of a rectangle are congruent.
QS = RT = 10
We know that,
The diagonals of a rectangle are congruent and bisect each other.
According to the rectangle diagonals Theorem,
RP = PT
RP = $$\frac{10}{2}$$
RP = 5

In Exercises 49-54. the diagonals of square LMNP intersect at K. Given that LK = 1. find the indicated measure.

Question 49.
m∠MKN
m∠MKN = 90°

Explanation:

Question 50.
m∠LMK
m∠LMK is 90°

Explanation:
We know that, the diagonals of a square are congruent and they are perpendicular to each other.
∠LMK = 90°

Question 51.
m∠LPK
m∠LPK = 45°

Explanation:

Question 52.
KN
KN = 1

Explanation:
We know that, the diagonals of a square are congruent and bisect each other.
According to the Square Diagonals Congruent Theorem,
LN = LK + KN
LK = KN
KN = 1
Hence, the length of KN is 1.

Question 53.
LN
LN = 2
Explanation:

Question 54.
MP
MP = 2

Explanation:
We know that, the diagonals of a square are congruent and are perpendicular to each other.
According to the Square Diagonals Congruent Theorem,
LN = MP
MP = 2
Hence, the length of MP is 2.

In Exercises 55-69. decide whetherJKLM is a rectangle, a rhombus. or a square. Give all names that apply. Explain your reasoning.

Question 55.
J(- 4, 2), K(0, 3), L(1, – 1), M(- 3, – 2)
The parallelogram JKLM is a rhombus, rectangle and square.

Explanation:
The given vertices are :
J(- 4, 2), K(0, 3), L(1, – 1), M(- 3, – 2)

Question 56.
J(- 2, 7), K(7, 2), L(- 2, – 3), M(- 11, 2)
The parallelogram JKLM is a rhombus.

Explanation:
The given vertices are :
J (-2, 7), K (7, 2), L (-2, -3), M (-11, 2)
The diagonals of the parallelogram JKLM are: JL and KM
JL = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(2 – 2)² + (7 + 3)²}$$
= $$\sqrt{(0)² + (10)²}$$
= 10
KM = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(2 – 2)² + (7 + 11)²}$$
= $$\sqrt{(0)² + (18)²}$$
= 18
JK = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(7 – 2)² + (7 + 2)²}$$
= $$\sqrt{(5)² + (9)²}$$
= 10.29
KL = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(2 + 3)² + (7 + 2)²}$$
= $$\sqrt{(5)² + (9)²}$$
= 10.29
LM = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(11 – 2)² + (2 + 3)²}$$
= $$\sqrt{(9)² + (5)²}$$
= 10.29
MJ = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(11 – 2)² + (7 – 2)²}$$
= $$\sqrt{(9)² + (5)²}$$
= 10.29
JL = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{-3 – 7}{2 – 2}$$
= -10
KM = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{2 – 2}{-11 – 7}$$
= -18
Hence, the parallelogram JKLM is a rhombus.

Question 57.
J(3, 1), K(3, – 3), L(- 2, – 3), M(- 2, 1)
The parallelogram JKLM is a rectangle.

Explanation:
The given vertices are :
J(3, 1), K(3, – 3), L(- 2, – 3), M(- 2, 1)

Question 58.
J(- 1, 4), K(- 3, 2), L(2, – 3), M(4, – 1)
The parallelogram JKLM is a rectangle.

Explanation:
The given vertices are :
J (-1, 4), K (-3, 2), L (2, -3), M (4, -1)
The diagonals of the parallelogram JKLM are: JL and KM
JL = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(2 + 1)² + (4 + 3)²}$$
= $$\sqrt{(3)² + (7)²}$$
= 7.61
KM = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(4 + 3)² + (2 + 1)²}$$
= $$\sqrt{(7)² + (3)²}$$
= 7.61
JK = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(3 – 1)² + (4 – 2)²}$$
= $$\sqrt{(2)² + (2)²}$$
= 2.82
KL = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(2 + 3)² + (2 + 3)²}$$
= $$\sqrt{(5)² + (5)²}$$
= 7.07
LM = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(4 – 2)² + (1 – 3)²}$$
= $$\sqrt{(2)² + (2)²}$$
= 2.82
MJ = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(4 + 1)² + (4 + 1)²}$$
= $$\sqrt{(5)² + (5)²}$$
= 7.07
JL = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{-3 – 4}{2 + 1}$$
= –$$\frac{7}{3}$$
KM = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{-1 – 2}{4 + 3}$$
= –$$\frac{3}{7}$$
Hence, the parallelogram JKLM is a rectangle.

Question 59.
J(5, 2), K(1, 9), L(- 3, 2), M(1, – 5)
The parallelogram JKLM is a rhombus.

Explanation:
The given vertices are:
J(5, 2), K(1, 9), L(- 3, 2), M(1, – 5)

Question 60.
J(5, 2), K(2, 5), L(- 1, 2), M(2, – 1)
The parallelogram JKLM is a square.

Explanation:
The given vertices are :
J (5, 2), K (2, 5), L (-1, 2), M (2, -1)
The diagonals of the parallelogram JKLM are: JL and KM
JL = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(2 – 2)² + (5 + 1)²}$$
= $$\sqrt{(0)² + (6)²}$$
= 6
KM = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(2 – 2)² + (5 + 1)²}$$
= $$\sqrt{(0)² + (6)²}$$
= 6
JK = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(5 – 2)² + (5 – 2)²}$$
= $$\sqrt{(3)² + (3)²}$$
= 4.24
KL = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(2 + 1)² + (5 – 2)²}$$
= $$\sqrt{(3)² + (3)²}$$
= 4.24
LM = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(-1 – 2)² + (2 + 1)²}$$
= $$\sqrt{(3)² + (3)²}$$
= 4.24
MJ = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
= $$\sqrt{(5 – 2)² + (2 + 1)²}$$
= $$\sqrt{(3)² + (3)²}$$
= 4.24
KM = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{-1 – 5}{2 – 2}$$
= Undefined
JL = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{2 – 2}{-1 – 5}$$
= 0
So, JL ⊥ KM
Hence, the parallelogram JKLM is a square.

MATHEMATICAL CONNECTIONS
In Exercises 61 and 62, classify the quadrilateral. Explain your reasoning. Then find the values of x and y.

Question 61.

x = 76 and y = 4

Explanation:

Question 62.

The given figure is square.
x = 9 and y = 2

Explanation:

From the given figure,
We observe that all the angles are 90° and the diagonals are perpendicular.
Hence the given figure is a Square.
We know that, the opposite angles are congruent in a square.
5x° = (3x + 18)°
5x° – 3x° = 18°
2x° = 18°
x° = $$\frac{18}{2}$$
x° = 9°
2y° = 10
y° = $$\frac{10}{2}$$
y° = 5°
So,  the given figure is a Square.
The values of x and y are: 9 and 2 respectively.

Question 63.
DRAWING CONCLUSIONS
In the window, $$\overline{B D}$$ ≅ $$\overline{D F}$$ ≅ $$\overline{B H}$$ ≅ $$\overline{H F}$$. Also, ∠HAB, ∠BCD, ∠DEF, and ∠FGH are right angles.

a. Classify HBDF and ACEG. Explain your reasoning.
HBDF is a rhombus.
ACEG is a square.

Explanation:

b. What can you conclude about the lengths of the diagonals $$\overline{A E}$$ and $$\overline{G C}$$? Given that these diagonals intersect at J, what can you conclude about the lengths of $$\overline{A J}$$, $$\overline{J E}$$, $$\overline{C J}$$ and $$\overline{J G}$$? Explain.
AJ = JE and CJ = JG

Explanation:
From part (a), we can observe that ACEG is a rectangle.
We know that, the diagonals of a rectangle are congruent and bisect each other.
AE and GC are the diagonals in the rectangle ACEG
So, AE = GC
Therefore, AJ = JE and CJ = JG

Question 64.
ABSTRACT REASONING
Order the terms in a diagram so that each term builds off the previous term(s). Explain why each figure is in the location you chose.

Quadrilateral, Parallelogram, Rectangle, Square and Rhombus.

Explanation:
The figure that has 4 sides is called a “Quadrilateral”
Ex:
Parallelogram, Square etc.,
The order of the terms in a diagram follows as,
so that each term builds off the previous term is:
a. Quadrilateral – No equal sides
b. Parallelogram – The parallel sides are equal and the angles are not 90°
c. Rectangle – The parallel sides are equal and all the angles are 90°
d. Square – All the sides are equal and all the angles are 90°
e. Rhombus – All the sides are equal but only one angle is 90°

CRITICAL THINKING
In Exercises 65-70, complete each statement with always, sometimes, or never. Explain your reasoning.
Question 65.
A square is ____________ a rhombus.
always a rhombus.

Explanation:

Question 66.
A rectangle is __________ a square.
sometimes a square.

Explanation:
A rectangle is sometimes a square,
because a rectangle has the congruent opposite sides whereas a square has all the congruent sides.

Question 67.
A rectangle _____________ has congruent diagonals.
always congruent diagonals.

Explanation:

Question 68.
The diagonals of a square _____________ bisect its angles.
always bisect its angles.

Explanation:
The diagonals of a square always bisect its angles,
according to the Square Diagonals Congruent Theorem.

Question 69.
A rhombus __________ has four congruent angles.
some has 4 congruent angles.

Explanation:

Question 70.
A rectangle ____________ has perpendicular diagonals.
some times perpendicular diagonals.

Explanation:
A rectangle sometimes has perpendicular diagonals,
because the diagonals of a rectangle bisect each other but not perpendicular to each other.
Whereas a square has the perpendicular diagonals.

Question 71.
USING TOOLS
You want to mark off a square region for a garden at school. You use a tape measure to mark off a quadrilateral on the ground. Each side of the quadrilateral is 2.5 meters long. Explain how you can use the tape measure to make sure that the quadrilateral is a square.

Question 72.
PROVING A THEOREM
Use the plan for proof below to write a paragraph proof for one part of the Rhombus Diagonals Theorem (Theorem 7. 11).

Parallelogram ABCD is a rhombus.

Explanation:
Given ABCD is a Parallelogram.
$$\overline{A C}$$ ⊥ $$\overline{B D}$$
Prove: ABCD is a rhombus.
Because ABCD is a parallelogram.
Its diagonals bisect each other at X.
Use $$\overline{A C}$$ ⊥ $$\overline{B D}$$ to show that ∆BXC ≅ ∆DXC. Then show that $$\overline{B C}$$ ≅ $$\overline{D C}$$.
Use the properties of a parallelogram to show that ABCD is a rhombus.

PROVING A THEOREM
In Exercises 73 and 74, write a proof for parts of the Rhombus Opposite Angles Theorem (Theorem 7.12).

Question 73.
Given: PQRS is a parallelogram.
$$\overline{P R}$$ bisects ∠SPQ and ∠QRS.
$$\overline{S Q}$$ bisects ∠PSR and ∠RQP.
Prove: PQRS is a rhombus.

Question 74.
Given: WXYZ is a rhombus
Prove: $$\overline{W Y}$$ bisects ∠ZWX and ∠XYZ.
$$\overline{Z X}$$ bisects ∠WZY and ∠YXW.

Question 75.
ABSTRACT REASONING
Will a diagonal of a square ever divide the square into two equilateral triangles? Explain your reasoning.

Question 76.
ABSTRACT REASONING
Will a diagonal of a rhombus ever divide the rhombus into two equilateral triangles? Explain your reasoning.
It is possible that a diagonal of a rhombus divides the rhombus into two equilateral triangles.

Explanation:
We know that, the diagonals of a rhombus are not congruent.
Sometimes, the interior angles of a rhombus are 120°, 120°, 60°, and 60°
We know that, the interior angles of an equilateral triangle are 60°
Hence, it is possible that a diagonal of a rhombus divides the rhombus into two equilateral triangles.

Question 77.
CRITICAL THINKING

Question 78.
HOW DO YOU SEE IT?
What other information do you need to determine whether the figure is a rectangle?

Rectangle.

Explanation:
From the given figure,
We observe that the opposite sides are congruent and all the interior angles of the given figure are 90°.
A quadrilateral that has the congruent opposite sides and all the interior angles are 90° is called a “Rectangle”
Hence, the given figure is a rectangle.

Question 79.
REASONING
Are all rhombuses similar? Are all squares similar? Explain your reasoning.
No, all rhombuses similar.
Yes, all squares similar.

Explanation:

Question 80.
THOUGHT PROVOKING
Use the Rhombus Diagonals Theorem (Theorem 7. 1I) to explain why every rhombus has at least two lines of symmetry.

PROVING A COROLLARY
In Exercises 81-83, write the corollary as a conditional statement and its converse. Then explain why each statement is true.

Question 81.
Rhombus Corollary (Corollary 7.2)

Question 82.
Rectangle Corollary (Corollary 7.3)
The quadrilateral should be either a rectangle or a square.

Explanation:
Conditional statement:
If a quadrilateral is a rectangle, then it has four right angles, four sides and four corners.
Converse:
If a quadrilateral has four right angles, then it is a rectangle.
The conditional statement is true since a quadrilateral is a rectangle, it has 4 right angles.
The corollary is not right because by having 4 right angles,
the quadrilateral should be either a rectangle or a square.

Question 83.
Square Corollary (Corollary 7.4)

Question 84.
MAKING AN ARGUMENT
Your friend claims a rhombus will never have congruent diagonals because it would have to be a rectangle. Is your friend correct? Explain your reasoning.
Yes, friend is correct.

Explanation:
We know that,
If a rhombus has congruent diagonals, then it would have to be a square only when all the angles will be 90°.
But, it is not possible,
if a rhombus with non-congruent diagonals will never be a rectangle,
because a rhombus won’t have all the angles 90°.

Question 85.
PROOF
Write a proof in the style of your choice.
Gien ∆XYZ ≅ ∆XWZ, ∠XYW ≅ ∠ZWY
Prove WVYZ is a rhombus.

Question 86.
PROOF
Write a proof in the style of your choice.
Given
Prove ABCD is a rectangle.

PROVING A THEOREM
In Exercises 87 and 88. write a proof for part of the Rectangle Diagonals Theorem (Theorem 7.13).

Question 87.
Given PQRS is a rectangle.
Prove $$\overline{P R} \cong \overline{S Q}$$

Question 88.
Given PQRS is a parallelogram.
$$\overline{P R} \cong \overline{S Q}$$
Prove PQRS is a rectangle.

Maintaining Mathematical Proficiency

$$\overline{D E}$$ is a midsegment of ∆ABC. Find the values of x and y.

Question 89.

x = 10 and y = 8

Explanation:
Given, $$\overline{D E}$$ is a midsegment of ∆ABC.

Question 90.

x = 14 and y = 6

Explanation:
The given figure is:

Given that $$\overline {D E}$$ is a midsegment of ΔABC.
From the above figure, AD = DB
y = 6
BC = 2 (DE)
BC = 2 (7)
BC = 14
x = 14
Hence, the values of x and y are 14 and 6 respectively.

Question 91.

x = 9 and y = 26

Explanation:

### 7.5 Properties of Trapezoids and Kites

Exploration 1

Sample

Work with a partner. Use dynamic geometry software.

a. Construct a trapezoid whose base angles are congruent. Explain your process.

c. Repeat parts (a) and (b) for several other trapezoids. Write a conjecture based on your results.
PERSEVERE IN SOLVING PROBLEMS
To be proficient in math, you need to draw diagrams of important features and relationships, and search for regularity or trends.

Exploration 2

Discovering a Property of Kites

Work with a partner. Use dynamic geometry software.

Sample

a. Construct a kite. Explain your process.

b. Measure the angles of the kite. What do you observe?

c. Repeat parts (a) and (b) for several other kites. Write a conjecture based on your results.

Question 3.
What are some properties of trapezoids and kites?

Question 4.
Is the trapezoid at the left isosceles? Explain.

Question 5.
A quadrilateral has angle measures of 7o, 70°, 1100, and 110°, Is the quadrilateral a kite? Explain.

### Lesson 7.5 Properties of Trapezoids and Kites

Monitoring progress

Question 1.
The points A(- 5, 6), B(4, 9) C(4, 4), and D(- 2, 2) form the vertices of a quadrilateral. Show that ABCD is a trapezoid. Then decide whether it is isosceles.

In Exercises 2 and 3, use trapezoid EFGH.

Question 2.
If EG = FH, is trapezoid EFGH isosceles? Explain.

Question 3.
If m∠HEF = 70° and ,m∠FGH = 110°, is trapezoid EFGH isosceles? Explain.

Question 4.
In trapezoid JKLM, ∠J and ∠M are right angles, and JK = 9 centimeters. The length of midsegment $$\overline{N P}$$ of trapezoid JKLW is 12 centimeters. Sketch trapezoid JKLM and its midsegment. Find ML. Explain your reasoning.

Question 5.
Explain another method you can use to find the length of $$\overline{Y Z}$$ in Example 4.

Question 6.
In a kite. the measures of the angles are 3x° 75°, 90°, and 120°. Find the value of x. What are the measures of the angles that are congruent?

Question 7.
Quadrilateral DEFG has at least one pair of opposite sides congruent. What types of quadrilaterals meet this condition?

Question 8.

Question 9.

Question 10.

### Exercise 7.5 Properties of Trapezoids and Kites

Vocabulary and Core Concept Check

Question 1.
WRITING
Describe the differences between a trapezoid and a kite.

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.

Is there enough information to prove that trapezoid ABCD is isosceles?

Is there enough information to prove that $$\overline{A B}$$ ≅ $$\overline{D C}$$?

Is there enough information to prove that the non-parallel sides of trapezoid ABCD are congruent?

Is there enough information to prove that the legs of trapezoid ABCD are congruent?

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, show that the quadrilateral with the given vertices is a trapezoid. Then decide whether it is isosceles.

Question 3.
W(1, 4), X(1, 8), Y(- 3, 9), Z(- 3, 3)

Question 4.
D(- 3, 3), E(- 1, 1), F(1, – 4), G(- 3, 0)

Question 5.
M(- 2, 0), N(0, 4), P(5, 4), Q(8, 0)

Question 6.
H(1, 9), J(4, 2), K(5, 2), L(8, 9)

In Exercises 7 and 8, find the measure of each angle in the isosceles trapezoid.

Question 7.

Question 8.

In Exercises 9 and 10. find the length of the midsegment of the trapezoid.

Question 9.

Question 10.

In Exercises 11 and 12, find AB.

Question 11.

Question 12.

In Exercises 13 and 14, find the length of the midsegment of the trapezoid with the given vertices.

Question 13.
A(2, 0), B(8, – 4), C(12, 2), D(0, 10)

Question 14.
S(- 2, 4), T(- 2, – 4), U(3, – 2), V(13, 10)

In Exercises 15 – 18, Find m ∠ G.

Question 15.

Question 16.

Question 17.

Question 18.

Question 19.
ERROR ANALYSIS
Describe and correct the error in finding DC.

Question 20.
ERROR ANALYSIS
Describe and correct the error in finding m∠A.

In Exercises 21 – 24. given the most specific name for the quadrilateral. Explain your reasoning.

Question 21.

Question 22.

Question 23.

Question 24.

REASONING
In Exercises 25 and 26, tell whether enough information is given in the diagram to classify the quadrilateral by the indicated name. Explain.

Question 25.
rhombus

Question 26.
Square

MATHEMATICAL CONNECTIONS
In Exercises 27 and 28, find the value of x.

Question 27.

Question 28.

Question 29.
MODELING WITH MATHEMATICS
In the diagram, NP = 8 inches, and LR = 20 inches. What is the diameter of the bottom layer of the cake?

Question 30.
PROBLEM SOLVING
You and a friend arc building a kite. You need a stick to place from X to Wand a stick to place from W to Z to finish constructing the frame. You want the kite to have the geometric shape of a kite. How long does each stick need to be? Explain your reasoning.

REASONING
In Exercises 31 – 34, determine which pairs of segments or angles must be congruent so that you can prove that ABCD is the indicated quadrilateral. Explain our reasoning. (There may be more than one right answer.)

Question 31.
isosceles trapezoid

Question 32.
Kite

Question 33.
Parallelogram

Question 34.
square

Question 35.
PROOF
Write a proof
Given $$\overline{J L} \cong \overline{L N}$$, $$\overline{K M}$$ is a midsegment of ∆JLN
Prove Quadrilateral JKMN is an isosceles trapezoid.

Question 36.
PROOF
Write a proof
Given ABCD is a kite.
$$\overline{A B} \cong \overline{C B}$$, $$\overline{A D} \cong \overline{C D}$$
Prove $$\overline{C E} \cong \overline{A E}$$

Question 37.
ABSTRACT REASONING
Point U lies on the perpendicular bisector of $$\overline{R T}$$. Describe the set of points S for which RSTU is a kite.

Question 38.
REASONING
Determine whether the points A(4, 5), B(- 3, 3), C(- 6, – 13), and D(6, – 2) are the vertices of a kite. Explain your reasoning.

PROVING A THEOREM
In Exercises 39 and 40, use the diagram to prove the given theorem. In the diagram, $$\overline{E C}$$’ is drawn parallel to $$\overline{A B}$$.

Question 39.
Isosceles Trapezoid Base Angles Theorem (Theorem 7.14)
Given ABCD is an isosceles trapezoid.
$$\overline{B C}$$ || $$\overline{A D}$$
Prove ∠A ≅ ∠D, ∠B ≅ ∠BCD

Question 40.
Isosceles Trapezoid Base Angles Theorem (Theorem 7.15)
Given ABCD is a trapezoid
∠A ≅ ∠D, $$\overline{B C}$$ || $$\overline{A D}$$
Prove ABCD is an isosceles trapezoid.

Question 41.
MAKING AN ARGUMENT
Your cousin claims there is enough information to prove that JKLW is an isosceles trapezoid. Is your cousin correct? Explain.

Question 42.
MATHEMATICAL CONNECTIONS
The bases of a trapezoid lie on the lines y = 2x + 7 and y = 2x – 5. Write the equation of the line that contains the midsegment of the trapezoid.

Question 43.
CONSTRUCTION
$$\overline{A C}$$ and $$\overline{B D}$$ bisect each other.
a. Construct quadrilateral ABCD so that $$\overline{A C}$$ and $$\overline{B D}$$ are congruent. hut not perpendicular. Classify the quadrilateral. Justify your answer.

b. Construct quadrilateral ABCD so that $$\overline{A C}$$ and $$\overline{B D}$$ are perpendicular. hut not congruent. Classify the quadrilateral. Justify your answer.

Question 44.
PROOF Write a proof.
Given QRST is an isosceles trapezoid.
Prove ∠TQS ≅ ∠SRT

Question 45.
MODELING WITH MATHEMATICS
A plastic spiderweb is made in the shape of a regular dodecagon (12-sided polygon). $$\overline{A B}$$ || $$\overline{P Q}$$, and X is equidistant from the vertices of the dodecagon.

a. Are you given enough information to prove that ABPQ is an isosceles trapezoid?
b. What is the measure of each interior angle of ABPQ

Question 46.
ATTENDING TO PRECISION
In trapezoid PQRS, $$\overline{P Q}$$ || $$\overline{R S}$$ and $$\overline{M N}$$ is the midsegment of PQRS. If RS = 5 . PQ. what is the ratio of MN to RS?
(A) 3 : 5
(B) 5 : 3
(C) 1 : 2
(D) 3 : 1

Question 47.
PROVING A THEOREM
Use the plan for proof below to write a paragraph proof of the Kite Opposite Angles Theorem (Theorem 7.19).
Given EFGH is a Kite.
, 
Prove ∠E ≅ ∠G, ∠F ∠H

Plan for Proof: First show that ∠E ≅ ∠G. Then use an indirect argument to show that ∠F ∠H.

Question 48.
HOW DO YOU SEE IT?
One of the earliest shapes used for cut diamonds is called the table cut, as shown in the figure. Each face of a cut gem is called a facet.

a. $$\overline{B C}$$ || $$\overline{A D}$$, and $$\overline{A B}$$ and $$\overline{D C}$$ are not parallel. What shape is the facet labeled ABCD?
b. $$\overline{D E}$$ || $$\overline{G F}$$, and $$\overline{D G}$$ and $$\overline{E F}$$ are congruent but not parallel. What shape is the facet labeled DEFG?

Question 49.
PROVING A THEOREM
In the diagram below, $$\overline{B G}$$ is the midsegment of ∆ACD. and $$\overline{G E}$$ is the midsegment of ∆ADF Use the diagram to prove the Trapezoid Midsegment Theorem (Theorem 7.17).

Question 50.
THOUGHT PROVOKING

Question 51.
PROVING A THEOREM
To prove the biconditional statement in the Isosceles Trapezoid Diagonals Theorem (Theorem 7.16), you must prove both Parts separately.
a. Prove part of the Isosceles Trapezoid Diagonals Theorem (Theorem 7. 16).
Given JKLM is an isosecles trapezoid.
$$\overline{K L}$$ || $$\overline{J M}$$, $$\overline{J L} \cong \overline{K M}$$
Prove $$\overline{J L} \cong \overline{K M}$$

b. Write the other parts of the Isosceles Trapezoid Diagonals Theorem (Theorem 7. 16) as a conditional. Then prove the statement is true.

Question 52.
PROOF
What special type of quadrilateral is EFGH? Write a proof to show that your answer is Correct.
Given In the three-dimensional figure, $$\overline{J L} \cong \overline{K M}$$. E, F, G, and H arc the midpoints of $$\overline{J L}$$. $$\overline{K l}$$, $$\overline{K M}$$, and $$\overline{J M}$$. respectively.
Prove EFGH is a ____________ .

Maintaining Mathematical Proficiency

Describe a similarity transformation that maps the blue preimage to the green image.

Question 53.

Explanation:
Transformation with scale factor 2.
If the size of a shape is increased or reduced then,
the image of the shape will be similar to the pre-image.
The similar figures have dimensions equal in proportion.

Question 54.

### 7.1 Angles of Polygons

Question 1.
Find the sum of the measures of the interior angles of a regular 30-gon. Then find the measure of each interior angle and each exterior angle.

Find the va1ue of x.

Question 2.

Question 3.

Question 4.

### 7.2 Properties of Parallelograms

Find the value of each variable in the parallelogram.

Question 5.

Question 6.

Question 7.

Question 8.
Find the coordinates of the intersection of the diagonals of QRST with vertices Q(- 8, 1), R(2, 1). S(4, – 3), and T(- 6, – 3).

Question 9.
Three vertices of JKLM are J(1, 4), K(5, 3), and L(6, – 3). Find the coordinates of vertex M.

### 7.3 Proving that a Quadrilateral is a Parallelogram

State which theorem you can use to show that the quadrilateral is a parallelogram.

Question 10.

Question 11.

Question 12.

Question 13.
Find the values of x and y that make the quadrilateral a parallelogram.

Question 14.
Find the value of x that makes the quadrilateral a parallelogram.

Question 15.
Show that quadrilateral WXYZ with vertices W(- 1, 6), X(2, 8), Y(1, 0), and Z(- 2, – 2) is a parallelogram.

### 7.4 Properties of Special Parallelograms

Question 16.

Question 17.

Question 18.

Question 19.
Find the lengths of the diagonals of rectangle WXYZ where WY = – 2y + 34 and XZ = 3x – 26.

Question 20.
Decide whether JKLM with vertices J(5, 8), K(9, 6), L(7, 2), and M(3, 4) is a rectangle. a rhombus, or a square. Give all names that apply. Explain.

### 7.5 Properties of Trapezoids and Kites

Question 21.
Find the measure of each angle in the isosceles trapezoid WXYZ.

Question 22.
Find the length of the midsegment of trapezoid ABCD.

Question 23.
Find the length of the midsegment of trapezoid JKLM with vertices J(6, 10), K(10, 6), L(8, 2), and M(2, 2).

Question 24.
A kite has angle measures of 7x°, 65°, 85°, and 105°. Find the value of x. What are the measures of the angles that are congruent?

Question 25.
Quadrilateral WXYZ is a trapezoid with one pair of congruent base angles. Is WXYZ all isosceles trapezoid? Explain your reasoning.

Question 26.

Question 27.

Question 28.

### Quadrilaterals and Other Polygons Test

Find the value of each variable in the parallelogram.

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.
In a convex octagon. three of the exterior angles each have a measure of x°. The other five exterior angles each have a measure of (2x + 7)°. Find the measure of each exterior angle.

Question 8.
Quadrilateral PQRS has vertices P(5, 1), Q(9, 6), R(5, 11), and 5(1, 6), Classify quadrilateral PQRS using the most specific name.

Determine whether enough information is given to show that the quadrilateral is a parallelogram. Explain your reasoning.

Question 9.

Question 10.

Question 11.

Question 12.
Explain why a parallelogram with one right angle must be a rectangle.

Question 13.
Summarize the ways you can prove that a quadrilateral is a square.

Question 14.
Three vertices of JKLM are J(- 2, – 1), K(0, 2), and L(4, 3),
a. Find the coordinates of vertex M.

b. Find the coordinates of the intersection of the diagonals of JKLM.

Question 15.
You are building a plant stand with three equally-spaced circular shelves. The diagram shows a vertical cross section of the plant stand. What is the diameter of the middle shell?

Question 16.
The Pentagon in Washington. D.C., is shaped like a regular pentagon. Find the measure of each interior angle.

Question 17.
You are designing a binocular mount. If $$\overline{B C}$$ is always vertical, the binoculars will point in the same direction while they are raised and lowered for different viewers. How can you design the mount so $$\overline{B C}$$ is always vertical? Justify your answer.

Question 18.
The measure of one angle of a kite is 90°. The measure of another angle in the kite is 30°. Sketch a kite that matches this description.

### Quadrilaterals and Other Polygons Cummulative Assessment

Question 1.
Copy and complete the flowchart proof of the Parallelogram Opposite Angles Theorem (Thm. 7.4).
Given ABCD is a parallelogram.
Prove ∠A ≅ ∠C, ∠B ≅ ∠D

Question 2.
Use the steps in the construction to explain how you know that the circle is inscribed within ∆ABC.

Question 3.
Your friend claims that he can prove the Parallelogram Opposite Sides Theorem (Thm. 7.3) using the SSS Congruence Theorem (Thm. 5.8) and the Parallelogram Opposite Sides Theorem (Thin. 7.3). Is your friend correct? Explain your reasoning.

Question 4.
Find the perimeter of polygon QRSTUV Is the polygon equilateral? equiangular? regular? Explain your reasoni ng.

Question 5.
Choose the correct symbols to complete the proof of the Converse of the Hinge Theorem (Theorem 6. 13).

Given
Prove m ∠ B > m ∠ E
Step 1 Assume temporarily that m ∠ B m ∠ E. Then it follows that either m∠B____ m∠E or m∠B ______ m ∠ E.

Step 2 If m ∠ B ______ m∠E. then AC _____ DF by the Hinge Theorem (Theorem 6. 12). If, m∠B _______ m ∠ E. then ∠B _____ ∠E. So. ∆ABC ______ ∆DEF by the SAS Congruence Theorem (Theorem 5.5) and AC _______ DF.

Step 3 Both conclusions contradict the given statement that AC _______ DF. So, the temporary assumption that m ∠ B > m ∠ E Cannot be true. This proves that m ∠ B ______ m ∠ E.
>     <    =    ≠    ≅

Question 6.
Use the Isosceles Trapctoid Base Angles Conersc (Thm. 7.15) to prove that ABCD is an isosceles trapezoid.
Given $$\overline{B C}$$ || $$\overline{A D}$$. ∠EBC ≅ ∠¿ECB, ∠ABE ≅ ∠DCE
Prove ABCD is an isoscelcs trapezoid.

Question 7.
One part of the Rectangle Diagonals Theorem (Thm. 7.13) says. “If the diagonals of a parallelogram are congruent, then it is a rectangle.” Using the reasons given. there are multiple ways to prove this part of the theorem. Provide a statement for each reason to form one possible proof of this part of the theorem.
Given QRST is a parallelogram
$$\overline{Q S} \cong \overline{R T}$$
Prove QRST is a rectangle

 Statements Reasons 1. $$\overline{Q S} \cong \overline{R T}$$ 1. Given 2. __________________________ 2. Parallelogram Opposite Sides Theorem (Thm. 7.3) 3. __________________________ 3. SSS Congruence Theorem (Thm. 5.8) 4. __________________________ 4. Corresponding parts of congruent triangles are congruent. 5. __________________________ 5. Parallelogram Consecutive Angles Theorem (Thm. 7.5 6. __________________________ 6. Congruent supplementary angles have the same measure. 7. __________________________ 7. Parallelogram Consecutive Angles Theorem (Thm. 7.5) 8. __________________________ 8. Subtraction Property of Equality 9. __________________________ 9. Definition of a right angle 10. __________________________ 10. Definition of a rectangle

## Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles

Find step-by-step solutions for all the Questions from our Big Ideas Math Geometry Answers Chapter 6. The concepts included in Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles are Perpendicular and Angle Bisectors, Bisectors of Triangles, Midsegment Theorem, Indirect Proof and Inequalities in one triangle, etc. Get Free Access to Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles from here. So, Download Ch 6 Relationships Within Triangles Big Ideas Math Geometry Answers pdf. With the help of BIM Geometry Answers Chapter 6 Relationships Within Triangles both students and teachers can find simple methods to solve the problems.

## Big Ideas Math Book Geometry Answer Key Chapter 6 Relationships Within Triangles

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### Relationships Within Triangles Maintaining Mathematical Proficiency

Write an equation of the line passing through point P that is perpendicular to the given line.

Question 1.
P(3, 1), y = $$\frac{1}{3}$$x – 5
Given equation is y = $$\frac{1}{3}$$x – 5
m = $$\frac { 1 }{ 3 }$$
The slope of the perpendicular line = -3
Substitute the values in y = mx + c
1 = -3(3) + c
1 = -9 + c
c = 1 + 9
c = 10
Substitute the values of m and c in the slope-intercept form of a linear equation.
y = -3x + 10

Question 2.
P(4, – 3), y = – x – 5
Given equation is y = -3x – 5
m = -3
The slope of the perpendicular line = $$\frac { 1 }{ 3 }$$
Substitute the values (4, – 3) and m = $$\frac { 1 }{ 3 }$$ in y = mx + c
-3 = –$$\frac { 1 }{ 3 }$$(4) + c
c = -3 + $$\frac { 4 }{ 3 }$$ = $$\frac { -9 + 4 }{ 3 }$$ = $$\frac { -5 }{ 3 }$$
Substitute the values of m and c in the slope-intercept form of a linear equation.
y = $$\frac { 1 }{ 3 }$$x + $$\frac { -5 }{ 3 }$$
y = $$\frac { 1 }{ 3 }$$x – $$\frac { 5 }{ 3 }$$

Question 3.
P(- 1, – 2), y = – 4x + 13
Given equation is y = – 4x + 13
m = -4
The slope of the perpendicular line = $$\frac { 1 }{ 4 }$$
Substitute the values (- 1, – 2) and m = $$\frac { 1 }{ 4 }$$ in y = mx + c
-2 = $$\frac { 1 }{ 4 }$$(-1) + c
c = -2 + $$\frac { 1 }{ 4 }$$ = $$\frac { -8 + 1 }{ 4 }$$ = $$\frac { -7 }{ 4 }$$
Substitute the values of m and c in the slope-intercept form of a linear equation.
y = $$\frac { 1 }{ 4 }$$x + $$\frac { -7 }{ 4 }$$
y = $$\frac { 1 }{ 4 }$$x – $$\frac { 7 }{ 4 }$$

Write the sentence as an inequality.
Question 4.
A number w is at least – 3 and no more than 8.
Given, A number w is at least – 3 and no more than 8.
We have to write the given sentence as inequality.
w ≥ -3 and w < 8
-3 ≤ w < 8

Question 5.
A number m is more than 0 and less than 11.
Given, A number m is more than 0 and less than 11.
We have to write the given sentence as inequality.
more than means > and less than means <
m > 0 and m < 11
0 < m < 11

Question 6.
A number s is less than or equal to 5 or greater than 2.
Given, A number s is less than or equal to 5 or greater than 2.
We have to write the given sentence as inequality.
less than or equal to means ≤ and greater than means >
s ≤ 5 or s > 2
2 < s ≤ 5

Question 7.
A number d is fewer than 12 or no less than – 7.
Given, A number d is fewer than 12 or no less than – 7.
We have to write the given sentence as inequality.
fewer than means < no less than means >
d < 12 or d > -7
-7 < d < 12

Question 8.
ABSTRACT REASONING
Is it possible for the solution of a compound inequality to be all real numbers? Explain your reasoning.
Yes, it is possible that the solution to an inequality can include all real numbers.

### Relationships Within Triangles Mathematical Practices

Monitoring Progress

Refer to the figures at the top of the page to describe each type of line, ray, or segment in a triangle.
Question 1.
perpendicular bisector
A perpendicular bisector can be defined as a line segment that bisects another line segment at 90 degrees. In other words, a perpendicular bisector intersects another line segment at 90° and divides it into two equal parts.

Question 2.
angle bisector
An angle bisector of an angle is a ray in a triangle that divides the angle into two congruent angles or equal parts.

Question 3.
median
A median is a line segment joining a vertex of the triangle to the midpoint of the opposite side.

Question 4.
altitude
The altitude of a triangle is the perpendicular drawn from the vertex of the triangle to the opposite side. It is also known as the height of the triangle.

Question 5.
midsegment
A midsegment is the line segment connecting the midpoints of two sides of a triangle. Since a triangle has three sides, each triangle has three midsegments. A midsegment is parallel to the third side of the triangle and is half of the length of the third side.

### 6.1 Perpendicular and Angle Bisectors

Exploration 1

Points on a Perpendicular Bisector
Work with a partner. Use dynamic geometry software.
a. Draw any segment and label it $$\overline{A B}$$. Construct the perpendicular bisector of $$\overline{A B}$$.

b. Label a point C that is on the perpendicular bisector of $$\overline{A B}$$ but is not on $$\overline{A B}$$.

c. Draw $$\overline{C A}$$ and $$\overline{C B}$$ and find their lengths. Then move point C to other locations on the perpendicular bisector and note the lengths of $$\overline{C A}$$ and $$\overline{C B}$$.

d. Repeat parts (a) – (c) with other segments. Describe any relationships(s) you notice.

Exploration 2

Points on an Angle Bisector
Work with a partner. Use dynamic geometry software.
a. Draw two rays $$\vec{A}$$B and $$\vec{A}$$C to form ∠BAC. Construct the bisector of ∠BAC.

b. Label a point D on the bisector of ∠BAC.

c. Construct and find the lengths of the perpendicular segments from D to the sides of ∠BAC. Move point D along the angle bisector and note how the lengths change.

d. Repeat parts (a)-(c) with other angles. Describe a relationship(s) you notice.
USING TOOLS STRATEGICALLY
To be proficient in math, you need to visualize the results of varying assumptions, explore consequences, and compare predictions with data.

Question 3.
What conjectures can you make about a point on the perpendicular bisector of a segment and a point on the bisector of an angle?

• At every point on the perpendicular bisector will be equidistant from both ends of the line it is bisecting.
• A point is equidistant from the sides of the angle if the point is on the bisector of an angle.

Question 4.
In Exploration 2. what is the distance from point D to $$\vec{A}$$B when the distance from D to $$\vec{A}$$C is 5 units? Justify your answer.

### Lesson 6.1 Perpendicular and Angle Bisectors

Use the diagram and tile given information to find the indicated measure.

Question 1.
is the perpendicular bisector of $$\overline{W Y}$$, and $$\overline{y Z}$$ = 13.75. Find WZ.
Using the perpendicular bisector theorem YZ = WZ
13.75 = WZ

Question 2.
is the perpendicular bisector of $$\overline{W Y}$$, WZ = 4n – 13, and YZ = n + 17. Find YZ.
Using the perpendicular bisector theorem YZ = WZ
n + 17 = 4n – 13
4n – n = 17 + 13
3n = 30
n = 10
YZ = 10 + 17 = 27

Question 3.
Find WX when WZ = 20.5. WY = 14.8. and YZ = 20.5.
Because WZ = ZY and ZX is the perpendicular bisector of WY by the converse of the perpendicular bisector theorem. By the definition of segment sector, WY = 2WX
14.8 = 2WX
WX = 14.8/2
WX = 7.4

Use the diagram and the given information to find the indicated measure.

Question 4.
$$\vec{B}$$D bisects ∠ABC, and DC = 6.9, Find DA.
By using the angle bisector theorem
DC = DA
DA = 6.9

Question 5.
$$\vec{B}$$D bisects ∠ABC, AD = 3z + 7, and CD = 2z + 11. Find CD.
By using the angle bisector theorem
DC = DA
2z + 11 = 3z + 7
3z – 2z = 11 – 7
z = 4

Question 6.
Find m∠ABC when AD = 3.2, CD = 3.2, and m∠DBC = 39°.
m∠DBC = 39 = m∠DBA
m∠ABC = m∠DBC + m∠DBA
= 39 + 39 = 78°
∴m∠DBC = 78°

Question 7.
Do you have enough information to conclude that $$\vec{Q}$$S bisects ∠PQR? Explain.

Yes, $$\vec{Q}$$S bisects ∠PQR
Because PS = SR.

Question 8.
Write an equation of the perpendicular bisector of the segment with endpoints (- 1, – 5) and (3, – 1).
Midpoint = ($$\frac { -1 + 3 }{ 2 }$$, $$\frac { -5 – 1 }{ 2 }$$) = ($$\frac { 2 }{ 2 }$$, $$\frac { -6 }{ 2 }$$) = (1, -3)
Slope of the line = $$\frac { -1 + 1 }{ 3 + 5 }$$ = 0
The slope of the perpendicular line = 0
Equation of the perpendicular bisector is (y + 5) = 0(x + 1)
y + 5 = 0

### Exercise 6.1 Perpendicular and Angle Bisectors

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
Point C is in the interior of ∠DEF. If ∠DEC and ∠CEF are congruent, then $$\vec{E}$$C is the ________ of ∠DEF.

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.

Is point B the same distance from both X and Z?
Yes

Is point B equidistant from X and Z?
Yes

Is point B collinear with X and Z?
No.

Is point B on the perpendicular bisector of $$\overline{X Z}$$?
Yes

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, find the indicated measure. Explain your reasoning.

Question 3.
GH

Question 4.
QR

Given RS = 1.3
QR = RS
So, QR = 1.3

Question 5.
AB

Question 6.
UW

UV = WU
9x + 1 = 7x + 13
9x – 7x = 13 – 1
2x = 12
x = 6
UW = 7(6) + 13 = 42 + 13 = 55

In Exercises 7-10. tell whether the information in the diagram allows you to conclude that point P lies on the perpendicular bisector of $$\overline{L M}$$. Explain your reasoning.

Question 7.

Question 8.

PN ⊥ LM.
So, LN – MN
Yes, point P lies on the LM.

Question 9.

Question 10.

Yes, you would need to know that PN ⊥ ML.

In Exercises 11-14. find the indicated measure. Explain your reasoning.

Question 11.
m∠ABD

Question 12.

RS = 12
PS = RS (Angle Bisector Theorem)
So, PS = 12

Question 13.
m∠KJL

Question 14.
FG

FG = x + 11
GH = 3x + 1
FG = GH
x + 11 = 3x + 1
3x – x = 11 – 1
2x = 10
x = 5
FG = 5 + 11 = 16

In Exercises 15 and 16, tell whether the information in the diagram allows you to conclude that $$\vec{E}$$H bisects ∠FEG. Explain your reasoning.

Question 15.

Question 16.

Yes, because H is equidistant from EF and EG, EH bisects ∠FEG by the Angle Bisector Theorem.

In Exercises 17 and 18, tell whether the information in the diagram allows you to conclude that DB = DC. Explain your reasoning.

Question 17.

Question 18.

Yes, because AD is perpendicular to BD and DC
DB = DC (perpendicular bisector theorem)

In Exercises 19-22, write an equation of the perpendicular bisector of the segment with the given endpoints.

Question 19.
M(1, 5), N(7, – 1)

Question 20.
Q(- 2, 0), R(6, 12)
Given, Q(- 2, 0), R(6, 12)
Slope of QR = $$\frac { 12 + 2 }{ 6 – 0 }$$ = $$\frac { 7 }{ 3 }$$
The slope of the perpendicular line m = –$$\frac { 3 }{ 7 }$$
Midpoint of QR = ($$\frac { -2 + 6 }{ 2 }$$, $$\frac { 0 + 12 }{ 2 }$$
= (-2, 6)
y = mx + b
6 = –$$\frac { 3 }{ 7 }$$(-2) + b
6 = $$\frac { 6 }{ 7 }$$ + b
b = $$\frac { 36 }{ 7 }$$
y = –$$\frac { 3 }{ 7 }$$x + $$\frac { 36 }{ 7 }$$

Question 21.
U(- 3, 4), V(9, 8)

Question 22.
Y( 10, – 7), Z(- 4, 1)
Given, Y( 10, – 7), Z(- 4, 1)
Slope of YZ = $$\frac { 1 + 7 }{ -4 – 10 }$$ = $$\frac { -4 }{ 7 }$$
The slope of the perpendicular line m = $$\frac { 7 }{ 4 }$$
Midpoint of YZ = ($$\frac { 10 – 4 }{ 2 }$$, $$\frac { -7 + 1 }{ 2 }$$)
= (3, -3)
y = mx + b
-3 = $$\frac { 7 }{ 4 }$$(3) + b
b = -3 – $$\frac { 21 }{ 4 }$$ = $$\frac { -12 – 21 }{ 4 }$$ = $$\frac { -33 }{ 4 }$$
y = $$\frac { 7 }{ 4 }$$x – $$\frac { 33 }{ 4 }$$

ERROR ANALYSIS
In Exercises 23 and 24, describe and correct the error in the student’s reasoning.

Question 23.

Question 24.

CP is not perpendicular to BP.
So, x = 5 is wrong.

Question 25.
MODELING MATHEMATICS
In the photo, the road is perpendicular to the support beam and $$\overline{A B} \cong \overline{C B}$$. Which theorem allows you to conclude that $$\overline{A D} \cong \overline{C D}$$?

Question 26.
MODELING WITH MATHEMATICS
The diagram shows the position of the goalie and the puck during a hockey game. The goalie is at point G. and the puck is at point P.

a. What should be the relationship between $$\vec{P}$$G and ∠APB to give the goalie equal distances to travel on each side of $$\vec{P}$$G?

PG should be the perpendicular bisector of ∠APB.

b. How does m∠APB change as the puck gets closer to the goal? Does this change make it easier or more difficult for the goalie to defend the goal? Explain your reasoning.

Question 27.
CONSTRUCTION
Use a compass and straightedge to construct a copy of $$\overline{X Y}$$. Construct a perpendicular bisector and plot a point Z on the bisector so that the distance between point Z and $$\overline{X Y}$$ is 3 centimeters. Measure $$\overline{X Z}$$ and $$\overline{Y Z}$$. Which theorem does this construction demonstrate?

Question 28.
WRITING
Explain how the Converse of the Perpendicular Bisector Theorem (Theorem 6.2) is related to the construction of a perpendicular bisector.
Converse of the Perpendicular Bisector Theorem:
If a point is equidistant from both the endpoints of the line segments in the same plane, then that point is on the perpendicular bisector of the line segment.

Question 29.
REASONING
What is the value of x in the diagram?

(A) 13
(B) 18
(C) 33
(D) not enough information

Question 30.
REASONING
Which point lies on the perpendicular bisector of the segment with endpoints M(7, 5) and m(- 1, 5)?
(A) (2, 0)
(B) (3, 9)
(C) (4, 1)
(D) (1, 3)
Given, M(7, 5) and m(- 1, 5)
The distance from M to (3, 9) is = √(7 – 3)² + (5 – 9)² = √16 + 16 = √32
The distance from m to (3, 9) = √(-1 – 3)² + (5 – 9)² = √16 + 16 = √32

Question 31.
MAKING AN ARGUMENT
Your friend says it is impossible for an angle bisector of a triangle to be the same line as the perpendicular bisector of the opposite side. Is your friend correct? Explain your reasoning.

Question 32.
PROVING A THEOREM
Prove the Converse of the Perpendicular Bisector Theorem (Thm. 6.2). (Hint: Construct a line through point C perpendicular to $$\overline{A B}$$ at point P.)

Given CA = CB
Prove Point C lies on the perpendicular bisector of $$\overline{A B}$$.
Given CA = CB
Prove Point C lies on the perpendicular bisector of $$\overline{A B}$$.

Compare △ACP, △CPB
AP = BP
CP = CP
∠APC = ∠BPC = 90°
So, △ACP ≅ △CPB (SAS congruence theorem)
So, CA = CB

Question 33.
PROVING A THEOREM
Use a congruence theorem to prove each theorem.
a. Angle Bisector Theorem (Thin. 6.3)
b. Converse of the Angle Bisector Theorem (Thm. 6.4)

Question 34.
HOW DO YOU SEE IT?
The figure shows a map of a city. The city is arranged so each block north to south is the same length and each block east to west is the same length.

a. Which school is approximately equidistant from both hospitals? Explain your reasoning.

b. Is the museum approximately equidistant from Wilson School and Roosevelt School? Explain your reasoning.

Question 35.
MATHEMATICAL CONNECTIONS
Write an equation whose graph consists of all the points in the given quadrants that are equidistant from the x- and y-axes.
a. I and III
b. II and IV
c. I and II

Question 36.
THOUGHT PROVOKING
The postulates and theorems in this book represent Euclidean geometry. In spherical geometry, all points are on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. In spherical geometry, is it possible for two lines to be perpendicular but not bisect each other? Explain your reasoning.

Question 37.
PROOF
Use the information in the diagram to prove that $$\overline{A B} \cong \overline{C B}$$ if and onI if points D, E, and B are collinear.

Question 38.
PROOF
prove the statement in parts (a) – (c)

Given Plane P is a perpendicular bisector of $$\overline{X Z}$$ at point Y.
Prove
a. $$\overline{X W} \cong \overline{Z W}$$
b. $$\overline{X V} \cong \overline{Z V}$$
c. ∠VXW ≅ ∠VZW
a. By using the perpendicular bisector theorem $$\overline{X W} \cong \overline{Z W}$$
b. By using the perpendicular bisector theorem $$\overline{X V} \cong \overline{Z V}$$

Maintaining Mathematical Proficiency

Classify the triangle by its sides.

Question 39.

Question 40.

The triangle is scalene as all the sides have different lengths.

Question 41.

Classify the triangles by its angles.

Question 42.

The triangle is scalene as all angles are different.

Question 43.

Question 44.

The triangle is an obtuse triangle.

### 6.2 Bisectors of Triangles

Exploration 1

Properties of the Perpendicular Bisectors of a Triangle

Work with a partner: Use dynamic geometry software. Draw any ∆ABC.
a. Construct the perpendicular bisectors of all three sides of ∆ABC. Then drag the vertices to change ∆ABC. ‘What do you notice about the perpendicular bisectors?

b. Label a point D at the intersection of the perpendicular bisectors.

c. Draw the circle with center D through vertex A of ∆ABC. Then drag the vertices to change ∆ABC. What do you notice?

Exploration 2

Properties of the Angle Bisectors of a Triangle

Work with a partner. Use dynamic geometry software. Draw any ∆ABC.

a. Construct the angle bisectors of all three angles of ∆ABC, Then drag the vertices to change ∆ABC. What do you notice about the angle bisectors?

b. Label a point D at the intersection of the angle bisectors.

c. Find the distance between D and $$\overline{A B}$$. Draw the circle with center D and this distance as a radius. Then drag the vertices to change ∆ABC. What do you notice?
LOOKING FOR STRUCTURE
To be proficient in math, you need to see complicated things as single objects or as being composed of several objects.

Question 3.
What conjectures can you make about the perpendicular bisectors and the angle bisectors of a triangle?

### Lesson 6.2 Bisectors of Triangles

Monitoring Progress

Question 1.
Three snack carts sell hot pretzels horn points A, B, and E. What is the location of the pretzel distributor if it is equidistant from the three carts? Sketch the triangle and show the location.

Find the coordinates of the circumcenter of the triangle with the given vertices.

Question 2.
R(- 2, 5), S(- 6, 5), T(- 2, – 1)
R(- 2, 5), S(- 6, 5), T(- 2, – 1)
PR = PS = PT
PR² = PS² = PT²
PR² = PS²
(x + 2)² + (y – 5)² = (x + 6)² + (y – 5)²
x² + 4x + 4 + y² – 10x + 25 = x² + 12x + 36 + y² -10y + 25
4x + 4 = 12x + 36
12x – 4x = 36 – 4
8x = 32
x = 4
PS² = PT²
(x + 6)² + (y – 5)² = (x + 2)² + (y + 1)²
x² + 12x + 36 + y² -10y + 25 = x² + 4x + 4 + y² + 2y + 1
12x – 4x – 10y – 2y + 61 = 5
8x – 12y = -56
8(4) – 12y = -56
32 – 12y = -56
-12y = -56 – 32
-12y = -88
y = 7.33
The circumcenter is (4, 7.33)

Question 3.
W(- 1, 4), X(1, 4), Y( 1, – 6)
W(- 1, 4), X(1, 4), Y( 1, – 6)
PW = PX = PY
PW² = PY² = PX²
PX² = PY²
(x – 1)² + (y – 4)² = (x – 1)² + (y + 6)²
x² -2x + 1 + y² – 8y + 16 = x² -2x + 1 + y² + 12y + 36
-8y – 12y = 36 – 16
-20y = 20
y = -1
PW² = PX²
(x + 1)² + (y – 4)² = (x – 1)² + (y – 4)²
x² + 2x + 4 + y² – 8y + 16 = x² – 2x + 1 + y² – 8y + 16
4x = 1
x = 0.25
The circumcenter is (0.25, -1)

Question 4.
In the figure shown, QM = 3x + 8 and QN = 7x + 2. Find QP.

QM = QN
3x + 8 = 7x + 2
7x – 3x = 8 – 2
4x = 6
x = $$\frac { 3 }{ 2 }$$
QP = QN
= 7($$\frac { 3 }{ 2 }$$) + 2
= $$\frac { 23 }{ 2 }$$

Question 5.
Draw a sketch to show the location L of the lamppost in Example 4.

### Exercise 6.2 Bisectors of Triangles

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
When three or more lines, rays, or segments intersect in the same Point. they are called _____________ lines, rays, or segments.
When three or more lines, rays, or segments intersect in the same Point. they are called concurrent lines, rays, or segments.

Question 2.
WHICH ONE DOESNT BELONG?
Which triangle does not belong with the other three? Explain your reasoning.

The third triangle does not belong with the other three. Because the point P in the remaining triangles is the circumcenter. But P is not circumcenter in the third triangle.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, the perpendicular bisectors of ∆ABC intersect at point G and are shown in blue. Find the indicated measure.

Question 3.
Find BG

Question 4.
Find GA

Because G is the circumcenter of ∆ABC, AG = BG = CG
AG = BG = 11
AG = 11

In Exercises 5 and 6, the angle bisectors of ∆XYZ intersect at point P and are shown in red. Find the indicated measure.

Question 5.
Find PB.

Question 6.
Find HP.

Because P is the incenter of ∆XYZ, PH = PF = PK
PK = 15
HP = 15

In Exercises 7-10. find the coordinates of the circumcenter of the triangle with the given vertice

Question 7.
A(2, 6), B(8, 6), C(8, 10)

Question 8.
D(- 7, – 1), E(- 1, – 1), F(- 7, – 9)
D(- 7, – 1), E(- 1, – 1), F(- 7, – 9) be the vertices of the given triangle.
PD = PE = PF
PD² = PE² = PF²
PD² = PE²
(x + 7)² + (y + 1)² = (x + 1)² + (y + 1)²
x² + 14x + 49 + y² + 2y +1 = x² + 2x + 1 + y² + 2y + 1
14x – 2x = 1 – 49
12x = -48
x = -4
PD² = PF²
(x + 7)² + (y + 1)² = (x + 7)² + (y + 9)²
x² + 14x + 49 + y² + 2y + 1 = x² + 14x + 49 + y² + 18y + 81
18y – 2y = 1 – 81
16y = -80
y = -5
The circumcenter is (-4, -5)

Question 9.
H(- 10, 7), J(- 6, 3), K(- 2, 3)

Question 10.
L(3, – 6), M(5, – 3) , N (8, – 6)
L(3, – 6), M(5, – 3), N (8, – 6) be the vertices of the given triangle.
PL = PM = PN
PL² = PM² = PN²
PL² = PN²
(x – 3)² + (y + 6)² = (x – 8)² + (y + 6)²
x² – 6x + 9 + y² + 12y + 36 = x² -16x + 64 + y² + 12y + 36
-16x + 6x = 9 – 64
-10x = -55
x = 5.5
PL² = PM²
(x – 3)² + (y + 6)² = (x – 5)² + (y + 3)²
x² – 6x + 9 + y² + 12y + 36 = x² – 10x + 25 + y² + 6y + 9
-6x + 10x + 45 = 6y – 12y + 34
4x = -6y -11
4(5.5) = -6y – 11
22 + 11 = -6y
33 = -6y
y = -5.5
The circumcenter is (5.5, -5.5)

In Exercises 11-14, N is the incenter of ∆ABC. Use the given information to find the indicated measure.

Question 11.
ND = 6x – 2
NE = 3x + 7
Find NF.

Question 12.
NG = x + 3
NH = 2x – 3
Find NJ.

NG = NH = NJ
x + 3 = 2x – 3
2x – x = 3 + 3
x = 6
By Incenter theorem, NG = NH = NJ
NJ = 6 + 3 = 9
NJ = 9

Question 13.
NK = 2x – 2
NL = – x + 10
Find NM

Question 14.
NQ = 2x
NR = 3x – 2
Find NS.

NQ = NR
2x = 3x – 2
3x – 2x = 2
x = 2
NQ = 2 (2) = 4
By the Incenter theorem NS = NR = NQ
NS = 4

Question 15.
P is the circumcenter of ∆XYZ. Use the given information to find PZ.
PX = 3x + 2
PY = 4x – 8

Question 16.
P is the circumcenter of ∆XYZ. Use the given information to find PY.
PX = 4x + 3
PZ = 6x – 11

PX = PZ
4x + 3 = 6x – 11
6x – 4x = 3 + 11
2x = 14
x = 7
PZ = 6(7) – 11 = 42 – 11 = 31
By the incenter theorem, PX = PZ = PY
PY = 31

CONSTRUCTION
In Exercises 17-20. draw a triangle of the given type. Find the circumcenter. Then construct the circumscribed circle.

Question 17.
right

Question 18.
obtuse

Question 19.
acute isosceles

Question 20.
equilateral

CONSTRUCTION
In Exercises 21-24, copy the triangle with the given angle measures. Find the incenter. Then construct the inscribed circle.

Question 21.

Question 22.

Question 23.

Question 24.

ERROR ANALYSIS
In Exercises 25 and 26. describe and correct the error in identifying equal distances inside the triangle.

Question 25.

Question 26.

The point T is the intersection of the angle bisectors, it is the incenter. Because TV and TF are not necessarily perpendicular to a side of the triangle, there is not sufficient evidence to conclude that TV and TZ are congruent.
Point T is equidistance from the sides of the triangle.

Question 27.
MODELING WITH MATHEMATICS
You and two friends plan to meet to walk your dogs together. You want the meeting place to be the same distance from each person’s house. Explain how you can use the diagram to locate the meeting place.

Question 28.
MODELING WITH MATHEMATICS
You are placing a fountain in a triangular koi pond. YOU want the foutain to be the same distance from each edge of the Pond. Where should you place the fountain? Explain your reasoning. Use a sketch to support your answer.

A, B, C are the edges of the pond.
Fountain to be the same distance from each edge of the Pond
Find the circumcenter of the triangle.

CRITICAL THINKING
In Exercises 29-32, complete the statement with always, sometimes, or never. Explain your reasoning.

Question 29.
The circumenter of a scalene triangle is ______________ inside the triangle.

Question 30.
If the perpendicular bisector of one side of a triangle intersects the opposite vertex. then the triangle is ______________ isosceles.
If the perpendicular bisector of one side of a triangle intersects the opposite vertex. then the triangle is always isosceles.

Question 31.
The perpendicular bisectors of a triangle intersect at a point that is ______________ equidistant from the midpoints of the sides of the triangle.

Question 32.
The angle bisectors of a triangle intersect at a point that is ______________ equidistant from the sides of the triangle.
The angle bisectors of a triangle intersect at a point that is circumcenter equidistant from the sides of the triangle.

CRITICAL THINKING
In Exercises 33 and 34, find the coordinates of the circumcenter of the triangle with the given vertices.

Question 33.
A(2, 5), B(6, 6). C( 12. 3)

Question 34.
D(- 9, – 5), E(- 5, – 9), F(- 2, – 2)
D(- 9, – 5), E(- 5, – 9), F(- 2, – 2) be the vertices of the given triangle.
PD = PE = PF
PD² = PE² = PF²
PD² = PE²
(x + 9)² + (y + 5)² = (x + 5)² + (y + 9)²
x² + 18x + 81 + y² + 10y + 25 = x² + 10x + 25 + y² + 18y + 81
18x – 10x = 18y – 10y
8x = 8y
x = y
PE² = PF²
(x + 5)² + (y + 9)² = (x + 2)² + (y + 2)²
x² + 10x + 25 + y² + 18y + 81 = x² + 4x + 4 + y² + 2y + 4
10x – 4x + 106 = 8 + 2y – 18y
6x + 106 = 8 – 16y
6x + 98 = -16x
22x = -98
x = 4.45
y = 4.45
The circumcenter is (4.45, 4.45)

MATHEMATICAL CONNECTIONS
In Exercises 35 and 36. find the a1ue of x that makes N the incenter of the triangle.

Question 35.

Question 36.

25² = 24² + (14x)²
625 = 576 + 196x²
49 = 196x²
x = $$\frac { 7 }{ 14 }$$ = $$\frac { 1 }{ 2 }$$
The value of x will make N the incenter is $$\frac { 12 }{ 2 }$$

Question 37.
PROOF
Where is the circumcenter located in any right triangle? Write a coordinate proof of this result.

Question 38.
PROVING A THEOREM
Write a proof of the Incenter Theorem (Theorem 6.6).
Given ∆ABC, $$\overline{A D}$$ bisects∠CAB,
$$\overline{B D}$$ bisects ∠CBA, $$\overline{D E}$$ ⊥ $$\overline{A B}$$, $$\overline{D F}$$ ⊥ $$\overline{B C}$$, and $$\overline{D G}$$ ⊥ $$\overline{C A}$$
Prove The angle bisectors intersect at D, which is equidistant from $$\overline{A B}$$, $$\overline{B C}$$, and $$\overline{C A}$$

Question 39.
WRITING
Explain the difference between the circumcenter and the incenter of a triangle.

Question 40.
REASONING
Is the incenter of a triangle ever located outside the triangle? Explain your reasoning.
No; because the incenter is the center of an inscribed circle, it must always be inside the triangle.

Question 41.
MODELING WITH MATHEMATICS
You are installing a circular pool in the triangular courtyard shown. You want to have the largest pool possible on the site without extending into the walkway.

a. Copy the triangle and show how to install the pool so that it just touches each edge. Then explain how you can he sure that you could not fit a larger pool on the site.
b. You want to have the largest pool possible while leaving at least I foot of space around the pool. Would the center of the pool be in the same position as in part (a)? Justify your answer.

Question 42.
MODELING WITH MATHEMATICS
Archaeologists find three stones. They believe that the stones were once pail of a circle of stones with a community fire pit at its center. They mark the locations of stones A, B, and C on a graph. where distances are measured in feet.

a. Explain how archaeologists can use a sketch to estimate the center of the circle of stones.
b. Copy the diagram and find the approximate coordinates of the point at which the archaeologists should look for the fire pit.

To get the center of the circle P, we need to find the circumcenter of the triangle ABC.
A(2, 10), B(13, 6), C(6, 1) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle.
PA = PB = PC
PA² = PB² = PC²
PA² = PB²
(x – 2)² + (y – 10)² = (x – 13)² + (y – 6)²
x² – 4x + 4 + y² – 20y + 100 = x² – 26x + 169 + y² – 12y + 36
-4x + 26x – 20y + 12y = 205 – 104
22x – 8y = 101
PB² = PC²
(x – 13)² + (y – 6)² = (x – 8)² + (y – 1)²
x² – 26x + 169 + y² – 12y + 36 = x² – 16x + 64 + y² – 2y + 1
-26x + 16x – 12y + 2y = 65 – 205
-10x – 10y = -140
x + y = 14
x = 14 – y
22(14 – y) – 8y = 101
308 – 22y – 8y = 101
308 – 101 = 30y
207 = 30y
y = 6.9
Substitute y = 6.9
x = 14 – 6.9
x = 7.1
The circumcenter is (7.1, 6.9)

Question 43.
REASONING
Point P is inside ∆ABC and is equidistant from points A and B. On which of the following segments must P be located?
(A) $$\overline{A B}$$
(B) the perpendicular bisector of $$\overline{A B}$$
(C) $$\overline{A C}$$
(D) the perpendicular bisector of $$\overline{A C}$$

Question 44.
CRITICAL THINKING
A high school is being built for the four towns shown on the map. Each town agrees that the school should be an equal distance from each of the tourist towns. Is there a single point where they could agree to build the school? If so, find it. If not, explain why not. Justify your answer with a diagram.

Question 45.
MAKING AN ARGUMENT
Your friend says that the circumcenter of an equilateral triangle is also the incenter of the triangle. Is your friend correct? Explain in your reasoning.

Question 46.
HOW DO YOU SEE IT?
The arms of the windmill are the angle bisectors of the red triangle. What point of concurrency is the point that Connects the three arms?

The arms of a windmill are equal. So, this means that the incenter is equidistant from the vertices of the red triangle.
A point that is equidistant from the vertices of a triangle is known as the circumcenter of the triangle.
The only case where the incenter and circumcenter coincide is the case of the equilateral triangle.

Question 47.
ABSTRACT REASONING
You are asked to draw a triangle and all its perpendicular bisectors and angle bisectors.
a. For which type of triangle would you need the fewest segments? What is the minimum number of segments you would need? Explain.
b. For which type of triangle would you need the most segments? What is the maximum number of segments you would need? Explain.

Question 48.
THOUGHT PROVOKING
The diagram shows an official hockey rink used by the National Hockey League. Create a triangle using hockey players as vertices in which the center circle is inscribed in the triangle. The center dot should he the incenter of your triangle. Sketch a drawing of the locations of your hockey players. Then label the actual lengths of the sides and the angle measures in your triangle.

AB = 60 ft
BD = 75 ft
∠A = 90°
∠B = 37°
∠D = 53°

COMPARING METHODS
In Exercises 49 and 50. state whether you would use perpendicular bisectors or angle bisectors. Then solve the problem.

Question 49.
You need to cut the largest circle possible from an isosceles triangle made of paper whose sides are 8 inches, 12 inches, and 12 inches. Find the radius of the circle.

Question 50.
On a map of a camp. You need to create a circular walking path that connects the pool at (10, 20), the nature center at (16, 2). and the tennis court at (2, 4). Find the coordinates of the center of the circle and the radius of the circle.

Let the center of the circle be at O (x, y)
Slope of AB = $$\frac { 20 – 4 }{ 10 – 2 }$$ = 2
The slope of XO must be $$\frac { -1}{ 2 }$$ the negative reciprocal of the slope of AB as the 2 lines are perpendicular
Slope of XO = $$\frac { y – 12 }{ x – 6 }$$ = $$\frac { -1 }{ 2 }$$
y – 12 = -0.5x + 3
0.5x + y = 12 + 3 = 15
x + 2y = 30
The slope of BC = $$\frac { 2 – 20 }{ 16 – 10 }$$ = -3
The slope of XO must be $$\frac { 1 }{ 3 }$$ = $$\frac { 11 – y }{ 13 – x }$$
33 – 3y = 13 – x
x – 3y = -33 + 13 = -20
Solve two equations
x + 2y – x + 3y = 30 + 20
y = 10
x – 30 = -20
x = 10
r = √(10 – 2)² + (10 – 4)²
r = 10

Question 51.
CRITICAL THINKING
Point D is the incenter of ∆ABC. Write an expression for the length x in terms of the three side lengths AB, AC, and BC.

Maintaining Mathematical Proficiency

The endpoints of $$\overline{A B}$$ are given. Find the coordinates of the midpoint M. Then find AB.
Question 52.
A(- 3, 5), B(3, 5)
Midpoint of AB = ($$\frac { -3 + 3 }{ 2 }$$, $$\frac { 5 + 5 }{ 2 }$$) = (0, 5)
AB = √(3 + 3)² + (5 – 5)² = 6

Question 53.
A(2, – 1), B(10, 7)

Question 54.
A(- 5, 1), B(4, – 5)
Given two points A(- 5, 1), B(4, – 5)
Midpoint of AB = ($$\frac { -5 + 4 }{ 2 }$$, $$\frac { 1 – 5 }{ 2 }$$) = ($$\frac { -1 }{ 2 }$$, -2)
AB = √(4 + 5)² + (-5 – 1)² = √81 + 36 =  10.81

Question 55.
A(- 7, 5), B(5, 9)

Write an equation of the line passing through point P that is perpendicular to the given line.
Graph the equations of the lines to check that they are perpendicular.
Question 56.
P(2, 8), y = 2x + 1
The slope of the given line m = 2
The slope of the perpendicular line M = $$\frac { -1 }{ 2 }$$
The perpendicular line passes through the given point P(2, 8) is
8 = $$\frac { -1 }{ 2 }$$(2) + b
b = 9
So, y = $$\frac { -1 }{ 2 }$$x + 9

Question 57.
P(6, -3), y = – 5

Question 58.
P(- 8, – 6), 2x + 3y = 18
The given line is 2x + 3y = 18
y = $$\frac { -2 }{ 3 }$$x + 6
The slope of the given line m = $$\frac { -2 }{ 3 }$$
The slope of the perpendicular line is M = $$\frac { 3 }{ 2 }$$
The perpendicular line passing through the point P(-8, -6) is
-6 = $$\frac { 3 }{ 2 }$$(-8) + b
b = 6
The perpendicular equation is y = $$\frac { 3 }{ 2 }$$x + 6

Question 59.
P(- 4, 1), y + 3 = – 4(x + 3)

### 6.3 Medians and Altitudes of Triangles

Exploration 1

Finding Properties of the Medians of a Triangle
Work with a partner. Use dynamic geometry software. Draw any ∆ABC.

a. Plot the midpoint of $$\overline{B C}$$ and label it D, Draw $$\overline{A D}$$, which is a median of ABC. Construct the medians to the other two sides of ∆ABC.

b. What do you notice about the medians? Drag the vertices to change ∆ABC. Use your observations to write a conjecture about the medians of a triangle.

c. In the figure above, point G divides each median into a shorter segment and a longer segment. Find the ratio of the length of each longer segment to the length of the whole median. Is this ratio always the same? Justify your answer.

Exploration 2

Finding Properties of the Altitudes of a Triangle

Work with a partner. Use dynamic geometry software. Draw any ∆ABC.

a. Construct the perpendicular segment from vertex A to $$\overline{B C}$$. Label the endpoint D. $$\overline{A D}$$ is an altitude of ∆ABC.

b. Construct the altitudes to the other two sides of ∆ABC. What do you notice?

c. Write a conjecture about the altitudes of a triangle. Test your conjecture by dragging the vertices to change ∆ABC.
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.

Question 3.
What conjectures can you make about the medians and altitudes of a triangle?

The median of the triangle is a line segment that joins the vertex and the midpoint of the opposite side of the triangle.
f is the median of ∆ABC.

The altitude of the triangle is a line segment that perpendicularly joins the vertex and the opposite side of the triangle.
g is the altitude of ∆ABC.

Question 4.
The length of median $$\overline{R U}$$ in ∆RST is 3 inches. The point of concurrency of the three medians of ∆RST divides $$\overline{R U}$$ into two segments. What are the lengths of these two segments?
Given,
The length of median $$\overline{R U}$$ in ∆RST is 3 inches.
The point of concurrency of the three medians of ∆RST divides $$\overline{R U}$$ into two segments.

SU = UT
RW = WS
RV = VT
Draw a line from U to VT such that it is parallel to OV and name it UX.

UT/SU = TX/VX
SU = UT
UT = TX
TX = VX
Also VT = 2XT
RO/OU = RV/VX
RV = 2XT
RO/OU = 2VX/VX
= 1/2
RU = RO + OU
RU = 2OU + OU = 3OU
So, OU = RU/3
OU = 3/3 = 1 inch.
RO = RU – OU
= 3 – 1 = 2 inches
So, RO = 2 inches and OU = 1 inch

### Lesson 6.3 Medians and Altitudes of Triangles

Monitoring Progress

There are three paths through a triangular park. Each path goes from the midpoint of one edge to the opposite corner. The paths meet at point P.

Question 1.
Find PS and PC when SC = 2100 feet.
Given,
SC = 2100 feet
PC = $$\frac { 2 }{ 3 }$$SC
SC = PS + PC
SC = PS + $$\frac { 2 }{ 3 }$$SC
PS = $$\frac { 1 }{ 3 }$$SC
So, PS = $$\frac { 1 }{ 3 }$$(2100) = 700
PC = $$\frac { 2 }{ 3 }$$(2100) = 1400
PS = 700, PC = 2100

Question 2.
Find TC and BC when BT = 1000 feet.
Given,
BT = 1000 feet
BT = TC
So, TC = 1000 ft
BC = BT + TC
BC = 1000 + 1000 = 2000 ft
BC = 2000 ft, TC = 1000 ft

Question 3.
Find PA and TA when PT = 800 feet.
Given,
PT = 800 feet
PT = $$\frac { 1 }{ 3 }$$PA
PA = 3PT
= 3 • 800 = 2400
TA = $$\frac { 2 }{ 3 }$$PA
TA = $$\frac { 2 }{ 3 }$$(2400)
= 1600
PA = 2400 ft, TA = 1600 ft

Find the coordinates of the centroid of the triangle with the given vertices.

Question 4.
F(2, 5), G(4, 9), H(6, 1)
Given three points F(2, 5), G(4, 9), H(6, 1)
The centroid of the triangle FGH is O = ($$\frac { 2 + 4 + 6 }{ 3 }$$, $$\frac { 5 + 9 + 1 }{ 3 }$$)
= ($$\frac { 12 }{ 3 }$$, $$\frac { 15 }{ 3 }$$)
= (4, 5)
The centroid of the given triangle is (4, 5).

Question 5.
X(- 3, 3), Y(1, 5), Z(- 1, – 2)
Given three points X(- 3, 3), Y(1, 5), Z(- 1, – 2)
The centroid of the triangle XYZ is O = ($$\frac { -3 + 1 – 1 }{ 3 }$$, $$\frac { 3 + 5 – 2 }{ 3 }$$)
= ($$\frac { -3 }{ 3 }$$, $$\frac { 6 }{ 3 }$$)
= (-1, 2)
The centroid of the given triangle is (-1, 2)

Tell whether the orthocenter of the triangle with the given vertices is inside, on, or outside the triangle. Then find the coordinates of the orthocenter.

Question 6.
A(0, 3), B(0, – 2), C(6, -3)
Given,
A(0, 3), B(0, – 2), C(6, -3)
The slope of the line BC = $$\frac { -3 + 2 }{ 6 – 0 }$$ = $$\frac { -1 }{ 6 }$$
The slope of the perpendicular line = 6
The perpendicular line is (y – 3) = 6(x – 0)
y – 3 = 6x
y = 6x + 3
The slope of AC = $$\frac { -3 – 3 }{ 6 – 0 }$$ = $$\frac { -6 }{ 6 }$$ = -1
The slope of the perpendicular line = 1
(y + 2) = 1(x – 0)
y + 2 = x
Substitute y = 6x + 3 in above equation
6x + 3 + 2 = x
5x = -5
x = -1
y + 2 = x
y + 2 = -1
y = -3
The orthocenter is (-1, -3)
The orthocenter lies outside the triangle.

Question 7.
J(- 3, – 4), K(- 3, 4), L(5, 4)
Given,
J(- 3, – 4), K(- 3, 4), L(5, 4)
The slope of JL = $$\frac { 4 + 4 }{ 5 + 3 }$$ = 2
The slope of perpendicular line is $$\frac { -1 }{ 2 }$$
The equation of perpendicular line is (y – 4) = $$\frac { -1 }{ 2 }$$(x + 3)
2y – 8 = -x – 3
x + 2y – 5 = 0
The slope of JK = $$\frac { 4 + 4 }{ -3 + 3 }$$ = 0
The slope of the perpendicular line is 0
y – 4 = 0
y = 4
x + 2(4) – 5 = 0
x + 3 = 0
x = -3
So, the orthocenter is (-3, 4)
It lies on the vertex of the triangle.

Question 8.
WHAT IF?
In Example 4, you want to show that median $$\overline{B D}$$ is also an angle bisector. How would your proof be different?
Given ABC is an isosceles triangle and BD is the median.
AB = BC
BD is the common sides
ΔABD ≅ ΔBDC
∠ABD = ∠CBD
So BD bisects ∠ABC
BD is the angle bisector.
Hence proved

### Exercise 6.3 Medians and Altitudes of Triangles

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Name the four types of points of concurrency. Which lines intersect to form each of the points?

Question 2.
COMPLETE THE SENTENCE
The length of a segment from a vertex to the centroid is ______________ the length of the median from that vertex.
The length of a segment from a vertex to the centroid is one-third of the length of the median from that vertex.

Monitoring progress and Modeling with Mathematics

In Exercises 3-6, point P is the centroid of ∆LMN. Find PN and QP.

Question 3.
QN = 9

Question 4.
QN = 21

PN = $$\frac { 2 }{ 3 }$$QN
PN = $$\frac { 2 }{ 3 }$$(21)
PN = 14
QP = $$\frac { 1 }{ 3 }$$QN
= $$\frac { 1 }{ 3 }$$(21)
= 7

Question 5.
QN = 30

Question 6.
QN = 42

PN = $$\frac { 2 }{ 3 }$$QN
PN = $$\frac { 2 }{ 3 }$$(42)
PN = 28
QP = $$\frac { 1 }{ 3 }$$QN
= $$\frac { 1 }{ 3 }$$(42)
= 14

In Exercises 7-10. point D is the centroid of ∆ ABC. Find CD and CE.

Question 7.
DE = 5

Question 8.
DE = 11

DE = $$\frac { 1 }{ 3 }$$CE
11 = $$\frac { 1 }{ 3 }$$ CE
CE = 33
CD = $$\frac { 2 }{ 3 }$$ CE
CD = $$\frac { 2 }{ 3 }$$(33)
CD = 22

Question 9.
DE = 9

Question 10.
DE = 15

DE = $$\frac { 1 }{ 3 }$$CE
15 = $$\frac { 1 }{ 3 }$$ CE
CE = 45
CD = $$\frac { 2 }{ 3 }$$ CE
CD = $$\frac { 2 }{ 3 }$$(45)
CD = 30

In Exercises 11-14. point G is the centroid of ∆ABC. BG = 6, AF = 12, and AE = 15. Find the length of the segment.

Question 11.
$$\overline{F C}$$

Question 12.
$$\overline{B F}$$

All the medians have the same length.
so, AE = 15 = BF

Question 13.
$$\overline{A G}$$

Question 14.
$$\overline{G E}$$
AE = AG + GE
15 = 10 + GE
GE = 5

In Exercises 15-18. find the coordinates of the centroid of the triangle with the given vertices.

Question 15.
A(2, 3), B(8, 1), C(5, 7)

Question 16.
F(1, 5), G( – 2, 7), H(- 6, 3)
Given,
F(1, 5), G( – 2, 7), H(- 6, 3)
The centroid of the triangle = ($$\frac { 1 – 2 – 6 }{ 3 }$$, $$\frac { 5 + 7 + 3 }{ 3 }$$)
= ($$\frac { -7 }{ 3 }$$, 5)
The centroid is ($$\frac { -7 }{ 3 }$$, 5)

Question 17.
S(5, 5), T(11, – 3), U(- 1, I)

Question 18.
X(1, 4), Y(7, 2), Z(2, 3)
Given,
X(1, 4), Y(7, 2), Z(2, 3)
The centroid of the triangle = ($$\frac { 1 + 7 + 2 }{ 3 }$$, $$\frac { 4 + 2 + 3 }{ 3 }$$)
= ($$\frac { 10 }{ 2 }$$, 3)
The centroid of the triangle is ($$\frac { 10 }{ 2 }$$, 3)

In Exercises 19-22. tell whether the orthocenter is inside, on, or outside the triangle. Then find the coordinates of the orthocenter.

Question 19.
L(0, 5), M(3, 1), N(8, 1)

Question 20.
X(- 3, 2), Y(5, 2), Z(- 3, 6)
Given,
X(- 3, 2), Y(5, 2), Z(- 3, 6)
The slope of YZ = $$\frac { 6 – 2 }{ -3 – 5 }$$ = $$\frac { -1 }{ 2 }$$
The slope of the perpendicular line is 2
The equation of a perpendicular line is (y – 2) = 2(x + 3)
y – 2 = 2x + 6
2x – y + 8 = 0
The slope of XZ = $$\frac { 6 – 2 }{ -3 + 3 }$$ = 0
The equation of the perpendicular line is (y – 2) = 0
y = 2
Substitute y = 2 in 2x – y + 8 = 0
2x – 2 + 8 = 0
2x + 6 = 0
x = -3
the orthocenter is (-3, 2)
The orthocenter lies on the vertex of the triangle.

Question 21.
A(- 4, 0), B(1, 0), C(- 1, 3)

Question 22.
T(-2, 1), U( 2, 1), V(0, 4)
The slope of UV = $$\frac { 4 – 1 }{ 0 – 2 }$$ = $$\frac { -3 }{ 2 }$$
The slope of the perpendicular line is $$\frac { 2 }{ 3 }$$
The equation of the perpendicular line is (y – 1) = $$\frac { 2 }{ 3 }$$(x + 2)
3(y – 1) = 2(x + 2)
3y – 3 = 2x + 2
2x – 3y + 5 = 0
The slope of TV = $$\frac { 4 – 1 }{ 0 + 2 }$$ = $$\frac { 3 }{ 2 }$$
The slope of the perpendicular line is $$\frac { -2 }{ 3 }$$
The equation of the perpendicular line is (y – 1) = $$\frac { -2 }{ 3 }$$(x – 2)
3(y – 1) = -2(x – 2)
3y – 3 = -2x + 4
2x + 3y – 7 = 0
2x – 3y + 5 + 2x + 3y – 7 = 0
4x – 2 = 0
x = 0.5
2x – 1.5 + 5 = 0
x = -1.75
So, the orthocenter is (0, 2.33)
The orthocenter lies inside the triangle ABC.

CONSTRUCTION
In Exercises 23-26, draw the indicated triangle and find its centroid and orthocenter.
Question 23.
isosceles right triangle

Question 24.
obtuse scalene triangle

Question 25.
right scalene triangle

Question 26.
acute isosceles triangle

ERROR ANALYSIS
In Exercises 27 and 28, describe and correct the error in finding DE. Point D is the centroid of ∆ABC.

Question 27.

Question 28.

The length of DE should be 1/3 of the length of AE because it is the shorter segment from the centroid
DE = $$\frac { 1 }{ 3 }$$AE
= $$\frac { 1 }{ 3 }$$(24)
= 8
So, DE = 8

PROOF
In Exercises 29 and 30, write a proof of the statement.

Question 29.
The angle bisector from the vertex angle to the base of an isosceles triangle is also a median.

Question 30.
The altitude from the vertex angle to the base of an isosceles triangle is also a perpendicular bisector.

CRITICAL THINKING
In Exercises 31-36, complete the statement with always, sometimes, or never. Explain your reasoning.

Question 31.
The centroid is _____________ on the triangle.

Question 32.
The orthocenter is _____________ outside the triangle.
The orthocenter is always outside the triangle.

Question 33.
A median is _____________ the same line segment as a perpendicular bisector.

Question 34.
An altitude is ______________ the same line segment as an angle bisector.
Answer: An altitude is sometimes the same line segment as an angle bisector.

Question 35.
The centroid and orthocenter are _____________ the same point.

Question 36.
The centroid is ______________ formed by the intersection of the three medians.
The centroid is a point formed by the intersection of the three medians.

Question 37.
WRITING
Compare an altitude of a triangle with a perpendicular bisector of a triangle.

Question 38.
WRITING
Compare a median. an altitude, and an angle bisector of a triangle.
1. Median is a line segment joining a vertex of a triangle with the midpoint of the opposite side.

2. An angle bisector is a line segment joining a vertex of a triangle with the opposite side such that the angle at the vertex is split into two equal parts.

3. An altitude is also a line segment joining a vertex of a triangle with an opposite side such that the segment is perpendicular to the opposite side.

In general, medians, angle bisectors, and altitudes drawn from the same vertex of a triangle are different line segments. In an equilateral triangle, altitude, median, and angle bisector are drawn from the same vertex. In an isosceles triangle, the altitude drawn to the base is the median and the angle bisector, the median drawn to the base is the altitude and the angle bisector, and the bisector of the angle opposite to the base is the altitude and the median.

Question 39.
MODELING WITH MATHEMATICS
Find the area of the triangular part of the paper airplane wing that is outlined in red. Which special segment of the triangle did you use?

Question 40.
ANALYZING RELATIONSHIPS
Copy and complete the statement for ∆DEF with centroid K and medians
$$\overline{D H}$$, $$\overline{E J}$$, and $$\overline{F G}$$.

a. EJ = ____ KJ

Since KJ is the shorter distance between the two parts of the median, its length is 1/3 of the EJ.

b. DK = ____ KH
DK = 2 KH
Since KH is the shorter distance between the two parts of the median, its length is 1/3 of the DH.

c. FG = ___ KF
FG = $$\frac { 2 }{ 3 }$$KF
Since KF is the longest distance between two parts, its length is $$\frac { 2 }{ 3 }$$ of FG

d. KG = ___ FG
KG = 2FG
Since KG is the shorter distance between the two parts of the median, its length is 1/3 of the FG.

MATHEMATCAL CONNETIONS
In Exercises 41-44, point D is the centroid of ∆ABC. Use the given information to find the value of x.

Question 41.
BD = 4x + 5 and BF = 9x

Question 42.
GD = 2x – 8 and GC = 3x + 3
GD = $$\frac { 1 }{ 3 }$$GC
2x – 8 = $$\frac { 1 }{ 3 }$$(3x + 3)
2x – 8 = x + 1
x = 9

Question 43.
AD = 5x and DE = 3x – 2

Question 44.
DF = 4x – 1 and BD = 6x + 4
DF = 4x – 1 and BD = 6x + 4
BD = 2DF
6x + 4 = 2(4x – 1)
6x + 4 = 8x – 2
2x = 6
x = 3

Question 45.
MATHEMATICAL CONNECTIONS
Graph the lines on the same coordinate plane. Find the centroid of the triangle formed by their intersections.
y1 = 3x – 4
y2 = $$\frac{3}{4}$$x + 5
y2 = – $$\frac{3}{2}$$x – 4

Question 46.
CRITICAL THINKING
In what types of triangles can a vertex be one of the points of concurrency of the triangle? Explain your reasoning.

Question 47.
WRITING EQUATIONS
Use the numbers and symbols to write three different equations for PE.

Question 48.
HOW DO YOU SEE IT?
Use the figure.

a. What type of segment is $$\overline{K M}$$? Which point of concurrency lies on $$\overline{K M}$$?
KM is the median drawn from K to JL.

b. What type of segments is $$\overline{K N}$$? Which point of concurrency lies on $$\overline{K N}$$?
KN is the altitude drawn from K to JL.

c. Compare the areas of ∆JKM and ∆KLM. Do you think the areas of the triangles formed by the median of any triangle will always compare this way? Explain your reasoning.

Question 49.
MAKING AN ARGUMENT
Your friend claims that it is possible for the circumcenter, incenter, centroid, and orthocenter to all be the same point. Do you agree? Explain your reasoning.

Question 50.
DRAWING CONCLUSIONS
The center of gravity of
a triangle, the point where a triangle can balance on the tip of a pencil, is one of the four points of concurrency. Draw and cut out a large scalene triangle on a piece of cardboard. Which of the four points of concurrency is the center of gravity? Explain.
The median is the balancing line of the triangle i.e median divides the triangle into two triangles of equal areas.
The area of ABC = 0.5bh
The median divides the base into two equal parts, each forming a smaller triangle with the same height as the original one.
AD is the median of base BC
So, bD = CD = $$\frac { 1 }{ 2 }$$ BC
The area of ABC = $$\frac { 1 }{ 2 }$$ x BC x h
Area of ABD = $$\frac { 1 }{ 2 }$$ x BD x h = $$\frac { 1 }{ 2 }$$ x ABC area
Area of ACD = $$\frac { 1 }{ 2 }$$ x CD x h = $$\frac { 1 }{ 2 }$$ x ABC area
The centroid is the concurrency point of three medians.
So, the centroid is the gravity center of the triangle.

Question 51.
PROOF
Prose that a median of an equilateral triangle is also an angle bisector, perpendicular bisector, and altitude.

Question 52.
THOUGHT PROVOKING
Construct an acute scalene triangle. Find the orthocenter, centroid, and circumcenter. What can you conclude about the three points of concurrency?
Below is the scalene triangle with the centroid, orthocenter, and circumcenter
Point G is the centroid, H is the orthocenter and J is the circumcenter.

If we draw a line through these points, we can see all the points lie on the same line.
Therefore, we can conclude that the centroid, orthocenter and circumcenter are collinear.
Below mentioned is the construction part.

Question 53.
CONSTRUCTION
Follow the steps to construct a nine-point circle. Why is it called a nine-point circle?
Step 1 Construct a large acute scalene triangle.
Step 2 Find the orthocenter and circumcenter of the triangle.
Step 3 Find the midpoint between the orthocenter and circumcenter.
Step 4 Find the midpoint between each vertex and the orthocenter.
Step 5 Construct a circle. Use the midpoint in Step 3 as the center of the circle, and the distance from the center to the midpoint of a side of the triangle as the radius.

Question 54.
PROOF
Prove the statements in parts (a)-(c).
Given $$\overline{L P}$$ and $$\overline{M Q}$$ are medians of scaling ∆LMN.
Point R is on $$\vec{L}$$P such that $$\overline{L P} \cong \overline{P R}$$. Point S is on $$\vec{M}$$Q such that $$\overline{M Q} \cong \overline{Q S}$$.
Prove
a. $$\overline{N S} \cong \overline{N R}$$
b. $$\overline{N S}$$ and $$\overline{N R}$$ are both parallel to $$\overline{L M}$$.
c. R, N, and S are collinear.

Maintaining Mathematical Proficiency

Determine whether $$\overline{A B}$$ is parallel to $$\overline{C D}$$.

Question 55.
A(5, 6), B (- 1, 3), C(- 4, 9), D(- 16, 3)

Question 56.
A(- 3, 6), B(5, 4), C(- 14, – 10), D(- 2, – 7)
Slope of AB = $$\frac { 4 – 6 }{ 5 + 3 }$$ = $$\frac { -1 }{ 4 }$$
Slope of CD = $$\frac { -7 + 10 }{ -2 + 14 }$$ = $$\frac { 1 }{ 4 }$$
The slopes of AB and CD are not equal. So AB and CD are not parallel.

Question 57.
A (6, – 3), B(5, 2), C(- 4, – 4), D(- 5, 2)

Question 58.
A(- 5, 6), B(- 7, 2), C(7, 1), D(4, – 5)
Slope of AB = $$\frac { 2 – 6 }{ -7 + 5 }$$ = 2
Slope of CD = $$\frac { -5 – 1 }{ 4 – 7 }$$ = 2
The slopes of AB and CD are equal. So, AB is parallel to CD.

### 6.1 and 6.3 Quiz

Find the indicated measure. Explain your reasoning.

Question 1.
UV

SV = VU
2x + 11 = 8x – 1
8x – 2x = 11 + 1
6x = 12
x = 2
UV = 8(2) – 1 = 15

Question 2.
QP

QP = QR
6x = 3x + 9
3x = 9
x = 3
QP = 6(3) = 18

Question 3.
m∠GJK

5x – 4 = 4x + 3
x = 7
∠JGK = 4(7) + 3 = 31
m∠GJK = 180 – (31 + 90) = 180 – 121 = 59
m∠GJK = 59°

Find the coordinates of the circumcenter of the triangle with the given vertices.

Question 4.
A(- 4, 2), B(- 4, – 4), C(0, – 4)
A(- 4, 2), B(- 4, – 4), C(0, – 4) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle.
PA = PB = PC
PA² = PB² = PC²
PA² = PB²
(x + 4)² + (y – 2)² = (x + 4)² + (y + 4)²
x² + 8x + 16 + y² – 4y + 4 = x² + 8x + 16 + y² + 8y + 16
12y = -12
y = -1
PB² = PC²
(x + 4)² + (y + 4)² = (x – 0)² + (y + 4)²
x² + 8x + 16 + y² + 8y + 16 = x² + y² + 8y + 16
8x = -16
x = -2
The circumcenter is (-2, -1)

Question 5.
D(3, 5), E(7, 9), F(11, 5)
D(3, 5), E(7, 9), F(11, 5) be the vertices of the given triangle
P(x,y) be the circumcentre of this triangle
PD = PE = PF
PD² = PE² = PF²
PD² = PE²
(x – 3)² + (y – 5)² = (x – 7)² + (y – 9)²
x² – 6x + 9 + y² – 10y + 25 = x² – 14x + 49 + y² – 18y + 81
-6x + 14x – 10y + 18y = 130 – 34
8x + 8y = 96
x + y = 12 — (i)
PE² = PF²
(x – 7)² + (y – 9)² = (x – 11)² + (y – 5)²
x² – 14x + 49 + y² – 18y + 81 = x² – 22x + 121 + y² – 10y + 25
-14x + 22x – 18y + 10y = 146 – 130
8x – 8y = 16
x – y = 2 — (ii)
x + y + x – y = 12 + 2
2x = 14
x = 7
Put x = 7 in (i)
7 + y = 12
y = 5
The circumcenter is (7, 5)

The incenter of ∆ABC is point N. Use the given information to find the indicated measure.

Question 6.
NQ = 2x + 1, NR = 4x – 9
Find NS.

NQ = NR = NS
2x + 1 = 4x – 9
4x – 2x = 10
2x = 10
x = 5
NQ = 10 + 1 = 11
NS = 11

Question 7.
NU = – 3x + 6, NV = – 5x
Find NT.

NU = NV = NT
-3x + 6 = -5x
-3x + 5x = -6
2x = -6
x = -3
NT = -5(-3) = 15

Question 8.
NZ = 4x – 10, NY = 3x – 1
Find NW.

NZ = NY = NW
4x – 10 = 3x – 1
x = 9
NZ = 4(9) – 10 = 36 – 10 = 26
NW = 26

Find the coordinates of the centroid of the triangle wilt the given vertices.
Question 9.
J(- 1, 2), K(5, 6), L(5, – 2)
Centroid of the triangle = ($$\frac { -1 + 5 + 5 }{ 3 }$$, $$\frac { 2 + 6 – 2 }{ 3 }$$)
= (3, 2)

Question 10.
M(- 8, – 6), N(- 4, – 2), P(0, – 4)
Centroid of the triangle = ($$\frac { -8 – 4 + 0 }{ 3 }$$, $$\frac { -6 – 2 – 4 }{ 3 }$$)
= (-4, -4)

Tell whether the orthocenter is inside, on, or outside the triangle. Then find its coordinates.

Question 11.
T(- 2, 5), U(0, 1), V(2, 5)
The slope of TU = $$\frac { 1 – 5 }{ 0 + 2 }$$ = -2
The slope of the perpendicular line is $$\frac { 1 }{ 2 }$$
The perpendicular line is y – 5 = $$\frac { 1 }{ 2 }$$(x – 2)
2y – 10 = x – 2
x – 2y + 8 = 0
The slope of UV = $$\frac { 5 – 1 }{ 2 – 0 }$$ = 2
The slope of the perpendicular line is $$\frac { -1 }{ 2 }$$
The perpendicular line is y – 5 = $$\frac { -1 }{ 2 }$$(x + 2)
2y – 10 = -x – 2
x + 2y – 8 = 0
equate both equations
x – 2y + 8 = x + 2y – 8
-4y = -16
y = 4
x – 2(4) + 8 = 0
x = 0
So, the orthocenter is (0, 4)
The orthocenter lies inside the triangle TUV

Question 12.
X(- 1, – 4), Y(7, – 4), Z(7, 4)
The slope of XY = $$\frac { -4 + 4 }{ 7 + 1 }$$ = 0
The slope of the perpendicular line is 0
The perpendicular line is y – 4 = 0
y = 4
The slope of XZ =$$\frac { 4 + 4 }{ 7 + 1}$$ = 1
The slope of the perpendicular line is -1
The perpendicular line is y + 4 = -1(x – 7)
y + 4 = -x + 7
x + y – 3 = 0
x + 4 – 3 = 0
x = -1
So, the orthocenter is (-1, 4)
The orthocenter lies on the vertex.

Question 13.
A woodworker is culling the largest wheel possible from a triangular scrap of wood. The wheel just touches each side of the triangle, as shown.

a. Which point of concurrency is the center of the circle? What type of segments are $$\overline{B G}$$, $$\overline{C G}$$, and $$\overline{A G}$$?
The center of the circle is the incenter of the triangle ABC and BC, CG and AG are the angle bisectors of the triangle ABC.

b. Which theorem can you use to prove that ∆BGF ≅ ∆BGE?
AG, BG, and CG are the angle bisectors of the triangle.
As BG is the angle bisector of ∠EBF.
∠FBG = ∠EBG
∠GFC = ∠GEA = 90°
As ∠GFC forms a linear pair with ∠GFB and ∠GEA forms a linear pair with ∠GEB
∠GFB = 180° – ∠GFC
∠GEB = 180° – ∠GEA
∠GFB = ∠GEB = 90°
In ΔBGF and ΔBGE
∠GFB ≅ ∠GEB
∠FBG ≅ ∠EBG
BG ≅ BG

c. Find the radius of the wheel to the nearest tenth of a centimeter. Justify your answer.
ΔBGF ≅ ΔBGE
BF = BE
the length of BE is also 3 cm
BE + EA = 10
3 + EA = 10
EA = 10 – 3
EA = 7
Apply the Pythagoras theorem in ΔAEG
EA² + GE² = AG²
7² + GE² = 8²
GE² = 8² – 7²
GE² = 64 – 49
GE² = 15
GE = √15 = 3.88
So, the length of GE is 3.88 cm.
The radius of the wheel to the nearest tenth is 3.9 cm.

Question 14.
The Deer County Parks Committee plans to build a park at point P, equidistant from the three largest cities labeled X, Y, and Z. The map shown was created b the committee.

a. Which point of concurrency did the commIttee use as the location of the Park?

• That committee plans to build a park at point P such that it is at an equal distance from three big cities.
• As the three big cities are vertices X, Y, and Z, and point P is equidistant to them, point P must be the point where the perpendicular bisectors of the triangle meet.
• There the committee used the point of concurrency of the perpendicular bisectors.

b. Did the committee use the best point of concurrency for the location of the park? Ii not, which point would be better to use? Explain.

• A point of intersection of perpendicular bisectors is the only point that is at equidistance from all three vertices.
• It is the best point for the location of the park.

### 6.4 The Triangle Midsegment Theorem

Exploration 1

Midsegments of a Triangle

Work with a partner. Use dynamic geometry software. Draw any ∆ABC.

a. Plot midpoint D of $$\overline{A B}$$ and midpoint E of $$\overline{B C}$$. Draw $$\overline{D E}$$, which is a midsegment of ∆ABC.

b. Compare the slope and length of $$\overline{D E}$$ with the slope and length of $$\overline{A C}$$.

c. Write a conjecture about the relationships between the midsegments and sides of a triangle. Test your conjecture by drawing the other midsegments of ∆ABC, dragging vertices to change ∆ABC. and noting whether the relationships hold.

Exploration 2

Midsegments of a Triangle

Work with a partner. Use dynamic geometry software. Draw any ∆ABC.

a. Draw all three midsegments of ∆ABC.

b. Use the drawing to write a Conjecture about the triangle formed by the midsegments of the original triangle.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to make conjectures and build a logical progression of statements to explore the truth of your conjectures.

Question 3.
How are the midsegments of a triangle related to the sides of the triangle?
A midsegment of a triangle has its endpoints as the midpoints of the two sides of the triangle. It is parallel to the third side and is half as long as that side.

Question 4.
In ∆RST. $$\overline{U V}$$ is the midsegment connecting the midpoints of $$\overline{R S}$$ and $$\overline{S T}$$. Given
UV = 12, find RT.

### Lesson 6.4 The Triangle Midsegment Theorem

Monitoring progress

Use the graph of △ABC.

Question 1.
In △ABC, show that midsegments $$\overline{D E}$$ is parallel to $$\overline{A C}$$ and that DE = $$\frac { 1 }{ 2 }$$AC.
AB midpoint = D($$\frac { 1 – 1 }{ 2 }$$, $$\frac { 4 – 6 }{ 2 }$$) = D(0, -1)
Midpoint of BC = E($$\frac { 5 – 1 }{ 2 }$$, $$\frac { 4 + 0 }{ 2 }$$) = E(2, 2)
Slope of AC = $$\frac { 0 + 6 }{ 5 – 1 }$$ =$$\frac { 3 }{ 2 }$$
Slope of DE = $$\frac { 2 + 1 }{ 2 – 0 }$$ = $$\frac { 3 }{ 2 }$$
DE = √(2 – 0)² + (2 + 1)² = √13
AC = √(5 – 1)² + (0 + 6)² = √16 + 36 = √52
So, DE = $$\frac { 1 }{ 2 }$$AC

Question 2.
Find the coordinates of the endpoints of midsegments $$\overline{E F}$$, Which opposite $$\overline{A B}$$. show that $$\overline{E F}$$ is parallel to $$\overline{A B}$$ and that EF = $$\frac { 1 }{ 2 }$$AB.
The coordinates of E = ($$\frac { -1 + 5 }{ 2 }$$, $$\frac { 0 + 4 }{ 2 }$$) = (2, 2)
The coordinates of F = ($$\frac { 1 + 5 }{ 2 }$$, $$\frac { 0 + 6 }{ 2 }$$) = (3, 3)
The slope of AB = $$\frac { 4 + 6 }{ -1 – 1 }$$ = -5
The slope of EF = $$\frac { 3 – 2 }{ 3 – 2 }$$ = 1
The slopes are different. So the lines are not parallel
EF = √(3 – 2)² + (3 – 2)² = √2
AB = √(-1 – 1)² + (4 + 6)² = √4 + 100 = √104

Question 3.
In Example 2, find the coordinates of F, the midpoint of $$\overline{O C}$$. Show that $$\overline{F E}$$ || $$\overline{O B}$$ and FE = $$\frac { 1 }{ 2 }$$OB.
The coordinates of F = ($$\frac { 0 + 2p }{ 2 }$$, $$\frac { 0 + 0 }{ 2 }$$)
= (p, 0)
The slope of OB = $$\frac { 2r – 0 }{ 2q – 0 }$$ = $$\frac { r }{ q }$$
The slope of FE = $$\frac { r – 0 }{ q + p – p }$$ = $$\frac { r }{ q }$$
The slopes are equal. So, the lines are parallel.
FE = √(q + p – p)² + (r – 0)² = √q² + r²
OB = √(2q – 0)² + (2r – 0)² = √4q² + 4r²  = 2√q² + r²
OB = 2(FE)
FE = $$\frac { 1 }{ 2 }$$OB.

Question 4.
Copy the diagram in Example 3. Draw and name the third midsegment.
Then find the length of $$\overline{V S}$$ when the length of the third midsegment is 81 inches.

The length of the third midsegment = 81 in
The length of VS = $$\frac { 81 }{ 2 }$$ = 40.5

Question 5.
In Example 4. if F is the midpoint of $$\overline{C B}$$, what do you know about $$\overline{D F}$$?
DF || AE

Question 6.
WHAT IF?
In Example 5, you jog down Peach Street to Plum Street, over Plum Street to Cherry Street. up Cherry Street to Pear Street. over Pear Street to Peach Street. and then back home up Peach Street. Do you jog more miles in Example 5? Explain.
The distance you jog = 2.25 + 1.4 + 1.3 + .65 = 5.6

### Exercise 6.4 The Triangle Midsegment Theorem

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
The ___________ of a triangle is a segment that connects the midpoints of two sides of the triangle.

Question 2.
COMPLETE THE SENTENCE
If $$\overline{D E}$$ is the midsegment opposile $$\overline{A C}$$ in ∆ABC, then $$\overline{D E}$$ || $$\overline{A C}$$ and DE = ________ AC by the Triangle Midsegrnent Theorem
If $$\overline{D E}$$ is the midsegment opposile $$\overline{A C}$$ in ∆ABC, then $$\overline{D E}$$ || $$\overline{A C}$$ and DE = $$\frac { 1 }{ 2 }$$ AC by the Triangle Midsegrnent Theorem (Theorem 6.8).

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, use the graph of ∆ABC with midsegments $$\overline{D E}$$, $$\overline{E F}$$, and $$\overline{D F}$$.

Question 3.
Find the coordinates of points D, E, and F.

Question 4.
Show that $$\overline{D E}$$ is parallel to $$\overline{C B}$$ and that DE = $$\frac{1}{2}$$CB.
The slope of DE = $$\frac { 0 + 2 }{ -2 + 4 }$$ = 1
The slope of CB = $$\frac { -2 + 6 }{ 1 + 3 }$$  = 1
$$\overline{D E}$$ is parallel to $$\overline{C B}$$
DE = √(-2 + 4)² + (0 + 2)² = √4 + 4 = √8
CB = √(1 + 3)² + (-2 + 6)² = √16 + 16 = √32
So, DE = $$\frac{1}{2}$$CB

Question 5.
Show that $$\overline{E F}$$ is parallel to $$\overline{A C}$$ and that EF = $$\frac{1}{2}$$AC.

Question 6.
Show that $$\overline{D F}$$ is parallel to $$\overline{A B}$$ and that DF = $$\frac{1}{2}$$AB.
The slope of DF = $$\frac { -4 + 2 }{ -1 + 4 }$$ = $$\frac { -2 }{ 3 }$$
The slope of AB = $$\frac { -2 – 2 }{ 1 + 5 }$$ = $$\frac { -4 }{ 6 }$$ = $$\frac { -2 }{ 3 }$$
The slopes are equal. So, the lines are parallel.
DF = √(-1 + 4)² + (-4 + 2)² = √9 + 4 = √13
AB = √(1 + 5)² + (-2 – 2)² = √36 + 16 = √52
DF = $$\frac{1}{2}$$AB

In Exercises 7-10, $$\overline{D E}$$ is a midsegment of ∆ABC Find the value of x.

Question 7.

Question 8.

DE = $$\frac { 1 }{ 2 }$$AB
5 = $$\frac { 1 }{ 2 }$$x
x = 10

Question 9.

Question 10.

BE = EC
x = 8

In Exercise 11-16, $$\overline{X J} \cong \overline{J Y}$$, $$\overline{Y L} \cong \overline{L Z}$$, and $$\overline{X K} \cong \overline{K Z}$$. Copy and complete the statement.

Question 11.
$$\overline{J K}$$ || __________

Question 12.
$$\overline{X Y}$$ || __________
$$\overline{X Y}$$ || KL

Question 13.
$$\overline{J L}$$ || __________

Question 14.
$$\overline{J L}$$ ≅ __________ ≅ __________
$$\overline{J L}$$ ≅ JK ≅ KL

Question 15.
$$\overline{J Y}$$ ≅ __________ ≅ __________

Question 16.
$$\overline{J K}$$ ≅ __________ ≅ __________
$$\overline{J K}$$ ≅ JL ≅ KL

MATHEMATICAL CONNECTIONS
In Exercises 17-19. use ∆GHJ, where A, B, and C are midpoints of the sides.

Question 17.
When AB = 3x + 8 and GJ = 2x + 24, what is AB?

Question 18.
When AC = 3y – 5 and HJ = 4y + 2, what is HB?
AC = 0.5(HJ)
3y – 5 = 0.5(4y + 2)
3y – 5 = 2y + 1
3y – 2y = 1 + 5
y = 6
HB = 0.5(HJ)
= 0.5(4(6) + 2) = 0.5(26) = 13
HB = 13

Question 19.
When GH = 7 – 1 and CB = 4z – 3. what is GA?

Question 20.
ERROR ANALYSIS
Describe and correct the error.

BC = $$\frac { 1 }{ 2 }$$DE
10 = $$\frac { 12 }{ 2 }$$(5)
According to the triangle midsegment theorem, AD = DB and AE = EC.

Question 21.
MODELING WITH MATHEMATICS
The distance between consecutive bases on a baseball held is 90 feet. A second baseman stands halfway between first base and second base, a shortstop stands hallway between second base and third base, and a pitcher stands halfway between first base and third base. Find the distance between the shortstop and the pitcher.

Question 22.
PROVING A THEOREM
Use the figure from Example 2 to prose the Triangle Midsegment Theorem (Theorem 6.8) for midsegment $$\overline{D F}$$, where F is the midpoint of $$\overline{O C}$$.

Question 23.
CRITICAL THINKING
$$\overline{X Y}$$ is a midsegment of ∆LMN. Suppose $$\overline{D E}$$ is called a “quarter segment” of ∆LMN. What do you think an “eighth segment” would be? Make conjectures about the properties of a quarter segment and an eighth segment. Use variable coordinates to verify your conjectures.

Question 24.
THOUGHT PROVOKING
Find a real-life object that uses midsegments as part of its structure. Print a photograph of the object and identify the midsegments of one of the triangles in the structure.

• When we look around, we will see that many structures like swings, rooftops of houses, bridges, etc, use midsegments in their design.
• One such structure is the swing.

Question 25.
ABSTRACT REASONING
To create the design shown. shade the triangle formed by the three midsegments of the triangle. Then repeat the process for each unshaded triangle.

a. What is the perimeter of the shaded triangle in Stage 1?
b. What is the total perimeter of all the shaded triangles in Stage 2?
c. What is the total perimeter of all the shaded triangles in Stage 3?

Question 26.
HOW DO YOU SEE IT?
Explain how you know that the yellow triangle is the midsegment triangle of the red triangle in the pattern of floor tiles shown.

Yellow triangle vertices are located at the midpoints of the red triangle sides.

Question 27.
ATTENDING TO PRECISION
The points P(2, 1), Q(4, 5), and R(7, 4) are the midpoints of the sides of a triangle. Graph the three midsegments. Then show how to use your graph and the properties of midsemeriLs to draw the original triangle. Give the coordinates of each vertex.

Maintaining Mathematical Proficiency

Find a counter example to show that the conjecture is false.

Question 28.
The difference of two numbers is always less than the greater number.
When we subtract -3 from 4 i.e 4 -(-3) = 7.
Subtraction of a negative number from a positive number gives an answer greater than both numbers.

Question 29.
An isosceles triangle is always equilateral.

### 6.5 Indirect Proof and Inequalities in One Triangle

Exploration 1

Comparing Angle Measures and Side Lengths

Work with a partner: Use dynamic geometry software. Draw any scalene ∆ABC

a. Find the side lengths and angle measures of the triangle.

b. Order the side lengths. Order the angle measures. What do you observe?

c. Drug the vertices of ∆ABC to form new triangles. Record the side lengths and angle measures in a table. Write a conjecture about your findings.

Exploration 2

A Relationship of the Side Lengths of a Triangle

Work with a partner. Use dynamic geometry software. Draw any ∆ABC.

a. Find the side lengths of the triangle.

b. Compare each side length with the sum of the other two side lengths.
ATTENDING TO PRECISION
To be proficient in math, you need to express numerical answers with a degree of precision appropriate for the content.

c. Drag the vertices of ∆ABC to form new triangles and repeat parts (a) and (b). Organize your results in a table. Write a conjecture about your findings.

Question 3.
How are the sides related to the angles of a triangle? How are any two sides of a triangle related to the third side?
Scalene triangle – A triangle in which all of its three sides have different side measures. The relation between the sides and the angles of the triangle is that in any triangle the largest side and the largest angle of the triangle are always opposite to each other and the shortest side and the shortest angle are always opposite to each other.
The relation of the two sides of a triangle to the third side is that the sum of the lengths of any two sides of a triangle is greater than the length of its third side. This is called the triangle inequality theorem.

Question 4.
Is it possible for a triangle to have side lengths of 3, 4, and 10? Explain.
Given side lengths are 3, 4, and 10.
3 + 4 > 10
7 > 10
Hence, It is not possible to construct a triangle.

### Lesson 6.5 Indirect Proof and Inequalities in One Triangle

Monitoring Progress

Question 1.
Write in indirect proof that a scalene triangle cannot have two congruent angles.
According to the property of the scalene triangle, all the angles are unequal and all the sides are of different lengths which contradict our assumption.
∠PQR is not congruent to ∠PRQ.
The assumption that scalene triangles have two congruent angles is wrong. So, by the contradictory method, it is proved that the scalene triangle cannot have two congruent angles.

Question 2.
List the angle of ∆PQR in order from smallest to largest.

The angles from smallest to the largest are ∠Q, ∠P, ∠R
The sides from the largest to smallest are PQ, QR, PR
The angles opposite to the sides are ∠R, ∠P, ∠Q

Question 3.
List the sides of ∆RST in order from shortest to longest.

The angles from shortest to longest are ∠R, ∠T, ∠S

Question 4.
A triangle has one side of length 12 inches and another side of length 20 inches. Describe the possible lengths of the third side.
Let the length of the third side be x.
Draw diagrams to help visualize the small and large values of x.
The smallest value of x is x + 12 > 20. So, x > 8
The largest value of x is 12 + 20 > x
32 > x
So, 32 > x or 32 < x
The length of the third side must be greater than 8 and lesser than 32.

Decide Whether it is possible to construct a triangle with the given side lengths. Explain your reasoning.

Question 5.
4 ft, 9 ft, 10 ft
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.
9 + 4 = 13
13 < 10
13 is not greater than 10.
It is not possible to construct a triangle.

Question 6.
8 ft, 9 ft, 18 ft
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.
8 + 9 = 17 < 18
17 is less than 18.
It is possible to construct a triangle.

Question 7.
5 cm, 7 cm, 12 cm
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.
7 + 5 = 12
12 = 12
It is not possible to construct a triangle.

### Exercise 6.5 Indirect Proof and Inequalities in One Triangle

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Why is an indirect proof also called a proof by contradiction?

Question 2.
WRITING
How can you tell which side of a triangle is the longest from the angle measures of the triangle? How can you tell which side is the shortest?
The shortest side is always opposite to the smallest interior angle and the longest side is opposite to the largest interior angle.

Monitoring progress and Modeling with Mathematics

In Exercises 3-6, write the first step in an indirect proof of the statement.

Question 3.
If WV + VU ≠ 12 inches and VU = 5 inches, then WV ≠ 7 inches.

Question 4.
If x and y are odd integers. then xy is odd.
Assume x = 5, y = 3
xy = 5(3) = 15
xy = 15 that is odd.

Question 5.
In ∆ABC. if m∠A = 100°, then ∠B is not a right angle.

Question 6.
In ∆JKL, if M is the midpoint of $$\overline{K L}$$, then $$\overline{J M}$$ is a median.
Assume temporarily JM is a median.

In Exercises 7 and 8, determine which two statements contradict each other. Explain your reasoning.

Question 7.
(A) ∆LMN is a right triangle.
(B) ∠L ≅∠V
(C) ∆LMN is equilateral.

Question 8.
(A) Both ∠X and ∠Y have measures greater than 20°.
(B) Both ∠X and ∠Y have measures less than 30°.
(C) m∠X + m∠Y = 62°
B and C. If ∠X and ∠Y are less than 30, then m∠X + m∠Y ≠ 62.

In Exercises 9 and 10, use a ruler and protractor to draw the given type of triangle. Mark the largest angle and longest side in red and the smallest angle and shortest side in blue. What do you notice?

Question 9.
acute scalene

Question 10.
right scalene
In order to construct a scalene triangle draw a segment and label it AB.
The three sides have different lengths.
Draw an arc with center A and AC as the radius draw an arc and draw another arc taking BC as the radius and B as the center.
Label the point of intersection of two arcs as C.

The largest side is BC. So, the largest angle is ∠A.
The smallest side is AB and the smallest angle is ∠C.

In Exercises 11 and 12, list the angles of the given triangle from smallest to largest.

Question 11.

Question 12.

The sides of △JKL are from smallest to largest: KL, JL, and JK.
The angles from smallest to largest are ∠J, ∠K, ∠and L.

In Exercises 13-16, list the sides of the given triangle from shortest to longest.

Question 13.

Question 14.

The angles of the triangle from smallest to largest are ∠Z, ∠X, ∠and Y.
The sides from shortest to largest are XY, YZ, and XZ.

Question 15.

Question 16.

∠F = 90°
∠G = 33°
Sum of angles = 180°
∠D = 180° – (90 + 33)° = 180° – 57°
The angles from smallest to largest are ∠G, ∠D, ∠F.
So by the triangle larger angle theorem, the sides from shortest to longest are FD, FG, and DG.

In Exercises 17-20, describe the possible lengths of the third side of the triangle given the lengths of the other to sides.

Question 17.
5 inches, 12 inches

Question 18.
12 feet, 18 feet
x + 12 > 18
x > 6
12 + 18 > x
30 > x or x < 30
The possible lengths of the third side are greater than 6 and less than 30 ft

Question 19.
2 feet, 40 inches

Question 20.
25 meters, 25 meters
x + 25 > 25
x > 0
25 + 25 > x
50 > x or x < 50
The possible lengths of the third side are greater than 0 and less than 50 m.

In Exercises 21-23, is it possible to construct a triangle with the given side lengths? If not, explain why not.

Question 21.
6, 7, 11

Question 22.
3, 6, 9
3 + 6 = 9 → 9 = 9
6 + 9 = 15 → 15 < 9
9 + 3 = 12 → 12 < 9
No, the sum of any two side lengths of a triangle is less than the length of the third side.

Question 23.
28, 17, 46

Question 24.
35, 120, 125
No; 35 + 120 = 155 → 155 < 125

Question 25.
ERROR ANALYSIS
Describe and correct the error in writing the first step of an indirect proof.

Question 26.
ERROR ANALYSIS
Describe and correct the error in labeling the side lengths 1, 2, and √3 on the triangle.

Number 1 is opposite to 90° and 2 is opposite to 60° and the remaining is √3 on the triangle.

Question 27.
REASONING
You are a lawyer representing a client who has been accused of a crime. The crime took place in Los Angeles, California. Security footage shows your client in New York at the time of the crime. Explain how to use indirect reasoning to prove your client is innocent.

Question 28.
REASONING
Your class has fewer than 30 students. The teacher divides your class into two groups. The first group has 15 students. Use indirect reasoning to show that the second group must have fewer than 15 students.
The number of students in a class N < 30
The number of students in the first group N1 = 15
The number of students in the second group N2
N = N1 + N2
Assume temporarily that the number of students in the second group has 15 or more students
N2 ≥ 15
N1 + N2 ≥ 15 + 15
30 ≥ 30
So, the assumption that N2 ≥ 15 is false.

Question 29.
PROBLEM SOLVING
Which statement about ∆TUV is false?

(A) UV > TU
(B) UV + TV > TU
(C) UV < TV
(D) ∆TUV is isosceles.

Question 30.
PROBLEM SOLVING
In ∆RST. which is a possible side length for ST? Select all that apply.

(A) 7
(B) 8
(C) 9
(D) 10
∠R = 180 – (65 + 56) = 180 – 121 = 59, which indicates that ∆RST is isosceles.
By the triangle longer side theorem, ST is 7
Option A is the correct answer.

Question 31.
PROOF
Write an indirect proof that an odd number is not divisible by 4.

Question 32.
PROOF
Write an indirect proof of the statement
“In ∆QRS, if m∠Q + m∠R = 90°, then m∠S = 90°.”

Question 33.
WRITING
Explain why the hypotenuse of a right triangle must always be longer than either leg.

Question 34.
CRITICAL THINKING
Is it possible to decide if three side lengths form a triangle without checking all three inequalities shown in the Triangle Inequality Theorem (Theorem 6. 11)? Explain your reasoning.
Yes, it is possible to decide if three side lengths form a triangle without checking all three inequalities shown in the Triangle Inequality Theorem.
If the sum of the lengths of the two shortest sides is greater than the length of the longest side. then the other two inequalities will also be true.

Question 35.
MODELING WITH MATHEMATICS
You can estimate the width of the river from point A to the tree at point B by measuring the angle to the tree at several locations along the riverbank. The diagram shows the results for locations C and D.

a. Using ∆BCA and ∆BDA, determine the possible widths of the river. Explain your reasoning.
b. What could you do if you wanted a closer estimate?

Question 36.
MODELING WITH MATHEMATICS
You travel from Fort Peck Lake to Glacier National Park and from Glacier National Park to Granite Peak.

a. Write two inequalities to represent the possible distances from Granite Peak back to Fort Peck Lake.
x < 565 + 489
x < 1054
x + 489 > 565
x > 565 – 489
x > 76
x + 489 > 565
x > 565 – 489
x > 76
Combining these two inequalities to find the possible distance from Granite Peak back to Fork Peck Lake
x < 1054
x > 76
76 < x < 1054

b. How is your answer to part (a) affected if you know that m∠2 < m∠1 and m∠2 < m∠3?
∠WXY is smaller than ∠YXZ, ∠YXZ is smaller than ∠WYX, ∠WYX is smaller than ∠XZY and ∠XZY is smaller than ∠XYZ.

Question 37.
REASONING
In the figure. $$\overline{X Y}$$ bisects ∠WYZ. List all six angles of ∆XYZ and ∆WXY in order from smallest to largest. Explain ‘our reasoning.

Question 38.
MATHEMATICAL CONNECTIONS
In ∆DEF, m∠D = (x + 25)°. m∠E = (2x – 4)°, and in m∠F = 63°. List the side lengths and angle measures of the triangle in order from least to greatest.
The sum of angles is 180 degrees
x + 25 + 2x – 4 + 63 = 180
3x + 84 = 180
3x = 96
x = 32
m∠D = 32 + 25 = 57
m∠E = 2(32) – 4 = 60
m∠F = 63
The angles from least to greatest are m∠D, m∠E, m∠F, and the side lengths are EF, DF, DE.

Question 39.
ANALYZING RELATIONSHIPS
Another triangle inequality relationship is given by the Exterior Angle Inequality Theorem. It states:
The measure of an exterior angle of a triangle is greater than the measure of either of the nonadjacent interior angles.

Explain how you know that m∠1 > m∠A and m∠1 > m∠B in ∆ABC with exterior angle ∠1.

MATHEMATICAL CONNECTIONS
In Exercises 40 and 41, describe the possible values of x.

Question 40.

JK + JL > KL
x + 11 + 5x – 9 > 2x +10
6x + 2 > 2x + 10
4x > 8
x > 2
JK + KL > JL
x + 11 + 2x + 10 > 5x – 9
3x + 21 > 5x – 9
30 > 2x
15 > x
JL + KL > JK
5x – 9 + 2x + 10 > x + 11
7x + 1 > x + 11
6x > 10
x > $$\frac { 5 }{ 3 }$$
The possible values of x are x > $$\frac { 5 }{ 3 }$$ and x < 15

Question 41.

Question 42.
HOW DO YOU SEE IT?
Your house is on the corner of Hill Street and Eighth Street. The library is on the corner of View Street and Seventh Street. What is the shortest route to et from your house to the library? Explain your reasoning.

Washington ave is the shortest route to reach from the house to the library.

Question 43.
PROVING A THEOREM
Use the diagram to prove the Triangle Longer Side Theorem (Theorem 6.9).

Given BC > AB, BD = BA
Prove m∠BAC > m∠C

Question 44.
USING STRUCTURE
The length of the base of an isosceles triangle is l. Describe the possible lengths for each leg. Explain our reasoning.
The length of the base of an isosceles triangle is l.
Let the other lengths are x and x
Then, x + x > l
2x > l
x > l/2
x + l > x
The possible lengths of each leg are l l/2.

Question 45.
MAKING AN ARGUMENT
Your classmate claims to have drawn a triangle with one side length of 13 inches and a perimeter of 2 feet. Is this possible? Explain your reasoning.

Question 46.
THOUGHT PROVOKING
Cut two pieces of string that are each 24 centimeters long. Construct an isosceles triangle out of one string and a scalene triangle out of the other. Measure and record the side lengths. Then classify each triangle by its angles.
Cut two pieces of string that are each 24 centimeters long.

9 + 9 + 6 = 24

6 + 8 + 10 = 24

Question 47.
PROVING A THEOREM
Prove the Triangle Inequality Theorem (Theorem 6. 11).
Given ∆ABC
Prove AB + BC > AC, AC + BC > AB, and AB + AC > BC

Question 48.
ATTENDING TO PRECISION
The perimeter of ∆HGF must be between what two integers? Explain your reasoning.

1 < GH < 7
1 < FH > 9
2 < GF < 8
4 < GH + FH + GF < 24
4 < Δ GHF < 24

Question 49.
PROOF
Write an indirect proof that a perpendicular segment is the shortest segment from a point to a plane.

Given $$\overline{P C}$$ ⊥ palne M
Prove $$\overline{P C}$$ is the shortest from P to plane M.

Maintaining Mathematical Proficiency

Name the indicated angle between the pair of sides given.

Question 50.
$$\overline{A E}$$ and $$\overline{B E}$$
The included angle between $$\overline{A E}$$ and $$\overline{B E}$$ is ∠AEB

Question 51.
$$\overline{A C}$$ and $$\overline{D C}$$

Question 52.
$$\overline{A D}$$ and $$\overline{D C}$$
The included angle between $$\overline{A D}$$ and $$\overline{D C}$$ is ∠ADC

Question 53.
$$\overline{C E}$$ and $$\overline{B E}$$

### 6.6 Inequalities in Two Triangles

Exploration 1

Comparing Measures in Triangles

Work with a partner. Use dynamic geometry software.

a. Draw ∆ABC, as shown below.

b. Draw the circle with center C(3, 3) through the point A(1, 3).

c. Draw ∆DBC so that D is a point on the circle.

d. Which two sides of ∆ABC are congruent to two sides of ∆DBC? Justify your answer.

e. Compare the lengths of $$\overline{A B}$$ and $$\overline{D B}$$. Then compare the measures of ∠ACB and ∠DCB. Are the results what you expected? Explain.

f. Drag point D to several locations on the circle. At each location, repeat part (e). Copy and record your results in the table below.

g. Look for a pattern of the measures in our table. Then write a conjecture that summarizes your observations.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to make conjectures and build a logical progression of statements to explore the truth of your conjectures.

Question 2.
If two sides of one triangle are congruent to two sides of another triangle, what can you say about the third sides of the triangles?

Answer: If two sides of one triangle are congruent to two sides of another triangle, the third side of the first triangle is greater than the third side of the second triangle, then the included angle of the first triangle is larger than the included angle of the second.

Question 3.
Explain how you can use the hinge shown at the left to model the concept described in Question 2.

### Lesson 6.6 Inequalities in Two Triangles

Monitoring Progress

Use the diagram

Question 1.
If PR = PS and m∠QPR > m∠QPS, which is longer, $$\overline{S Q}$$ or $$\overline{R Q}$$?
PR = PS and PQ ≅ PQ by the reflexive property of congruence.
m∠QPR > m∠QPS.
So, two sides of △PSQ are congruent to two sides of △PRQ, and included angle in △PRQ is larger.
RQ > SQ

Question 2.
If PR = PS and RQ < SQ, which is larger, ∠RPQ or ∠SPQ?
PR = PS, RQ < SQ
PQ ≅ PQ (Reflexive property of congruence theorem)
Two sides of △PSQ are congruent to two sides of △PRQ, and the third side of △PSQ is longer.
By the converse of the hinge theorem, ∠SPQ > ∠RPQ

Question 3.
Write a temporary assumption you can make to prove the Hinge Theorem indirectly. What two cases does that assumption lead to?
According to the Hinge theorem, If two sides of the triangle are congruent to two sides of another triangle and the included angle of the first is larger than the included angle of the second, then the third side of the first triangle is longer than the third side of the second triangle.

Question 4.
WHAT IF?
In Example 5, Group C leaves camp and travels 2 miles due north. then turns 40° towards east and travels 1.2 miles. Compare the distances from camp for all three groups.

group A = 180 – 45 = 135
group B = 180 – 30 = 150
Group C = 180 – 40 = 140
Because 150 > 140 > 135, the distance between Group B from the camp is greater than the distance between Group C, and A
The distances are Group B > group C > group A

### Exercise 6.6 Inequalities in Two Triangles

Vocabulary and Core Concept Check

Question 1.
WRITING
Explain why Theorem 6.12 is named the “Hinge Theorem.”

Question 2.
COMPLETE THE SENTENCE
In ∆ABC and ∆DEF, $$\overline{A B} \cong \overline{D E}$$, $$\overline{B C} \cong \overline{E F}$$. and AC < DF. So m∠_______ > m ∠ _______ by the Converse of the Hinge Theorem
In ∆ABC and ∆DEF, $$\overline{A B} \cong \overline{D E}$$, $$\overline{B C} \cong \overline{E F}$$. and AC < DF. So m∠E > m ∠B by the Converse of the Hinge Theorem

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, copy and complete the statement with <, >, or = Explain your reasoning.

Question 3.
m∠1 ________ m∠2

Question 4.
m∠1 ________ m∠2

m∠1 < m∠2 by the converse of the hinge theorem.

Question 5.
m∠1 ________ m∠2

Question 6.
m∠1 ________ m∠2

m∠1 > m∠2 by the converse of the hinge theorem.

In Exercises 7-10. copy and complete the statement with <, >, or =. Explain your reasoning.

Question 7.

Question 8.
MN ________ LK

MN < KL by the converse of the hinge theorem.

Question 9.
TR ________ UR

Question 10.
AC ________ DC

AC > DC (If two triangles have two congruent sides, then the one with the longer third side has the larger included angle).

PROOF
In Exercises 11 and 12, write a proof.

Question 11.
Given $$\overline{X Y} \cong \overline{Y Z}$$, m∠WYZ > m∠WYX
Prove WZ > WX

Question 12.
Given $$\overline{B C} \cong \overline{D A}$$, DC < AB
Prove m∠BCA > m∠DAC

AC ≅ AC by the reflexive property of congruence theorem
DC < AB is given
So, m∠BCA > m∠DAC by the hinge theorem.

In Exercises 13 and 14, you and your friend leave on different flights from the same airport. Determine which flight is farther from the airport. Explain your reasoning.

Question 13.
Your flight: Flies 100 miles due west, then turns 20° toward north and flies 50 miles.
Friend’s flight: Flies 100 miles due north, then turns 30° toward east and flies 50 miles.

Question 14.
Your flight: Flies 21 miles due south, then turns 70° toward west and flies 80 miles.
Friend’s flight: Flies 80 miles due north, then turns 50° toward east and flies 210 miles.
The figure below represents the situation.

I flew 210 miles north to point B then ran 70° to the west and flew 80 miles to point C.
My friend flew 80 miles north to point D then turned 50° east and flew 210 miles to point F.
We should take into consideration that 70° to west means 110° and 50° to east means 130°
AB = DF
So, AC < AF
My friend’s flight is farther from the airport.

Question 15.
ERROR ANALYSIS
Describe and correct the error in using the Hinge Theorem (Theorem 6.12).

Question 16.
REPEATED REASONING
Which is possible measure for ∠JKM? Select all that apply.

(A) 15°
(B) 22°
(C) 25°
(D) 35°
KL ≅ KJ is given
KM ≅ KM (reflexive property of congruence)
LM > MJ
So, m∠LKM > m∠MKJ
25 > 35

Question 17.
DRAWING CONCLUSIONS
The path from E to F is longer than the path from E to D. The path from G to D is the same length as the path from G to F. What can you conclude about the angles of the paths? Explain your reasoning.

Question 18.
ABSTRACT REASONING
In ∆EFG, the bisector of ∠F intersects the bisector of ∠G at point H. Explain why $$\overline{F G}$$ must be longer than $$\overline{F H}$$ or $$\overline{H G}$$.
The sum of the interior angles of a triangle is 180°
The sum of angles F and G cannot be larger than 180°
∠HFG is half of the measurement of ∠F. ∠HGF is half of the measurement of ∠G. thus, ∠HFG + ∠HGF < 90°
∠FHG = 180° – ∠HGF – ∠HFG > 90°

Question 19.
ABSTRACT REASONING
$$\overline{N R}$$ is a median of ∆NPQ, and NQ > NP Explain why ∠NRQ is obtuse.

MATHEMATICAL CONNECTIONS
In Exercises 20 and 21, write and solve an inequality for the possible values of x.

Question 20.

3x + 2 = x + 3
3x – x = 3 – 2
2x = 1
x = $$\frac { 1 }{ 2 }$$

Question 21.

Question 22.
HOW DO YOU SEE IT?
In the diagram, triangles are formed by the locations of the players on the basketball court. The dashed lines represent the possible paths of the basketball as the players pass. How does m∠ACB compare with m∠ACD?

BC ≅ CD
AC ≅ AC (Reflexive property of congruence theorem)
So, m∠ACB < m∠ACD (Converse of the hinge theorem)

Question 23.
CRITICAL THINKING
In ∆ABC, the altitudes from B and C meet at point D, and m∠BDC. What is true about ∆ABC? Justify your answer.

Question 24.
THOUGHT PROVOKING
The postulates and theorems in this book represent Euclidean geometry. In spherical geometry, all points are on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. In spherical geometry, state an inequality involving the sum of the angles of a triangle. Find a formula for the area of a triangle in spherical geometry.
The sum of the interior angles of a triangle is equal to 180°. If we include in the spherical geometry i.e., we will connect the vertices of the triangle that is created by the arcs of the great circles.

In Euclidean Triangle, the angle measures of the spherical triangle are greater than the angle measures. This tells us that the sum ‘S’ of the angle measures of a triangle in spherical geometry is greater than 180°.

L(x)/x = 4πr²/360
L(x) = xπr²/90
Let x, y, z be the interior angles of the spherical triangles
L(y) = yπr²/90
L(z) = zπr²/90
2L(x) + 2L(y) + 2L(z) – 4 = 4πr²
2(xπr²/90) + 2(yπr²/90) + 2(zπr²/90) – 4A = 4πr²
A = (x/180 + y/180 + z/180 – 1)πr²

Maintaining Mathematical proficiency

Find the value of x.

Question 25.

Question 26.

36° + x° + x° = 180°
36° + 2x° = 180°
2x° = 144°
x = 72°

Question 27.

Question 28.

The sum of interior angles in  a triangle is 180 degrees
44° + 64° + y = 180°
y = 180° – 108°
y = 72°
x + y = 180°
x + 72° = 180°
x = 108°

### 6.1 Perpendicular and Angle Bisectors

Find the indicated measure. Explain your reasoning.

Question 1.
DC

20 = 7x – 15
7x = 35
x = 5
DC = 7(5) – 15 = 20

Question 2.
RS

PS = SR
6x + 5 = 9x – 4
9x – 6x = 9
3x = 9
x = 3
RS = 9(3) – 4 = 23

Question 3.
m∠JFH

m∠JFH = m∠JFG
m∠JFH = 47°

### 6.2 Bisectors of Triangles

Find the coordinates of the circumcenter of the triangle with the given vertices.

Question 4.
T(- 6, – 5), U(0, – 1), V(0, – 5)
T(- 6, – 5), U(0, – 1), V(0, – 5)
PT = PU = PV
PT² = PU² = PV²
PT² = PU²
(x + 6)² + (y + 5)² = (x – 0)² + (y + 1)²
x² + 12x + 36 + y² + 10y + 25 = x² + y² + 2y + 1
12x + 8y + 61 = 1
12x + 8y + 60 = 0
3x + 2y + 15 = 0
PU² = PV²
(x – 0)² + (y + 1)² = (x – 0)² + (y + 5)²
x² + y² + 2y + 1 = x² + y² + 10y + 25
8y + 24 = 0
8y = -24
y = -3
Substitute y = -3 in 3x + 2y + 15 = 0
3x + 2(-3) + 15 = 0
3x – 6 + 15 = 0
3x = -9
x = -3
The circumcenter is (-3, -3)

Question 5.
X(- 2, 1), Y(2, – 3), Z(6, – 3)
Given points are X(- 2, 1), Y(2, – 3), Z(6, – 3)
PX = PY = PZ
PX² = PY² = PZ²
PX² = PY²
(x + 2)² + (y – 1)² = (x – 2)² + (y + 3)²
x² + 4x + 4 + y² – 2y + 1 = x² – 4x + 4 + y² + 6y + 9
8x – 8y + 5 = 13
8x – 8y – 8 = 0
x – y = 1
PY² = PZ²
(x – 2)² + (y + 3)² = (x – 6)² + (y + 3)²
x² – 4x + 4 + y² + 6y + 9 = x² – 12x + 36 + y² + 6y + 9
8x = 32
x = 4
Substitue x = 4 in x – y = 1
4 – y = 1
-y = -3
y = 3
The circumcenter is (4, 3)

Question 6.
Point D is the incenter of ∆LMN. Find the value of x.
x = 13

### 6.3 Medians and Altitudes of Triangles

Find the coordinates of the centroid of the triangle with the given vertices.

Question 7.
A(- 10, 3), B(- 4, 5), C(- 4, 1)
The centroid = ($$\frac { -10 – 4 – 4 }{ 3 }$$, $$\frac { 3 + 5 + 1 }{ 3 }$$)
= (-6, 3)

Question 8.
D(2, – 8), E(2, – 2), F(8, – 2)
The centroid = ($$\frac { 2 + 2 + 8 }{ 3 }$$, $$\frac { -8 – 2 – 2 }{ 3 }$$)
= (4, -4)

Tell whether the orthocenter of the triangle with the given vertices is inside, on, or outside the triangle. Then find the coordinates of the orthocenter.

Question 9.
G(1, 6), H(5, 6), J(3, 1)
The slope of the line HJ = $$\frac { 1 – 6 }{ 3 – 5}$$ = $$\frac { 5 }{ 2 }$$
The slope of the perpendicular line = $$\frac { -2 }{ 5 }$$
The perpendicular line is (y – 6) = $$\frac { -2 }{ 5 }$$(x – 1)
5(y – 6) = -2(x – 1)
5y – 30 = -2x + 2
2x + 5y – 32 = 0 — (i)
The slope of GJ = $$\frac { 1 – 6 }{ 3 – 1 }$$ = $$\frac { -5 }{ 2 }$$
The slope of the perpendicular line = $$\frac { 2 }{ 5 }$$
The equation of perpendicular line (y – 6) = $$\frac { 2 }{ 5 }$$(x – 5)
5(y – 6) = 2(x – 5)
5y – 30 = 2x – 10
2x – 5y + 20 = 0 — (ii)
Equate both equations
2x + 5y – 32 = 2x – 5y + 20
10y = 52
y = 5.2
Substitute y = 5.2 in (i)
2x + 5(5.2) – 32 = 0
2x + 26 – 32 = 0
2x = 6
x = 3
The orthocenter is (3, 5.2)
The orthocenter lies inside the triangle.

Question 10.
K(-8, 5), L(- 6, 3), M(0, 5)
The slope of LM = $$\frac { 5 – 3}{ 0 + 6}$$ = $$\frac { 1 }{ 3 }$$
The slope of the perpendicular line = -3
The perpendicular line is (y – 5) = -3(x + 8)
y – 5 = -3x – 24
3x + y + 19 = 0 —- (ii)
The slope of KL = $$\frac { 3 – 5 }{ -6 + 8 }$$ = -1
The slope of the perpendicular line = $$\frac { 1 }{ 2 }$$
The equation of perpendicular line (y – 5) = $$\frac { 1 }{ 2 }$$(x – 0)
2y – 10 = x —- (ii)
Substitute (ii) in (i)
3(2y – 10) + y + 19 = 0
6y – 30 + y + 19 = 0
7y – 11 = 0
y = $$\frac { 11 }{ 7 }$$
x = -6
The orthocenter is (-6, -1)
The orthocenter lies outside of the triangle

### 6.4 The Triangle Midsegment Theorem

Find the coordinates of the vertices of the midsegment triangle for the triangle with the given vertices.

Question 11.
A(- 6, 8), B(- 6, 4), C(0, 4)
The midsegment of AB = ($$\frac { -6 – 6 }{ 2 }$$, $$\frac { 8 + 4 }{ 2 }$$)
= (-6, 6)
The midsegment of AB = (-6, 6)
The midsegment of BC = ($$\frac { -6 + 0 }{ 2 }$$, $$\frac { 4 + 4 }{ 2 }$$)
= (-3, 4)
The midsegment of BC = (-3, 4)
The midsegment of AC = ($$\frac { -6 + 0 }{ 2 }$$, $$\frac { 8 + 4 }{ 2 }$$)
= (-3, 6)
The midsegment of AC = (-3, 6)

Question 12.
D(- 3, 1), E(3, 5), F(1, – 5)
The midsegment of DE = ($$\frac { -3 + 3 }{ 2 }$$, $$\frac { 1 + 5 }{ 2 }$$)
= (0, 3)
The midsegment of DE = (0, 3)
The midsegment of EF = ($$\frac { 3 + 1 }{ 2 }$$, $$\frac { 5 – 5 }{ 2 }$$)
= (2, 0)
The midsegment of EF = (2, 0)
The midsegment of DF = ($$\frac { -3 + 1 }{ 2 }$$, $$\frac { 1 – 5 }{ 2 }$$)
= (-1, -2)
The midsegment of DF = (-1, -2)

### 6.5 Indirect Proof and Inequalities in One Triangle

Describe the possible lengths of the third side of the triangle given the lengths of the other two sides.

Question 13.
4 inches, 8 inches
4 + 8 > x
12 > x
4 + x > 8
x > 4
8 + x > 4
x > -4
4 < x < 12

Question 14.
6 meters, 9 meters
6 + 9 > x
15 > x
6 + x > 9
x > 3
9 + x > 6
x > -3
3 < x < 15

Question 15.
11 feet, 18 feet
11 + 18 > x
29 > x
11 + x > 18
x > 7
18 + x > 11
x > -7
7 < x < 29

Question 16.
Write an indirect proof 0f the statement “In ∆XYZ, if XY = 4 and XZ = 8. then YZ > 4.”
4 + 8 > x
12 > x
4 + x > 8
x > 4
8 + x > 4
x > -4
12 > x > 4
YZ > 4

### 6.6 Inequalities in Two Triangles

Use the diagram.

Question 17.
If RQ = RS and m∠QRT > m∠SRT, then how does $$\overline{Q T}$$ Compare to $$\overline{S T}$$?
RQ = RS and m∠QRT > m∠SRT
RT = RT by the reflexive property of the congruence theorem
TQ > ST by the converse of the hinge theorem.

Question 18.
If RQ = RS and QT > ST, then how does ∠QRT compare to ∠SRT?
RQ = RS and QT > ST
RT = RT (Reflexive property of the congruence theorem)
So, m∠QRT > m∠SRT (converse of the hinge theorem)

### Relationships Within Triangles Chapter Test

In Exercise 1 and 2, $$\overline{M N}$$ is a midsegment of ∆JKL. Find the value of x.

Question 1.

KL = 0.5(MN)
12 = 0.5x
x = 6

Question 2.

JM = ML
x = 9

Find the indicated measure. Identify the theorem you use.

Question 3.
ST

RS = ST by the perpendicular bisector theorem
3x + 8 = 7x – 4
7x – 3x = 12
4x = 12
x = 3
ST = 7(3) – 4 = 21 – 4 = 17

Question 4.
WY

6x + 2 = 9x – 13
3x = 15
x = 5
WY = 6(5) + 2 = 32

Question 5.
BW

WC = BW = AW (The incenter of a triangle is the intersection point of all the three interior angle bisectors of the triangle)
BW = 20

Copy and complete the statement with <, >, or =.

Question 6.
AB _____ CB
AB > CB

Question 7.
m∠1 _____ m∠2
m∠1 < m∠2

Question 8.
m∠MNP ________ m∠NPM
m∠MNP < m∠NPM

Question 9.
Find the coordinates of the circumcenter, orthocenter, and centroid of the triangle with vertices A(0, – 2), B(4, – 2), and C(0, 6).
Graph the triangle
Draw the bisector of every angle and their intersection to get the circumcenter
Draw the altitudes and their intersections to get the orthocenter
Find the midpoint of all sides and then connect it with the opposite vertex. The intersection of these medians is the centroid.

The circumcenter is (1.5, -0.5)
orthocenter is (0, -2)
centroid is (1.3, 0.7)

Question 10.
Write an indirect proof of the Corollary to the Base Angles Theorem (Corollary 5.2): If ∆PQR is equilateral, then it is equiangular.
AB = BC = AC
If two sides of a triangle are congruent, then the angles opposite them are congruent.
Since AB = AC, the base angle theorem tells that m∠B = m∠C
Since AB = BC, the base angle theorem tells that m∠A = m∠C
By the transitive property of equality m∠A = m∠B = m∠C, which contradicts the assumption that triangle ABC is equiangular.
The corollary is true.

Question 11.
∆DEF is a right triangle with area A. Use the area for ∆DEF to write an expression for the area of ∆GEH. Justify your answer.

The area of ∆DEF = A
As HG is the midsegment of ∆DEF.
Area of ∆GEH = A/2

Question 12.
Two hikers start at a visitor center. The first hikes 4 miles hikes due west, then turns 40° toward south and hikes 1.8 miles. The second hikes 4 miles due east, then turns 52° toward north and hikes 1 .8 miles. Which hiker is farther from the visitor center? Explain how you know.
The second hiker is farther away from the visitor center as all things being equal cos 128 < cos 140 from distance² = 4² + 1.8² — 2 x 4 x 1.8 cos (angle turned)

In Exercises 13-15, use the map.

Question 13.
Describe the possible lengths of Pine Avenue.
7 + 9 > x
16 > x
7 + x > 9
x > 2
9 + x > 7
x > -2
2 < x < 16
The possible lengths of the pine avenue are between 2 and 16.

Question 14.
You ride your bike along a trail that represents the shortest distance from the beach to Main Street. You end up exactly halfway between your house and the movie theatre. How long is Pine Avenue? Explain.
halfway between your house and the movie theatre = 7/2 = 3.5
Pine Avenue is 3.5 mi

Question 15.
A market is the same distance from your house, the movie theater, and the beach. Copy the map and locate the market.
Centroid is the market.

### Relationships Within Triangles Cumulative Assessment

Question 1.
Which definitions(s) and/or theorem(s) do you need to use to prove the Converse of the Perpendicular Bisector Theorem (Theorem 6.2)? Select all that apply.
Given CA = CB
Prove Point C lies on the perpendicular bisector of AB.

 Definition of perpendicular bisector Definition of angle bisector Definition of segment congruence Definition of angle congruence Base Angles Theorem (Theorem 5.6) Converse of the Base Angles Theorem (Theorem 5.7) ASA Congruence Theorem (Theorem 5.10) AAS Congruence Theorem (Theorem 5.11)

The definition of perpendicular bisector is used to prove the Converse of the Perpendicular Bisector Theorem.

Question 2.
Use the given information to write a two-column proof.
Given $$\overline{Y G}$$ is the perpendicular bisector of $$\overline{D F}$$.
Prove ∆DEY ≅ ∆FEY

Question 3.
What are the coordinates of the centroid of ∆LMN?

(A) (2, 5)
(B) (3, 5)
(C) (4, 5)
(D) (5, 5)
The coordinates of L(3, 8), M (1, 5), N(5, 2)
Centroid = ($$\frac { 3 + 1 + 5 }{ 3 }$$, $$\frac { 8 + 5 + 2 }{ 3 }$$)
= (3, 5)
Option B is the correct answer.

Question 4.
Use the steps in the construction to explain how you know that the circle is circumscribed about ∆ABC.

Draw a perpendicular bisector for AB
Draw a perpendicular bisector for BC
Extend those bisectors to meet at one point.
By taking the point of intersection of perpendicular bisectors as the radius, draw a circle that connects three vertices of the triangle.

Question 5.
Enter the missing reasons in the proof of the Base Angles Theorem (Theorem 5.6).

Given $$\overline{A B} \cong \overline{\Lambda C}$$
Prove ∠B = ∠C

 Statements Reasons 1. Draw $$\overline{A D}$$, the angle bisector of ∠CAB 1. Construction of angle bisector 2. ∠CAD ≅ ∠BAD 2. ________________________ 3. $$\overline{\Lambda B} \cong \overline{A C}$$ 3. ________________________ 4. $$\overline{D A} \cong \overline{D A}$$ 4. ________________________ 5. ∆ADB ≅ ∆ADC 5. ________________________ 6. ∠B ≅ ∠C 6. ________________________

 Statements Reasons 1. Draw $$\overline{A D}$$, the angle bisector of ∠CAB 1. Construction of angle bisector 2. ∠CAD ≅ ∠BAD 2. Angle Bisector Congruence Theorem 3. $$\overline{\Lambda B} \cong \overline{A C}$$ 3. Given 4. $$\overline{D A} \cong \overline{D A}$$ 4. Reflexive property of congruence theorem 5. ∆ADB ≅ ∆ADC 5. SAS congruence theorem 6. ∠B ≅ ∠C 6. Angle congruence theorem

Question 6.
Use the graph of ∆QRS.

a. Find the coordinates of the vertices of the midsegment triangle. Label the vertices T, U, and V.
The coordinates of Q(-3, 8), R(3, 6), S(1, 2)
The midpoint of QR = T = ($$\frac { -3 +3 }{ 2 }$$, $$\frac { 8 + 6 }{ 2 }$$) = (0, 7)
Midpoint of RS = U = ($$\frac { 3 + 1 }{ 2 }$$, $$\frac { 6 + 2 }{ 2 }$$) = (2, 4)
Midpoint of SR = V = ($$\frac { 1 – 3 }{ 2 }$$, $$\frac { 2 + 8 }{ 2 }$$) = (-1, 5)

b. Show that each midsegment joining the midpoints of two sides is parallel to the third side and is equal to half the length of the third side.
The slope of RS = $$\frac { 2 – 6 }{ 1 – 3 }$$ = 2
The slope of TV = $$\frac { 5 – 7 }{ -1 – 0 }$$ = 2
As the slopes are same. The lines are parallel.
TV = √(-1 – 0)² + (5 – 7)² = √1 + 4 = √5
RS = √(1 – 3)² + (2 – 6)² = √4 + 16 = √20
TV = 0.5(RS)

Question 7.
A triangle has vertices X(- 2, 2), Y(1, 4), and Z(2, – 2). Your friend claims that a translation of (x, y) → (x + 2, y – 3) and a translation by a scale factor of 3 will produce a similarity transformation. Do you support our friend’s claim? Explain our reasoning.
A triangle has vertices X(- 2, 2), Y(1, 4), and Z(2, – 2).
The slope of XY
m1 = 1-(-2)/4-2 = 3/2
The slope of the side AB,
m2 = 3-0/(1-(-1)
m2 = 3/2 = 3/2
The slope of the side YZ,
m3 = 2-1/(-2-4) = -1/6
The slope of the side BC
m4 = 4-3/-5-1 = -1/6
The slope of the side XZ
m5 = 4-(-2)/(-2-2) = 4/-4 = -1
The slope of the side AC
m6 = 4-0/-5-(-1) = 4/-4 = -1
m1 = m2
m3 = m4
m5 = m6
XY || AB, YZ || BC and XZ || AC
ΔABC is a similar transformation of the triangle ΔXYZ.

Question 8.
The graph shows a dilation of quadrilateral ABCD b a scale factor of 2. Show that the line containing points B and D is parallel to the line Containing points B’ and D’.

The coordinates of B(4, 1), D(1, -1), B'(8, 2), D'(2, -2)
The slope of BD = $$\frac { -1 – 1 }{ 1 – 4 }$$ = $$\frac { 2 }{ 3 }$$
The slope of B’D’ = $$\frac { -2 – 2 }{ 2 – 8 }$$ = $$\frac { -4 }{ -6 }$$ = $$\frac { 2 }{ 3 }$$
The slopes are equal. So, BD is parallel to B’D’.

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### Parallel and Perpendicular Lines Maintaining Mathematical Proficiency

Find the slope of the line.

Question 1.

The given points are A (-1, 2), and B (3, -1)
Compare the given points with A (x1, y1), B (x2, y2)
m = $$\frac{y2 – y1}{x2 – x1}$$
Substitute A (-1, 2), and B (3, -1) in the formula.
m = $$\frac{-1 – 2}{3 + 1}$$ = $$\frac{-3}{4}$$
So, slope of the given line is $$\frac{-3}{4}$$

Question 2.

The given points are A (-2, 2), and B (-3, -1)
Compare the given points with A (x1, y1), B (x2, y2)
Substitute A (-2, 2), and B (-3, -1) in the formula.
m = $$\frac{y2 – y1}{x2 – x1}$$
m = $$\frac{-1 – 2}{-3 + 2}$$
m = $$\frac{-3}{-1}$$
m = 3
So, the slope of the given line is 3.

Question 3.

The given points are A (-3, -2), and B (1, -2)
Compare the given points with A (x1, y1), B (x2, y2)
m = $$\frac{y2 – y1}{x2 – x1}$$
m = $$\frac{-2 + 2}{3 + 1}$$
m = $$\frac{0}{4}$$
m = 0
So, the slope of the given line is 0.

Write an equation of the line that passes through the given point and has the given slope.

Question 4.
(6, 1); m = – 3
Given, (6, 1) and m = -3
The equation of a line y = mx + b
Where,
m is the slope
b is the y-intercept
Substitute the values in the equation.
y = -3x + b
To find the value of b,
1 = -3 (6) + b
1 = -18 + b
1 + 18 = b
b = 19
So, the equation of the line along with the y-intercept is y = -3x + 19

Question 5.
(-3, 8); m = – 2
Given (-3, 8) and m = -2
The equation of a line is y = mx + b
Where,
m is the slope
b is the y-intercept
y = -2x + b
Substitute the given point in the equation.
x = -3 and y = 8
8 = -2 (-3) + b
8 = 6 + b
8 – 6 = b
b = 2
So, the equation of the line along with the y-intercept is y = -2x + 2

Question 6.
(- 1, 5); m = 4
Given (-1, 5) and m = 4
The equation of a line is y = mx + b
Where,
m is the slope
b is the y-intercept
y = 4x + b
Substitute the given point in the equation
5 = 4 (-1) + b
5 = -4 + b
5 + 4 = b
b = 9
So, the equation of the line along with the y-intercept is y = 4x + 9

Question 7.
(2, – 4); m = $$\frac{1}{2}$$
Given (2, -4) and m = $$\frac{1}{2}$$
The equation of a line is y = mx + b
Where,
m is the slope
b is the y-intercept
y = $$\frac{1}{2}$$x + b
Substitute the given point in the equation.
-4 = $$\frac{1}{2}$$ (2) + b
-4 = 1 + b
-4 – 1 = b
b = -5
So, the equation of the line along with the y-intercept is y = $$\frac{1}{2}$$x – 5

Question 8.
(- 8, – 5); m = –$$\frac{1}{4}$$
Given (-8, -5) and m = –$$\frac{1}{4}$$
The equation of a line is y = mx + b
Where,
m is the slope
b is the y-intercept
y = –$$\frac{1}{4}$$x + b
Substitute the given point in the equation
-5 = –$$\frac{1}{4}$$ (-8) + b
-5 = 2 + b
-5 – 2 = b
b = -7
So, the equation of the line along with the y-intercept is y = –$$\frac{1}{4}$$x – 7

Question 9.
(0, 9); m = $$\frac{2}{3}$$
Given (0, 9) and m = $$\frac{2}{3}$$
The equation of a line is y = mx + b
Where,
m is the slope
b is the y-intercept
y = $$\frac{2}{3}$$x + b
Substitute the given point in the equation
9 = $$\frac{2}{3}$$ (0) + b
9 = 0 + b
9 – 0 = b
b = 9
So, the equation of the line along with the y-intercept is y = $$\frac{2}{3}$$x + 9

Question 10.
ABSTRACT REASONING
Why does a horizontal line have a slope of 0, but a vertical line has an undefined slope?
m = $$\frac{y2 – y1}{x2 – x1}$$
For a horizontal line,
The coordinates of y are the same. i.e.,
y1 = y2 = y3  ………
For a vertical line,
The coordinates of x are the same. i.e.,
x1 = x2 = x3 …….
The slope of the horizontal line (m) = $$\frac{y2 – y2}{x2 – x1}$$
The slope of the horizontal line (m) = 0
The slope of vertical line (m) = $$\frac{y2 – y1}{x2 – x1}$$
The slope of the vertical line (m) = Undefined

### Parallel and Perpendicular Lines Mathematical Practices

Use a graphing calculator to graph the pair of lines. Use a square viewing window. Classify the lines as parallel, perpendicular, coincident, or non-perpendicular intersecting lines. Justify your answer.

Question 1.
x + 2y = 2
2x – y = 4
The pair of lines are
x + 2y = 2
2x – y = 4
x + 2y = 2 ⇒ 2y = -x + 2 ⇒ y = -1/2 x + 1
2x – y = 4 ⇒ y = 2x – 4
The representation of the given pair of lines in the coordinate plane is:

The product of the slope of the first line and the slope of the second line will be equal to -1
So,
By comparing the given pair of lines with
y = mx + b
We get
The slope of first line (m1) = –$$\frac{1}{2}$$
The slope of the second line (m2) = 2
m1 × m2 = –$$\frac{1}{2}$$ × 2
m1m2 = -1
The given pair of lines are perpendicular lines.

Question 2.
x + 2y = 2
2x + 4y = 4
The given pair of lines are:
x + 2y = 2 ⇒ 2y = -x + 2 ⇒ y = -1/2 x + 1
2x + 4y = 4 ⇒ y = -1/2 x + 1

y = mx + b
m1 = –$$\frac{1}{2}$$, b1 = 1
m2 = –$$\frac{1}{2}$$, b2 = 1
So, the given pair of lines are coincident lines.

Question 3.
x + 2y = 2
x + 2y = – 2
The given pair of lines are:
x + 2y = 2 ⇒ y = -1/2 x + 1
x + 2y = -2 ⇒ y = -1/2 x – 1

y = mx + b
We get
m1 = –$$\frac{1}{2}$$, b1 = 1
m2 = $$\frac{1}{2}$$, b2 = -1
From the graph, we observe that the given pair of lines are parallel lines

Question 4.
x – 2y = 2
x – y = – 4
The given pair of lines are:
x – 2y = 2 ⇒ y = 1/2 x – 1
x – y = -4 ⇒ y = x + 4
Hence,
The representation of the given pair of lines in the coordinate plane is:

y = mx + b
The slope of first line (m1) = $$\frac{1}{2}$$
The slope of the second line (m2) = 1
m1m2  = $$\frac{1}{2}$$
From the above graph we observe that the given pair of lines are non-perpendicular lines

### 3.1 Pairs of Lines and Angles

Exploration 1

Points of intersection

work with a partner: Write the number of points of intersection of each pair of coplanar lines.

a. The points of intersection of parallel lines:
The Parallel lines have the same slope but have different y-intercepts
We can say that any parallel line does not intersect at any point.
Parallel lines are those lines that are equidistant from each other and never meet.
We can say that the number of points of intersection of parallel lines is 0

b. The points of intersection of intersecting lines:
When two or more lines cross each other in a plane, they are called intersecting lines.
The number of points of intersection of intersecting lines is: 1

c. The points of intersection of coincident lines:
Intersecting lines have one point in common and coincident lines have infinitely many points in common.
Any coincident line does not intersect at any point or intersect at 1 point.
The number of points of intersection of coincident lines is: 0 or 1

Exploration 2

Classifying Pairs of Lines
Work with a partner: The figure shows a right rectangular prism. All its angles are right angles. Classify each of the following pairs of lines as parallel, intersecting, coincident, or skew. Justify your answers. (Two lines are skew lines when they do not intersect and are not coplanar.)

The given rectangular prism is:

Parallel lines are those lines that are equidistant from each other and never meet.
Intersecting lines are those lines that intersect with each other and are in the same plane.
Coincident lines are the lines that lie on each other and in the same plane.
Skew lines are lines that do not intersect in the same plane and do not intersect.

Exploration 3
Identifying Pairs of Angles
Work with a partner: In the figure, two parallel lines are intersected by a third line called a transversal.

a. Identify all the pairs of vertical angles. Explain your reasoning.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results.
The angles that are opposite to each other when two lines cross are called “Vertical angles”
From the figure,
∠1 and ∠3 are vertical angles
∠2 and ∠4 are vertical angles
∠5 and ∠7 are vertical angles
∠6 and ∠8 are vertical angles

b. Identify all the linear pairs of angles. Explain your reasoning.
A “Linear pair” is a pair of adjacent angles formed when two lines intersect
From the figure,
∠1 and ∠2 linear pair of angles
∠4 and ∠3 linear pair of angles
∠5 and ∠6 linear pair of angles
∠8 and ∠7 linear pair of angles

Question 4.
What does it mean when two lines are parallel, intersecting, coincident, or skew?

• Parallel: The two lines are said to be Parallel when they do not intersect each other and are coplanar.
• Intersecting: The two lines are said to be Intersecting lines when they intersect each other and are coplanar.
• Coincident: The two lines are said to be Coincident lines when they lie on each other and are coplanar. They are neither parallel nor perpendicular lines.
• Skew: The two lines are said to be Skewed when they do not intersect each other and are not parallel.

Question 5.
In Exploration 2. find more pairs of lines that are different from those given. Classify the pairs of lines as parallel, intersecting, coincident, or skew. Justify your answers.
The given rectangular prism of Exploration 2 is:

The pair of lines that are different from the given pair of lines in Exploration 2 are:
a. $$\overline{C D}$$ and $$\overline{A E}$$
b. $$\overline{D H}$$ and $$\overline{F G}$$
a. $$\overline{C D}$$ and $$\overline{A E}$$ are “Skew lines” because they are not intersecting and are non-coplanar
b. $$\overline{D H}$$ and $$\overline{F G}$$ are “Skew lines” because they are not intersecting and are non-coplanar

### Lesson 3.1 Pairs of Lines and Angles

Monitoring Progress

Question 1.
Look at the diagram in Example 1. Name the line(s) through point F that appear skew to .
The line that passes through point F that appears skewed to is: $$\overline{F C}$$

Question 2.
In Example 2, can you use the Perpendicular Postulate to show that is not perpendicular to ? Explain why or why not.
Perpendicular Postulate:
Perpendicular postulate states that if a line and a point are not on the line, then there is exactly one line through the point perpendicular to the given line.
$$\overline{A C}$$ is not perpendicular to $$\overline{B F}$$ because as per the perpendicular Postulate, $$\overline{A C}$$ will be a straight line but it is not a straight line.
Hence we can use the “Perpendicular Postulate” to show that $$\overline{A C}$$ is not perpendicular to $$\overline{B F}$$

Classify the pair of numbered angles.

Question 3.

Two angles that share a common vertex and side are called adjacent angles.
From the figure, we observe that ∠1 and ∠5 are the adjacent angles.

Question 4.

Vertical angles are the angles that are opposite to each other where two lines cross.
From the figure, we observe that ∠2 and ∠7 are the “Vertical angles”

Question 5.

Vertical angles are the angles that are opposite to each other where two lines cross.
From the figure, we observe that ∠4 and ∠5 are the “Vertical angles”

### Exercise 3.1 Pairs of Lines and Angles

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
Two lines that do not intersect and are also not parallel are ________ lines.
Answer: Two lines that do not intersect and are also not parallel are skew lines.

Question 2.
WHICH ONE did DOESN’T BELONG?
Which angle pair does not belong with the other three? Explain our reasoning.

∠2 and ∠3
∠4 and ∠5
∠1 and ∠8
∠2 and∠7
Answer: ∠4 and ∠5 angle-pair does not belong with the other three

Explanation:
The angles that have a common side are called “Adjacent angles”
The angles that are opposite to each other when 2 lines cross are called “Vertical angles”
From the figure, we observe that,
∠2 and ∠3 are vertical angles
∠4 and ∠5 are adjacent angles
∠1 and ∠8 are vertical angles
∠2 and ∠7 are vertical angles
Thus, ∠4 and ∠5 angle-pair does not belong with the other three

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, think of each segment in the diagram as part of a line. All the angles are right angles. Which line(s) or plane(s) contain point B and appear to fit the description?

Question 3.
line(s) parallel to .

Question 4.
line(s) PerPendicular to .
The perpendicular lines are two lines that intersect each other and the angle formed between the two lines should be equal to 90 degrees.
The line that is perpendicular to $$\overline{C D}$$ is: $$\overline{A D}$$ and $$\overline{C B}$$

Question 5.
line(s) skew to

Question 6.
plane(s) parallel to plane CDH
The plane parallel to plane CDH is Plane BAE

In Exercises 7-10, Use the diagram.

Question 7.
Name a pair of parallel lines.

Question 8.
Name a pair of perpendicular lines.
The perpendicular lines are two lines that intersect each other and the angle formed between the two lines should be equal to 90 degrees.
$$\overline{N P}$$ and $$\overline{P O}$$ are perpendicular lines

Question 9.

Question 10.

The perpendicular lines are two lines that intersect each other and the angle formed between the two lines should be equal to 90 degrees.
$$\overline{P R}$$ and $$\overline{P O}$$ are not perpendicular lines.

In Exercises 11-14, identify all pairs of angles of the given type.

Question 11.
corresponding

Question 12.
alternate interior
Alternate interior angles are the angles formed when a transversal intersects two coplanar lines.
So, the alternate interior angles are: ∠4 and ∠5; ∠3 and ∠6

Question 13.
alternate exterior

Question 14.
consecutive interior
The pair of angles on one side of the transversal and inside the two lines are called the Consecutive interior angles.
So, the consecutive interior angles are: ∠3 and ∠5; ∠4 and ∠6

USING STRUCTURE
In Exercises 15-18, classify the angle pair as corresponding. alternate interior, alternate exterior, or consecutive interior angles.

Question 15.
∠5 and ∠1

Question 16.
∠11 and ∠13
The pair of angles on one side of the transversal and inside the two lines are called the Consecutive interior angles.
So, the consecutive interior angles are: ∠11 and ∠13

Question 17.
∠6 and ∠13

Question 18.
∠2 and ∠11
∠2 and ∠11 are the Vertical angles
The angles that are opposite to each other when 2 lines cross are called “Vertical angles”

ERROR ANALYSIS
In Exercises 19 and 20. describe and correct the error in the conditional statement about lines.

Question 19.

Question 20.

Perpendicular postulate states that if a line and a point are not on the line, then there is exactly one line through the point perpendicular to the given line.
The statement is false.

Question 21.
MODELING WITH MATHEMATICS
Use the photo to decide whether the statement is true or false. Explain Your reasoning.

a. The plane containing the floor of the treehouse is parallel to the ground.
b. The lines containing the railings of the staircase, such as , are skew to all lines in the plane containing the ground.
c. All the lines containing the balusters. such as , are perpendicular to the plane containing the floor of the treehouse.

Question 22.
THOUGHT-PROVOKING
If two lines are intersected by a third line, is the third line necessarily a transversal? Justify your answer with a diagram.
Parallel lines do not intersect each other
Perpendicular lines intersect at each other at right angles
The third intersecting line can intersect at the same point that the two lines have intersected as shown below:

Question 23.
MATHEMATICAL CONNECTIONS
Two lines are cut by a transversal. Is it possible for all eight angles formed to have the same measure? Explain your reasoning.

Question 24.
HOW DO YOU SEE IT?
Think of each segment in the figure as part of a line.

a. Which lines are parallel to ?
The lines that do not intersect each other are called Parallel lines.
The line parallel to $$\overline{N Q}$$ is $$\overline{M P}$$

b. Which lines intersect ?
The lines that are coplanar and any two lines that have a common point are called Intersecting lines.
The lines that intersect are: $$\overline{Q P}$$, $$\overline{N K}$$, $$\overline{N M}$$

c. Which lines are skew to ?
The lines that do not intersect or not parallel and are non-coplanar are called “Skew lines”
The lines that are skew to are $$\overline{K L}$$, $$\overline{L M}$$, and $$\overline{L S}$$

d. Should you have named all the lines on the cube in parts (a)-(c) except $$\overline{N Q}$$? Explain.
No, we did not name all the lines on the cube in parts (a) – (c) except $$\overline{N Q}$$

In exercises 25-28. copy and complete the statement. List all possible correct answers.

Question 25.
∠BCG and __________ are corresponding angles.

Question 26.
∠BCG and __________ are consecutive interior angles.
∠BCG and ∠FCA and ∠BCA are consecutive interior angles.

Question 27.
∠FCJ and __________ are alternate interior angles.

Question 28.
∠FCA and __________ are alternate exterior angles.
∠FCA and ∠JCB are alternate exterior angles.

Question 29.
MAKING AN ARGUMENT
Your friend claims the uneven parallel bars in gymnastics are not really Parallel. She says one is higher than the other. so they cannot be on the same plane. Is she correct? Explain.

Maintaining Mathematical Proficiency

Use the diagram to find the measure of all the angles.

Question 30.
m∠1 = 76°
Sum of the adjacent angles is: 180°
The adjacent angles are: ∠1 and ∠2; ∠2 and ∠3; ∠3 and ∠4; and ∠4 and ∠1
The vertical angles are: ∠1 and ∠3; ∠2 and ∠4
∠1 + ∠2 = 180°
∠2 = 180° – ∠1
= 180° – 76° = 104°
∠2 = 104°
∠3 = 76° and ∠4 = 104°
∠1 = 76°, ∠2 = 104°, ∠3 = 76°, and ∠4 = 104°

Question 31.
m∠2 = 159°

### 3.2 Parallel Lines and Transversals

Exploration 1

Exploring parallel Lines

Work with a partner: Use dynamic geometry software to draw two parallel lines. Draw a third line that intersects both parallel lines. Find the measures of the eight angles that are formed. What can you conclude?

We observe that,
∠3 = 53.7° and ∠4 = 53.7°
The angle measures of the vertical angles are congruent
∠1 = 53.7° and ∠5 = 53.7°
Hence, ∠1 = ∠2 = ∠3 = ∠4 = ∠5 = ∠6 = ∠7 = 53.7°

Exploration 2

Writing conjectures

Work with a partner. Use the results of Exploration 1 to write conjectures about the following pairs of angles formed by two parallel lines and a transversal.
ATTENDING TO PRECISION
To be proficient in math, you need to communicate precisely with others.
a. corresponding angles

When two lines are crossed by another line, the angles in matching corners are called “Corresponding angles”
From the figure,
∠1 and ∠5 are corresponding angles
∠4 and ∠8 are corresponding angles

b. alternate interior angles

Alternate Interior Angles are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal.
From the figure,
∠3 and ∠5 are alternate interior angles
∠2 and ∠8 are alternate interior angles

c. alternate exterior angles

Alternate exterior angles are the pair of angles that lie on the outer side of the two parallel lines but on either side of the transversal line.
From the figure,
∠1 and ∠7 alternate exterior angles
∠6 and ∠4 alternate exterior angles

d. consecutive interior angles

When two lines are cut by a transversal, the pair of angles on one side of the transversal and inside the two lines are called the Consecutive interior angles.
∠2 and ∠5 consecutive interior angles
∠3 and ∠8 consecutive interior angles

Question 3.
When two parallel lines are cut by a transversal, which of the resulting pairs of angles are congruent?
If two parallel lines are cut by a transversal, then the pairs of “Alternate exterior angles” are congruent.
If two parallel lines are cut by a transversal, then the pairs of “Corresponding angles” are congruent.
If two parallel lines are cut by a transversal, then the pairs of “Alternate interior angles” are congruent.

Question 4.
In Exploration 2. m∠1 = 80°. Find the other angle measures.
Given,
m∠1 = 80°
All the angle measures are equal
∠1 = ∠2 = ∠3 = ∠4 = ∠5 = ∠6 = ∠7 = ∠8 = 80°

### Lesson 3.2 Parallel Lines and Transversals

Monitoring Progress

Use the diagram

Question 1.
Given m∠1 = 105°, find m∠4, m∠5, and m∠8. Tell which theorem you use in each case.

Given m∠1 = 105°
To find ∠4:
Verticle angle theorem:
Vertical Angles Theorem states that vertical angles, angles that are opposite each other and formed by two intersecting straight lines, are congruent
m∠1 = m∠4
m∠4 = 105°
To find ∠5:
Alternate Interior angles theorem:
The Alternate Interior Angles Theorem states that, when two parallel lines are cut by a transversal, the resulting alternate interior angles are congruent
m∠4 = m∠5
m∠5 = 105°
To find ∠8:
Verticle angle theorem:
Vertical Angles Theorem states that vertical angles, angles that are opposite each other and formed by two intersecting straight lines, are congruent.
m∠5 = m∠8
m∠8 = 105°

Question 2.
Given m∠3 = 68° and m∠8 = (2x + 4)°, what is the value of x? Show your steps.
Given m∠3 = 68° and m∠8 = (2x + 4)°
∠3 and ∠8 are consecutive exterior angles.
If parallel lines are cut by a transversal line, then consecutive exterior angles are supplementary.
∠3 + ∠8 = 180°
68° + (2x + 4)° = 180°
2x + 72° = 180°
2x° = 180° – 72°
2x° = 108°
x = $$\frac{108}{2}$$
x = 54°

Question 3.
In the proof in Example 4, if you use the third statement before the second statement. could you still prove the theorem? Explain.
If you even interchange the second and third statements, you could still prove the theorem as the second line before interchange is not necessary.
If you use the third statement before the second statement, you could still prove the theorem

Question 4.
WHAT IF?
In Example 5. yellow light leaves a drop at an angle of m∠2 = 41°. What is m∠1? How do you know?
m∠2 = 41°
∠1 and ∠2, then they are alternate interior angles
Now,
According to the Alternate interior angle theorem,
∠1 = ∠2
∠2 = 41°
∠1 = 41°

### Exercise 3.2 Parallel Lines and Transversals

Vocabulary and Core Concept Check

Question 1.
WRITING
How are the Alternate Interior Angles Theorem (Theorem 3.2) and the Alternate Exterior
Angles Theorem (Theorem 3.3) alike? How are they different?

Question 2.
WHICH ONE did DOESN’T BELONG?
Which pair of angle measures does not belong with the other three? Explain.

m∠1 and m∠3
m∠2 and m∠4
m∠2 and m∠3
m∠1 and m∠5

From the figure we can find the vertical, alternate and consecutive angles
∠1 and ∠3 are vertical angles
∠2 and ∠4 are alternate interior angles
∠2 and ∠3 are consecutive interior angles
∠1 and ∠5 are alternate exterior angles
All the angles except ∠1 and ∠3 are the interior and exterior angles
∠1 and ∠3 pair does not belong with the other three.

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, find m∠1 and m∠2. Tell which theorem you use in each case.

Question 3.

Question 4.

From the figure,
∠1 = ∠2 (Vertical Angles theorem)
∠2 = 150° (Alternate exterior angles theorem)
∠1 = ∠2 = 150°

Question 5.

Question 6.

From the figure,
∠1 + ∠2 = 180° (consecutive interior angles theorem)
∠2 = 140° (Vertical angles theorem)
∠1 = 180° – 140°
∠1 = 40°
∠1 = 40° and ∠2 = 140°

In Exercises 7-10. find the value of x. Show your steps.

Question 7.

Question 8.

From the figure,
72° + (7x + 24)° = 180° (Consecutive interior angles theory)
(7x + 24)° = 180° – 72°
(7x + 24)° = 108°
7x° = 108° – 24°
7x° = 84°
x° = $$\frac{84}{7}$$
x° = 12°

Question 9.

Question 10.

From the figure,
(8x + 6)° = 118° (Vertical Angles theorem)
8x° = 118° – 6°
8x° = 112°
x° = $$\frac{112}{8}$$
x° = 14°

In Exercises 11 and 12. find m∠1, m∠2, and m∠3. Explain our reasoning.

Question 11.

Question 12.

From the figure,
∠3 + 133° = 180° (Consecutive Interior angles theorem)
∠3 = 180° – 133°
∠3 = 47°
∠2 + ∠3 = 180°
∠2 = 180° – ∠3
∠2 = 180° – 47°
∠2 = 133°
∠1 = ∠2
∠1 = ∠2 = 133° and ∠3 = 47°

Question 13.
ERROR ANALYSIS
Describe and correct the error in the students reasoning

Question 14.
HOW DO YOU SEE IT?
Use the diagram

a. Name two pairs of congruent angles when $$\overline{A D}$$ and $$\overline{B C}$$ are parallel? Explain your reasoning.
Let the congruent angle be ∠P
From the figure,
The pair of angles when $$\overline{A D}$$ and $$\overline{B C}$$ are parallel is ∠APB and ∠DPB

b. Name two pairs of supplementary angles when $$\overline{A B}$$ and $$\overline{D C}$$ are parallel. Explain your reasoning.
The two pairs of supplementary angles when $$\overline{A B}$$ and $$\overline{D C}$$ are parallel is ∠ACD and ∠BDC

PROVING A THEOREM
In Exercises 15 and 16, prove the theorem.

Question 15.
Alternate Exterior Angles Theorem (Thm. 3.3)

Question 16.
Consecutive Interior Angles Theorem (Thm. 3.4)
Statement: If two parallel lines are cut by a transversal, then the pairs of consecutive interior angles formed are supplementary
Proof:

Question 17.
PROBLEM-SOLVING
A group of campers ties up their food between two parallel trees, as shown. The rope is pulled taut. forming a straight line. Find m∠2. Explain our reasoning.

Question 18.
DRAWING CONCLUSIONS
You are designing a box like the one shown.

a. The measure of ∠1 is 70°. Find m∠2 and m∠3.
b. Explain why ∠ABC is a straight angle.
c. If m∠1 is 60°, will ∠ABC still he a straight angle? Will the opening of the box be more steep or less steep? Explain.

Question 19.
CRITICAL THINKING
Is it possible for consecutive interior angles to be congruent? Explain.

Question 20.
THOUGHT-PROVOKING
The postulates and theorems in this book represent Euclidean geometry. In spherical geometry, all points are points on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. In spherical geometry, is it possible that a transversal intersects two parallel lines? Explain your reasoning.
Answer: It is not possible that a transversal intersects two parallel lines

Explanation:
According to Euclidean geometry,
For a parallel line, there will be no intersecting point
But, In spherical geometry, even though there is some resemblance between circles and lines, there is no possibility to form parallel lines as the lines will intersect at least at 1 point on the circle which is called a tangent.
Hence, it is not possible that a transversal intersects two parallel lines.

MATHEMATICAL CONNECTIONS
In Exercises 21 and 22, write and solve a system of linear equations to find the values of x and y.

Question 21.

Question 22.

From the figure,
2y° + 4x° = 180°
(2x + 12)° + (y + 6)° = 180°
2x° + y° + 18° = 180°
2x° + y° = 180° – 18°
2x° + y° = 162°———(1)
4x° + 2y° = 180°——–(2)
Solve eq. (1) & eq. (2)
2x° = 18°
x° = $$\frac{18}{2}$$
x° = 9°
Substitute the value of x to find y.
y° = 162° – 2 (9°)
y° = 162° – 18°
y° = 144°

Question 23.
MAKING AN ARGUMENT
During a game of pool. your friend claims to be able to make the shot Shown in the diagram by hitting the cue ball so that m∠1 = 25°. Is your friend correct? Explain your reasoning.

Question 24.
REASONING
In diagram. ∠4 ≅∠5 and $$\overline{S E}$$ bisects ∠RSF. Find m∠1. Explain your reasoning.

Given,
∠4 ≅∠5 and $$\overline{S E}$$ bisects ∠RSF
∠FSE = ∠ESR
From ΔESR,
Sum of the angle measures = 180°
∠3 + ∠4 + ∠5 = 180°
∠3 = 60° (∠4 ≅ ∠5 and the triangle is not a right triangle)
From the figure,
∠1 = ∠3 (Corresponding angles theorem)
∠1 = 60°

Maintaining Mathematical Proficiency

Write the converse of the conditional statement. Decide whether it is true or false.

Question 25.
If two angles are vertical angles. then they are congruent.

Question 26.
If you go to the zoo, then you will see a tiger.
Statement: If you go to the zoo, then you will see a tiger
Converse: If you will see a tiger, then you go to the zoo (false).

Question 27.
If two angles form a linear pair. then they are supplementary.

Question 28.
If it is warm outside, then we will go to the park.
Statement: If it is warm outside, then we will go to the park
Converse: If you will go to the park, then it is warm outside (False).

### 3.3 Proofs with Parallel Lines

Exploration 1

Exploring Converses

Work with a partner: Write the converse of each conditional statement. Draw a diagram to represent the converse. Determine whether the converse is true. Justify your conclusion.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to make conjectures and build a logical progression of statements to explore the truth of your conjectures.

a. Corresponding Angles Theorem (Theorem 3.1): If two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent.

Converse:
If the pairs of corresponding angles are congruent, then the two parallel lines are cut by a transversal.
The Converse of the Corresponding Angles Theorem says that if two lines and a transversal form congruent corresponding angles, then the lines are parallel.
When we compare the converses we obtained from the given statement and the actual converse,
The given statement is true.
Proof of Converse of Corresponding Angles Theorem:
Consider the 2 lines L1 and L2 intersected by a transversal line L3 creating 2 corresponding angles 1 and 2 which are congruent
We want to prove L1 and L2 are parallel and we will prove this by using “Proof of Contradiction”
Let us assume L1 is not parallel to L2
According to the parallel line axiom, there is a different line than L2 that passes through the intersection point of L2 and L3, which is parallel to L1.
Let’s draw that line, and call it P. Let’s also call the angle formed by the traversal line and this new line angle 3, and we see that if we add some other angle, call it angle 4, to it, it will be the same as ∠2.
P || L1
∠1 ≅ ∠3,
m∠1=m∠3
∠1 and ∠2 are congruent
This contradiction means our assumption that L1 is not parallel to L2 is false, and so L1 must be parallel to L2.
The representation of the Converse of Corresponding Angles Theorem is:

b. Alternate Interior Angles Theorem (Theorem 3.2): If two parallel lines are cut by a transversal, then the pairs of alternate interior angles are congruent.

Converse:
If the pairs of alternate interior angles are congruent, then the two parallel lines are cut by a transversal.
The converse of the Alternate Interior Angles Theorem states that if two lines are cut by a transversal and the alternate interior angles are congruent, then the lines are parallel.
When we compare the actual converse and the converse according to the given statement,
The statement is false.

c. Alternate Exterior Angles Theorem (Theorem 3.3): If two parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent.

Converse:
If the pairs of alternate exterior angles are congruent, then the two parallel lines are cut by a transversal.
The converse of the Alternate Exterior Angles Theorem states that if alternate exterior angles of two lines crossed by a transversal are congruent, then the two lines are parallel.
When we compare the converses we obtained from the given statement and the actual converse,
The given statement is true
Proof of Alternate exterior angles Theorem:
Given: ∠1 ≅ ∠2
Prove: l || m
The flow proof for the Converse of Alternate exterior angles Theorem is:

Converse of the Exterior angles Theorem:

d. Consecutive Interior Angles Theorem (Theorem 3.4): If two parallel lines are cut by a transversal. then the pairs of consecutive interior angles are supplementary.

Converse:
If the pairs of consecutive interior angles are supplementary, then the two parallel lines are cut by a transversal
Converse of the consecutive Interior angles Theorem” states that if the consecutive interior angles on the same side of a transversal line intersecting two lines are supplementary, then the two lines are parallel.
When we compare the conversations we obtained from the given statement and the actual converse
We can conclude that the converse we obtained from the given statement is true
Proof of the Converse of the Consecutive Interior angles Theorem:
a.  m∠5 + m∠4 = 180° (From the given statement)
b.  m∠1 + m∠4 = 180° (Linear pair of angles are supplementary)
c. m∠5=m∠1 // (1), (2), transitive property of equality
d.  AB||CD (Converse of the Corresponding Angles Theorem)

Question 2.
For which of the theorems involving parallel lines and transversals is the converse true?
The theorems involving parallel lines and transversals that the converse is true are:
a. Corresponding Angles Theorem
b. Alternate Exterior angles Theorem
c. Consecutive Interior angles Theorem

Question 3.
In Exploration 1, explain how you would prove any of the theorems that you found to be true.
For the proofs of the theorems that you found to be true, refer to Exploration 1

### Lesson 3.3 Proofs with Parallel Lines

Monitoring Progress

Question 1.
Is there enough information in the diagram to conclude that m || n? Explain.

The given angles are the consecutive exterior angles
m || n
Converse of the Consecutive Exterior angles Theorem:
a.  m∠1 + m∠8 = 180° (From the given statement)
b.  m∠1 + m∠4 = 180° (Linear pair of angles are supplementary)
c. m∠5=m∠1 // (1), (2), transitive property of equality
d.  AB||CD (Converse of the Corresponding Angles Theorem)
The representation of the Converse of the Consecutive Interior angles Theorem is:

Question 2.
Explain why the Corresponding Angles Converse is the converse of the Corresponding Angles Theorem (Theorem 3.1).
Corresponding Angles Postulate states that, when two parallel lines are cut by a transversal, the resulting corresponding angles are congruent
Converse: When the corresponding angles are congruent, the two parallel lines are cut by a transversal
The Converse of Corresponding Angles Theorem:
If the corresponding angles formed are congruent, then two lines l and m are cut by a transversal.
When we observe the Converse of the Corresponding Angles Theorem we obtained and the actual definition, both are the same.
The corresponding Angles Converse is the converse of the Corresponding Angles Theorem

Question 3.
If you use the diagram below to prove the Alternate Exterior Angles Converse. what Given and Prove statements would you use?

It is given that the given angles are the alternate exterior angles.
Alternate Exterior angle Theorem:
If two parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent
Converse of the Alternate Exterior Angles Theorem:
Converse of the Alternate Exterior Angles Theorem states that if alternate exterior angles of two lines crossed by a transversal are congruent, then the two lines are parallel.
For the Converse of the alternate exterior angles Theorem,
The given statement is: ∠1 ≅ 8
To prove: l || k

Question 4.
Copy and complete the following paragraph proof of the Alternate Interior Angles Converse using the diagram in Example 2.
It is given that ∠4 ≅∠5. By the _______ . ∠1 ≅ ∠4. Then by the Transitive Property of Congruence (Theorem 2.2), _______ . So, by the _______ , g || h.
It is given that ∠4 ≅∠5. By the Vertical Angles Congruence Theorem (Theorem 2.6). ∠1 ≅ ∠4. Then by the Transitive Property of Congruence (Theorem 2.2), ∠1 ≅∠5. So, by the Corresponding Angles Converse, g || h.

Question 5.
Each step is parallel to the step immediately above it. The bottom step is parallel to the ground. Explain why the top step is parallel t0 the ground.

The steps are intersecting each other
In the same way, when we observe the floor from any step,
They are also parallel.
The top step is also parallel to the ground since they do not intersect each other at any point.

Question 6.
In the diagram below. p || q and q || r. Find m∠8. Explain your reasoning.

The given angles are the consecutive exterior angles
∠8 + 115° = 180°
∠8 = 180° – 115°
∠8 = 65°

### Exercise 3.3 Proofs with Parallel Lines

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Two lines are cut by a transversal. Which angle pairs must be congruent for the lines to be parallel?

Question 2.
WRITING
Use the theorems from Section 3.2 and the converses of those theorems in this section to write three biconditional statements about parallel lines and transversals.

Monitoring Progress and Modeling with Mathematics

In Exercises 3-8. find the value of x that makes m || n. Explain your reasoning.

Question 3.

Question 4.

According to the Corresponding Angles Theorem,
(2x + 15)° = 135°
2x° = 135° – 15°
2x° = 120°
x° = $$\frac{120}{2}$$
x° = 60°

Question 5.

Question 6.

According to the Corresponding Angles Theorem,
(180 – x)° = x°
180° = x°  + x°
2x° = 180°
x° = $$\frac{180}{2}$$
x° = 90°

Question 7.

Question 8.

According to the Corresponding Angles Theorem,
(2x + 20)° = 3x°
20° = 3x°  – 2x°
x° = 20°

In Exercises 9 and 10, use a compass and straightedge to construct a line through point P that is parallel to line m.

Question 9.

Question 10.

Let A and B be two points on line m.
Draw $$\overline{A P}$$ and construct an angle ∠1 on n at P so that ∠PAB and ∠1 are corresponding angles

PROVING A THEOREM
In Exercises 11 and 12. prove the theorem.
Question 11.
Alternate Exterior Angles Converse (Theorem 3.7)

Question 12.
Consecutive Interior Angles Converse (Theorem 3.8)
Proof of the Converse of the Consecutive Interior angles Theorem:
Given: m∠5 + m∠4 = 180°
Prove: AB || CD

Now,
a.  m∠5 + m∠4 = 180° (From the given statement)
b.  m∠1 + m∠4 = 180° (Linear pair of angles are supplementary)
c. m∠5=m∠1 // (1), (2), transitive property of equality
d.  AB||CD (Converse of the Corresponding Angles Theorem)

In Exercises 13-18. decide whether there is enough information to prove that m || n. If so, state the theorem you would use.

Question 13.

Question 14.

Yes, there is enough information to prove m || n
The theorem we can use to prove that m || n is: Alternate Exterior angles Converse theorem

Question 15.

Question 16.

No, there is not enough information to prove m || n

Question 17.

Question 18.

Yes, there is enough information to prove m || n

ERROR ANALYSIS
In Exercises 19 and 20, describe and correct the error in the reasoning.

Question 19.

Question 20.

∠1 and ∠2 are Consecutive Interior angles
It also shows that a and b are cut by a transversal and they have the same length.
From the converse of the Consecutive Interior angles Theorem,
a || b

In Exercises 21-24. are and parallel? Explain your reasoning.

Question 21.

Question 22.

From the figure,
The sum of the given angle measures = 180°
The given angles are consecutive exterior angles
From the Consecutive Exterior angles Converse,
AC || DF

Question 23.

Question 24.

From the figure,
We can observe that the sum of the angle measures of all the pairs i.e., (115 + 65)°, (115 + 65)°, and (65 + 65)° is not 180°
The sum of the angle measures is not supplementary, according to the Consecutive Exterior Angles Converse,
AC is not parallel to DF

Question 25.
ANALYZING RELATIONSHIPS
The map shows part of Denser, Colorado, Use the markings on the map. Are the numbered streets parallel to one another? Explain your reasoning.

Question 26.
ANALYZING RELATIONSHIPS
Each rung of the ladder is parallel to the rung directly above it. Explain why the top rung is parallel to the bottom rung.

Given,
Each rung of the ladder is parallel to the rung directly above it.
The rungs are not intersecting at any point i.e., they have different points
The parallel lines do not have any intersecting points
We can conclude that the top rung is parallel to the bottom rung.

Question 27.
MODELING WITH MATHEMATICS
The diagram of the control bar of the kite shows the angles formed between the Control bar and the kite lines. How do you know that n is parallel to m?

Question 28.
REASONING
Use the diagram. Which rays are parallel? Which rays are not parallel? Explain your reasoning.

Question 29.
ATTENDING TO PRECISION
Use the diagram. Which theorems allow you to conclude that m || n? Select all that apply. Explain your reasoning.

(A) Corresponding Angles Converse (Thm 3.5)
(B) Alternate Interior Angles Converse (Thm 3.6)
(C) Alternate Exterior Angles Converse (Thm 3.7)
(D) Consecutive Interior Angles Converse (Thm 3.8)

Question 30.
MODELING WITH MATHEMATICS
One way to build stairs is to attach triangular blocks to angled support, as shown. The sides of the angled support are parallel. If the support makes a 32° angle with the floor, what must m∠1 so the top of the step will be parallel to the floor? Explain your reasoning.

It is given that the sides of the angled support are parallel and the support makes a 32° angle with the floor.
To make the top of the step where ∠1 is present to be parallel to the floor, the angles must be “Alternate Interior angles”
Alternate Interior angles are congruent.
∠1 = 32°

Question 31.
ABSTRACT REASONING
In the diagram, how many angles must be given to determine whether j || k? Give four examples that would allow you to conclude that j || k using the theorems from this lesson.

Question 32.
THOUGHT-PROVOKING
Draw a diagram of at least two lines cut by at least one transversal. Mark your diagram so that it cannot be proven that any lines are parallel. Then explain how your diagram would need to change in order to prove that lines are parallel.
The diagram that represents the figure that it can not be proven that any lines are parallel is:

From the above,
The diagram can be changed by the transformation of transversals into parallel lines and a parallel line into transversal

PROOF
In Exercises 33-36, write a proof.

Question 33.
Given m∠1 = 115°, m∠2 = 65°
Prove m||n

Question 34.
Given ∠1 and ∠3 are supplementary.
Prove m||n

Given: ∠1 and ∠3 are supplementary
To Prove: m || n

Question 35.
Given ∠1 ≅ ∠2, ∠3 ≅ ∠4
Prove $$\overline{A B} \| \overline{C D}$$

Question 36.
Given a||b, ∠2 ≅ ∠3
Prove c||d

Given: a || b, ∠2 ≅ ∠3
Prove: c || d
Hence,

Question 37.
MAKING AN ARGUMENT

Question 38.
HOW DO YOU SEE IT?
Are the markings on the diagram enough to conclude that any lines are parallel? If so. which ones? If not, what other information is needed?

∠1 and ∠4 are the pairs of corresponding angles
∠2 and ∠3 are the pairs of corresponding angles
According to the Converse of the Corresponding angles Theorem,
If the corresponding angles are congruent, then the two lines that are cut by a transversal are parallel lines
We can say that p and q; r and s are the pairs of parallel lines

Question 39.
PROVING A THEOREM
Use these steps to prove the Transitive Property of Parallel Lines Theorem
a. Cops the diagram with the Transitive Property of Parallel Lines Theorem on page 141.
b. Write the Given and Prove statements.
c. Use the properties of angles formed by parallel lines cut by a transversal to prove the theorem.

Question 40.
MATHEMATICAL CONNECTIONS
Use the diagram

a. Find the value of x that makes p || q.
From the figure,
p || q,
(2x + 2)° and (x + 56)°
We can observe that the given angles are corresponding angles
(2x + 2)° = (x + 56)°
2x – x = 56° – 2°
x° = 54°

b. Find the value of y that makes r || s.
r || s,
The angles are (y + 7)° and (3y – 17)°
We can observe that the given angles are corresponding angles
(y + 7)° = (3y – 17)°
y – 3y = -17° – 7°
-2y° = -24°
y = $$\frac{24}{2}$$
y = 12°

c. Can r be parallel to s and can p, be parallel to q at the same time? Explain your reasoning.
No, p ||q and r ||s will not be possible at the same time because when p || q, r, and s can act as transversal and when r || s, p, and q can act as transversal

Maintaining Mathematical Proficiency
Use the Distance Formula to find the distance between the two points.

Question 41.
(1, 3) and (- 2, 9)

Question 42.
(- 3, 7) and (8, – 6)
(-3, 7), and (8, -6)
Compare the given points with (x1, y1), and (x2, y2)
We know that,
d = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
d = $$\sqrt{(8 + 3)² + (7 + 6)²}$$
d = $$\sqrt{(11)² + (13)²}$$
d = $$\sqrt{290}$$
d = 17.02

Question 43.
(5, – 4) and (0, 8)

Question 44.
(13, 1) and (9, – 4)
(13, 1), and (9, -4)
Compare the given points with (x1, y1), and (x2, y2)
We know that,
d = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
d = $$\sqrt{(13 – 9)² + (1 + 4)²}$$
d = $$\sqrt{(4)² + (5)²}$$
d = $$\sqrt{41}$$
d = 6.40

### 3.1 – 3.3 Study Skills: Analyzing Your Errors

Mathematical Practices

Question 1.
Draw the portion of the diagram that you used to answer Exercise 26 on page 130.

Question 2.
In Exercise 40 on page 144. explain how you started solving the problem and why you started that way.
You started solving the problem by considering the 2 lines parallel and two lines as transversals
If p and q are the parallel lines, then r and s are the transversals
If r and s are the parallel lines, then p and q are the transversals

### 3.1 – 3.3 Quiz

Think of each segment in the diagram as part of a line. Which lines(s) or plane(s) contain point G and appear to fit the description?

Question 1.
line(s) parallel to .
The line parallel to is $$\overline{D H}$$

Question 2.
line(s) perpendicular to .
The lines perpendicular to are $$\overline{F B}$$ and $$\overline{F G}$$

Question 3.
line(s) skew to .
The lines skew to are $$\overline{C D}$$, $$\overline{C G}$$, and $$\overline{A E}$$

Question 4.
The plane parallel to plane ADE is plane GCB

Identify all pairs of angles of the given type.

Question 5.
consecutive interior
When two lines are cut by a transversal, the pair of angles on one side of the transversal and inside the two lines are known as consecutive interior angles.
The consecutive interior angles are 3 and 5, 4 and 6

Question 6.
alternate interior
Alternate Interior Angles are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal.
The alternate interior angles are 3 and 6, 4 and 5

Question 7.
corresponding
When two lines are crossed by another line, the angles in matching corners are called corresponding angles.
The corresponding angles are ∠1 and ∠5, ∠3 and ∠7, ∠2 and ∠4, ∠6 and ∠8.

Question 8.
alternate exterior
Alternate exterior angles are the pair of angles that lie on the outer side of the two parallel lines but on either side of the transversal line.
The alternate exterior angles are ∠1 and ∠8, ∠7 and ∠2

Find m∠1 and m∠2. Tell which theorem you use in each case.

Question 9.

From the figure,
By using the linear pair theorem,
∠1 + 138° = 180°
∠1 = 180° – 138°
∠1 = 42°
By using the Alternate Exterior Angles Theorem,
∠1 = ∠2
∠1 = ∠2 = 42°

Question 10.

From the figure,
By using the Vertical Angles Theorem,
∠2 = 123°
By using the vertical Angles Theorem,
∠1 = ∠2
∠1 = ∠2 = 123°

Question 11.

From the figure,
By using the linear pair theorem,
∠1 + 57° = 180°
∠1 = 180° – 57°
∠1 = 123°
By using the consecutive interior angles theorem,
∠1 + ∠2 = 180°
∠2 = 180° – 123°
∠2 = 57°
∠1 = 123° and ∠2 = 57°

Decide whether there is enough information to prove that m || n. If so, state the theorem you would use.

Question 12.

By using the Consecutive Interior angles Converse
If the angle measure of the angles is a supplementary angle, then the lines cut by a transversal are parallel
69° + 111° = 180°
Hence, m || n by using the Consecutive Interior angles Theorem.

Question 13.

By using the Corresponding Angles Theorem,
If the corresponding angles are congruent, then the lines cut by a transversal are parallel
Hence, m || n by using the Corresponding Angles Theorem

Question 14.

From the figure,
It is given that l || m and l || n
By using the parallel lines property
If a || b and b || c, then a || c
Hence m || n

Question 15.
Cellular phones use bars like the ones shown to indicate how much signal strength a phone receives from the nearest service tower. Each bar is parallel to the bar directly next to it.

a. Explain why the tallest bar is parallel to the shortest bar.
From the given bars,
We can observe that there is no intersection between any bars
If we represent the bars in the coordinate plane, we can observe that the number of intersection points between any bar is 0
The number of intersection points for parallel lines is 0
The tallest bar is parallel to the shortest bar.

b. Imagine that the left side of each bar extends infinitely as a line.
If m∠1 = 58°, then what is m∠2?
Given, m∠1 = 58°
∠1 and ∠2 are the consecutive interior angles
The sum of the angle measure between 2 consecutive interior angles is 180°
∠1 + ∠ 2 = 180°
58 + ∠ 2 = 180°
∠2 = 180° – 58°
∠2 = 122°

Question 16.
The diagram shows lines formed on a tennis court.

a. Identify two pairs of parallel lines so that each pair is in a different plane.
There are a total of 5 lines.
The two pairs of parallel lines so that each pair is in a different plane are q and p, k and m

b. Identify two pairs of perpendicular lines.
There are 2 perpendicular lines
The two pairs of perpendicular lines are l and n

c. Identify two pairs of skew line
There are 2 pairs of skew lines
The 2 pairs of skew lines are q and p, l and m

d. Prove that ∠1 ≅ ∠2.
∠1 and ∠2 are the alternate exterior angles
If the line cut by a transversal is parallel, then the corresponding angles are congruent
∠1 ≅ ∠2 (Alternate Exterior angles Theorem)

### 3.4 Proofs with Perpendicular Lines

Exploration 1

Writing Conjectures

Work with a partner: Fold a piece of pair in half twice. Label points on the two creases. as shown.

a. Write a conjecture about $$\overline{A B}$$ and $$\overline{C D}$$. Justify your conjecture.
The conjecture about $$\overline{A B}$$ and $$\overline{c D}$$ is “If two lines intersect to form a linear pair of congruent angles, then the lines are perpendicular.”

b. Write a conjecture about $$\overline{A O}$$ and $$\overline{O B}$$ Justify your conjecture.
The conjecture about $$\overline{A O}$$ and $$\overline{O B}$$ is “In a plane, if two lines are perpendicular to the same line, then they are parallel to each other.”

Exploration 2

Exploring a segment Bisector

Work with a partner: Fold and crease a piece of paper. as shown. Label the ends of the crease as A and B.

a. Fold the paper again so that point A coincides with point B. Crease the paper on that fold.

b. Unfold the paper and examine the four angles formed by the two creases. What can you conclude about the four angles?
When we unfold the paper and examine the four angles formed by the two creases, we can say that the four angles formed are the right angles i.e., 90°

Exploration 3

Writing a conjecture

Work with a partner.

a. Draw $$\overline{A B}$$, as shown.
b. Draw an arc with center A on each side of AB. Using the same compass selling, draw an arc with center B on each side $$\overline{A B}$$. Label the intersections of arcs C and D.
c. Draw $$\overline{C D}$$. Label its intersection with $$\overline{A B}$$ as O. Write a conjecture about the resulting diagram. Justify your conjecture.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to make conjectures and build a logical progression of statements to explore the truth of your conjectures.
The resultant diagram is:

The angles formed at all the intersection points are: 90°
The lengths of the line segments are equal i.e., AO = OB and CO = OD

Question 4.
What conjectures can you make about perpendicular lines?
1. If two lines intersect to form a linear pair of congruent angles, then the lines are said to be perpendicular lines.
2. If a transversal is perpendicular to one of two parallel lines, then it is perpendicular to the other line.
3. If two lines are perpendicular to the same line, then they are parallel to each other.

Question 5.
In Exploration 3. find AO and OB when AB = 4 units.
AB = AO + OB
AO = OB
AB = 4 units
The values of AO and OB are: 2 units

### Lesson 3.4 Proofs with Perpendicular Lines

Monitoring Progress

Question 1.
Find the distance from point E to

Given E is ⊥ to $$\overline{F H}$$
To find the distance between E and $$\overline{F H}$$, we need to find the distance between E and G i.e., EG
From the coordinate plane,
E (-4, -3), G (1, 2)
Compare the given points with
E (x1, y1), G (x2, y2)
d = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
d = $$\sqrt{(1 + 4)² + (2 + 3)²}$$
d = $$\sqrt{(5)² + (5)²}$$
d = $$\sqrt{50}$$
d = 7.07

Question 2.
Prove the Perpendicular Transversal Theorem using the diagram in Example 2 and the Alternate Exterior Angles Theorem (Theorem 3.3).
Answer:In a plane, if a line is perpendicular to one of the two parallel lines, then it is perpendicular to the other line also

Proof:
Given: k || l, t ⊥ k
Prove: t ⊥ l

Alternate Exterior Angles Theorem:
When two parallel lines are cut by a transversal, the resulting alternate exterior angles are congruent.

Proof:
Given: k || l
Prove: ∠1 ≅ ∠7 and ∠4 ≅ ∠6
Since k || l, by the Corresponding Angles Postulate,
∠1 ≅ ∠5
Also, by the Vertical Angles Theorem,
∠5 ≅ ∠7
Then, by the Transitive Property of Congruence,
∠1 ≅ ∠7
You can prove that 4 and 6 are congruent using the same method.

Use the lines marked in the photo.

Question 3.
Is b || a? Explain your reasoning.
a and b are nonintersecting lines.
b || a

Question 4.
Is b ⊥ c? Explain your reasoning.
The angle between b and c is 90°
Hence, b is perpendicular to c

### Exercise 3.4 Proofs with Perpendicular Lines

Vocabulary and core Concept Check

Question 1.
COMPLETE THE SENTENCE
The perpendicular bisector of a segment is the line that passes through the _______________ of the segment at a _______________ angle.

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.

Find the distance from point X to
X (-3, 3), Y (3, 1)
Compare the given points with
(x1, y1), (x2, y2)
We know that,
XY = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
XY = $$\sqrt{(3 + 3)² + (3 – 1)²}$$
XY = $$\sqrt{(6)² + (2)²}$$
XY = 6.32

Find XZ
The given points are:
X (-3, 3), Z (4, 4)
Compare the given points with
(x1, y1), (x2, y2)
We know that,
XZ = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
XZ = $$\sqrt{(4 + 3)² + (3 – 4)²}$$
XZ = $$\sqrt{(7)² + (1)²}$$
XZ = 7.07

Find the length of $$\overline{X Y}$$
The given points are:
X (-3, 3), Y (3, 1)
Compare the given points with
(x1, y1), (x2, y2)
We know that,
XY = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
XY = $$\sqrt{(3 + 3)² + (3 – 1)²}$$
XY = $$\sqrt{(6)² + (2)²}$$
XY = 6.32

Find the distance from line l to point X.
The given points are:
X (-3, 3), Y (3, 1)
Compare the given points with
(x1, y1), (x2, y2)
We know that,
XY = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
XY = $$\sqrt{(3 + 3)² + (3 – 1)²}$$
XY = $$\sqrt{(6)² + (2)²}$$
XY = 6.32

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4. find the distance from point A to .

Question 3.

Question 4.

The given points are:
X (3, 3), Y (2, -1.5)
Compare the given points with
(x1, y1), (x2, y2)
We know that,
XY = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
XY = $$\sqrt{(3 + 1.5)² + (3 – 2)²}$$
XY = $$\sqrt{(4.5)² + (1)²}$$
XY = 4.60

CONSTRUCTION
In Exercises 5-8, trace line m and point P. Then use a compass and straightedge to construct a line perpendicular to line m through point P.

Question 5.

Question 6.

1. Using P as the center, draw two arcs intersecting with line m.
2. Label the intersections as points X and Y.
3. Using X and Y as centers and an appropriate radius, draw arcs that intersect.
4. Label the intersection as Z. Draw $$\overline{P Z}$$

Question 7.

Question 8.

1. Using P as the center and any radius, draw arcs intersecting m and label those intersections as X and Y.
2. Using X as the center, open the compass so that it is greater than half of XP and draw an arc.
3. Using Y as the center and retaining the same compass setting, draw an arc that intersects with the first
4. Label the point of intersection as Z. Draw $$\overline{P Z}$$

CONSTRUCTION
In Exercises 9 and 10, trace $$\overline{A B}$$. Then use a compass and straightedge to construct the perpendicular bisector of $$\overline{A B}$$

Question 9.

Question 10.

Using a compass setting greater than half of AB, draw two arcs using A and B as centers
Connect the points of intersection of the arcs with a straight line

ERROR ANALYSIS
In Exercises 11 and 12, describe and correct the error in the statement about the diagram.
Question 11.

Question 12.

The distance from the perpendicular to the line is given as the distance between the point and the non-perpendicular line
The distance from point C to AB is the distance between point C and A i.e., AC
Hence, the distance from point C to AB is 12 cm

PROVING A THEOREM
In Exercises 13 and 14, prove the theorem.
Question 13.
Linear Pair Perpendicular Theorem (Thm. 3. 10)

Question 14.
Lines Perpendicular to a Transversal Theorem (Thm. 3.12)
In a plane, if a line is perpendicular to one of the two parallel lines, then it is perpendicular to the other line also

Proof:
Given: k || l, t ⊥ k
Prove: t ⊥ l

PROOF
In Exercises 15 and 16, use the diagram to write a proof of the statement.

Question 15.
If two intersecting lines are perpendicular. then they intersect to form four right angles.
Given a ⊥ b
Prove ∠1, ∠2, ∠3, and ∠4 are right angles.

Question 16.
If two sides of two adjacent acute angles are perpendicular, then the angles are complementary.
Given $$\overrightarrow{B A}$$ ⊥$$\vec{B}$$C
Prove ∠1 and ∠2 are complementary

In Exercises 17-22, determine which lines, if any, must be parallel. Explain your reasoning.

Question 17.

Question 18.

From the figure, we observe that a is perpendicular to both lines b and c
By using the Perpendicular transversal theorem, a is both perpendicular to b and c and b is parallel to c

Question 19.

Question 20.

From the figure, we can observe that a is perpendicular to d and b is perpendicular to c.
By using the Perpendicular transversal theorem, a is perpendicular to d and b is perpendicular to c.

Question 21.

Question 22.

From the figure,
w ⊥ v and w⊥ y
We can say that w and v are parallel lines by the “Perpendicular Transversal Theorem”
z ⊥ x and w ⊥ z
We can say that w and x are parallel lines by the “Perpendicular Transversal theorem”

Question 23.
USING STRUCTURE
Find all the unknown angle measures in the diagram. Justify your answer for cacti angle measure.

Question 24.
MAKING AN ARGUMENT
Your friend claims that because you can find the distance from a point to a line, you should be able to find the distance between any two lines. Is your friend correct? Explain your reasoning.
It is given your friend claims that because you can find the distance from a point to a line, you should be able to find the distance between any two lines.

Question 25.
MATHEMATICAL CONNECTIONS
Find the value of x when a ⊥ b and b || c.

Question 26.
HOW DO YOU SEE IT?
You are trying to cross a stream from point A. Which point should you jump to in order to jump the shortest distance? Explain your reasoning.

From the figure,
Point A is perpendicular to Point C.
According to Perpendicular Transversal Theorem,
The distance between the perpendicular points is the shortest.
In order to jump the shortest distance, you have to jump to point C from point A

Question 27.
ATTENDING TO PRECISION
In which of the following diagrams is $$\overline{A C}$$ || $$\overline{B D}$$ and $$\overline{A C}$$ ⊥ $$\overline{C D}$$? Select all that apply.
(A)
(B)
(C)
(D)
(E)

Question 28.
THOUGHT-PROVOKING
The postulates and theorems in this book represent Euclidean geometry. In spherical geometry, all points are points on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. In spherical geometry. how many right angles are formed by two perpendicular lines? Justify your answer.
Given,
The postulates and theorems in this book represent Euclidean geometry. In spherical geometry, all points are points on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere.
In Euclidean geometry, the two perpendicular lines form 4 right angles whereas, In spherical geometry, the two perpendicular lines form 8 right angles according to the Parallel lines Postulate in spherical geometry.
Hence 8 right angles are formed by two perpendicular lines in spherical geometry.

Question 29.
CONSTRUCTION
Construct a square of side length AB

Question 30.
ANALYZING RELATIONSHIPS
The painted line segments that brain the path of a crosswalk are usually perpendicular to the crosswalk. Sketch what the segments in the photo would look like if they were perpendicular to the crosswalk. Which type of line segment requires less paint? Explain your reasoning.

Question 31.
ABSTRACT REASONING
Two lines, a and b, are perpendicular to line c. Line d is parallel to line c. The distance between lines a and b is x meters. The distance between lines c and d is y meters. What shape is formed by the intersections of the four lines?

Question 32.
MATHEMATICAL CONNECTIONS
Find the distance between the lines with the equations y = $$\frac{3}{2}$$ + 4 and – 3x + 2y = – 1.
y = $$\frac{3}{2}$$ + 4 and -3x + 2y = -1
y = $$\frac{3}{2}$$ + 4 and y = $$\frac{3}{2}$$x – $$\frac{1}{2}$$
Compare the given equations with
y = mx + c
m1 = m2 = $$\frac{3}{2}$$
c1 = 4
c2= –$$\frac{1}{2}$$
The distance between the two parallel lines is:
d = | c1 – c2 |
The distance between the given 2 parallel lines = | c1 – c2 |
= | 4 + $$\frac{1}{2}$$ |
= $$\frac{9}{2}$$

Question 33.
WRITING
Describe how you would find the distance from a point to a plane. Can you find the distance from a line to a plane? Explain your reasoning.

Maintaining Mathematical Proficiency

Simplify the ratio.

Question 34.
$$\frac{6-(-4)}{8-3}$$
Given,
$$\frac{6 – (-4)}{8 – 3}$$
= $$\frac{6 + 4}{8 – 3}$$
= $$\frac{10}{5}$$
= 2

Question 35.
$$\frac{3-5}{4-1}$$

Question 36.
$$\frac{8-(-3)}{7-(-2)}$$
Given,
$$\frac{8 – (-3)}{7 – (-2)}$$
= $$\frac{8 + 3}{7 + 2}$$
= $$\frac{11}{9}$$

Question 37.
$$\frac{13-4}{2-(-1)}$$

Identify the slope and the y-intercept of the line.

Question 38.
y = 3x + 9
y = 3x + 9
y = mx + c
Where,
m is the slope
c is the y-intercept
The slope is: 3
The y-intercept is: 9

Question 39.
y = –$$\frac{1}{2}$$x + 7

Question 40.
y = $$\frac{1}{6}$$x – 8
y = $$\frac{1}{6}$$x – 8
y = mx + c
m = $$\frac{1}{6}$$ and c = -8
The slope is: $$\frac{1}{6}$$
The y-intercept is: -8

Question 41.
y = – 8x – 6

### 3.5 Equations of Parallel and Perpendicular Lines

Exploration 1

Writing Equations of Parallel and Perpendicular Lines

Work with a partner: Write an equation of the line that is parallel or perpendicular to the given line and passes through the given point. Use a graphing calculator to verify your answer. What is the relationship between the slopes?
a.

The given lines are parallel lines
The equation for another line is:
y = mx + c
y = $$\frac{3}{2}$$x + c
Substitute (0, 2) in the above equation
2 = 0 + c
c = 2
The equation for another parallel line is:
y = $$\frac{3}{2}$$x + 2
The lines that have the same slope and different y-intercepts are Parallel lines.
y = $$\frac{3}{2}$$x – 1
y = $$\frac{3}{2}$$x + 2

b.

The given lines are perpendicular lines
The equation for another line is:
y = $$\frac{3}{2}$$x + c
The slope of perpendicular lines is: -1
m1m2 = -1
$$\frac{3}{2}$$ . m2 = -1
m2 = –$$\frac{2}{3}$$
y = –$$\frac{2}{3}$$x + c
Substitute (0, 1) in the above equation
1 = 0 + c
c = 1
The equation for another perpendicular line is:
y = –$$\frac{2}{3}$$x + 1
We can observe that the product of the slopes are -1 and the y-intercepts are different
The lines that have the slopes product -1 and different y-intercepts are Perpendicular lines
y = $$\frac{3}{2}$$x – 1
y = –$$\frac{2}{3}$$x + 1

c.

The given lines are parallel lines
The equation for another line is:
y = mx + c
y = $$\frac{1}{2}$$x + c
Substitute (2, -2) in the above equation
-2 = $$\frac{1}{2}$$ (2) + c
-2 = 1 + c
c = 2 – 1
c = -3
The equation for another parallel line is:
y = $$\frac{1}{2}$$x – 3
When we compare the given equation with the obtained equation
The lines that have the same slope and different y-intercepts are Parallel lines
y = $$\frac{1}{2}$$x + 2
y = $$\frac{1}{2}$$x – 3

d.

From the figure,
The given lines are perpendicular lines
The equation for another line is:
y = $$\frac{1}{2}$$x + c
The slope of perpendicular lines is: -1
m1m2 = -1
$$\frac{1}{2}$$ . m2 = -1
m2 = -2
So,
y = -2x + c
Substitute (2, -3) in the equation
-3 = -2 (2) + c
-3 = -4 + c
c = 4 – 3
c = 1
The equation for another perpendicular line is:
y = -2x + 1
We can observe that the product of the slopes are -1 and the y-intercepts are different.
The lines that have the slopes product -1 and different y-intercepts are Perpendicular lines.
y = $$\frac{1}{2}$$x + 2
y = -2x + 1

e.

From the figure, the given lines are parallel lines
The equation for another line is:
y = -2x + c
Substitute (0, -2) in the equation
-2 = 0 + c
c = -2
The equation for another parallel line is:
y = -2x – 2
We can observe that the slopes are the same and the y-intercepts are different
The lines that have the same slope and different y-intercepts are Parallel lines
y = -2x + 2
y = -2x – 2

f.

From the figure, the given lines are perpendicular lines
The equation for another line is:
y = -2x + c
The slope of perpendicular lines is: -1
m1m2 = -1
-2 . m2 = -1
m2 = $$\frac{1}{2}$$
y = $$\frac{1}{2}$$x + c
Substitute (4, 0) in the equation
0 = $$\frac{1}{2}$$ (4) + c
0 = 2 + c
c = 0 – 2
c = -2
The equation for another perpendicular line is:
y = $$\frac{1}{2}$$x – 2
We can observe that the product of the slopes is -1 and the y-intercepts are different
The lines that have the slopes product -1 and different y-intercepts are Perpendicular lines.
y = $$\frac{1}{2}$$x – 2
y = -2x + 2

Exploration 2

Writing Equations of Parallel and Perpendicular Lines

Work with a partner: Write the equations of the parallel or perpendicular lines. Use a graphing calculator to verify your answers.

a.

The given lines are the parallel lines
The coordinates of the line of the first equation are: (-1.5, 0), and (0, 3)
The coordinates of the line of the second equation are: (1, 0), and (0, -2)
Compare the given coordinates with
A (x1, y1), and B (x2, y2)
Slope (m) = $$\frac{y2 – y1}{x2 – x1}$$
The slope of the line of the first equation is:
m = $$\frac{3 – 0}{0 + 1.5}$$
m = $$\frac{3}{1.5}$$
m = 2
The standard linear equation is:
y = mx + c
y = 2x + c
The slopes are the same but the y-intercepts are different
The given parallel line equations are:
y = 2x + c1
y = 2x + c2

b.

From the figure,
The given lines are perpendicular lines
The coordinates of the line of the first equation are: (0, -3), and (-1.5, 0)
The coordinates of the line of the second equation are: (-4, 0), and (0, 2)
Compare the given coordinates with
A (x1, y1), and B (x2, y2)
Slope (m) = $$\frac{y2 – y1}{x2 – x1}$$
The slope of the line of the first equation is:
m = $$\frac{0 + 3}{0 – 1.5}$$
m = $$\frac{3}{-1.5}$$
m = $$\frac{-30}{15}$$
m = -2
The standard linear equation is:
y = mx + c
y = -2x + c
The product of the slopes is -1 and the y-intercepts are different.
m1 × m2 = -1
-2 × m2 = -1
m2 = $$\frac{1}{2}$$
The given perpendicular line equations are:
y = -2x + c1
y = $$\frac{1}{2}$$x + c2

Question 3.
How can you write an equation of a line that is parallel or perpendicular to a given line and passes through a given point?
MODELING WITH MATHEMATICS
To be proficient in math, you need to analyze relationships mathematically to draw conclusions.
The standard form of a linear equation is:
y = mx + c
The slopes are the same and the y-intercepts are different
To find the y-intercept of the equation that is parallel to the given equation, substitute the given point and find the value of c.
The product of the slopes is -1
The slope of the equation that is perpendicular to the given equation is: –$$\frac{1}{m}$$
To find the y-intercept of the equation that is perpendicular to the given equation, substitute the given point and find the value of c.

Question 4.
Write an equation of the line that is (a) parallel and (b) perpendicular to the line y = 3x + 2 and passes through the point (1, -2).
Given,
y = 3x + 2
The given point is: (1, -2)
a) Parallel line equation:
The slope of the parallel equations is the same
The slope of the equation that is parallel t the given equation is: 3
The equation that is parallel to the given equation is
y = 3x + c
Substitute (1, -2) in the equation
-2 = 3 (1) + c
-2 – 3 = c
c = -5
The equation of the line that is parallel to the given equation is y = 3x – 5
b) Perpendicular line equation:
The product of the slope of the perpendicular equations is: -1
m1 m2 = -1
3m2 = -1
m2 = –$$\frac{1}{3}$$
The slope of the equation that is parallel t the given equation is –$$\frac{1}{3}$$
The equation that is perpendicular to the given equation is y = –$$\frac{1}{3}$$x + c
Substitute (1, -2) in the above equation
-2 = –$$\frac{1}{3}$$ (-2) + c
-2 – $$\frac{2}{3}$$ = c
c = –$$\frac{8}{3}$$
The equation of the line that is perpendicular to the given equation is y = –$$\frac{1}{3}$$x –$$\frac{8}{3}$$

### Lesson 3.5 Equations of Parallel and Perpendicular Lines

Monitoring Progress

Find the coordinates of point P along the directed line segment AB so that AP to PB is the given ratio.

Question 1.
A(1, 3), B(8, 4); 4 to 1
Given,
A (1, 3), and B (8, 4)
Compare the given points with
A (x1, y1), and B (x2, y2)
AP : PB = 4 : 1
Divide AB into 5 parts
Slope = $$\frac{y2 – y1}{x2 – x1}$$
Slope of AB = $$\frac{4 – 3}{8 – 1}$$
Slope of AB = $$\frac{1}{7}$$
To find the coordinates of P, add slope to AP and PB
P = (4 + (4 / 5) × 7, 1 + (4 / 5) × 1)
P = (22.4, 1.8)
The coordinates of P are (22.4, 1.8)

Question 2.
A(- 2, 1), B(4, 5); 3 to 7
Given, A (-2, 1), and B (4, 5)
Compare the given points with
A (x1, y1), and B (x2, y2)
AP : PB = 3 : 7
Divide AB into 10 parts
m = $$\frac{y2 – y1}{x2 – x1}$$
Slope of AB = $$\frac{5 – 1}{4 + 2}$$
Slope of AB = $$\frac{4}{6}$$
Slope of AB = $$\frac{2}{3}$$
To find the coordinates of P, add slope to AP and P
P = (3 + ($$\frac{3}{10}$$ × 3), 7 + ($$\frac{3}{10}$$ × 2))
P = (3.9, 7.6)
The coordinates of P are (3.9, 7.6)

Question 3.
Determine which of the lines are parallel and which of the lines are perpendicular.

From the figure,
The coordinates of line a are (0, 2), and (-2, -2)
The coordinates of line b are (2, 3), and (0, -1)
The coordinates of line c are (4, 2), and (3, -1)
The coordinates of line d are (-3, 0), and (0, -1)
Compare the given coordinates with (x1, y1), and (x2, y2)
The slope of line a (m) = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{-2 – 2}{-2 – 0}$$
= $$\frac{-4}{-2}$$
= 2
The slope of line b (m) = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{-1 – 3}{0 – 2}$$
= $$\frac{-4}{-2}$$
= 2
The slope of line c (m) = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{-1 – 2}{3 – 4}$$
= $$\frac{-3}{-1}$$
= 3
The slope of line d (m) = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{-1 – 0}{0 + 3}$$
= $$\frac{-1}{3}$$
= –$$\frac{1}{3}$$
The parallel lines have the same slopes
The perpendicular lines have the product of slopes equal to -1
Linea and Line b are parallel lines.
Line c and Line d are perpendicular lines.

Question 4.
Write an equation of the line that passes through the point (1, 5) and is
(a) parallel to the line y = 3x – 5 and
Given,
y = 3x – 5
(1, 5)
The parallel lines have the same slope
Compare the given equation with
y = mx + c
The slope of the parallel line that passes through (1, 5) is 3
y = 3x + c
To find the value of c, substitute (1, 5) in the above equation
5 = 3 (1) + c
c = 5 – 3
c = 2
The equation of the parallel line that passes through (1, 5) is y = 3x + 2

(b) perpendicular to the line y = 3x – 5.
Given,
y = 3x – 5
(1, 5)
The product of the slopes of the perpendicular lines is equal to -1
Compare the given equation with
y = mx + c
m = 3
The slope of the perpendicular line that passes through (1, 5) is m1m2 = -1
3m2 = -1
m2 = –$$\frac{1}{3}$$
The equation of the perpendicular line that passes through (1, 5) is y = –$$\frac{1}{3}$$x + c
To find the value of c, substitute (1, 5) in the above equation
5 = –$$\frac{1}{3}$$ + c
c = 5 + $$\frac{1}{3}$$
c = $$\frac{16}{3}$$
The equation of the perpendicular line that passes through (1, 5) is y = –$$\frac{1}{3}$$x + $$\frac{16}{3}$$

Question 5.
How do you know that the lines x = 4 and y = 2 are perpendicular?
The given lines are x = 4 and y = 2
The line x = 4 is a vertical line that has the right angle i.e., 90°
The line y = 4 is a horizontal line that has a straight angle i.e., 0°
The angle at the intersection of the 2 lines = 90° – 0° = 90°
The lines x = 4 and y = 2 are perpendicular lines

Question 6.
Find the distance from the point (6, 4) to the line y = x + 4.
Given
y = x + 4
Compare the given equation with
ax + by + c = 0
x – y + 4 = 0
a = 1, and b = -1
The distance from the point (6, 4)
The distance from the point (x, y) to the line ax + by + c = 0
d = | ax + by + c| /$$\sqrt{a² + b²}$$
d = | x – y + 4 | / $$\sqrt{1² + (-1)²}$$
d = | x – y + 4 | / $$\sqrt{2}$$}
Substitute (6, 4) in the above equation
So,
d = | 6 – 4 + 4 |/ $$\sqrt{2}$$}
d = 3√2

Question 7.
Find the distance from the point (- 1, 6) to the line y = – 2x.
Given,
y = -2x
ax + by + c = 0
2x + y = 0
a = 2, and b = 1
The given point is: (-1, 6)
The distance from the point (x, y) to the line ax + by + c = 0 is
d = | ax + by + c| /$$\sqrt{a² + b²}$$
d = | 2x + y | / $$\sqrt{2² + (1)²}$$
d = | 2x + y | / $$\sqrt{5}$$}
Substitute (-1, 6) in the above equation
So,
d = | -2 + 6 |/ $$\sqrt{5}$$
d = $$\frac{4}{√5}$$

### Exercise 3.5 Equations of Parallel and Perpendicular Lines

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
A _________ line segment AB is a segment that represents moving from point A to point B.
A directed line segment AB is a segment that represents moving from point A to point B.

Question 2.
WRITING
How are the slopes of perpendicular lines related?
The Perpendicular lines are lines that intersect at right angles.
If you multiply the slopes of two perpendicular lines in the plane, you get −1 i.e., the slopes of perpendicular lines are opposite reciprocals.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6. find the coordinates of point P along the directed line segment AB so that AP to PB is the given ratio.

Question 3.
A(8, 0), B(3, – 2); 1 to 4

Question 4.
A(- 2, – 4), B(6, 1); 3 to 2
A (-2, -4), and B (6, 1)
Compare the given points with
A (x1, y1), and B (x2, y2)
AP : PB = 3 : 2
Divide AB into 5 parts
Slope (m) = $$\frac{y2 – y1}{x2 – x1}$$
Slope of AB = $$\frac{1 + 4}{6 + 2}$$
Slope of AB = $$\frac{5}{8}$$
To find the coordinates of P, add slope to AP and PB
So,
P = (3 + (3 / 5) × 8, 2 + (3 / 5) × 5)
P = (7.8, 5)
The coordinates of P are (7.8, 5)

Question 5.
A(1, 6), B(- 2, – 3); 5 to 1

Question 6.
A(- 3, 2), B(5, – 4); 2 to 6
Given,
A (-3, 2), and B (5, -4)
Compare the given points with
A (x1, y1), and B (x2, y2)
AP : PB = 2 : 6
Divide AB into 8 parts
Slope (m) = $$\frac{y2 – y1}{x2 – x1}$$
Slope of AB = $$\frac{-4 – 2}{5 + 3}$$
Slope of AB = $$\frac{-6}{8}$$
To find the coordinates of P, add slope to AP and PB
P = (2 + (2 / 8) × 8, 6 + (2 / 8) × (-6))
P = (4, 4.5)
The coordinates of P are (4, 4.5)

In Exercises 7 and 8, determine which of the lines are parallel and which of the lines are perpendicular.

Question 7.

Question 8.

From the figure,
The coordinates of line a are (2, 2), and (-2, 3)
The coordinates of line b are (3, -2), and (-3, 0)
The coordinates of line c are (2, 4), and (0, -2)
The coordinates of line d are (0, 6), and (-2, 0)
Compare the given coordinates with (x1, y1), and (x2, y2)
The slope of line a (m) = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{3 – 2}{-2 – 2}$$
= $$\frac{1}{-4}$$
= –$$\frac{1}{4}$$
The slope of line b (m) = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{0 + 2}{-3 – 3}$$
= $$\frac{2}{-6}$$
= –$$\frac{1}{3}$$
The slope of line c (m) = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{-4 – 2}{0 – 2}$$
= $$\frac{-6}{-2}$$
= 3
The slope of line d (m) = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{6 – 0}{0 + 2}$$
= $$\frac{6}{2}$$
= 3
The parallel lines have the same slopes
The perpendicular lines have the product of slopes equal to -1.
Line c and Line d are parallel lines
Line b and Line c are perpendicular lines

In Exercises 9 – 12, tell whether the lines through the given points are parallel, perpendicular, or neither. justify your answer.

Question 9.
Line 1: (1, 0), (7, 4)
Line 2: (7, 0), (3, 6)

Question 10.
Line 1: (- 3, 1), (- 7, – 2)
Line 2: (2, – 1), (8, 4)
The coordinates of line 1 are (-3, 1), (-7, -2)
The coordinates of line 2 are (2, -1), (8, 4)
Compare the given coordinates with
(x1, y1), (x2, y2)
Slope (m) = $$\frac{y2 – y1}{x2 – x1}$$
Slope of line 1 = $$\frac{-2 – 1}{-7 + 3}$$
= $$\frac{-3}{-4}$$
= $$\frac{3}{4}$$
Slope of line 2 = $$\frac{4 + 1}{8 – 2}$$
= $$\frac{5}{6}$$
By comparing the slopes, we can say that the given lines are neither parallel nor perpendicular

Question 11.
Line 1: (- 9, 3), (- 5, 7)
Line 2: (- 11, 6), (- 7, 2)

Question 12.
Line 1: (10, 5), (- 8, 9)
Line 2: (2, – 4), (11, – 6)
The coordinates of line 1 are (10, 5), (-8, 9)
The coordinates of line 2 are (2, -4), (11, -6)
Compare the given coordinates with
(x1, y1), (x2, y2)
Slope (m) = $$\frac{y2 – y1}{x2 – x1}$$
Slope of line 1 = $$\frac{9 – 5}{-8 – 10}$$
= $$\frac{4}{-18}$$
= –$$\frac{2}{9}$$
Slope of line 2 = $$\frac{4 – 6}{11 – 2}$$
= $$\frac{-2}{9}$$
= –$$\frac{2}{9}$$
The given lines are parallel

In Exercises 13 – 16. write an equation of the line passing through point P that ¡s parallel to the given line. Graph the equations of the lines to check that they are parallel.

Question 13.
P(0, – 1), y = – 2x + 3

Question 14.
P(3, 8), y = $$\frac{1}{5}$$(x + 4)
Given,
y = $$\frac{1}{5}$$ (x + 4)
P (3, 8)
y = $$\frac{1}{5}$$x + $$\frac{4}{5}$$
The slopes are equal fot the parallel lines
The equation that is parallel to the given equation is y = $$\frac{1}{5}$$x + c
Substitute P (3, 8) in the above equation to find the value of c
8 = $$\frac{1}{5}$$ (3) + c
c = 8 – $$\frac{3}{5}$$
c = $$\frac{37}{5}$$
The parallel line equation that is parallel to the given equation is y = $$\frac{1}{5}$$x + $$\frac{37}{5}$$

Question 15.
P(- 2, 6), x = – 5

Question 16.
P(4, 0), – x + 2y = 12
Given,
-x + 2y = 12
The given point is: P (4, 0)
y = $$\frac{1}{2}$$x + 6
The slopes are equal fot the parallel lines
The equation that is parallel to the given equation is y = $$\frac{1}{2}$$x + c
Substitute P (4, 0) in the above equation to find the value of c
0 = $$\frac{1}{2}$$ (4) + c
c = 2 – 0
c = 2
The parallel line equation that is parallel to the given equation is y = $$\frac{1}{2}$$x + 2

In Exercises 17 – 20. write an equation of the line passing through point P that is perpendicular to the given line. Graph the equations of the lines to check that they are perpendicular.

Question 17.
P(0, 0), y = – 9x – 1

Question 18.
P(4, – 6); y = – 3
Given,
y = -3; P (4, -6)
The line that is perpendicular to y=n is
x = n
The line that is perpendicular to the given equation is:
x = n
Substitute P (4, -6) in the equation
x = 4
The equation that is perpendicular to y = -3 is:
x = 4

Question 19.
P(2, 3), y – 4 = – 2(x + 3)

Question 20.
P(- 8, 0), 3x – 5y = 6
Given,
3x – 5y = 6
P (-8, 0)
5y = 3x – 6
y = $$\frac{3}{5}$$x – $$\frac{6}{5}$$
The product of the slopes of perpendicular lines is equal to -1
m = –$$\frac{5}{3}$$
The equation that is perpendicular to the given line equation is y = –$$\frac{5}{3}$$x + c
Substitute P(-8, 0) in the above equation
0 = –$$\frac{5}{3}$$ ( -8) + c
c = $$\frac{40}{3}$$
The equation that is perpendicular to the given equation is y = –$$\frac{5}{3}$$x + $$\frac{40}{3}$$

In Exercises 21 – 24, find the distance from point A to the given line.

Question 21.
A(- 1, 7), y = 3x

Question 22.
A(- 9, – 3), y = x – 6
Given,
y = x – 6; A (-9, -3)
Compare the given equation with
y = mx + c
The product of the slopes of the perpendicular lines is equal to -1
m1m2 = -1
1 (m2) = -3
m2 = -1
The equation that is perpendicular to the given line equation is y = -x + c
To find c:
-3 = 9 + c