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## 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 – Page 357
- Quadrilaterals and Other Polygons Mathematical Practices – Page 358
- 7.1 Angles of Polygons – Page 359
- Lesson 7.1 Angles of Polygons – Page(360-366)
- Exercise 7.1 Angles of Polygons – Page(364-366)
- 7.2 Properties of Parallelograms – Page 367
- Lesson 7.2 Properties of Parallelograms – Page(368-374)
- Exercise 7.2 Properties of Parallelograms – Page(372-374)
- 7.3 Proving That a Quadrilateral is a Parallelogram – Page 375
- Lesson 7.3 Proving that a Quadrilateral is a Parallelogram – Page(376-384)
- Exercise 7.3 Proving that a Quadrilateral is a Parallelogram – Page(381-384)
- 7.1 – 7.3 Quiz – Page 386
- 7.4 Properties of Special Parallelograms – Page 387
- Lesson 7.4 Properties of Special Parallelograms – Page (388-396)
- Exercise 7.4 Properties of Special Parallelograms – Page(393-396)
- 7.5 Properties of Trapezoids and Kites – Page 397
- Lesson 7.5 Properties of Trapezoids and Kites – Page(398-406)
- Exercise 7.5 Properties of Trapezoids and Kites – Page(403-406)
- Quadrilaterals and Other Polygons Review – Page(408-410)
- Quadrilaterals and Other Polygons Test – Page 411
- Quadrilaterals and Other Polygons Cummulative Assessment – Page(412-413)

### 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

Answer:

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

Answer:

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

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Some quadrilaterals are squares.

Answer:

True.

Explanation:

The given statement is:

Some quadrilaterals are squares.

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.

Answer:

The below ae the three true statements based on the Venn diagram given,.

Some squares are part of rectangles.

No kites are parallelograms.

Some quadrilaterals are squares.

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.

d. Some quadrilaterals are kites.

e. Some quadrilaterals are trapezoids.

f. Some quadrilaterals are parallelograms.

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.

Answer:

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

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

\(\frac{180° (n – 2)}{n}\)

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.

Answer:

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.

Answer:

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

Communicate Your Answer

Question 3.

What is the sum of the measures of the interior angles of a polygon?

Answer:

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).

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

∠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.

Answer:

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?

Answer:

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?

Answer:

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?

Answer:

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 |

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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°

Answer:

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°

Answer:

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°

Answer:

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°

Answer:

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.

Answer:

x = 64°

Explanation:

Given that,

Question 12.

Answer:

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.

Answer:

x = 89°

Explanation:

Given that,

Question 14.

Answer:

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.

Answer:

x = 70°

Explanation:

Question 16.

Answer:

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.

Answer:

x = 150°

Explanation:

Question 18.

Answer:

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.

Answer:

m∠X = m∠Y = 92°

Explanation:

Question 20.

Answer:

∠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.

Answer:

m∠X = m∠Y = 100.5°

Explanation:

Question 22.

Answer:

∠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.

Answer:

X = 111°

Explanation:

The sum of the angle measures = 360°

Question 24.

Answer:

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.

Answer:

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.

Answer:

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

Answer:

x = 72°

Explanation:

Question 28.

18-gon

Answer:

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

Answer:

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

Answer:

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.

Answer:

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.

Answer:

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?

Answer:

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.

Answer:

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°.

Answer:

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°.

Answer:

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°.

Answer:

15°

Explanation:

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

Question 38.

Each interior angle has a measure of 165°.

Answer:

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°.

Answer:

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°.

Answer:

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°

Answer:

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.

Answer:

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).

Answer:

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.

Answer:

Yes, your friend is correct

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.

Answer:

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.

Answer:

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.

Answer:

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.

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?

Answer:

Question 50.

**HOW DO YOU SEE IT?**

Is the hexagon a regular hexagon? Explain your reasoning.

Answer:

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°.)

Answer:

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.

Answer:

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.

Answer:

101°

Explanation:

Sum of the angles = 180°

Question 54.

Answer:

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.

Answer:

16°

Explanation:

Sum of the angles = 180°

Question 56.

Answer:

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

Answer:

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?

Answer:

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?

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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?

Answer:

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.

Answer:

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.

Answer:

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

Communicate Your Answer

Question 3.

What are the properties of parallelograms?

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

∠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,

∠ADC + ∠BCD = 180°

∠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.

Answer:

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).

Answer:

(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.

Answer:

(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?

Answer:

Question 2.

**WRITING**

You are given one angle measure of a parallelogram. Explain how you can find the other angle measures of the parallelogram.

Answer:

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.

Answer:

x = 9; y = 15

Explanation:

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

Question 4.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

m∠B = 129°

Explanation:

The sum of the angles = 180°

m∠A = 51°; m∠B = x

Question 8.

Find m ∠ N.

Answer:

∠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

Answer:

Question 10.

LP

Answer:

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

Answer:

Question 12.

MQ

Answer:

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

Answer:

m∠LMN = 80°

Explanation:

Question 14.

m∠NQL

Answer:

∠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

Answer:

m∠MNQ = 100°

Explanation:

The given figure is:

Question 16.

m∠LMQ

Answer:

∠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.

Answer:

n = 110°; m = 35°

Explanation:

Sum of angles of parallelogram is 180°

Question 18.

Answer:

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.

Answer:

k = 7; m = 8

Explanation:

From the given figure,

we know that the diagonals of a parallelogram are equal.

Question 20.

Answer:

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.

Answer:

m∠v = 130°

The above answer is wrong.

Explanation:

Question 22.

Answer:

\(\overline{G F}\) = \(\overline{F J}\)

The above answer is wrong.

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 \(\overline{A B} \cong \overline{F E}\)

Answer:

Question 24.

Given ABCD, EBGF, and HJKD are parallelograms.

Prove ∠2 ≅∠3

Answer:

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)

Answer:

(1, 2.5)

Explanation:

Question 26.

Q(- 1, 3), R(5, 2), S(1, – 2), T(- 5, – 1)

Answer:

(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)

Answer:

F(3, 3)

Explanation:

Question 28.

D(- 2, – 4), F(0, 7), G(1, 0)

Answer:

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)

Answer:

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)

Answer:

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.

Answer:

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.

Answer:

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**

In quadrilateral ABCD.

m∠B = 124°, m∠A = 56°, and m∠C = 124°.

Your friend claims quadrilateral ABCD could be a parallelogram. Is your friend correct? Explain your reasoning.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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

Answer:

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°

Answer:

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

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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 = (x^{2})°, m∠TUV = 12x°, and ∠TUV is an acute angle. Find m∠USV.

Answer:

m∠USV = 16°

Explanation:

Given that,

In STUV m∠TSU = 32°, m∠USV = (x^{2})°, 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.

Answer:

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.

Answer:

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.)

Answer:

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.)

Answer:

Maintaining Mathematical Proficiency

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

Question 48.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

b. Is the quadrilateral a parallelogram? Justify your answer.

Answer:

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.

Answer:

If the opposite sides of a quadrilateral are equal, then the opposite angles of a quadrilateral are equal.

Explanation:

From the above quadrilateral,

The conjecture about the quadrilaterals is:

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.

Answer:

If the opposite sides of a quadrilateral are equal, then the opposite angles of a quadrilateral are equal.

Explanation:

From part (c),

The conjecture about quadrilaterals is:

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°,

AD = BC = 2.9

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.

Answer:

Explanation:

We know that the opposite angles of a quadrilateral are congruent.

So, ∠B = ∠D

b. Is the quadrilateral a parallelogram? Justify your answer.

Answer:

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.

Answer:

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.

Answer:

From part (c),

The conjecture about quadrilaterals are,

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.

Communicate Your Answer

Question 3.

How can you prove that a quadrilateral is a parallelogram?

Answer:

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.

Is the quadrilateral at the left a parallelogram? Explain your reasoning

Answer:

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.

Answer:

Yes, WXYZ a parallelogram.

Explanation:

Given that, quadrilateral WXYZ,

∠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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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

Answer:

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**

A quadrilateral has four congruent sides. Is the quadrilateral a parallelogram? Justify your answer.

Answer:

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. |

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

Yes, the given quadrilateral is a parallelogram.

The values of x = 3 and y = 4

Explanation:

Question 12.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

Yes, the given quadrilateral is a parallelogram.

The value of x = 7

Explanation:

The given figure quadrilateral is a parallelogram.

Question 16.

Answer:

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)

Answer:

Question 18.

E(- 3, 0), F(- 3, 4), G(3, – 1), H(3, – 5)

Answer:

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)

Answer:

Question 20.

N(- 5, 0), P(0, 4), Q(3, 0), R(- 2, – 4)

Answer:

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.

Answer:

Question 22.

Answer:

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**

What value of x makes the quadrilateral a parallelogram? Explain how you found your answer.

Answer:

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.

Answer:

Yes, your friend is correct

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)

Answer:

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)

Answer:

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)

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

Both are correct.

Explanation:

Given that the quadrilateral QRST is a parallelogram.

According to the Opposite Sides Parallel and Congruent Theorem,

In the quadrilateral QRST,

QR ≅TS and QT ≅ RS

QR || TS and QT || RS

According to your brother,

You have to show that

QR || TS and QT || RS

According t your sister,

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.

Answer:

Question 32.

Answer:

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?

d. Is quadrilateral EFGH a parallelogram? Explain your reasoning.

Answer:

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.

Answer:

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.

Answer:

∠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 \(\overline{J K}\) || \(\overline{P Q}\)?

Answer:

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.

Answer:

Question 36.

Answer:

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.

Answer:

Question 38.

**REASONING**

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

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

Question 42.

**PROOF**

Write the proof.

Given DEBF is a parallelogram.

AE = CF

Prove ABCD is a parallelogram.

Answer:

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.

Answer:

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?

Answer:

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.

Answer:

Question 46.

**THOUGHT-PROVOKING**

Create a regular hexagon using congruent parallelograms.

Answer:

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.

Answer:

Question 48.

**PROOF**

Write the proof.

Given ABCD is a parallelogram.

∠A is a right angle.

Prove ∠B, ∠C, and ∠D are right angles.

Answer:

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.)

Answer:

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.

Answer:

Maintaining Mathematical proficiency

Classify the quadrilateral.

Question 51.

Answer:

Question 52.

Answer:

Yes, the given quadrilateral is a rectangle.

Explanation:

The given figure is:

From the given quadrilateral,

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.

Answer:

Question 54.

Answer:

Yes, the given quadrilateral is a rhombus.

Explanation:

The given figure is:

From the given quadrilateral,

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.

Answer:

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.

Answer:

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.

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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.

AD

Answer:

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,

AD = BC

BC = 7

So, AD = 7

Question 10.

AE

Answer:

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

Answer:

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

Answer:

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

Answer:

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.

m∠ADC

Answer:

m∠ADC is 60°

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

Answer:

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.

Answer:

Parallelograms Opposite sides congruent and parallel Theorem.

Explanation:

The given quadrilateral is:

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.

Answer:

Diagonals Congruent Converse Theorem.

Explanation:

The given quadrilateral is:

We observe that the diagonals bisect each other.

So, according to the Diagonals Congruent Converse Theorem,

the given quadrilateral is a parallelogram.

Question 18.

Answer:

Parallelograms Opposite Angles Theorem.

Explanation:

The given quadrilateral is:

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)

Answer:

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)

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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?

Answer:

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.

Answer:

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**

Identifying Special Quadrilaterals

Work with a partner: Use dynamic geometry software.

Sample

a. Draw a circle with center A.

Answer:

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.

Answer:

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.

c. Draw quadrilateral BDCE.

Answer:

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.

Answer:

BDCE is a rhombus.

Explanation:

From the above quadrilateral,

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.

Answer:

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**

Identifying Special Quadrilaterals

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.

Answer:

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.

b. Draw quadrilateral AEBD.

Answer:

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.

Answer:

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.

Answer:

From parts (a) – (c),

the quadrilaterals formed from the perpendicular bisectors are only “Rhombus”.

Communicate Your Answer

Question 3.

What are the properties of the diagonals of rectangles, rhombuses, and squares?

Answer:

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.

Answer:

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?

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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?

Answer:

∠ADC = ∠BCD = 61°

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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?

Answer:

Question 2.

**WRITING**

What should you look for in a parallelogram to know if the parallelogram is also a rhombus?

Answer:

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

Answer:

Question 4.

∠K ≅∠M

Answer:

∠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.

\(\overline{J M} \cong \overline{K L}\)

Answer:

Question 6.

\(\overline{J K} \cong \overline{K L}\)

Answer:

\(\overline{J K} \cong \overline{K L}\) 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.

\(\overline{J L} \cong \overline{K M}\)

Answer:

Question 8.

∠JKM ≅ ∠LKM

Answer:

∠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.

In Exercises 9-12. classify the quadrilateral. Explain your reasoning.

Question 9.

Answer:

Question 10.

Answer:

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.

Answer:

Question 12.

Answer:

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.

Answer:

Question 14.

Answer:

Question 15.

Answer:

Question 16.

Answer:

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

Answer:

Question 18.

\(\overline{W X} \cong \overline{Y Z}\)

Answer:

\(\overline{W X} \cong \overline{Y Z}\) 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.

\(\overline{W X} \cong \overline{X Y}\)

Answer:

Question 20.

\(\overline{W Y} \cong \overline{X Z}\)

Answer:

\(\overline{W Y} \cong \overline{X Z}\) 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.

\(\overline{W Y}\) ⊥ \(\overline{X Z}\)

Answer:

Question 22.

∠WXZ ≅∠YXZ

Answer:

∠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.

Answer:

Question 24.

Answer:

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

Answer:

Question 26.

WY = 14x + 10

XZ = 11x + 22

Answer:

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

Answer:

WY = XZ = 4

Explanation:

Question 28.

WY = 16x – 2

XZ = 36x – 6

Answer:

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.

Answer:

Question 30.

It is equiangular and equilateral.

Answer:

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

Question 31.

The diagonals are perpendicular.

Answer:

Question 32.

The opposite sides are congruent.

Answer:

Parallelogram, Rectangle, Square, and Rhombus are the quadrilaterals,

where the opposite sides are congruent to each other.

Question 33.

The diagonals bisect each other.

Answer:

Question 34.

The diagonals bisect opposite angles.

Answer:

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.

Answer:

x° = 143°

Explanation:

Question 36.

**ERROR ANALYSIS**

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

Answer:

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

Answer:

m∠DAC = 53°

Explanation:

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

Question 38.

m∠AED

Answer:

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.

m∠ADC

Answer:

m∠ADC = 74°

Explanation:

Question 40.

DB

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

m∠SRT = 56°

Explanation:

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

Question 46.

QP

Answer:

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

Answer:

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

Answer:

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

Answer:

m∠MKN = 90°

Explanation:

Question 50.

m∠LMK

Answer:

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

Answer:

m∠LPK = 45°

Explanation:

Question 52.

KN

Answer:

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

Answer:

LN = 2

Explanation:

Question 54.

MP

Answer:

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)

Answer:

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)

Answer:

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)

Answer:

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)

Answer:

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)

Answer:

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)

Answer:

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.

Answer:

Given quadrilateral is rhombus.

x = 76 and y = 4

Explanation:

Question 62.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

always a rhombus.

Explanation:

Question 66.

A rectangle is __________ a square.

Answer:

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.

Answer:

always congruent diagonals.

Explanation:

Question 68.

The diagonals of a square _____________ bisect its angles.

Answer:

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.

Answer:

some has 4 congruent angles.

Explanation:

Question 70.

A rectangle ____________ has perpendicular diagonals.

Answer:

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.

Answer:

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).

Answer:

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.

Answer:

Question 74.

Given: WXYZ is a rhombus

Prove: \(\overline{W Y}\) bisects ∠ZWX and ∠XYZ.

\(\overline{Z X}\) bisects ∠WZY and ∠YXW.

Answer:

Question 75.

**ABSTRACT REASONING**

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

Answer:

Question 76.

**ABSTRACT REASONING**

Will a diagonal of a rhombus ever divide the rhombus into two equilateral triangles? Explain your reasoning.

Answer:

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**

Which quadrilateral could be called a regular quadrilateral? Explain your reasoning.

Answer:

Question 78.

**HOW DO YOU SEE IT?**

What other information do you need to determine whether the figure is a rectangle?

Answer:

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.

Answer:

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.

Answer:

**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)

Answer:

Question 82.

Rectangle Corollary (Corollary 7.3)

Answer:

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)

Answer:

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.

Answer:

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.

Answer:

Question 86.

**PROOF**

Write a proof in the style of your choice.

Given

Prove ABCD is a rectangle.

Answer:

**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 \(\overline{P R} \cong \overline{S Q}\)

Answer:

Question 88.

Given PQRS is a parallelogram.

\(\overline{P R} \cong \overline{S Q}\)

Prove PQRS is a rectangle.

Answer:

Maintaining Mathematical Proficiency

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

Question 89.

Answer:

x = 10 and y = 8

Explanation:

Given, \(\overline{D E}\) is a midsegment of ∆ABC.

Question 90.

Answer:

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.

Answer:

x = 9 and y = 26

Explanation:

### 7.5 Properties of Trapezoids and Kites

**Exploration 1**

Making a conjecture about Trapezoids

Sample

Work with a partner. Use dynamic geometry software.

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

Answer:

b. Is the trapezoid isosceles? Justify your answer.

Answer:

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.

Answer:

Exploration 2

Discovering a Property of Kites

Work with a partner. Use dynamic geometry software.

Sample

a. Construct a kite. Explain your process.

Answer:

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

Answer:

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

Answer:

Communicate Your Answer

Question 3.

What are some properties of trapezoids and kites?

Answer:

Question 4.

Is the trapezoid at the left isosceles? Explain.

Answer:

Question 5.

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

Answer:

### 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.

Answer:

In Exercises 2 and 3, use trapezoid EFGH.

Question 2.

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

Answer:

Question 3.

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

Answer:

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.

Answer:

Question 5.

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

Answer:

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?

Answer:

Question 7.

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

Answer:

Give the most specific name for the quadrilateral. Explain your reasoning.

Question 8.

Answer:

Question 9.

Answer:

Question 10.

Answer:

### 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.

Answer:

Question 2.

**DIFFERENT WORDS, SAME QUESTION**

Which is different? Find “both” answers.

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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)

Answer:

Question 4.

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

Answer:

Question 5.

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

Answer:

Question 6.

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

Answer:

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

Question 7.

Answer:

Question 8.

Answer:

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

Question 9.

Answer:

Question 10.

Answer:

In Exercises 11 and 12, find AB.

Question 11.

Answer:

Question 12.

Answer:

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)

Answer:

Question 14.

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

Answer:

In Exercises 15 – 18, Find m ∠ G.

Question 15.

Answer:

Question 16.

Answer:

Question 17.

Answer:

Question 18.

Answer:

Question 19.

**ERROR ANALYSIS**

Describe and correct the error in finding DC.

Answer:

Question 20.

**ERROR ANALYSIS**

Describe and correct the error in finding m∠A.

Answer:

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

Question 21.

Answer:

Question 22.

Answer:

Question 23.

Answer:

Question 24.

Answer:

**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

Answer:

Question 26.

Square

Answer:

**MATHEMATICAL CONNECTIONS**

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

Question 27.

Answer:

Question 28.

Answer:

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?

Answer:

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.

Answer:

**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

Answer:

Question 32.

Kite

Answer:

Question 33.

Parallelogram

Answer:

Question 34.

square

Answer:

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.

Answer:

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 \(\overline{C E} \cong \overline{A E}\)

Answer:

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.

Answer:

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.

Answer:

**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

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

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.

Answer:

Question 44.

PROOF Write a proof.

Given QRST is an isosceles trapezoid.

Prove ∠TQS ≅ ∠SRT

Answer:

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

Answer:

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

Answer:

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.

Answer:

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?

Answer:

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).

Answer:

Question 50.

**THOUGHT PROVOKING**

Is SSASS a valid congruence theorem be kites? Justify your answer.

Answer:

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.

Answer:

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 ____________ .

Answer:

Maintaining Mathematical Proficiency

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

Question 53.

Answer:

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.

Answer:

### Quadrilaterals and Other Polygons Review

### 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.

Answer:

Find the va1ue of x.

Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

### 7.2 Properties of Parallelograms

Find the value of each variable in the parallelogram.

Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

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).

Answer:

Question 9.

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

Answer:

### 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.

Answer:

Question 11.

Answer:

Question 12.

Answer:

Question 13.

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

Answer:

Question 14.

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

Answer:

Question 15.

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

Answer:

### 7.4 Properties of Special Parallelograms

Classify the special quadrilateral. Explain your reasoning.

Question 16.

Answer:

Question 17.

Answer:

Question 18.

Answer:

Question 19.

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

Answer:

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.

Answer:

### 7.5 Properties of Trapezoids and Kites

Question 21.

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

Answer:

Question 22.

Find the length of the midsegment of trapezoid ABCD.

Answer:

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).

Answer:

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?

Answer:

Question 25.

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

Answer:

Give the most specific name for the quadrilateral. Explain your reasoning.

Question 26.

Answer:

Question 27.

Answer:

Question 28.

Answer:

### Quadrilaterals and Other Polygons Test

Find the value of each variable in the parallelogram.

Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Give the most specific name for the quadrilateral. Explain your reasoning.

Question 4.

Answer:

Question 5.

Answer:

Question 6.

Answer:

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.

Answer:

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.

Answer:

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

Question 9.

Answer:

Question 10.

Answer:

Question 11.

Answer:

Question 12.

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

Answer:

Question 13.

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

Answer:

Question 14.

Three vertices of JKLM are J(- 2, – 1), K(0, 2), and L(4, 3),

a. Find the coordinates of vertex M.

Answer:

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

Answer:

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?

Answer:

Question 16.

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

Answer:

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.

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.

Answer:

### 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

Answer:

Question 2.

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

Answer:

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.

Answer:

Question 4.

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

Answer:

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.

> < = ≠ ≅

Answer:

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.

Answer:

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 |

Answer: