Sachin

Know the process to find Factorization by using identities. Based on the identities, we can simply factorize an algebraic equation. That means, depending on the identities or identity values, we can easily reduce the number of expressions into n number of terms. Generally, for the simplest factorization process, we have to follow some basic expressions. They are

(i) (x + y)^2 = (x)^2 + 2xy + (y)^2.
(ii) (x – y)^2 = (x)^2 – 2xy + (y)^2.
(iii) x^2 – y^2 = (x + y) (x – y).

Factoring Polynomials Identities Steps

Go through the below-listed guidelines on how to factor polynomials using algebraic identities. They are in the following fashion

• Note down the given expression and compare the expression with the basic expressions.
• If there are three terms and all are identified with the positive sign, then that is related to (x + y)^2.
• If the first and last terms are with the positive sign and the middle term is identified with the negative sign, then that is related to (x – y)^2.
• If there are only two terms with one positive sign and one negative sign, then that is related to x^2 – y^2.
• Now, compare the co-efficient values with the basic expressions.
• Find the values of coefficients.
• Based on the values, reduce the given expression into simple terms.

Factorization Using Identities Examples with Answers

1. Factorize using the formula of a square of the sum of two terms

(i) a² + 6a + 9

Solution: Given expression is a² + 6a + 9
There are three terms are identified with the positive sign. Then it is related to the expression (x)^2 + 2xy + (y)^2 = (x + y)^2.
Now, compare the co-efficient values of a² + 6a + 9 with the expression (x)^2 + 2xy + (y)^2.
Here, x = a, 2y = 6 then y = 3.
So, (x + y)^2 = (a + 3)^2.

Finally, our expression a^2 + 6a + 9 is reduced to (a + 3)^2.

(ii) a^2 + 20a + 100

Solution:
Given expression is a^2 + 20a + 100.
There are three terms that are identified with a positive sign. Then it is related to the expression (x)^2 + 2xy + (y)^2 = (x + y)^2.
Now, compare the co-efficient values of a^2 + 20a + 100 with the expression(x)^2 + 2xy + (y)^2.
Here, x = a, 2y = 20 then y = 10.
So, (x + y)^2 = (a + 10)^2.

Finally, our expression a^2 + 20a + 100 is reduced to (a + 10)^2.

2. Factorize using the formula of the square of the difference of two terms

(i) 4p^2 – 12pq + 9q^2

Solution:
Given expression is 4p^2 – 12pq + 9q^2.
There are three terms that are identified with the positive sign of the first and last terms and the negative sign of the middle term.
Then it is related to the expression (x)^2 – 2xy + (y)^2 = (x – y)^2.
Now, compare the co-efficient values of 4p^2 – 12pq + 9q^2with the expression (x)^2 – 2xy + (y)^2.
Here, x^2 = 4p^2 then x =2p, y^2 = 9q^2 then y = 3q.
So, (x – y)^2 = (2p – 3q)^2.

Finally, our expression 4p^2 – 12pq + 9q^2is reduced to (2p – 3q)^2.

(ii) a^2 – 10a + 25

Solution:
Given expression is a^2 – 10a + 25.
There are three terms that are identified with the positive sign of the first and last terms and the negative sign of the middle term.
Then it is related to the expression (x)^2 – 2xy + (y)^2 = (x – y)^2.
Now, compare the co-efficient values of a^2 – 10a + 25 with the expression (x)^2 – 2xy + (y)^2.
Here, x^2 = a^2 then x =a, y^2 = 25 then y = 5.
So, (x – y)^2 = (a – 5)^2.

Finally, our expression a^2 – 10a + 25 is reduced to (a – 5)^2.

3. Factorize using the formula of a difference of two squares:

(i) 49a^2 – 64

Solution:
Given expression is 49a^2 – 64
There are two terms identified with the positive sign of the first term and the negative sign of the last term.
Now, compare the co-efficient values of 49a^2 – 64 with the expression (x)^2 – (y)^2.
Here, x^2 = 49a^2 = (7a)^2 then x =7a, y^2 = 64 = (8)^2 then y = 8.
So, (x)^2 – (y)^2 = (7a)^2 – (8)^2.
(x + y) (x – y) = (7a + 8) (7a – 8)

Finally, our expression 49a^2 – 64is reduced to (7a + 8) (7a – 8).

(ii) 16a^2 – 36b^2

Solution:
Given expression is 16a^2 – 36b^2
There are two terms identified with the positive sign of the first term and the negative sign of the last term.
Now, compare the co-efficient values of 16a^2 – 36b^2 with the (x)^2 – (y)^2.
Here, x^2 = 16a^2 = (4a)^2 then x =4a, y^2 = 36b^2 = (6b)^2 then y = 6b.
So, (x)^2 – (y)^2 = (4a)^2 – (6b)^2.
(x + y) (x – y) = (4a + 6b) (4a – 6b)

Finally, our expression 16a^2 – 36b^2 is reduced to (4a + 6b) (4a – 6b).

(iii) 1 – [5(2p – 5q)]^2

Solution:
Given expression is 1 – [5(2p – 5q)]^2
We can write it as (1)^2 – [5(2p – 5q)]^2
There are two terms identified with the positive sign of the first term and the negative sign of the last term.
Now, compare the co-efficient values of (1)^2 – [5(2p – 5q)]^2 with the (x)^2 – (y)^2.
Here, x^2 = (1)^2 then x =1,
y^2 = [5(2p – 5q)]^2then y = 5(2p – 5q).
So, (x)^2 – (y)^2 = (1)^2 – [5(2p – 5q)]^2.
(x + y) (x – y) = [1 + 5(2p – 5q)] [1 – 5(2p – 5q)].

Finally, our expression 1 – 25(2p – 5q)^2 is reduced to [1 + 5(2p – 5q)] [1 – 5(2p – 5q)].

4. Factor completely using the formula of a difference of two squares

(i) p^4 –q^4

Solution:
Given expression is p^4 –q^4
We can write it as (p^2)^2 – (q^2)^2
There are two terms identified with the positive sign of the first term and the negative sign of the last term.
Now, compare the co-efficient values of (p^2)^2 – (q^2)^2with the (x)^2 – (y)^2.
Here, x^2 = (p^2)^2 then x =p^2,
y^2 = (q^2)^2 then y = q^2.
So, (x)^2 – (y)^2 = (p^2)^2 – (q^2)^2.
(x + y) (x – y) = (p^2 + q^2) (p^2 –q^2).

Finally, our expression p^4 – q^4 is reduced to (p^2 + q^2) (p^2 – q^2).

Parallelogram consists of a flat shape that has two opposite & parallel sides with equal length. The parallelogram has four sides and also it is called a quadrilateral. The pair of parallel sides are always equal in length of a parallelogram. Furthermore, the interior opposite angles also equal in measurement. While adding the adjacent angles of a parallelogram, you will get 180 degrees.

The area of the parallelogram always depends on its base and height. Also, the perimeter of the parallelogram depends on the length of its four sides.

List of Parallelogram Concepts

Have a glance at the list of Parallelogram Concepts available below and use them for your reference. All you need to do is simply tap on the quick links and avail the underlying concept within. Practice as much as you can and solve all the problems easily.

• Properties of a Rectangle Rhombus and Square
• Problems on Parallelogram
• Practice Test on Parallelogram

Parallelogram Definition

A parallelogram called a quadrilateral that has two pairs of parallel sides. The interior angles of the parallelogram on the same side of the transversal are supplementary. The sum of all the interior angles becomes 360 degrees in a parallelogram.

A rectangle and square also consist of similar properties of a parallelogram. If in case, the sides of the parallelogram become equal, then it treats as a rhombus. Or else, if a parallelogram has one parallel side and the other two sides are non-parallel, then it treats as a trapezium.

From the above figure, ABCD is a parallelogram, where CD || AB and BC || AD. Also, CD = AB and BC = AD.
And, ∠A = ∠C & ∠B = ∠D
Also, ∠A & ∠D are supplementary angles. Because ∠A & ∠D are interior angles present on the same side of the transversal. Similarly, ∠B & ∠C are supplementary angles.

Therefore, ∠B + ∠C = 180, ∠A + ∠D = 180.

Shape of Parallelogram

A parallelogram shape is a two-dimensional shape. It consists of four sides and two pairs of parallel sides. All the parallel sides of the Parallelogram are equal in length. Also, the interior angles of the parallelogram should always equal.

Angles of Parallelogram

The Parallelogram has four angles. Its opposite interior angles always equal and the angles on the same side of the transversal are always supplementary with each other. When you add the same side of the transversal angles, you can get 180 degrees. Furthermore, the sum of the interior angles is 360 degrees.

Properties of Parallelogram

Check the below properties of a parallelogram and solve the related problems easily by applying the same. They are as follows

• The opposite angles of a parallelogram are congruent.
• Also, the opposite sides are parallel and congruent
• The consecutive angles are supplementary
• Furthermore, The two diagonals bisect each other
• If one angle of a parallelogram is a right angle, then all other angles are right angles.
• Parallelogram law: The sum of squares of all the sides of a parallelogram is always equal to the sum of squares of its diagonals.
• Each diagonal bisects the parallelogram into two congruent triangles.

Area of Parallelogram

The area of the Parallelogram totally depends on the Base and height of the Parallelogram.
The area of the Parallelogram can be calculated as Area = Base × Height

Perimeter of Parallelogram

The perimeter of a parallelogram is calculated as the total distance of the boundaries of the parallelogram. By knowing the length and breadth of the parallelogram, we can get the perimeter of a parallelogram.

Perimeter = 2 (a+b) units where a and b are the length of the sides of the parallelogram.

Types of Parallelogram

Depends on the angles and sides of the Parallelogram, mainly four types of Parallelogram are considered.
Let ABCD is a Parallelogram,
1. If AB = BC = CD = DA are equal, then it is called a rhombus. The properties of the rhombus and parallelogram are equal.
2. Rectangle, Square are also types of the parallelogram.

Parallelogram Theorems

Theorem: Prove that in a parallelogram, the opposite sides are equal; the opposite angles are equal; diagonals bisect each other.

Proof:
Let ABCD be a parallelogram. Draw its diagonal AC.
In ∆ ABC and ∆ ACD, ∠1 = ∠4 (alternate angles)
∠3 = ∠2 (alternate angles)
and AC = CA (common)

Therefore, ∆ ABC ≅ ∆ ACD (by ASA congruence)
⇒ AB = CD, BC = AD and ∠B = ∠D.
Similarly, by drawing the diagonal BD, we can prove that
∆ ABD ≅ ∆ BCD
Therefore, ∠A = ∠C
Thus, AB = CD, BC = AD, ∠A = ∠C and ∠B = ∠D.
This proves (i) and (ii)
In order to prove (iii) consider parallelogram ABCD and draw its diagonals AC and BD, intersecting each other at O.
In ∆ OAB and ∆ OCD, we have
AB = CD [Opposite sides of a parallelogram]
∠AOB = ∠ COD [Vertically opposite angles]
∠OAB = ∠OCD [Alternate angles]
Therefore, ∆ OAB ≅ ∆ OCD [By ASA property]
⇒ OA = OC and OB = OD.
This shows that the diagonals of a parallelogram bisect each other.

Therefore, in a parallelogram, the opposite sides are equal; diagonals bisect each other; the opposite angles are equal.

Wanna learn all the basics of Integers concepts? Then start with the absolute value of an integer. Integer Absolute Value is an important and basic concept to learn other concepts of Integers. Refer to all the rules, definitions, solved examples, types, etc to understand the concept more clearly. Follow the below sections to get more idea on how to find the Absolute Value of an Integer in detail and Solved Problems for finding the Integer Absolute Value using different approaches.

Absolute Value of an Integer – Introduction

Integers Absolute Value is the distance of that integer value from zero irrespective of (positive or negative) direction. While considering the absolute value, its numerical value is taken without taking the sign into consideration.

An integer is any positive or negative whole number. Therefore, the positive integer has a negative sign and vice versa.

A positive number is a number that is greater than zero. It is represented by a “+” symbol and can be written with or without a symbol in front of it. The gain in some value or something is written with a positive number. A negative number is a number that is less than zero. It is represented by a “-” symbol and can be written with a “-” symbol in front of it. The loss in some value or something is written with a negative number.

A number line is a kind of diagram on which the numbers are marked at intervals. These are used to illustrate simple and easy numerical operations. Using the number line allows seeing a number is in relationship to other numbers and from zero. Zero, is present in the middle and separates positive and negative numbers. On the right side of zero, we can find numbers that are greater than zero (positive numbers) and on the left side, we can find numbers that are less than zero (negative numbers). The absolute value of the integer is the same as the distance from zero to a specific number.

Representation of Absolute Value of an Integer

Integer Absolute Value is represented with two vertical lines. i.e., | |, one on either side of the integer.

|a| = a, when a is the positive integer

|a| = -a, when a is the negative integer

Examples:

1. Absolute integer value of -15 is written as |-15| = 15 {here mod of -15 = 15}
2. Absolute integer value of 8 is written as |8| = 8 {here mod of 8 = 8}

Adding and Subtracting Absolute Value of Integers

To add 2 integers with the same (positive or negative) sign, add absolute values and assign the sum with the same sign as same both values.

Example:

(-7) + (-4) = -(7 + 4)= – 11.

To add 2 integers with different signs, find the difference in absolute integer values and give that product the same sign of the largest absolute value.

Example:

(-7)+(2)= -5

How to find the absolute value?

1. First of all, find the absolute values of (7 and 2)
2. Find the difference of numbers between 7 and 2 (7-2=5)
3. Find the sign of the largest absolute value i.e., negative, as 7 is of (-)sign
4. Add the sign to the difference we got in Step 2.
5. Hence the final solution is -5

When an integer is subtracted or added by another integer, the result will be an integer.

Multiplication of Absolute Value of Integers

To multiply the integer values, we have to multiply the absolute values. If the integers that are to be multiplied have the same sign, then the result will be positive. If both the integers have different signs, then the result will be negative.

When an integer is multiplied by another integer, then the result will be an integer itself.

Absolute Value of Integers Examples

Question 1:

Write the absolute value of each of the following?

(i) 15

(ii) -24

(iii) -375

(iv) 0

(v) +7

(vi) +123

Solution: 

(i) 15

(ii) 24

(iii) 375

(iv) 0

(v) 7

(vi) 123

Question 2: 

Evaluate the following integers

(i) |-7| + |+5| + |0|

(ii) |10| – |-15| + |+12|

(iii) – |+3| – |-3| + |-6|

(iv) |-8| – |17| + |-12|

Solution: 

(i) 12

(ii) 7

(iii) 0

(iv) 3

Question 3:

State whether the statements are true or false?

(i) The absolute value of -3 is 3.

(ii) The absolute value of an integer is always greater than the integer.

(iii) |+5| = +5

(iv) |-5| = -5

(v) – |+5| = 5

(vi) – |-5| = -5

Solution:

(i) True

(ii) False

(iii) True

(iv) False

(v) False

(vi) True

Methods to Compute the Absolute Value of Integer

There are 3 approaches to find the Integers Absolute value. These approaches are helpful while solving problems. Finding the absolute value is the first step in solving the problems.

Method 1: 

As the absolute value of the integer is always positive. For the positive integer, the absolute number is the number itself. For the negative number, the absolute number is multiplied by other negative numbers.

Method 2:

Negative numbers are saved in 2’s complement form. To get the absolute number, toggle bits of the number, then 1 to the result.

Method 3:

The built-in function library finds the absolute value of an integer.

To solve most of the problems, knowing the absolute value of the integer is important. This method is used in the real world and it has all sorts of math clues. Absolute values are often used in distance problems and also sometimes used for inequalities. Integers Absolute values are really helpful to get clarity on many things.

We have mentioned the important concept of Integers Absolute Value in the above article. Hope you got a clear idea of definitions, rules, and how to solve the problems. If you have any doubts, you can contact us from the below comment box. You can also directly ping us to know other information. Stay tuned to our page to get the latest and important information about all mathematical concepts.

Lines and Angles concepts are explained by experts in order to help students to get excellent marks in their examination. The line is a combination of infinite points that has both directions infinitely. There are different lines available such as intersecting lines, perpendicular lines, transversal lines, etc. An angle is nothing but a combination of two rays with a common endpoint.

Based on the operations performed on a line, then the lines are classified as Parallel Lines, Perpendicular Lines, Transversal, etc. In the same way, the angles are classified as Complementary Angles, Supplementary Angles, Adjacent Angles, Vertically Opposite Angles based on their operation.

Quick Links of Lines and Angles

Working on Lines and Angles problems will help students to get an idea about how to solve the problems. All the concept links are given below. Just check out the concepts and open the topic that is difficult for you. After that solve all the problems available on that topic to get a grip on that entire topic.

• Basic Geometrical Concepts
• Angles
• Classification of Angles
• Related Angles
• Some Geometric Terms and Results
• Complementary Angles
• Supplementary Angles
• Complementary and Supplementary Angles
• Adjacent Angles
• Linear Pair of Angles
• Vertically Opposite Angles
• Parallel Lines
• Transversal Line
• Parallel and Transversal Lines

Line Segment

The line segment is a line that has two endpoints with a fixed length.

Ray

A ray is a line that has only one endpoint and extends infinitely in another direction. From the below figure, point A is the and point.

Perpendicular Lines

Two lines are said to be Perpendicular Lines when they meet at a single point and forms a right angle. From the below figure, PQ and RS are two lines and they intersect at a point x. Also, the two lines PQ and RS form a right angle 90º.

Parallel Lines

Two lines are said to be Parallel Lines when they do not intersect each other and they have the same distance throughout the plane. From the below figure, A and B are two parallel lines.

Transversal Line

A transversal Line is a line that intersects two lines at distinct points. From the below figure, C is the Transversal Line, that intersects the lines A and B at two points D and E.

Properties of Lines

• The points those are not lie on the same line are called non-collinear points.
• The points that lie on the same line are called Collinear points

Acute Angle

The angle that is less than the right angle is called Acute Angle.

Obtuse Angle

The angle that is more than the right angle is called Acute Angle.

Right Angle

The angle with 90 degrees is called as Right Angle.

Straight Angle

The angle that is equal to 180 degrees is called Straight Angle.

Complementary Angles

If the sum of the two angles is 90 degrees, then they are called Complementary Angles.

Supplementary Angles

If the sum of the angles is 180 degrees, then those angles are called Supplementary Angles.

Properties of Angles

• An angle is formed at a point where two rays intersecting.
• If the angle is more than 180 degrees and less than 360 degrees, then it is called a reflex angle.
• If the two angles are adjacent angles and add up to 180 degrees then they form linear pair.

Quick Access to Lines and Angles

We have given solutions for all questions with step by step explanation in understandable language. Therefore, students can get great knowledge on the entire concepts of Lines and Angles. All the questions and answers that are present in the updated syllabus are included on our website with individual concept explanation. So, students who want to learn a single topic can go through the required link available below and learn it from the concerned link.

Are you searching for Lines and Angles Material? You have free access to Lines and Angles Material on our website. You can read and practice all problems either online or offline. So, immediately bookmark our website and practice all the math problems at your comfortable times. Do hard work and get complete knowledge on the Lines and Angles concept. Clear your doubts by contacting us through the comment section.

 

When two triangles lie on the same base and between the same parallels, then those triangles measure equal area. Check the proof for Triangles on the same base and between the same parallel are equal in Area. Get the relationship between the areas of triangles having the same base in the following sections. For the reference of students, we have also included the solved examples on Triangle on the Same Base and between Same Parallels.

Proof for Area of Triangles are Equal Having the Same Base

Get a detailed explanation for how the areas of triangles are equal when their bases are the same and between the same parallels.

In the following figure, ∆ABD and ∆DEF having equal base length and between the same parallels BF and AD.

In this case, Area of △ ABC = Area of △ DEF

To prove that Triangle on the Same Base and between Same Parallels have the same area, follow these instructions.

Let us take △ ABC and △ ABD on the same base AB and between the same parallels AB and CD.

Draw a parallelogram ABPQ with AB base and lying between the same parallels AB and Q.

The relationship between the area of the triangle and parallelogram says that when they lie on the same base and between the same parallels, then the triangle is half of the parallelogram area.

So, Area of ∆ ABC = 1/2 x Area of parallelogram ABPQ

In the same way, Area of ∆ ABD = 1/2 x Area of parallelogram ABPQ

Therefore, Area of ∆ ABC = Area of ∆ ABD.

Hence proved.

Questions on Triangles on the Same Base and between Same Parallels

Example 1.

AD is the median of ∆ABC and ∆ADC. E is any point on AD. Show that area of ∆ABE = area of ∆ACE.

Solution:

Here, AD is the median for △ ABC.

So, BD = CD

In the above figure, △ ABD and △ ADC are having equal bases BD and CD and they are between the same parallels BC and U.

So, Area of △ ADC = Area of △ ABD —– (i)

Since E lies on AD

Therefore, Ed is the median for △ BCE

Now, BDE and CDE are having the equal base length BD = CD and between the same paralleks BC and m, then

Area of △ CDE = Area of △ BDE —— (ii)

Subtracting equation (i) and equation (ii), we get

Area of △ ADC – Area of △ CDE = Area of ∆ABD – Area of ∆BED

Area of △ ABE = Area of △ ACE

Hence, shown.

Example 2.

Prove that the medians of the triangle divide it into two triangles having equal area?

Solution:

AD is the median of the ∆ BCA and AE is the altitude of ∆ BCA and also ∆ ADC.

And AE is perpendicular to BC

Here, AD is the median of ∆ ABC

Therefore, BD = CD

On multiplying both sides of the equation by AE, we get

BD x AE = CD x AE

Again multiply both sides by 1/2

1/2 x BD x AE = 1/2 x CD x AE

Area of ∆ABD = Area of ∆ADC

Hence shown.

If a triangle and parallelogram having the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram. Get the example problems and solutions on Triangle and Parallelogram on the Same Base and between Same Parallels. Also, find the relationship between them in the following sections explained step by step.

Proof for Area of Triangle Equal to Half of the Parallelogram having the Same Base

In the following figure, parallelogram ABCD and ∆ABD are on the same base AB and between the same parallels AF and DC. Get the area of the triangle ABD and area of the parallelogram ABCD and check their areas.

Area of the parallelogram ABCD = base x height

= AB x CF

Here, CF and DE area parallel lines CF = DE.

Area of the parallelogram = AB x DE = AB x CF.

Area of △ ABD = 1/2 x base x height

= 1/2 x AB x DE

= 1/2 x Area of the parallelogram ABCD

Therefore, Area of the parallelogram ABCD = 1/2 x Area of △ ABD

Hence proved.

Here are various case studies for the area of the parallelogram is half of the area of the triangle when they lie on the same base.

1. When a triangle and a parallelogram lies on the same base and have the same altitude, then the area of the triangle is equal to half of the parallelogram.
2. If they lie between the same parallels and have the same altitude, then the area of the triangle is equal to half of the area of the parallelogram.
3. When the triangle and rectangle have the same base and between the same parallels, then the area of the triangle is half of the area of the rectangle.

Example Questions for Triangle and Parallelogram on Same Base and between Same Parallels

Example 1:

∆ ABD and parallelogram ABCD are on the same base AB. If the base and altitude of the parallelogram are 10 cm and 6 cm, find the area of the triangle.

Solution:

Given that,

Base of the parallelogram = 10 cm

Altitude of the parallelogram = 6 cm

Area of the parallelogram = base x altitude

= 10 x 6 = 60 cm²

We know that when a triangle and parallelogram having the same base and between the same parallels then the area of the triangle is equal to half of the area of the parallelogram.

So, area of △ ABD = 1/2x Area of the parallelogram ABCD

= 1/2 x 60 = 30 cm²

Therefore, triangle area is 30 cm².

Example 2:

∆ ABD and parallelogram ABCD are on the same base AB. If the base and height of the triangle is 16 cm and 12 cm, find the area of the parallelogram.

Solution:

Given that,

The base length of the triangle = 16 cm

Height of the triangle = 12 cm

Area of ∆ ABD = 1/2 x base x height

= 1/2 x 16 x 12

= 1/2 x 192 = 96 cm²

We know that when a triangle and parallelogram having the same base and between the same parallels then the area of the parallelogram is equal to double the area of the parallelogram.

Area of the parallelogram ABCD = 2 x Area of ∆ ABD

= 2 x 96 = 192 cm²

Therefore, parallelogram area is 192 cm².

Example 3:

In the adjacent figure, BCDE is a rectangle and ABC is a triangle show that area of the triangle ABC is equal to half of the area of the rectangle BCDE.

Solution:

Given that,

Triangle ABC and rectangle BCDE lies on the same base BC and between the same parallels BC and ED>

Area of triangle ABC = 1/2 x base x height

= 1/2 x BC x AP

AP and Cd are parallel lines and having the same length. So, AP = CD.

△ ABC area = 1/2 x BC x AP = 1/2 x BC x CD

Area of the rectangle = length x breadth

= BC x CD

= 1/2 x Area of △ ABC.

Therefore, the Area of the rectangle BCDE = 1/2 x Area of △ ABC

Hence, shown.

When one parallelogram and a rectangle lies on the same base and between the same parallels then they measure equal area. Here we are providing the example questions, answers for Parallelograms, and Rectangles on Same Base and between the Same Parallels. We are also offering proof that shows parallelogram and rectangle on same between same parallels have the same area.

Proof for Rectangles and Parallelograms on the Same Base have Equal Area

Below-mentioned is the steps for the parallelogram and rectangles having the same base and between the same parallels having equal area.

In the following figure, ABCD is a parallelogram and ABEF is the rectangle that lies on the same base AB and the same parallels AB and DF.

Area of parallelogram ABCD = base x altitude

AF is the altitude of the parallelogram.

So, Parallelogram area = AB x AF.

Area of the rectangle ABEF = base x height

= AB x AF

Therefore, the area of the parallelogram ABCD = Area of the rectangle ABEF.

Solved Examples for Parallelograms and Rectangles on Same Base and between Same Parallels

Example 1:

Parallelogram ABCD and rectangle ABFE have the same AB and the length and breadth of the rectangle are 15 cm and 9 cm. Find the area of the parallelogram.

Solution:

Given that,

Length of the rectangle l = 15 cm

The breadth of the rectangle b = 9 cm

Parallelogram ABCD and rectangle ABFE have the same AB.

So, area of the parallelogram = Area of the rectangle.

Area of the rectangle = length x breadth

= 15 x 9 = 135 cm²

Therefore, area of the parallelogram is 135 cm².

Example 2:

In the adjacent figure, ABCD is a parallelogram and EFCD is a rectangle. Also, AL ⊥ DC. Prove that

(a) Area (ABCD) = Area (EFCD)

(b) Area (ABCD) = DC × AL

Solution:

(a)

Rectangle is CDEF.

Parallelogram is ABCD.

In the given figure, AL and DE are parallel and have the same length. So, DE = AL.

Area of ABCD = CD x AL

= CD x DE square units.

Area of EFCD = CD x DE

So, Area of EFCD = Area of ABCD

Hence proved.

(b)

As ABCD is a parallelogram.

Area of ABCD = base x altitude

The base is CD and the altitude is AL.

Therefore, Area of ABCD = CD x AL

Hence proved.

Example 3:

Parallelogram ABCD and rectangle ABFE have the same AB and the base and height of the parallelogram are 28 cm and 16 cm. Find the area of the rectangle.

Solution:

Given that,

Base of the parallelogram = 28 cm

Height of the parallelogram = 16 cm

Area of the parallelogram = base x height

= 28 x 16 = 448 cm²

We know, that when Parallelograms and Rectangles on Same Base and between Same Parallels, then the area of the parallelogram = area of the rectangle.

So, the area of the rectangle = 448 cm².

Students can check how parallelograms having the same base and between the same parallels are equal in area and proof for that. When two or more parallelogram lies on the same base and between same parallels then they have the same area. Also, find the solved example questions here explaining the concept and learn the concept behind them.

Proof for Theorem: Parallelogram on the Same Base & Between Same Parallels are Equal in Area

Let us draw a parallelogram ABCD on a cardboard sheet or thick sheet of paper.

Now, draw a line segment DE as shown in the figure.

Now, cut the triangle ADE congruent to triangle ADE in a seperate sheet with the help of a tracing paper and place the triangle ADE in such a way that AD coincides BC. The entire process is as shown below.

Note that there are two parallelograms ABCD and EE’CD on the same base CD and between the same parallel AE’ and CD.

∆ADE ≅ ∆ A’ D’ E’

Therefore, Area of ADE = Area of A’D’E’

Also, Area of ABCD = Area of ADE + Area of EBCE

= Area of A’D’E’ + Area of EBCD

= Area of EE’CD

So, the two parallelograms are in an equal area.

Parallelograms on Same Base and Between Same Parallels Examples

Example 1:

A farmer was having a field in the form of a parallelogram ABCD. She took any point P on AB and joined it to points C and D. In how many parts the fields are divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

Solution:

If DP and CP are joined parallelogram ABCD is divided into three triangles.

They are △ APD, △ CDP, △ BCP

△ CDP and parallelogram ABCD are on the same base i.e CD and between parallel sides CD ∥ AC

∴ Area of △ CDP = 1/2 x area of parallelogram ABCD —- (i)

∴ Area of △ APD + Area of △ BCP = 1/2 x area of parallelogram ABCD —- (ii)

From equation (i) and equation (ii)

Area of △ CDP = Area of △ APD + Area of △ BCP

∴ The farmer can use the part of the field to sow wheat, i.e. Area of △ APD + Area of △ BCP, and in the same area he can use the Area of △ CDP to sow pulses.

Example 2:

Parallelograms ABCD and ABEF are situated on the opposite sides of AB in such a way that D, A, F are not collinear. Prove that DCEF is a parallelogram, and parallelogram ABCD + parallelogram ABEF = parallelogram DCEF.

Solution:

AB and DC are two opposite sides of parallelogram ABCD,

Therefore, AB ∥ DC and AB = DC

Again, AB and EF are two opposite sides of parallelogram ABEF

Therefore, AB ∥ EF and AB ∥ EF

Therefore, DC ∥ EF and DC = EF

Therefore, DCEF is a parallelogram.

Therefore, ∆ADF and ∆BCE, we get

AD = BC (opposite sides of parallelogram ABCD)

AF = BE (opposite sides of parallelogram ABEF)

And DF = CE (opposite sides of parallelogram CDEF)

Therefore, ∆ADF ≅ ∆BCE (side – side – side)

Therefore, ∆ADF = ∆BCE

Therefore, polygon AFECD – ∆BCE = polygon AFCED – ∆ADF

Parallelogram ABCD + Parallelogram ABEF = Parallelogram DCEF