Class 9 AREAS OF PARALLELOGRAMS AND TRIANGLES FOR CBSE NCERT

### Topic covered

• Parallelograms on the same Base and Between the same Parallels

### Parallelograms on the same Base and Between the same Parallels

Now let us try to find a relation, if any, between the areas of two parallelograms on the same base
and between the same parallels. For this, let us perform the following activities:

color{green}("Activity 1 :") Let us take a graph sheet and draw two parallelograms ABCD and PQCD on it as shown in Fig. 9.9.

The above two parallelograms are on the same base DC and between the same parallels PB and DC. You may recall the method of finding the areas of these two parallelograms by counting the squares.

In this method, the area is found by counting the number of complete squares enclosed by the figure, the number of squares a having more than half their parts enclosed by the figure and the number of squares having half their parts enclosed by the figure.

The squares whose less than half parts are enclosed by the figure are ignored. You will find that areas of both the parallelograms are (approximately) 15cm^2.

Repeat this activity* by drawing some more pairs of parallelograms on the graph sheet. What do you observe? Are the areas of the two parallelograms different or equal? If fact, they are equal.

So, this may lead you to conclude that parallelograms on the same base and between the same parallels are equal in area. However, remember that this is just a verification.

color{blue}("Activity 2 :") Draw a parallelogram ABCD on a thick sheet of paper or on a cardboard sheet. Now, draw a line-segment DE as shown in Fig. 9.10.

*This activity can also be performed by using a Geoboard.

Next, cut a triangle A′ D′ E′ congruent to triangle ADE on a separate sheet with the help of a tracing paper and place DeltaA′D′E′ in such a way that A′D′ coincides with BC as shown in Fig 9.11.

"Note" that there are two parallelograms ABCD and EE′CD on the same base DC and between the same parallels A E′ and DC. What can you say about their areas?

color{green}("As" \ \ \ \ \ \ \ \ \ \ \ \ \Delta ADE ≅ D A′D′E′)
color{orange}("Therefore"\ \ \ \ \ \ \ \ \ar (ADE) = ar (A′D′E′))
color{red}("Also" \ \ \ \ \ \ \ \ \ \ \ \ \ ar (ABCD) = ar (ADE) + ar (EBCD))
color{red}(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ar (A′D′E′) + ar (EBCD))
color{navy}(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ar (E E′CD))

So, the two parallelograms are equal in area.

Let us now try to prove this relation between the two such parallelograms.

color{navy}("Theorem 9.1 :")

bb"Parallelograms on the same base and between the same parallels are equal in area."

color{red}("Proof :") Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels AF and DC are given (see Fig.9.12).

We need to prove that color{navy}("ar (ABCD) = ar (EFCD).")
In Delta ADE and Delta BCF,

∠DAE = ∠CBF (Corresponding angles from AD || BC and transversal AF) .......(1)

∠AED = ∠ BFC (Corresponding angles from ED || FC and transversal AF) .......(2)

Therefore, ∠ADE = ∠BCF (Angle sum property of a triangle) ..................(3)

Also, AD = BC (Opposite sides of the parallelogram ABCD)..........(4)

So, Delta ADE ≅ Delta BCF ................. [By ASA rule, using (1), (3), and (4)]

Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas).........(5)

Now, ar (ABCD) = ar (ADE) + ar (EDCB)

= ar (BCF) + ar (EDCB) ......... [From(5)]

 = ar (EFCD)

So, parallelograms ABCD and EFCD are equal in area.

Let us now take some examples to illustrate the use of the above theorem.

Q 3260178015

In Fig. 9.13, ABCD is a parallelogram and EFCD is a rectangle.
Also, AL ⊥ DC. Prove that
(i) ar (ABCD) = ar (EFCD)
(ii) ar (ABCD) = DC × AL
Class 9 Chapter 9 Example 1
Solution:

(i) As a rectangle is also a parallelogram,
therefore, ar (ABCD) = ar (EFCD) (Theorem 9.1)

(ii) From above result,
ar (ABCD) = DC × FC (Area of the rectangle = length × breadth) (1)
As AL ⊥ DC, therefore, AFCL is also a rectangle
So, AL = FC.(2)
Therefore, ar (ABCD) = DC × AL ........... [From (1) and (2)]
Can you see from the Result (ii) above that area of a parallelogram is the product of its any side and the coresponding altitude. Do you remember that you have studied this formula for area of a parallelogram in Class VII. On the basis of this formula, Theorem 9.1 can be rewritten as parallelograms on the same base or equal bases and between the same parallels are equal in area.
Can you write the converse of the above statement? It is as follows: Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels. Is the converse true? Prove the converse using the formula for area of the parallelogram.
Q 3210178019

If a triangle and a parallelogram are on the same base and between the same parallels, then prove that the area of the triangle is equal to half the area of the parallelogram.
Class 9 Chapter 9 Example 2
Solution:

Let Delta ABP and parallelogram ABCD be on the same base AB and between the same parallels AB and PC (see Fig. 9.14).

You wish to prove that ar (PAB) = 1/2 ar(ABCD)
Draw BQ || AP to obtain another parallelogram ABQP. Now parallelograms ABQP and ABCD are on the same base AB and between the same parallels AB and PC.

Therefore, ar (ABQP) = ar (ABCD) (By Theorem 9.1) (1)

But DeltaPAB ≅ Delta BQP (Diagonal PB divides parallelogram ABQP into two congruent triangles.)
So, ar (PAB) = ar (BQP) (2

Therefore, ar (PAB) = 1/2 ar(ABQD) [From (2)] (3)

This gives ar (PAB) = 1/2 ar(ABCD) [From (1) and (3)]