In Fig. 9.8, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = 1/2 ar (EFGD). Is the given statement true or false and justify your answer.

Solution:

Let us join PG

G is the midpoint of CD

In ∆ DPC, PG is the median which divides the triangle into two equal areas

ar (∆ DPG) = ar (∆ GPC) = 1/2 ar (∆ DPC) …. (1)

We know that

If a parallelogram and triangle lie on the same base and between the same parallels, then the area of triangle is half the area of parallelogram

As parallelogram EFGD and ∆ DPG lie on the same base and between the same parallels DG and EF

ar (∆ DPG) = 1/2 ar (EFGD) … (2)

From equation (1) and (2)

1/2 ar (∆ DPC) = 1/2 ar (EFGD)

ar (∆ DPC) = ar (EFGD)

Therefore, the statement is false.

✦ Try This: PQRS is a parallelogram whose area is 70 cm² and A is any point on the diagonal QS. Find the area of ∆ ASR.

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 9

NCERT Exemplar Class 9 Maths Exercise 9.2 Problem 5

In Fig. 9.8, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = 1/2 ar (EFGD). Is the given statement true or false and justify your answer.

Summary:

The statement “In Fig. 9.8, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = 1/2 ar (EFGD)” is false

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