Question

Question 5
Write True or False and justify your answer :

In the figure, ABCD and EFGD are two parallelograms and G is the midpoint of CD. Then, $ar\left(\Delta DPC\right)=ar\frac{1}{2}\left(EFGD\right)$.

Answer 1

False.

In the given figure, join PG.

G is the mid-point of CD.
Thus, PG is a median of $\Delta$ DPC and it divides the triangle into parts of equal areas.
$ar\left(\Delta DPG\right)=ar\left(\Delta GPC\right)=\frac{1}{2}ar\left(\Delta DPC\right)$ ...(i)

Also, we know that; if a parallelogram and a triangle lie on the same base and between the same parallels, then the area of the triangle is equal to half of the area of the parallelogram.

Here, parallelogram EFGD and DPG lie on the same base DG and between the same parallels DG and EF.
So, $ar(\Delta DPG=\frac{1}{2}ar(EFGD)$ ....(ii)

From equations (i) and (ii), we get,
$\frac{1}{2}ar\left(\Delta DPC\right)=\frac{1}{2}ar\left(EFGD\right)$
$\Rightarrow ar\left(\Delta DPC\right)=ar\left(EFGD\right)$