Question

In the given figure, ABCD and EFGD are two parallelograms and G is the mid point of CD. Then, ar(ΔDPC) : ar(EFGD) = ________.

Answer 1

Given:
ABCD and EFGD are two parallelograms.
G is the mid point of CD


Since, G is the mid point of CD
Therefore, DG = GC

Since, ΔDPG and ΔGPC have equal base and common height,
Thus, ar(ΔDPG) = ar(ΔGPC)
⇒ ar(ΔDPC) = 2 ar(ΔDPG) ...(1)

Also, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
Thus, ar(ΔDPG) = $\frac{1}{2}$ar(EFGD) ...(2)

From (1) and (2),
ar(ΔDPC) = 2 × $\frac{1}{2}$ar(EFGD)
⇒ ar(ΔDPC) = ar(EFGD)
​

Hence, ar(ΔDPC) : ar(EFGD) = 1 : 1.