Question 10
In the figure, ABCD and AEFD are two parallelograms. Prove that $a r (Δ P E A) = a r (Δ Q F D)$
.

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Solution

In quadrilateral PQDA,
AP || DQ [In parallelogram ABCD, AB || CD]
PQ || AD [in parallelogram AEFD, PE || AD]
then, Quadrilateral PQDA is parallelogram.
Also, parallelogram PQDA and AEFD are on the same base AD and between the same parallels AD and EQ.
$∴$ ar ( parallelogram PQDA) = ar (parallelogram AEFD)
On subtracting ar (quadrilateral APFD) from both sides, we get,
ar (parallelogram PQDA) - ar (quadrilateral APFD)
= ar (parallelogram AEFD) - ar (quadrilateral APFD)
$\Rightarrow a r (Δ Q F D) = a r (Δ P E A)$

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Question 10 In the figure, ABCD and AEFD are two parallelograms. Prove that ar(ΔPEA)=ar(ΔQFD) .

Question 10
In the figure, ABCD and AEFD are two parallelograms. Prove that $a r (Δ P E A) = a r (Δ Q F D)$
.