Question

In fig 12.45, ABCD and AEFD are two parallelograms. Prove that ar(∆PEA) = ar(∆QFD).

Answer 1

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Answer 2

$\textbf{\huge{ANSWER:}}$

As I don't have any symbol of triangle, let's take (◇) as the symbol of a triangle.

Given:

ABCD is a parallelogram

AEFD is a parallelogram

To prove:

Ar (◇PEA) = Ar (◇QFD)

Proof:

In quadrilateral APQD:

AP is parallel to QD (As ABCD is a parallelogram and opposite sides of parallelogram are parallel)

EF is parallel to AD (As AEFD is a parallelogram and opposite sides of a parallelogram are parallel)

This shows us that APQD is also a parallelogram as both pair of sides are parallel.

Now,

Ar (AEFD) = Ar (APQD) {As they both lie on the same base and between the same parallel lines}

=》 Ar (AEFD) - APFD = Ar (APQD) - APFD

=》 Ar (◇PEA) = Ar (◇QFD)

Hence Proved!!