In the adjoining figure, ABCD and BQSC are two parallelograms. Prove that ar(∆RSC) = ar(∆PQB).

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Solution

In ∆RSC and ∆PQB
$∠$ CRS = $∠$ BPQ (CD || AB) so, corresponding angles are equal)
$∠$ CSR = $∠$ BQP ( SC || QB so, corresponding angles are equal)
SC = QB (BQSC is a parallelogram)
So, ∆RSC $≅$ ∆PQB (AAS congruency)
Thus, ar(∆RSC) = ar(∆PQB)

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In the adjoining figure, ABCD and BQSC are two parallelograms. Prove that ar(∆RSC) = ar(∆PQB).

In ∆RSC and ∆PQB ∠CRS = ∠BPQ (CD || AB) so, corresponding angles are equal) ∠CSR = ∠BQP ( SC || QB so, corresponding angles are equal) SC = QB (BQSC is a parallelogram) So, ∆RSC ≅ ∆PQB (AAS congruency) Thus, ar(∆RSC) = ar(∆PQB)

In ∆RSC and ∆PQB
$∠$ CRS = $∠$ BPQ (CD || AB) so, corresponding angles are equal)
$∠$ CSR = $∠$ BQP ( SC || QB so, corresponding angles are equal)
SC = QB (BQSC is a parallelogram)
So, ∆RSC $≅$ ∆PQB (AAS congruency)
Thus, ar(∆RSC) = ar(∆PQB)