In the adjoining figure, ABCD and BQSC are two parallelograms.

Question:

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

 

Solution:

In $\triangle \mathrm{RSC}$ and $\triangle \mathrm{PQB}$

$\angle \mathrm{CRS}=\angle \mathrm{BPQ}$      (CD || AB) so, corresponding angles are equal)

$\angle \mathrm{CSR}=\angle \mathrm{BQP}$             ( SC || QB so, corresponding angles are equal)

SC = QB                                    (BQSC is a parallelogram)

So, $\Delta \mathrm{RSC} \cong \Delta \mathrm{PQB}$      (AAS congruency)

Thus, $\operatorname{ar}(\Delta \mathrm{RSC})=\operatorname{ar}(\Delta \mathrm{PQB})$

 

 

 

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