Question

iv) In PQRS (PQ) = 13, (SR) = 9cm , (PS) = 8 , find the area of
O PQRS.
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Answer 1

Answer:

Let ∠SPQ=2θ.

⇒∠SRQ=2θ and ∠TPQ=θ.

Also, PQ∥SR⇒∠TMR=∠TPQ=θ.

∠SRQ is an exterior angle of △TMR.

⇒∠SRQ=∠TMR+∠MTR⇒2θ=θ+∠MTR.

⇒∠MTR=θ.

⇒∠TPQ is isosceles.

⇒TQ=PQ=12.

⇒RT=TQ−QR=TQ−PS=12−9=3 cm.