Question

In the given figure,  ∠PQR = ∠PST = 90°, PQ = 5cm and PS =  2cm.

1. Prove that ΔPQR ~ ΔPST.
2. Find the Area of ΔPQR : Area of quadrilateral SRQT.

Answer 1

To prove ΔPQR ∼ ΔPST

In ΔPQR and ΔPST

∠PQR = ∠PST = 90°

∠P = ∠P       ...( common )

∴ ΔPQR ∼ ΔPST       ...{by AA axiom}

(ii) `("Area of PQR ")/("Area of Quadrilateral") = ("ar" (ΔPQR))/("ar" (ΔPST ))`

= `(1/2 xx PQ xx QR )/(1/2 xxPS xx ST) = 5/2 xx 5/2`

`("ar" (ΔPQR))/("ar" (ΔPST )) = 25/4      ...[ ∵ (PQ)/(PS) = (QR)/(ST) ]`

Taking the reciprocals on both sides,

`("ar" (ΔPST ))/("ar" (ΔPQR)) = 4/25`

Now deducting both sides by 1,

` therefore 1 - ("ar" (ΔPST ))/("ar" (ΔPQR)) = 1-4/25`

`("ar" (ΔPQR )- "ar" (ΔPST)) /("ar"(ΔPQR)) = 21/25`

`("Area of quadrilateral")/("ar"(Delta PQR)) = 21/25`

Hence, the Area of ΔPQR : Area of quadrilateral SRQT is 25 : 21.