Sum

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

(i) Prove that ΔPQR ~ ΔPST.

(ii) Find Area of ΔPQR : Area of quadrilateral SRQT.

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#### Solution

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`

Concept: Similarity of Triangles

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