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प्रश्न
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.
उत्तर
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`
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