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Question
The given figure shows a triangle PQR in which XY is parallel to QR. If PX : XQ = 1 : 3 and QR = 9 cm, find the length of XY.
Further, if the area of ΔPXY = x cm2; find, in terms of x the area of :
- triangle PQR.
- trapezium XQRY.
Solution
In ΔPXY and ΔPQR, XY is parallel to QR, so corresponding angles are equal.
∠PXY = ∠PQR
∠PYX = ∠PRQ
Hence, ΔPXY ∼ ΔPQR ...(By AA similarity criterion)
`(PX)/(PQ) = (XY)/(QR)`
`=> 1/4 = (XY)/(QR)` ...(PX : XQ = 1 : 3 `=>` PX : PQ = 1 : 4)
`=> 1/4 = (XY)/(9)`
`=>` XY = 2.25 cm
i. We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
`(Ar(ΔPXY))/(Ar(ΔPQR)) = ((PX)/(PQ))^2 `
`x/(Ar(ΔPQR)) = (1/4)^2 = 1/16`
Ar (ΔPQR) = 16x cm2
ii. Ar (trapezium XQRY) = Ar (ΔPQR) – Ar (ΔPXY)
= (16x – x) cm2
= 15x cm2
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