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 :

1. triangle PQR.
2. trapezium XQRY.

Answer 1

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