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 △PXY = x cm²; find, in terms of x, the area of :

(i) triangle PQR

(ii) trapezium XQRY.

(i) In △PXY and △PQR,

∠PXY = ∠PQR [Corresponding angles are equal]

∠XPY = ∠QPR [Common angles]

∴ △PXY ~ △PQR [By AA]

Given,

⇒ PX : XQ = 1 : 3

⇒ $\dfrac{PX}{XQ} = \dfrac{1}{3}$

Let PX = a and XQ = 3a

From figure,

⇒ PQ = PX + XQ = a + 3a = 4a.

Since, corresponding sides of similar triangle are proportional.

$\therefore \dfrac{XY}{QR} = \dfrac{PX}{PQ} \\[1em] \Rightarrow \dfrac{XY}{9} = \dfrac{a}{4a} \\[1em] \Rightarrow XY = \dfrac{9}{4} = 2.25 \text{ cm}.$

Hence, XY = 2.25 cm.

(i) We know that,

The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides.

$\therefore\dfrac{\text{Area of ∆PXY}}{\text{Area of ∆PQR}} = \Big(\dfrac{PX}{PQ}\Big)^2 \\[1em] \Rightarrow \dfrac{x}{\text{Area of ∆PQR}} = \Big(\dfrac{a}{4a}\Big)^2 \\[1em] \Rightarrow \dfrac{x}{\text{Area of ∆PQR}} = \Big(\dfrac{1}{4}\Big)^2 \\[1em] \Rightarrow \dfrac{x}{\text{Area of ∆PQR}} = \dfrac{1}{16} \\[1em] \Rightarrow \text{Area of ∆PQR} = 16x.$

Hence, Area of ∆PQR = 16x cm².

(ii) From figure,

Area of trapezium XQRY = Area of ∆PQR - Area of ∆PXY

= 16x - x = 15x.

Hence, Area of trapezium XQRY = 15x cm².