In the given figure, AX : XB = 3 : 5.

Find :

(i) the length of BC, if the length of XY is 18 cm.

(ii) the ratio between the areas of trapezium XBCY and triangle ABC.

(i) Given, AX : XB = 3 : 5

Let AX = 3a and XB = 5a.

From figure,

AB = AX + XB = 3a + 5a = 8a.

(i) In ΔAXY and ΔABC,

As XY || BC, corresponding angles are equal.

∠AXY = ∠ABC and ∠AYX = ∠ACB

∴ ∆AXY ~ ∆ABC [By AA]

Since, corresponding sides of similar triangle are proportional to each other.

$\therefore \dfrac{AX}{AB} = \dfrac{XY}{BC} \\[1em] \Rightarrow \dfrac{3a}{8a} = \dfrac{18}{BC} \\[1em] \Rightarrow BC = \dfrac{18 \times 8a}{3a} = 48 \text{ cm}.$

Hence, BC = 48 cm.

(ii) We know that,

The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\therefore \dfrac{\text{Area of ∆AXY}}{\text{Area of ∆ABC}} = \dfrac{AX^2}{AB^2} \\[1em] = \dfrac{(3a)^2}{(8a)^2} \\[1em] = \dfrac{9a^2}{64a^2} \\[1em] = \dfrac{9}{64}. \\[1em]$

Let area of ∆AXY = 9b and area of ∆ABC = 64b.

From figure,

Area of trapezium XBCY = Area of ∆ABC - Area of ∆AXY

= 64b - 9b = 55b.

$\therefore \dfrac{\text{Area of trap. XBCY}}{\text{Area of ∆ABC}} = \dfrac{55b}{64b} = \dfrac{55}{64}$

Hence, ratio of area of trapezium XBCY and triangle ABC = 55 : 64.