Question

Question 79

Ratio of the area of $\Delta WXY$ to the area of $\Delta WZY$ is 3 : 4 in the given figure. If the area of $\Delta WXZ$ is $56c{m}^2$ and WY = 8 cm, find the lengths of XY and YZ.

Answer 1

Given, area of $\Delta WXZ=56c{m}^2$

$\Rightarrow \frac{1}{2}\times WY\times XZ=56[\because \text{area of triangle}=\frac{1}{2}\times base\times height]$

$\Rightarrow \frac{1}{2}\times 8\times XZ=56\Rightarrow XZ=14cm[\because WY=8cm,given]$

$\therefore$ Area of $\Delta WXY$ : Area of $\Delta WZY=3:4$

$\Rightarrow \frac{Areaof\Delta WXY}{Areaof\Delta WZY}=\frac{3}{4}$

$\Rightarrow \frac{\frac{1}{2}\times WY\times XY}{\frac{1}{2}\times YZ\times WY}=\frac{3}{4}$

$\Rightarrow \frac{XY}{YZ}=\frac{3}{4}$

$\Rightarrow \frac{XY}{XZ-XY}=\frac{3}{4}[\because YZ=XZ-XY]$

$\Rightarrow \frac{XY}{14-XY}=\frac{3}{4}$ [by cross-mutiplication]

$\Rightarrow 4XY=42-3XY$

$7XY=42\Rightarrow XY=6cm$

So, YZ = XZ - XY = 14-6

YZ = 8 cm

Hence, XY = 6 cm and YZ = 8 cm.