Question

A man of height 60 cm is walking away from the base of a building at a speed of 2.2 m/s. If the height of the building is 7.2 m, then find the length of the shadow of man after 5s.

Answer 1

Given, AB is the height of building = 7.2 m
EC is the height of man = 0.6 m
Speed of man = 2.2 m/s
$\therefore$ Distance travelled in 5 s = Speed $\times$ Time = 2.2 m/s $\times$ 5s = 11 m
$\Rightarrow$ y = 11 m
In $\Delta ABD$ and $\Delta ECD$,
$\angle ABD=\angle ECD$ [each ${90}^{\circ }$]
$\angle ADB=\angle EDC$ [common]
$\therefore \Delta ABD\sim \Delta ECD$ [by AA similarity]
$\Rightarrow \frac{AB}{EC}=\frac{BD}{CD}\Rightarrow \frac{AB}{EC}-1=\frac{BD}{CD}-1\Rightarrow \frac{AB-EC}{EC}=\frac{BD-CD}{CD}\Rightarrow \frac{7.2-0.6}{0.6}=\frac{BC}{CD}\Rightarrow \frac{6.6}{0.6}=\frac{11}{CD}\Rightarrow \frac{6.6}{0.6\times 11}=\frac{1}{CD}\Rightarrow 1=\frac{1}{CD}\Rightarrow CD=1m$
So, the length of the shadow of the man after 5 seconds is 1 m.