Question

A vertical pole of a length $6m$ casts a shadow $4m$ long on the ground and at the same time, a tower casts a shadow $28m$ long. Find the height of the tower.

Answer 1

Length of the vertical pole $=6m$

Shadow of the pole $=4m$

Length of the shadow of the tower $=28m$

Let the height of the tower$=hm$

From $\Delta ABCand\Delta DEF,$

$\angle C=\angle E$ (angular elevation of the sum)

$\angle B=\angle F=90°$

$\therefore \Delta ABC~\Delta DEF$ (AA similarity criterion)

Hence, $\frac{AB}{DF}=\frac{BC}{EF}$ (If two triangles are similar corresponding sides are proportional)

So we can represent it in the equation as

$$\begin{gathered} \frac{6}{h}=\frac{4}{28} \\ \Rightarrow h=\frac{(6\times 28)}{4} \\ \Rightarrow h=6\times 7 \\ \Rightarrow h=42m \end{gathered}$$

Hence, the height of the tower is $42m.$