Question

A vertical pole of length 7.5 m casts a shadow 5 m long on the ground and at the same time a tower casts a shadow 24 m long. Find the height of the tower.

Answer 1

The height of the pole = 7.5 m
length of pole of shadow = BC = 5 m
Length of shadow of tower = EF = 24 cm

Height of tower, ie, ED = ?

in $\Delta ABC$ and $\Delta DEF$,

$\angle B=\angle E={90}^o$ (Both ${90}^o$ as both are vertical to ground)

$\angle C=\angle F$ ( Same elevation in both cases as both the sahdows are cast at the same time)

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

We know that if two triangles are similar, ratio of their sides are in proportion,

so, $\frac{AB}{DE}=\frac{BC}{EF}\frac{7.5}{DE}=\frac{5}{24}DE=\frac{24\times 7.5}{5}=36m$

Height of the tower = 36 m