# Ex.6.3 Q15 Triangles Solution - NCERT Maths Class 10

## Question

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

**Diagram**

## Text Solution

**Reasoning:**

The ratio of any two corresponding sides in two equiangular triangles is always the same.

**Steps:**

\(AB\) is the pole = \(6 \) \(\rm m\)

\(BC\) is the shadow of pole = \(4 \) \(\rm m\)

\(PQ\) is the tower =?

\(QR\) is the shadow of the tower = \(28 \) \(\rm m\)

In \(\Delta ABD\) and \(\Delta PQR\)

\(\angle \mathrm{ABC}=\angle \mathrm{PQR}=90^{\circ}\qquad \text{(The objects and shadow are perpendicular to each other)}\)

\(\angle BAC=\angle QPR\ \qquad\qquad\text{(Sunrays fall on the pole and tower at the same angle, at the same time)}\)

\(\Rightarrow \Delta \text {ABC} \sim \Delta \text {PQR}\qquad \text{(AA criterion)}\)

The ratio of any two corresponding sides in two equiangular triangles is always the same.

\[\begin{align} \Rightarrow\qquad \frac{A B}{B C}&=\frac{P Q}{Q R} \\ \frac{6 \rm {}m}{6 \rm{}m}&=\frac{P Q}{28\rm{} m} \\ \Rightarrow\quad P Q&=\frac{6 \times 28}{4}{\rm {m}} \\ P Q &=42\rm{} m \end{align}\]

Hence, the height of the tower is \(42 \rm{}\,m.\)