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

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## 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 $$42 \rm{}\,m.$$

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