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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

 Video Solution
Triangles
Ex 6.3 | Question 15

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 \(\Delta PQR\)

\[\begin{align}&\angle ABC=\angle PQR= 90^{\circ} \\ &\begin{bmatrix}\text{The objects and shadow are } \\ \text{perpendicular to each other}\end{bmatrix}\end{align}\]

\[\begin{align}&\angle BAC=\angle QPR \\ &\begin{bmatrix} \text{Sunrays fall on the }\\\text{pole and tower at the }\\ \text{same angle, at the same time} \end{bmatrix} \end{align}\]

\[\begin{align}\Rightarrow \quad \Delta ABC \sim \Delta PQR  \\ \text{(AA Criterion)} \end{align}\]

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 \(42 \rm{}\,m.\)