A girl of height $ 90 \mathrm{~cm} $ is walking away from the base of a lamp-post at a speed $ 1.2 \mathrm{~m} / \mathrm{sec} $. If the lamp is $ 3.6 \mathrm{~m} $ above the ground, find the length of her shadow after 4 seconds.

Given:

A girl of height \( 90 \mathrm{~cm} \) is walking away from the base of a lamp-post at a speed \( 1.2 \mathrm{~m} / \mathrm{sec} \).

To do:

We have to find the length of her shadow after 4 seconds if the lamp is \( 3.6 \mathrm{~m} \) above the ground.

Solution:

Height of the girl$=90\ cm=0.9\ m$

Height of the lamp post$=3.6\ m$

Speed of the girl$=1.2\ m/s$

Time taken$=4\ s$

Distance covered in $4\ s=Speed \times Time$

$CY=1.2\times4=4.8\ m$.

Let the length of the shadow $AC$ be $x$.

In $\triangle ABC$ and $\triangle AXY$,

$\angle ACB=\angle AYX=90^o$

$\angle BAC=\angle XAY$    (Common angle)

Therefore,

$\triangle ABC \sim\ \triangle AXY$  (By AA similarity)

This implies,

$\frac{AC}{AY}=\frac{BC}{XY}$   (Corresponding sides of similar triangles are proportional)

$\frac{x}{x+4.8}=\frac{0.9}{3.6}$

$\frac{x}{x+4.8}=\frac{1}{4}$

$x(4)=1(x+4.8)$

$4x=x+4.8$

$4x-x=4.8$

$3x=4.8$

$x=\frac{4.8}{3}$

$x=1.6\ m$

The length of her shadow after 4 sec is 1.6 m.

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Updated on: 10-Oct-2022

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