Draw a right triangle in which sides (other than the hypotenuse) are of lengths $ 8 \mathrm{~cm} $ and $ 6 \mathrm{~cm} $. Then construct another triangle whose sides are $ 3 / 4 $ times the corresponding sides of the first triangle.

Given:

A triangle of sides (other than the hypotenuse) are of lengths \( 8 \mathrm{~cm} \) and \( 6 \mathrm{~cm} \).

To do:

We have to draw a right triangle in which sides (other than the hypotenuse) are of lengths \( 8 \mathrm{~cm} \) and \( 6 \mathrm{~cm} \). Then construct another triangle whose sides are \( 3 / 4 \) times the corresponding sides of the first triangle.

Solution:

Steps of construction:

(i) Draw right angled triangle $ABC$ with right angle at $B$ and $BC = 8\ cm$ and $BA = 6\ cm$.

(ii) Draw a line $BY$ making an acute angle with $BC$ and cut off four equal parts.

(iii) Join $B_4C$ and draw $B_3C’\ \parallel B_4C$ and $C’A’$ parallel to $CA$.

$BC’A’$ is the required triangle.

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Updated on: 10-Oct-2022

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