Construct a triangle similar to a given $ \triangle A B C $ such that each of its sides is $ (5 / 7)^{\text {th }} $ of the corresponding sides of $ \triangle A B C $. It is given that $ A B=5 \mathrm{~cm}, B C=7 \mathrm{~cm} $ and $ \angle A B C=50^{\circ} . $

Given:

A \( \triangle A B C \) of sides \( A B=5 \mathrm{~cm}, B C=7 \mathrm{~cm} \) and \( \angle A B C=50^{\circ} . \) 

To do:

We have to construct a triangle similar to a given \( \triangle A B C \) such that each of its sides is \( (5 / 7)^{\text {th }} \) of the corresponding sides of \( \triangle A B C \).

Solution:

Steps of construction:

(i) Draw a line segment $BC = 7\ cm$.

(ii) Draw a ray $BX$ making an angle of $50^o$ and cut off $BA = 5\ cm$.

(iii) Join $AC$.

$ABC$ is the required triangle.

(iv) Draw a ray $BY$ making an acute angle with $BC$ and cut off seven equal parts making $BB_1 =B_1B_2=B_2B_3=B_3B_4=B_4B_5=B_5B_6=B_6B_7$

(v) Join $B_7$ and $C$

(vi) Draw $B_5C^{'}$ parallel to $B_7C$ and $C^{'}A^{’}$ parallel to $CA$.

$A^{’}BC^{'}$ is the required triangle.

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Updated on: 10-Oct-2022

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