Construct a triangle similar to a given $ \triangle A B C $ such that each of its sides is $ (2 / 3)^{\text {rd }} $ of the corresponding sides of $ \triangle A B C $. It is given that $ B C=6 \mathrm{~cm}, \angle B=50^{\circ} $ and $ \angle C=60^{\circ} $.

Given:

A \( \triangle A B C \) of sides \( B C=6 \mathrm{~cm}, \angle B=50^{\circ} \) and \( \angle C=60^{\circ} \).

To do:

We have to construct a triangle similar to a given \( \triangle A B C \) such that each of its sides is \( (2 / 3)^{\text {rd }} \) of the corresponding sides of \( \triangle A B C \).

Solution:

Steps of construction:

(i) Draw a line segment $BC = 6\ cm$.

(ii) Draw a ray $BX$ making an angle of $50^o$ and $CY$ making $60^o$ with $BC$ which intersect each other at $A$.

$ABC$ is the required triangle.

(iii) From $B$, draw another ray $BZ$ making an acute angle below $BC$ and intersect three equal parts making $BB_1 =B_1B_2 = B_2B_3$

(iv) Join $B_3C$.

(v) From $B_2$, draw $B_2C^{’}$ parallel to $B_3C$ 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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