Draw a $ \triangle A B C $ with side $ B C=6 \mathrm{~cm}, A B=5 \mathrm{~cm} $ and $ \angle A B C=60^{\circ} $. Then, construct a triangle whose sides are $ (3 / 4)^{\text {th }} $ of the corresponding sides of the $ \triangle A B C $.

Given:

A \( \triangle A B C \) with side \( B C=6 \mathrm{~cm}, A B=5 \mathrm{~cm} \) and \( \angle A B C=60^{\circ} \).

To do:

We have to draw a \( \triangle A B C \) with side \( B C=6 \mathrm{~cm}, A B=5 \mathrm{~cm} \) and \( \angle A B C=60^{\circ} \). Then, construct a triangle whose sides are \( (3 / 4)^{\text {th }} \) of the corresponding sides of the \( \triangle A B C \). 

Solution:

Steps of construction:

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

(ii) At $B$, draw a ray $BX$ making an angle of $60^o$ with $BC$ and cut off $BA = 5\ cm$.

(iii) Join $AC$.

$ABC$ is the triangle.

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

(v) Join $B_4$ and $C$.

(vi) From $B_3$, draw $B_3C’$ parallel to $B_4C$ 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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