Draw a right triangle $ A B C $ in which $ A C=A B=4.5 \mathrm{~cm} $ and $ \angle A=90^{\circ} . $ Draw a triangle similar to $ \triangle A B C $ with its sides equal to $ (5 / 4) $ th of the corresponding sides of $ \triangle A B C $.

Given:

A triangle with sides \( A C=A B=4.5 \mathrm{~cm} \) and \( \angle A=90^{\circ} . \)

To do:

We have to draw a right triangle \( A B C \) in which \( A C=A B=4.5 \mathrm{~cm} \) and \( \angle A=90^{\circ} \) and draw a triangle similar to \( \triangle A B C \) with its sides equal to \( (5 / 4) \) th of the corresponding sides of \( \triangle A B C \).

Solution:

Steps of construction:

(i) Draw a line segment $AB = 4.5\ cm$.

(ii) At $A$, draw a ray $AX$ perpendicular to $AB$ and cut off $AC = AB = 4.5\ cm$.

(iii) Join $BC$. 

$ABC$ is the triangle.

(iv) Draw a ray $AY$ making an acute angle with $AB$ and cut off five equal parts making $AA_1 = A_1A_2 = A_2A_3 =A_3A_4 = A_4A_5$

(v) Join $A_4$ and $B$.

(vi) From $A_5$, draw $A_5B’$ parallel to  $A_4B$ and  $B’C’$ parallel to $BC$.

$AB’C’$ is the required triangle.

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Updated on: 10-Oct-2022

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