Draw a \( \triangle A B C \) in which \( B C=6 \mathrm{~cm}, A B=4 \mathrm{~cm} \) and \( A C=5 \mathrm{~cm} \). Draw a triangle similar to \( \triangle A B C \) with its sides equal to \( (3 / 4)^{\text {th }} \) of the corresponding sides of \( \triangle A B C \).

Given:

A \( \triangle A B C \) of sides \( B C=6 \mathrm{~cm}, A B=4 \mathrm{~cm} \) and \( A C=5 \mathrm{~cm} \).

To do:

We have to draw a \( \triangle A B C \) in which \( B C=6 \mathrm{~cm}, A B=4 \mathrm{~cm} \) and \( A C=5 \mathrm{~cm} \) and draw a triangle similar to \( \triangle A B C \) with its sides equal to \( (3 / 4)^{\text {th }} \) of the corresponding sides of \( \triangle A B C \).

Solution:

Steps of construction:

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

(ii) With centre $B$ and radius $4\ cm$ and with centre $C$ and radius $5\ cm$, draw arcs intersecting each other at $A$.

(iii) Join $AB$ and $AC$.

$ABC$ is the triangle,

(iv) Draw a ray $BX$ 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 $B_3C$ draw $C_3C’$ parallel to $B_4C$ and from $C’$, draw $C’A’$ parallel to $CA$.

$A’BC’$ is the required triangle.

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Updated on: 10-Oct-2022

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