# Construct a triangle with sides $5 \mathrm{~cm}, 6 \mathrm{~cm}$ and $7 \mathrm{~cm}$ and then another triangle whose sides are $\frac{5}{7}$ of the corresponding sides of the first triangle.

Given:

A triangle with sides $5 \mathrm{~cm}, 6 \mathrm{~cm}$ and $7 \mathrm{~cm}$.

To do:

We have to construct a triangle with sides $5 \mathrm{~cm}, 6 \mathrm{~cm}$ and $7 \mathrm{~cm}$ and then another triangle whose sides are $\frac{5}{7}$ of the corresponding sides of the first triangle.

Solution:

Steps of construction:

(i) Draw a line segment $BC = 5\ cm$.

(ii) With centre $B$ and radius $6\ cm$ and with centre $C$ and radius $7\ cm$, draw arcs intersecting each other at $A$.

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

$ABC$ is the required triangle.

(iv) Draw a ray $BX$ 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) From $B_5$, 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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