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# Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are $\frac{2}{3}$ of the corresponding sides of the first triangle.

Given:

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

To do:

We have to construct a triangle of sides \( 4 \mathrm{~cm}, 5 \mathrm{~cm} \) and \( 6 \mathrm{~cm} \) and then a triangle similar to it whose sides are \( (2 / 3) \) of the corresponding sides of it.

Solution:

__Steps of construction:__

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

(ii) With centre $B$ and radius $4\ cm$ and with centre $C$ and radius $6\ cm$, 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 three equal parts making $BB_1 = B_1B_2= B_2B_3$.

(v) Join $B_3C$.

(vi) Draw $B^{\'}C^{\'}$ parallel to $B_3C$ and $C^{’}A^{’}$ parallel to $CA$.

$A^{’}BC^{’}$ is the required triangle.

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