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

Given:

A triangle with sides 5 cm, 6 cm and 7 cm.

To do:

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

Solution:

__Steps of construction:__

(i) Draw a $\triangle ABC$ with sides $AB = 5\ cm, BC = 7\ cm$ and $AC = 6\ cm$.

(ii) Draw an acute angle $CBX$ below $BC$ at point $B$.

(iii) Mark the ray $BX$ as $B_1, B_2, B_3, B_4, B_5, B_6$ and $B_7$ such that $BB_1= B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5 = B_5B_6 = B_6B_7$.

(iv) Join $B_5$ and $C$.

(v) Draw $B_7C’$ parallel to $B_5C$, where $C’$ is a point on extended line $BC$.

(vi) Draw $A’C’ \| AC$, where $A’$ is a point on extended line $BA$.

Therefore,

$A’BC’$ is the required triangle.

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