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Construct a triangle, the lengths of whose sides are 5 cm, 6 cm and 7 cm. Now construct another triangle whose sides are $\frac{5}{7}$ times the corresponding sides of the first triangle.

Answers (1)

Steps of Construction:

Construct a $\mathrm{\Delta ABC}$ in which AB = 9cm, AC = 5 cm, AC = 6 cm.
Draw a ray from A by making an acute angle with AB opposite to C.
Locate seven points $A_1, A_2,A_3,A_4, A_5, A_6$ and $A_7$ such that $AA_1 =A_1A_2= A_2A_3 = ... = A_6A_7$
Join $A_7B$
From A₅ draw $A_5B' \parallel A_7B$ meeting produced part of AB at B'.
From B' draw $B'C' \parallel BC$ intersecting the line segement AC at C'

This AB'C' is the required triangle.

Posted by

Safeer PP

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