Construct a triangle similar to a given ΔABC such that each of its sides is (5/7)th of the corresponding sides of Δ ABC.

Solution:

Given that

$A B=5 \mathrm{~cm}, B C=7 \mathrm{~cm}$ and $\angle A B C=50^{0}$

Construct a triangle similar to a triangle $A B C$ such that each of sides is $(5 / 7)^{\text {th }}$ of the corresponding sides of triangle $A B C$.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment.

Step: II- With B as centre and draw an angle.

Step: III- With B as centre and radius, draw an arc, cut the line BY drawn in step II at C.

Step: IV- Joins AC to obtain.

Step: V- Below AB, makes an acute angle.

Step: VI- Along $A X_{t}$ mark off seven points $A_{1}, A_{2}, A_{3}, A_{4,} A_{5} A_{6}$ and $A_{7}$ such that $A A_{1}=A_{1} A_{2}=A_{2} A_{3}=A_{3} A_{4}=A_{4} A_{3}=A_{3} A_{6}=A_{0} A_{7}$

Step: VII-Join.

Step: VIII- Since we have to construct a triangle each of whose sides is $(5 / 7)^{\text {th }}$ of the corresponding sides of $\triangle A B C$.

So, we take five parts out of seven equal parts on $A X$ from point $A_{5}$ draw $A_{5} B^{\prime} \| A_{7} B$, and meeting $A B$ at $B$ '.

Step: IX- From $B^{\prime}$ draw $B^{\prime} C \| B C$, and meeting $A C$ at $C^{\prime}$

Thus, $\triangle A B^{\prime} C^{\prime}$ is the required triangle, each of whose sides is $(5 / 7)^{\text {th }}$ of the corresponding sides of $\triangle A B C$.