Construct a triangle similar to a given $\vartriangle ABC$ such that each of its sides is $(5/7)$ of the corresponding sides of $\vartriangle ABC$. It is given that $AB=5cm$, $BC=7cm$ and $\angle ABC={{50}^{\circ }}$

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Construct a triangle similar to a given $\vartriangle ABC$ such that each of its sides is $(5/7)$ of the corresponding sides of $\vartriangle ABC$ . It is given that $AB=5cm$ , $BC=7cm$ and $\angle ABC={{50}^{\circ }}$

Construct a triangle similar to a given $\vartriangle ABC$ such that each of its sides is $(5/7)$ of the corresponding sides of $\vartriangle ABC$ . It is given that $AB=5cm$ , $BC=7cm$ and $\angle ABC={{50}^{\circ }}$

Steps of construction:

1. Make a line segment $BC=7cm$ .

2. Construct a ray BX making an angle of ${{50}^{\circ }}$ and cut off $BA=5cm$ .

3. Now, join AC. Then ABC will the triangle.

4. Construct a ray BY making an acute angle with BC and cut off 7 equal parts making $B{{B}_{1}}={{B}_{1}}{{B}_{2}}={{B}_{2}}{{B}_{3}}={{B}_{3}}{{B}_{4}}={{B}_{4}}{{B}_{5}}={{B}_{5}}{{B}_{6}}={{B}_{6}}{{B}_{7}}$
5. Then, join ${{B}_{7}}$ and C