Question

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

Answer 1

Steps of construction:
1. Draw a line segment $BC=7cm$.
2. Draw a ray $BX$ making an angle of ${50}^{\circ }$ and cut off $BA=5cm$.
3. Join $AC$. Then $ABC$ is the triangle.
4.Draw a ray BY making an acute angle with $BC$ and cut off 7 equal parts making $B_1=B_1{B}_2=B_2{B}_3=B_3{B}_4=B_4{B}_5=B_5{B}_6=B_6{B}_7$ 5. Now, join $B_7$ and $C$. Draw $B_5{C}^{\prime }$ parallel to $B_{7}C$ and $C^{\prime }{A}^{\prime }$ parallel to CA.
Then, $△A^{\prime }{BC}^{\prime }$ is the required triangle.