Question

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

Answer 1

Let $\Delta ABC$ be the given triangle having $AB=5cm;BC=7cm;\angle ABC=50°$.

Steps of construction:

$\left(1\right).$ Draw a ray $BX$ making an acute angle with $BC$ on the opposite side of vertex $A$.

$\left(2\right)$. Mark seven points on $BX$ as $B_1,B_2,B_3,B_4,B_5,B_6,B_7$ such that $B{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4=B_4{B}_5=B_5{B}_6=B_6{B}_7$.

$\left(3\right)$. Join $B_{7}C$ and draw a line through $B_5$ parallel to $B_{7}C$ intersect $BC$ at $C^{\prime }$.

$\left(4\right).$ Draw a line $C^{\prime }$ parallel to $AC$ intersect $AB$ at $A^{\prime }$.

Thus, $\Delta A^{\prime }B{C}^{\prime }$ is required triangle.