Question

Construct a triangle similar to a given $△ABC$ such that each of its sides is $(3/4{)}^{th}$ of the corresponding sides of $△ABC$. It is given that $BC=6cm,\angle B={50}^{\circ }$ and $\angle C={60}^{\circ }$.

Answer 1

Let $\Delta ABC$ be a triangle having $BC=6cm;\angle B=50°;\angle C=60°$.

Steps of construction of triangle $\left(3/4\right)th$ of $△ABC$:

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

$\left(2\right)$. Mark four points on $BX$ as $B_1,B_2,B_3,B_4$ such that $B{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4$.

$\left(3\right)$. Join $B_{4}C$ and draw a line through $B_3$ parallel to $B_{4}C$ 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.