Question

Construct a triangle similar to a given triangle $ABC$ with its sides equal to $\frac{3}{4}$ of corresponding sides of $△ABC$.

Answer 1

Steps of construction

$\left(1\right).$Draw any ray $PX$ making an acute angle with $PQ$ on the side opposite to the verterx $Q$.

$\left(2\right).$Mark four points on PX $P_1,P_2,P_3$ and $P_4$ on PX so that $P{P}_1=P_1{P}_2=P_2{P}_3=P_3{P}_4$

$\left(3\right).$Join $P_{4}Q$ and draw a line through $P_3$ parallel to $P_{4}Q,$ to intersects $PQ$ at $Q^{\prime }$.

$\left(4\right).$Draw a line through $Q^{\prime }$ parallel to the line $QR$ to intersect $PR$ at $R^{\prime }$.

Thus the required triangle.

Hence, this is the answer.