Question

Construct a triangle similar to a given triangle $PQR$ with its sides equal to $\frac{7}{3}$ of the corresponding sides of the triangle $PQR$ (scale factor $\frac{7}{3}>1$ )

Answer 1

Given triangle PQR, we are required to construct another triangle whose sides are $\frac{7}{3}$ of the corresponding sides of the $△PQR$.
Steps of construction:
(i) Construct a $△PQR$ with any measurement.
(ii) Draw a ray $QX$ making an acute angle with $QR$ on the side opposite to the vertex $P$.
(iii) Locate 7 points $Q_1,Q_2,Q_3,Q_4,Q_5,Q_6,Q_7$ on $QX$.
So that ${QQ}_1=Q_1{Q}_2=Q_2{Q}_3=Q_3{Q}_4=Q_5{Q}_6=Q_6{Q}_7$

(iv) Join $Q_3$ to $R$ and draw a line through $Q_7$ parallel to $Q_{3}R$ intersecting the extended line segment $QR$ at $R^{\prime }$.
(v) Draw a line parallel to RP. Intersecting the extended line segment QP at $P^{\prime }$.
$\therefore △P^{\prime }{QR}^{\prime }$ is the required triangle.