### Theory:

A scale factor is defined as the ratio of the corresponding sides of similar figures.
Example:
Scale factor greater than $$1$$.

Construct a triangle similar to a given triangle $$PQR$$ with its sides equal to $$\frac{3}{2}$$ of the corresponding sides of the triangle $$PQR$$.

Solution:

Given a triangle $$PQR$$. We are required to construct an another triangle whose sides are $$\frac{3}{2}$$ of the corresponding sides of the triangle $$PQR$$.

Steps of construction:

Step 1: Construct a triangle $$PQR$$ with any measurement.

Step 2: Draw a ray $$QX$$ making an acute angle with $$QR$$ on the side opposite to vertex $$P$$.

Step 3: Locate $$3$$ (the greater of $$2$$ and $$3$$ in $$\frac{3}{2}$$) points. $$Q_1$$, $$Q_2$$, and $$Q_3$$ on $$QX$$ so that $$QQ_1 = Q_1Q_2 = Q_2Q_3$$.

Step 4: Join $$Q_2$$ ($$2$$ being smaller of $$2$$ and $$3$$ in $$\frac{3}{2}$$) to $$R$$ and draw a line through $$Q_3$$ parallel to $$Q_2R$$, intersecting the extended line segment $$QR$$ at $$R'$$.

Step 5: Draw a line through $$R'$$ parallel to the line $$RP$$ intersecting the extended line segment $$QP$$ at $$P'$$. Then, $$P'QR'$$ is the required triangle, each of whose side is $$\frac{3}{2}$$ of the corresponding sides of $$\triangle PQR$$.

JUSTIFICATION:

In the figure, $$QX$$ is a transversal for lines $$Q_3R'$$ and $$Q_2R$$.

$$\angle QQ_3R' = \angle QQ_2R$$

If corresponding angles are equal, then the lines are parallel.

Therefore, $$Q_3R'$$ is parallel to $$Q_2R$$.

$$\frac{QQ_2}{Q_2Q_3} = \frac{QR}{RR'}$$

$$\frac{QQ_2}{Q_2Q_3} = \frac{2}{1}$$

$$\frac{QR}{RR'} = \frac{2}{1}$$

Now, $$\frac{QR'}{QR} = \frac{QR + RR'}{QR}$$

$$= \frac{QR}{QR} + \frac{RR'}{QR}$$

$$= 1 + \frac{RR'}{QR}$$

$$= 1 + \frac{1}{2} = \frac{3}{2}$$

Therefore, $$\frac{QR'}{QR} = \frac{3}{2}$$.

By construction, $$R'P'$$ is parallel to $$RP$$.

In $$\Delta P'QR'$$ and $$\Delta PQR$$:

$$\angle P'QR' = \angle PQR$$ [common angle]

$$\angle QR'P' = \angle QRP$$ [corresponding angles]

Thus, $$\Delta P'QR' \sim \Delta PQR$$ [by AA similarity].

In similar triangles, corresponding sides are in the same ratio.

Therefore, $$\frac{P'Q}{PQ} = \frac{P'R'}{PR} = \frac{QR'}{QR} = \frac{3}{2}$$.

Let us consider the construction of triangle $$ABC$$ using the scale factor $$\frac{5}{3}$$.