### Theory:

A scale factor is defined as the ratio of the corresponding sides of similar figures.
We dealt with similar triangles theoretically in earlier classes. Now, let us discuss how to construct a similar triangle using the concept of the scale factor. There are $$2$$ cases.

Case 1: The scale factor is lesser than $$1$$.

Case 2: The scale factor is greater than $$1$$.

Let us understand the cases using examples.
Example:
Scale factor lesser than $$1$$:

Construct a triangle similar to the given triangle $$PQR$$ with its sides equal to $$\frac{2}{5}$$ of the corresponding sides of the triangle $$PQR$$.

Solution:

Given a triangle $$PQR$$. We are required to construct an another triangle whose sides are $$\frac{2}{5}$$ of the corresponding sides of the triangle $$PQR$$.

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 $$5$$ (the greater of $$2$$ and $$5$$ in $$\frac{2}{5}$$) points $$Q_1$$, $$Q_2$$, $$Q_3$$, $$Q_4$$ and $$Q_5$$ on $$QX$$ so that $$QQ_1$$ $$= Q_1Q_2$$ $$= Q_2Q_3$$ $$= Q_3Q_4$$ $$= Q_4Q_5$$.

Step 4: Join $$Q_5R$$ and draw a line through $$Q_2$$ (the second point, $$2$$ being smaller of $$2$$ and $$5$$ in $$\frac{2}{5}$$) parallel to $$Q_5R$$ to intersect $$QR$$ at $$R'$$.

Step 5: Draw a line through $$R'$$ parallel to the line $$RP$$ to intersect $$QP$$ at $$P'$$. Then, $$P'QR'$$ is the required triangle, each of whose side is two - fifths of the corresponding sides of $$\triangle PQR$$.

JUSTIFICATION:

In the figure, $$QX$$ is a transversal for lines $$Q_5R$$ and $$Q_2R'$$.

$$\angle QQ_5R = \angle QQ_2R'$$

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

Therefore, $$Q_5R$$ is parallel to $$Q_2R'$$.

$$\frac{QQ_2}{Q_2Q_5} = \frac{QR'}{R'R}$$

$$\frac{QQ_2}{Q_2Q_5} = \frac{2}{3}$$

$$\frac{QR'}{R'R} = \frac{2}{3}$$

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

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

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

$$= 1 + \frac{3}{2} = \frac{5}{2}$$

$$\frac{QR}{QR'} = \frac{5}{2}$$

Therefore, $$\frac{QR'}{QR} = \frac{2}{5}$$.

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.

$$\Rightarrow \frac{P'Q}{PQ} = \frac{P'R'}{PR} = \frac{QR'}{QR} = \frac{2}{5}$$.

Hence, justified.

Let us consider the construction of triangle $$ABC$$ using the scale factor $$\frac{3}{4}$$.