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 \(PQR\).
 
10.PNG
 
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 \(\frac{3}{4}\).