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 \(PQR\).
 
11.PNG
 
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 \(\frac{5}{3}\).