LEARNATHON

III

Competition for grade 6 to 10 students! Learn, solve tests and earn prizes!

Learn more### Theory:

A scale factor is defined as the ratio of the corresponding sides of similar figures.

In the above theoretical material, we deal with similar triangles theoretically. Let us discuss how to construct a similar triangle using the concept of the scale factor. There are \(2\) cases.

Let us understand the cases using examples.

Example:

**Case 1**: If the scale factor is less 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 another triangle whose sides are \(\frac{2}{5}\) of the corresponding sides of the triangle \(PQR\).

**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 \(5\) (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 \(\triangle PQR\).

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

**Case 2**: If the scale factor is 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 \(PQR\).

**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 \(Q_2\) (the 2nd point, 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 three - twos of the corresponding sides of \(\triangle PQR\).

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