NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 10: Construction - Exercise 10.4

Question 7

Step-by-Step Construction

1. Draw the given triangle:
   • First, draw a line segment \(AB=4\,\text{cm}\).
   • At point \(A\), use a compass to draw an arc of radius \(9\,\text{cm}\).
   • At point \(B\), draw another arc of radius \(6\,\text{cm}\). The arcs meet at \(C\).
   • Join \(AC\) and \(BC\). Now, \(\triangle ABC\) is ready.
2. Prepare for enlargement: From \(A\), draw a ray \(AX\) making an acute angle with \(AB\).
3. Mark equal divisions: Since the scale factor is \(\tfrac{3}{2}\), divide the ray into 2 equal parts using a compass. Mark them as \(A_1\) and \(A_2\). Then, extend one more equal part to get \(A_3\). (So total = 3 equal divisions on \(AX\)).
4. Connect and draw parallel: Join \(A_2\) with \(B\). Through \(A_3\), draw a line parallel to \(A_2B\). This line meets the extension of \(AB\) at \(B'\).
5. Locate the third vertex: Through \(B'\), draw a line parallel to \(BC\). Through \(A\), draw a line parallel to \(AC\). The two lines meet at \(C'\).
6. Complete the enlarged triangle: Join \(B'C'\). The new triangle \(\triangle AB'C'\) is similar to \(\triangle ABC\).

Justification

By construction, each side is enlarged by the scale factor \(\dfrac{3}{2}\): \(AB' = \dfrac{3}{2} \times AB = 6\,\text{cm}\), \(BC' = \dfrac{3}{2} \times BC = 9\,\text{cm}\), \(AC' = \dfrac{3}{2} \times AC = 13.5\,\text{cm}\).

Thus, \(\dfrac{AB'}{AB} = \dfrac{AC'}{AC} = \dfrac{B'C'}{BC} = \dfrac{3}{2}\). This proves that \(\triangle AB'C' \sim \triangle ABC\) by AA similarity.

Congruence Check

For congruence, scale factor must be exactly 1 (sides must be equal). Here, the scale factor is \(\dfrac{3}{2}\ne1\). So, the triangles are similar but not congruent.