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

Question 5

Step-by-Step Construction (reduction by \(\dfrac{5}{7}\))

1. Draw the given triangle.
   - First, draw a line segment \(AB = 5\,\text{cm}\).
   - At point \(B\), use a protractor to make an angle of \(60^\circ\).
   - From this angle line, mark point \(C\) such that \(BC = 6\,\text{cm}\).
   - Join \(A\) to \(C\). Now, \(\triangle ABC\) is drawn.
2. Draw a helper ray.
   - From point \(B\), draw a ray \(BX\) making an acute angle (less than \(90^\circ\)) with side \(BC\).
3. Mark equal divisions.
   - On \(BX\), use a compass or scale to cut 7 equal parts. Name them \(B_1, B_2, ..., B_7\) starting from \(B\).
   - So, \(B = B_0\), then \(B_1, B_2,...,B_7\) are equally spaced points.
4. Connect the last point to C.
   - Join \(B_7\) to \(C\).
5. Use parallel lines to reduce.
   - Locate \(B_5\) (because the scale factor is \(\tfrac{5}{7}\)).
   - Draw a line from \(B_5\) parallel to \(B_7C\). Let this line meet the extended side \(BC\) at \(C'\).
6. Construct the reduced A'.
   - Through \(C'\), draw a line parallel to \(AC\).
   - This new line meets the line through \(B\) parallel to \(BA\) at point \(A'\).
7. Complete the reduced triangle.
   - Join \(A'\) and \(C'\). The new triangle \(A'BC'\) is the required reduced triangle.

Justification (Why this works)

From the Basic Proportionality Theorem (Thales’ theorem):

\[ \dfrac{BB_5}{BB_7} = \dfrac{5}{7} \]

This ratio is carried forward to the sides of the new triangle:

• \(BA' / BA = 5/7\)
• \(BC' / BC = 5/7\)
• \(A'C' / AC = 5/7\)

So, each side of \(\triangle A'BC'\) is \(\tfrac{5}{7}\) of the corresponding side of \(\triangle ABC\). Hence, the triangles are similar with scale factor \(\tfrac{5}{7}\).