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

Question 1

Step-by-Step Explanation (Beginner Level)

1. Draw the triangle: Start by drawing \(\triangle ABC\). - Make \(AB = 5\,\text{cm}\). - Make \(AC = 7\,\text{cm}\). - Keep the angle between them, \(\angle BAC = 60^\circ\).
2. Mark point P on AB: We are told \(AP = \dfrac{3}{4}AB\). - Since \(AB = 5\,\text{cm}\), \[ AP = \tfrac{3}{4} \times 5 = 3.75\,\text{cm}. \] So, put point P at a distance of 3.75 cm from A on line AB.
3. Mark point Q on AC: We are told \(AQ = \dfrac{1}{4}AC\). - Since \(AC = 7\,\text{cm}\), \[ AQ = \tfrac{1}{4} \times 7 = 1.75\,\text{cm}. \] So, put point Q at a distance of 1.75 cm from A on line AC.
4. Join PQ: Now draw a straight line between P and Q. We need to find its length.
5. Use coordinate geometry to calculate: To make calculation easier, we place triangle ABC on the coordinate plane. - Take A at (0, 0). - Take B at (5, 0) because AB = 5 cm along x-axis. - Take C at \((7\cos60^\circ, 7\sin60^\circ)\). Since \(\cos60^\circ = 0.5\) and \(\sin60^\circ = \tfrac{\sqrt3}{2}\): \[ C = (3.5, 3.5\sqrt3). \]
6. Coordinates of P: P divides AB in ratio 3:1 (because AP = 3/4 of AB). So, P = (3.75, 0).
7. Coordinates of Q: Q divides AC in ratio 1:3 (because AQ = 1/4 of AC). So, Q = (0.25 × 3.5, 0.25 × 3.5√3) = (0.875, 0.875√3).
8. Find PQ using distance formula: Distance formula: \[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. \] Here, - P = (3.75, 0), - Q = (0.875, 0.875√3). Substituting: \[ PQ^2 = (3.75 - 0.875)^2 + (0 - 0.875\sqrt3)^2. \] Simplify: \[ PQ^2 = \left(\tfrac{23}{8}\right)^2 + \left(\tfrac{7\sqrt3}{8}\right)^2 = \tfrac{529}{64} + \tfrac{147}{64} = \tfrac{676}{64}. \] \[ PQ = \sqrt{\tfrac{676}{64}} = \tfrac{13}{4} = 3.25\,\text{cm}. \]

Therefore, the length of PQ is 3.25 cm.