In the circle given below, let OA = 1 unit, OB = 13 unit and PQ ⊥ OB. Then, the area of the triangle PQB (in square units) is :

Solution

Maths sir · 4 steps

O and B both sit on the circle, so OB is a diameter, and the chord PQ is cut in half at A.

1. Read the figure. The circle passes through O and B, and both lie on the x-axis. Because PQ ⊥ OB at A, the diameter OB bisects the chord PQ, so PA = AQ. Also AB = OB - OA = 13 - 1 = 12.
2. Intersecting chords. Chords PQ and OB cross at A. The products of their pieces are equal.

   PA · AQ = OA · AB ⇒ PA² = 1 × 12 = 12

3. Base and height. So PA = 2√3 and PQ = 4√3. The height of triangle PQB from B onto PQ is AB = 12.
4. So. Area = 1/2 × 4√3 × 12 = 24√3 square units, which is option (b).

The idea to keep. When a chord is perpendicular to a diameter, the half-chord squared equals the product of the two pieces of the diameter. This saves you from writing the circle's equation.

Answer from the NTA final answer key. Solution by PrepWiser.