In the circle given below, let O A = 1 unit, O B = 13 unit and P Q ⊥ O B . Then, the area of the triangle P Q B (in square units) is :

Step-by-step solution

Since x-axis is normal to the circle at O , and normal to the circle passes through its centre. So, O B is a diameter. Now, perpendicular from centre of circle to chord bisects the chord, so P A = A Q = λ   ( L e t ) Also, product of segments of chords in a circle is equal i.e. O A · A B = A P · A Q ⇒ 1 · 12 = λ · λ ⇒ λ = 2 3 units Now, area of Δ P Q B = 1 2 × P Q × A B = 1 2 × 2 λ × A B i.e. Δ = 1 2 · 4 3 × 12 sq.