Question

In the given figure O is the centre of the circle. ∠AOD = 120°. If the radius of the circle be 'r', then find the sum of the areas of quadrilateral AODP and OBQC.

(a) \(\frac{\sqrt3}{2}\) r²

(b) 3√3 r² 

(c) √3 r²

(d) 2√3 r²

Answer 1

(c) r² √3 cm² .

Let OQ = OB = OC = r          (radius of the circle) 

Given, ∠AOD = ∠BOC = 120°

∴ ∠BOQ = ∠COQ = 60° 

⇒ \(\frac{SB}{OB}\) = sin 60°  = \(\frac{\sqrt3}{2}\) 

⇒ SB = \(\frac{r\sqrt3}{2}\) 

∴ BC = 2 SB = r√3 

∴ Area of quadrilateral BQCO 

= \(\frac{1}{2}\) × product of diagonals 

= \(\frac{1}{2}\)× BC × OQ 

= \(\frac{1}{2}\) x r√3 x r = \(\frac{r^2\sqrt3}{2}\) cm² 

∴ Area of both the quadrilaterals 

= 2 x \(\frac{r^2\sqrt3}{2}\) = r² √3 cm² .