#### Question

In the given figure, O is the centre of the circle.* m* ( arc PQR) = 60^{° }OP = 10 cm. Find the area of the shaded region.( \[\pi\]= 3.14, \[\sqrt{3}\]= 1.73)

#### Solution

Radius of the circle, *r* = 10 cm*m*(arc PQR) = ∠POR = *θ* = 60º

∴ Area of the shaded region = Area of segment PQR

\[= r^2 \left( \frac{\pi\theta}{360°} - \frac{\sin\theta}{2} \right)\]

\[ = \left( 10 \right)^2 \left( \frac{3 . 14 \times 60°}{360°} - \frac{\sin60°}{2} \right)\]

\[ = 100 \times \left( \frac{3 . 14}{6} - \frac{\sqrt{3}}{4} \right)\]

\[ = \frac{314}{6} - \frac{173}{4} \left( \sqrt{3} = 1 . 73 \right)\]

\[ = 52 . 33 - 43 . 25\]

\[ = 9 . 08 {cm}^2\]

Thus, the area of the shaded region is 9.08 cm^{2}.

Is there an error in this question or solution?

Solution In the Given Figure, O is the Centre of the Circle. M ( Arc Pqr) = 60° Op = 10 Cm. Find the Area of the Shaded Region.( π = 3.14, √ 3 = 1.73) Concept: Areas of Sector and Segment of a Circle.