# In the Following Figure, If Abc is an Equilateral Triangle, Then Shaded Area is Equal to - Mathematics

MCQ

In the following figure, If ABC is an equilateral triangle, then shaded area is equal to

#### Options

• $\left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right) r^2$

• $\left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} \right) r^2$

• $\left( \frac{\pi}{3} + \frac{\sqrt{3}}{4} \right) r^2$

• $\left( \frac{\pi}{3} + \sqrt{3} \right) r^2$

#### Solution

We have given that ABC is an equilateral triangle.

∴ ∠A=60°

As we know that,∠BCA=1/2 m (∠BOC)

∴ 60=1/2 m (BOC)

m(∠BOC)=120°

Area of the shaded region = area of the segment BC.

Let  ∠BOC=θ

∴ Area of the segment= (piθ/360-sin  θ/2 cos  θ/2)

Substituting the values we get,

Area of the segment= ((pixx120)/360-sin60cos60)r^2

∴ Area of the segment=(pi/3-sin60cos60)r^2

Substituting sin 60=sqrt3/2 and 60=1/2we get,

∴ Area of the segment=(pi/3-1/2xxsqrt3/2)r^2

∴ Area of the segment=(pi/3-sqrt3/4)r^2

Therefore, area of the shaded region is (pi/3-sqrt3/4)r^2.

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 13 Areas Related to Circles
Q 37 | Page 72