Question

In the following figure, If ABC is an equilateral triangle, then shaded area is equal to
(a)$(\frac{\pi }{3}- \frac{\sqrt3} {4})r^{2}$
(b)$(\frac{\pi }{3}- \frac{\sqrt3} {2})r^{2}$
(c)$(\frac{\pi }{3}+\frac{\sqrt3} {4})r^{2}$
(d)$(\frac{\pi }{3}+ {\sqrt3} )r^{2}$
​

Answer 1

Answer:

The area of the shaded region is {π/3  - √3/4 } × r².

Among the given options option (a) {π/3  - √3/4 } × r² is the correct answer.

Step-by-step explanation:

Given :

Radius of a circle = r  

∆ABC is an equilateral triangle.

In an equilateral triangle all the three angles are 60° each.

∠A = ∠B = ∠C = 60°  

∠BAC = ½ ∠BOC

[Angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle]

60° = ½ ×∠BOC

60° × 2 = ∠BOC

∠BOC = 120°  

Angle at the centre of a circle, θ = 120°  

Area of the segment , A = {πθ/360 - sin θ /2 cos θ/2 }r²

A = {120°π/360° - sin 120°/2 cos 120°/2 }× r²

A = {π/3  - sin 60°cos 60°} × r²

A = {π/3  - √3/2 × 1/2} × r²

A = {π/3  - √3/4 } × r²

Area of the segment = {π/3 - √3/4 } × r²

Hence, the area of the shaded region is {π/3  - √3/4 } × r².

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Hey !

______________________________

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

(a)$(\frac{\pi }{3}- \frac{\sqrt3} {4})r^{2}$

(b)$(\frac{\pi }{3}- \frac{\sqrt3} {2})r^{2}$

(c)$(\frac{\pi }{3}+\frac{\sqrt3} {4})r^{2}$

(d)$(\frac{\pi }{3}+ {\sqrt3} )r^{2}$

______________________________

Thanks !