Question

In the following figure, the area of segment ACB is
(a)$(\frac{\pi }{3}- \frac{\sqrt3} {2})r^{2}$
(b)$(\frac{\pi }{3}-+\frac{\sqrt3} {2})r^{2}$
(c)$(\frac{\pi }{3}- \frac{\sqrt2} {3})r^{2}$
(d) None of these
​

Answer 1

Answer:

The area of the segment , ACB is {π/3  - √3/4 } × r².

Among the given options option (d) None of these is the correct answer.

Step-by-step explanation:

Given :

Radius of a circle = r  

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

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

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

HOPE THIS ANSWER WILL HELP YOU….

 

Answer 2

$\huge\underline\mathcal\blue{Answer}$

Area of the segment,ACB

$= (\frac{\pi}{3} - \frac{ \sqrt{3} }{4} ) \times {r}^{2}$

The correct option is option(d) none of these .

$\huge\underline\mathcal\blue{Explanation}$

Given :

Radius of a circle = r  

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

Solve :

Area of the segment , ACB ,

$a = ( \frac{\pi \: theta}{360} - \frac{ \sin(theta) }{2} \times \frac{ \cos(theta) }{2}) \times {r}^{2}$

$a = ( \frac{120\pi}{360} - \frac{ \sin(120) }{2} \times \frac{ \cos(120) }{2} ) \times {r}^{2}$

$a = ( \frac{\pi}{3} - \sin(60) \times \cos(60) ) \times {r}^{2}$

$a = ( \frac{\pi}{3} - \frac{ \sqrt{3} }{2} \times \frac{1}{2} ) \times {r}^{2}$

$a = ( \frac{\pi}{3} - \frac{ \sqrt{3} }{4} ) \times {r}^{2}$

Area of the segment,ACB

$= (\frac{\pi}{3} - \frac{ \sqrt{3} }{4} ) \times {r}^{2}$

Hence, the area of the segment , ACB is

$(\frac{\pi}{3} - \frac{ \sqrt{3} }{4} ) \times {r}^{2}$