Question

a chord pq of a circle of radius 10 cm subtends an angle of 60° at the centre of circle.Find the area of major and minor segments of the circle....diagram please

Answer 1

I hope it will help you.

Answer 2

Answer:  

The area of minor segments of the circle is 9.06 cm².  The area of major segment of the circle is 305.1cm².

Step-by-step explanation:  

The radius of the circle is 10 cm. The subtends angle is 60 degree at the center.  

The formula to find the area of minor segment is  

$A_1=\pi r^2\times \frac{\theta}{360}-\frac{1}{2}\times r^2\sin \theta$  

$A_1=\pi (10)^2\times \frac{60}{360}-\frac{1}{2}\times (10)^2\sin (60)$  

$A_1=\pi (10)^2\times \frac{1}{6}-\frac{1}{2}\times (10)^2\frac{\sqrt{3}}{2}$  

$A_1=9.06$  

The area of minor segments of the circle is 9.06 cm².

The area of circle is  

$A_2=\pi r^2$  

$A_2=\pi (10)^2$  

$A_2=314.16$  

The area of major segment of the circle is  

$A=A_2-A_1=314.16-9.06=305.1$  

Therefore the area of major segment of the circle is 305.1cm².