Question

Find the area of the minor segment of a circle of radius 14 cm, when its central angle is 60°­. Find the area of the corresponding major segment. (Use π = $\frac{22}{7}$)

Answer 1

Radius of the cirle with centre 'O' is r = 14 cm = OA = OB

The angle subtended at centre, θ = 60°
Thus, triangle AOB is an equilateral triangle
Length of the question AB = 14 cm

Now, angle of sector $\theta =60°Areaofsector=\frac{\theta }{360°}\times {\mathrm{\pi r}}^2=\frac{60°}{360°}\times \pi \times {14}^2=102.66{\mathrm{cm}}^2\mathrm{And},\mathrm{Area}\mathrm{of}\mathrm{equilateral}\mathrm{triangle}\mathrm{AOB}=\frac{\sqrt{3}}{4}\times {\left(\mathrm{side}\right)}^2=\frac{\sqrt{3}}{4}\times {\left(14\right)}^2=84.8658{\mathrm{cm}}^2\mathrm{Therefore},\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{minor}\mathrm{segment}\mathrm{ACB}=\left(\mathrm{area}\mathrm{of}\mathrm{sector}\right)-\left(\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{AOB}\right)=102.66-84.8658=17.79{\mathrm{cm}}^2\mathrm{Lastly},\mathrm{we}\mathrm{have}\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{major}\mathrm{segment}\mathrm{BDE}=\mathrm{Area}\mathrm{of}\mathrm{circle}-\mathrm{Area}\mathrm{of}\mathrm{minor}\mathrm{segment}\mathrm{ACB}={\mathrm{\pi r}}^2-17.79=\frac{22}{7}\times {\left(14\right)}^2-17.79=798.11{\mathrm{cm}}^2$