Question

AB is a chord of a circle centred at O such that ∠AOB = 60°. If OA = 14 cm, then find the area of the minor segment. ${\displaystyle (\text{take}\sqrt{3}=1.73)}$

Answer 1

Let ACB be given the subtending angle of 60° at the centre.

∵ θ = 60° and OA = OB

∴ ΔOAB is an equilateral triangle.

Here, r = 14 cm and θ = 60°

Area of minor Segment = Area of sector OACBO − Area of ΔOAB

= ${\displaystyle \frac{\theta }{360°}\times \pi r^2-\frac{\sqrt{3}}{4}{r}^2}$

= ${\displaystyle \frac{60°}{360°}\times \frac{22}{7}\times {(14 cm)}^2-\frac{\sqrt{3}}{4}\times {(14 cm)}^2}$

= ${\displaystyle \frac{60}{360}\times \frac{22}{7}\times 14 cm\times 14 cm-\frac{\sqrt{3}}{4}\times 14 cm\times 14 cm}$

= ${\displaystyle \frac{22\times 14}{3} c{m}^2-49\sqrt{3} c{m}^2}$

= ${\displaystyle \frac{308}{3} c{m}^2-49\times 1.732 c{m}^2}$

= 102.67 cm² − 84.87 cm²

= 17.8 cm²