A chord of a circle of radius r subtends an angle of 60° at the centre of the circle. Show that the area Class 9

A chord of a circle of radius r subtends an angle of 60° at the centre of the circle. Show that the area Class 9

Question 1.

A chord of a circle of radius r subtends an angle of 60° at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to $r^2\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right)$. Class 9

Soliution:

Central angle (θ) = 60°

Area of sector

$\begin{aligned} & =\frac{\theta}{360^{\circ}} \times \pi r^2 \\ & =\frac{60}{360} \times \pi r^2=\frac{1}{6} \pi r^2\end{aligned}$

∵ In ∆AOB, OA = OB = r units and ∠AOB = 60°

∴ ∆AOB is equilateral triangle

∴ Area of triangle = $\frac{\sqrt{3}}{4} r^2$

∵ Area of minor segment

= Area of sector - Area of triangle

$\begin{aligned} & =\frac{1}{6} \pi r^2-\frac{\sqrt{3}}{4} r^2 \\ & =r^2\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right)\end{aligned}$

∴ Area of minor segment = $r^2\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right)$

[Note: The question has been modified]