A chord of a circle of radius 15 cm subtends an angle of 60° at the centre of the circle. Find the areas Class 9

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre of the circle. Find the areas Class 9

Question 1.

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. Class 9

Use π ≈ 3.14 and $\sqrt{3}$ ≈ 1.73.)

Solution:

Radius (r) = 15 cm Angle at centre of circle (θ) = 60°

Area of sector

$\begin{aligned} & =\frac{\theta}{360^{\circ}} \times \pi r^2 \\ & =\frac{60}{360} \times 3.14 \times(15)^2 \\ & =\frac{1}{6} \times 3.14 \times 225\end{aligned}$

= 117.75 cm² ...(i)

In ∆OAB OA = OB = 15 cm and ∠AOB = 60

∴ ∆OAB is an equilateral triangle

∵ Area of equilateral triangle

$\begin{aligned} & =\frac{\sqrt{3}}{4} \times(\text { side })^2 \\ & =\frac{1.73}{4} \times(15)^2 \\ & =\frac{1.73}{4} \times 225\end{aligned}$

= 97.31 cm² ....(ii)

∵ Area of minor segment

= Area of minor sector - Area of triangle AOB

= 117.75 - 97.31 ...[From (i), (ii)]

= 20.44 cm²

∴ Area of minor segment = 20.44 cm²

Area of circle = πr² = 3.14 ×(15)²

= 706.5 cm²

∵ Area of major segment = Area of circle - Area of minor segment

= 706.5 - 20.44 = 686.06 cm²

∴ Area of major segment = 686.06 cm²