Question

A chord of a circle of radius 15 cm subtends an angle of \(60^\circ\) at the centre of the circle. Find the area of the corresponding minor segment. \((\pi = 3.14, \, \sqrt{3} = 1.73)\)

Hint:

Remember: Area of minor segment = Area of sector – Area of triangle. Use \(\sin \theta\) for the triangle part when \(\theta\) is known.

Answer 1

Step 1: Given data.
Radius \(r = 15 \, \text{cm}\), central angle \(\theta = 60^\circ\).
Step 2: Area of the minor segment.
\[ \text{Area of minor segment} = \text{Area of sector} - \text{Area of triangle} \]
Step 3: Find the area of the sector.
\[ \text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2 \] \[ = \frac{60}{360} \times 3.14 \times 15^2 = \frac{1}{6} \times 3.14 \times 225 = 117.75 \, \text{cm}^2 \]
Step 4: Find the area of the triangle.
For an isosceles triangle with sides \(r, r\) and included angle \(60^\circ\), \[ \text{Area of triangle} = \frac{1}{2} r^2 \sin \theta \] \[ = \frac{1}{2} (15)^2 \sin 60^\circ = \frac{1}{2} \times 225 \times \frac{\sqrt{3}}{2} \] \[ = \frac{225 \times 1.73}{4} = 97.3 \, \text{cm}^2 \]
Step 5: Find the area of the minor segment.
\[ \text{Area of segment} = 117.75 - 97.3 = 20.45 \, \text{cm}^2 \] Step 6: Conclusion.
Hence, the area of the minor segment is \(\boxed{20.45 \, \text{cm}^2}\).