Question

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

Hint:

Always use $\dfrac{\theta}{360^\circ}$ for finding the fractional part of the circle corresponding to the sector’s angle.

Answer 1

Step 1: Write the formula for the area of a sector.
\[ \text{Area of sector} = \dfrac{\theta}{360^\circ} \times \pi r^2 \] Step 2: Substitute the given values for the minor sector.
\[ \theta = 60^\circ, \quad r = 15 \, \text{cm} \] \[ \text{Area of minor sector} = \dfrac{60}{360} \times 3.14 \times 15^2 \] \[ = \dfrac{1}{6} \times 3.14 \times 225 = 117.75 \, \text{cm}^2 \] Step 3: Find the area of the major sector.
Since the total area of the circle is: \[ \pi r^2 = 3.14 \times 15^2 = 706.5 \, \text{cm}^2 \] \[ \text{Area of major sector} = 706.5 - 117.75 = 588.75 \, \text{cm}^2 \] Step 4: Conclusion.
\[ \boxed{\text{Minor sector area} = 117.75 \, \text{cm}^2} \] \[ \boxed{\text{Major sector area} = 588.75 \, \text{cm}^2} \]