Chapter 11 · Question 4

Find the area of the sector of a circle with radius $4\text{ cm}$ and of angle $30^\circ$ . Also, find the area of the corresponding major sector. (Use $\pi = 3.14$ )

Answer Revealed

Direct Answer:

Area of minor sector

= \frac{30}{360} \times 3.14 \times 4 \times 4 = \frac{1}{12} \times 3.14 \times 16 = \frac{50.24}{12} \approx 4.19\text{ cm}^2

. Area of major sector

= \pi r^2 - \text{minor sector} = 3.14 \times 16 - 4.19 = 50.24 - 4.19 = 46.05\text{ cm}^2

(approx

46.1\text{ cm}^2

The minor (

30^\circ

) sector has area about

4.19\text{ cm}^2

. The rest of the circle (the major sector of angle

330^\circ

) has area about

46.05\text{ cm}^2

Given: Radius

r = 4\text{ cm}

, sector angle

\theta = 30^\circ

\pi = 3.14

. Area of minor sector:

\text{Area} = \frac{\theta}{360} \times \pi r^2 = \frac{30}{360} \times 3.14 \times 4 \times 4 = \frac{1}{12} \times 3.14 \times 16 = \frac{50.24}{12} \approx 4.19\text{ cm}^2

. Area of major sector:

\text{Area} = \pi r^2 - \text{Area of minor sector} = 3.14 \times 16 - 4.19 = 50.24 - 4.19 = 46.05\text{ cm}^2

. Alternatively, using the angle of the major sector (

360^\circ - 30^\circ = 330^\circ

\text{Area} = \frac{330}{360} \times 3.14 \times 16 = \frac{11}{12} \times 50.24 = 46.05\text{ cm}^2

. Answer: Minor sector area

\approx 4.19\text{ cm}^2

, major sector area

\approx 46.05\text{ cm}^2

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