Question:medium

Find the area of the sector of a circle of radius 42 cm and of central angle \(30^\circ\). Also, find the area of the corresponding major sector. [Use \(\pi = \frac{22}{7}\)]

Show Hint

Using \(A_{\text{major}} = \text{Total Area} - A_{\text{minor}}\) is much faster and less prone to calculation errors than calculating with large fraction angles like \(\frac{330^\circ}{360^\circ}\).
Always calculate the total area first if both sectors are required!
Updated On: Jul 7, 2026
  • Minor Area = 462 \(\text{cm}^2\), Major Area = 5082 \(\text{cm}^2\)
  • Minor Area = 462 \(\text{cm}^2\), Major Area = 5544 \(\text{cm}^2\)
  • Minor Area = 231 \(\text{cm}^2\), Major Area = 5082 \(\text{cm}^2\)
  • Minor Area = 154 \(\text{cm}^2\), Major Area = 5082 \(\text{cm}^2\)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Work in radians instead of using the degree-out-of-360 fraction, and anchor the answer to the full circle's area.
First convert the central angle to radians:
\[ 30^\circ = \frac{\pi}{6} \text{ radians} \]
The sector area formula in radians is:
\[ A_{\text{sector}} = \frac{1}{2} r^2 \theta \]

Step 2: Compute the minor sector area with $r = 42$ cm and $\theta = \frac{\pi}{6}$, using $\pi = \frac{22}{7}$.
\[ A_{\text{minor}} = \frac{1}{2} \times 42^2 \times \frac{\pi}{6} = \frac{1}{2} \times 1764 \times \frac{22}{7 \times 6} \]
\[ A_{\text{minor}} = 882 \times \frac{22}{42} = 882 \times \frac{11}{21} = 42 \times 11 = 462 \text{ cm}^2 \]

Step 3: Compute the area of the whole circle as an anchor value.
\[ A_{\text{circle}} = \pi r^2 = \frac{22}{7} \times 1764 = 22 \times 252 = 5544 \text{ cm}^2 \]

Step 4: Get the major sector by subtracting the minor sector from the whole circle.
The major sector is simply everything left over once the minor sector is removed from the full circle:
\[ A_{\text{major}} = A_{\text{circle}} - A_{\text{minor}} = 5544 - 462 = 5082 \text{ cm}^2 \]

Step 5: Final Answer.
The minor sector area is 462 cm$^2$ and the major sector area is 5082 cm$^2$, so option (A) is correct. \[ \boxed{462 \text{ cm}^2 \text{ and } 5082 \text{ cm}^2} \]
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