Question

A chord PQ of a circle of diameter 28 cm subtends an angle of \(90^\circ\) at the centre O. Find the area of the sector OPCQ, where C is a point on minor arc PQ. Also, find the area of segment PCQ.

• A. Sector Area = 154 \(\text{cm}^2\), Segment Area = 56 \(\text{cm}^2\)
• B. Sector Area = 154 \(\text{cm}^2\), Segment Area = 98 \(\text{cm}^2\)
• C. Sector Area = 308 \(\text{cm}^2\), Segment Area = 154 \(\text{cm}^2\)
• D. Sector Area = 77 \(\text{cm}^2\), Segment Area = 28 \(\text{cm}^2\)

Hint:

For a central angle of \(90^\circ\), the sector is exactly one-quarter of the circle.
The segment area can be written directly using the simplified formula:
\[ A_{\text{segment}} = r^2 \left( \frac{\pi}{4} - \frac{1}{2} \right) \]
Substituting \(r = 14\):
\[ A_{\text{segment}} = 196 \left( \frac{22}{28} - \frac{14}{28} \right) = 196 \times \frac{8}{28} = 7 \times 8 = 56\ \text{cm}^2 \]
This direct formula is very useful for verifying your calculations!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a circle with a diameter of 28 cm. A chord \(PQ\) subtends a right angle (\(90^\circ\)) at the center \(O\) of the circle. We need to find:
1. The area of the sector \(OPCQ\).
2. The area of the minor segment \(PCQ\).

Step 2: Key Formula or Approach:
1. Radius \(r\) of the circle is half of the diameter:
\[ r = \frac{\text{Diameter}}{2} \]
2. The area of a sector with a central angle \(\theta\) is given by:
\[ A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 \]
3. The area of the minor segment \(PCQ\) is calculated by subtracting the area of the right-angled triangle \(OPQ\) from the area of the sector \(OPCQ\):
\[ A_{\text{segment}} = A_{\text{sector}} - A_{\Delta OPQ} \]
Since \(\angle POQ = 90^\circ\), the area of \(\Delta OPQ\) is:
\[ A_{\Delta OPQ} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times r \times r = \frac{1}{2} r^2 \]

Step 3: Detailed Explanation:
1. Calculate the radius \(r\) of the circle:
\[ r = \frac{28}{2} = 14\ \text{cm} \]
2. Calculate the area of the sector \(OPCQ\) with \(\theta = 90^\circ\):
\[ A_{\text{sector}} = \frac{90^\circ}{360^\circ} \times \pi \times 14^2 \]
\[ A_{\text{sector}} = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 \]
\[ A_{\text{sector}} = \frac{1}{4} \times 22 \times 2 \times 14 \]
\[ A_{\text{sector}} = 11 \times 14 = 154\ \text{cm}^2 \]
3. Calculate the area of triangle \(OPQ\):
Since the central angle is \(90^\circ\), \(\Delta OPQ\) is a right-angled triangle with base and height equal to the radius (14 cm):
\[ A_{\Delta OPQ} = \frac{1}{2} \times 14 \times 14 = 98\ \text{cm}^2 \]
4. Calculate the area of the segment \(PCQ\):
\[ A_{\text{segment}} = A_{\text{sector}} - A_{\Delta OPQ} \]
\[ A_{\text{segment}} = 154 - 98 = 56\ \text{cm}^2 \]

Step 4: Final Answer:
The area of the sector is \(154\ \text{cm}^2\) and the area of the segment is \(56\ \text{cm}^2\), which matches option (A).