Question

In the given figure, chord AB subtends an angle of \(120^\circ\) at the centre of the circle with radius 7 cm. Find (i) perimeter of major sector OACB, and (ii) area of the shaded segment, if area of \(\Delta OAB = 21.2\text{ cm}^2\).

Hint:

The perimeter of a sector includes the two straight boundary radii in addition to the curved arc length.
Do not make the common mistake of only calculating the arc length.

Answer 1

Step 1: Understanding the Question:
We are given a circle with a radius of \(7\text{ cm}\).
A chord \(AB\) subtends an angle of \(\theta = 120^\circ\) at the center \(O\).
We need to find:
(i) The perimeter of the major sector \(OACB\).
(ii) The area of the minor shaded segment.
We are also given the area of triangle \(\Delta OAB\) as \(21.2\text{ cm}^2\).

Step 2: Key Formula or Approach:
1. The angle of the major sector is \(\theta_{\text{major}} = 360^\circ - 120^\circ = 240^\circ\).
2. The perimeter of a sector with angle \(\phi\) and radius \(r\) is given by:
\[ P = 2r + \frac{\phi}{360^\circ} \times 2\pi r \] 3. The area of the minor segment is given by:
\[ \text{Area of Segment} = \text{Area of minor sector } OAB - \text{Area of } \Delta OAB \] 4. The area of a sector with angle \(\theta\) is:
\[ A = \frac{\theta}{360^\circ} \times \pi r^2 \]

Step 3: Detailed Explanation:
1. Part (i): Perimeter of major sector OACB:
- The central angle of the major sector is:
\[ \phi = 360^\circ - 120^\circ = 240^\circ \]
- Calculate the arc length of the major sector:
\[ \text{Arc Length} = \frac{240^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 7 = \frac{2}{3} \times 44 = \frac{88}{3} \approx 29.33\text{ cm} \]
- Calculate the perimeter of the major sector:
\[ P = 2r + \text{Arc Length} = 2(7) + 29.33 = 14 + 29.33 = 43.33\text{ cm} \]
2. Part (ii): Area of the shaded segment:
- Calculate the area of the minor sector \(OAB\):
\[ \text{Area of minor sector} = \frac{120^\circ}{360^\circ} \times \pi r^2 = \frac{1}{3} \times \frac{22}{7} \times 7^2 \]
\[ \text{Area of minor sector} = \frac{154}{3} \approx 51.33\text{ cm}^2 \]
- Given that the area of triangle \(\Delta OAB = 21.2\text{ cm}^2\):
\[ \text{Area of shaded segment} = 51.33 - 21.2 = 30.13\text{ cm}^2 \]

Step 4: Final Answer:
(i) The perimeter of the major sector \(OACB\) is \(43.33\text{ cm}\).
(ii) The area of the shaded segment is \(30.13\text{ cm}^2\).