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 cm\(^2\).
Show Hint
Remember that "perimeter of a sector" is not just the arc length.
It includes the two straight boundaries (radii) as well: \( \text{Perimeter} = L + 2r \).
For segment areas, subtracting the given triangle area directly from the sector area is extremely straightforward, so keep track of units and rounding decimals.
Step 1: Understanding the Question:
The topic of this question is Areas Related to Circles.
We are given a circle of radius \( r = 7\text{ cm} \).
A chord \( AB \) subtends an angle \( \theta = 120^{\circ} \) at the center \( O \).
We need to calculate:
- (i) The perimeter of the major sector \( OACB \).
- (ii) The area of the shaded segment (which is a minor segment).
Step 2: Key Formula or Approach:
- The perimeter of a sector of a circle is:
\[ \text{Perimeter} = \text{Arc length} + 2r \]
- The angle of the major sector is \( 360^{\circ} - \theta \).
- Length of an arc of a sector with angle \( \alpha \):
\[ \text{Arc Length} = \frac{\alpha}{360^{\circ}} \times 2\pi r \]
- Area of the shaded minor segment is:
\[ \text{Area of segment} = \text{Area of minor sector } OAB - \text{Area of } \Delta OAB \]
- Area of a sector with angle \( \theta \):
\[ \text{Area} = \frac{\theta}{360^{\circ}} \times \pi r^2 \]
Step 3: Detailed Explanation:
1. Part (i): Find perimeter of major sector OACB:
The angle of the major sector is:
\[ \alpha = 360^{\circ} - 120^{\circ} = 240^{\circ} \]
Calculate the length of the major arc \( ACB \):
\[ \text{Arc Length} = \frac{240^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 7 \]
\[ \text{Arc Length} = \frac{2}{3} \times 44 = \frac{88}{3}\text{ cm} \approx 29.33\text{ cm} \]
The perimeter of the major sector is the sum of the major arc length and the two radii:
\[ \text{Perimeter} = \text{Arc Length} + 2r \]
\[ \text{Perimeter} = 29.33 + 2(7) \]
\[ \text{Perimeter} = 29.33 + 14 = 43.33\text{ cm} \]
2. Part (ii): Find area of the shaded segment:
Calculate the area of the minor sector \( OAB \):
\[ \text{Area of minor sector} = \frac{120^{\circ}}{360^{\circ}} \times \pi r^2 \]
\[ \text{Area of minor sector} = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \]
\[ \text{Area of minor sector} = \frac{154}{3}\text{ cm}^2 \approx 51.33\text{ cm}^2 \]
We are given that the area of \( \Delta OAB = 21.2\text{ cm}^2 \).
Calculate the area of the shaded segment:
\[ \text{Area of segment} = \text{Area of minor sector} - \text{Area of } \Delta OAB \]
\[ \text{Area of segment} = 51.33 - 21.2 = 30.13\text{ cm}^2 \]
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 \(30.13\text{ cm}^2\).
