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:

Do not forget that the perimeter of a sector must include the two straight radii that bound it ($2r$), not just the curved arc length! This is a very common oversight.

Answer 1

Step 1: Understanding the Question:
This question is from "Areas Related to Circles".
We are given a circle of radius $r = 7\text{ cm}$ with a chord $AB$ subtending an angle of $\theta = 120^\circ$ at the center $O$.
We need to find two quantities:
1. The perimeter of the major sector $OACB$.
2. The area of the shaded segment (the minor segment formed by chord $AB$), given that the area of triangle $OAB$ is $21.2\text{ cm}^2$.

Step 2: Key Formula or Approach:
1.

Perimeter of Major Sector: The angle of the major sector is $360^\circ - \theta$. The perimeter includes the length of the major arc plus the two bounding radii:
\[ \text{Perimeter} = \frac{360^\circ - \theta}{360^\circ} \times 2\pi r + 2r \] 2.

Area of Segment: The area of the minor segment is calculated as the area of the minor sector minus the area of the triangle $OAB$:
\[ \text{Area of Segment} = \text{Area of Minor Sector} - \text{Area of } \Delta OAB \] \[ \text{Area of Minor Sector} = \frac{\theta}{360^\circ} \times \pi r^2 \]

Step 3: Detailed Explanation:

• Part (i): Perimeter of Major Sector OACB:
- Given radius $r = 7\text{ cm}$ and minor sector angle $\theta = 120^\circ$.
- The angle for the major sector is $360^\circ - 120^\circ = 240^\circ$.
- Calculate the length of the major arc:
\[ \text{Arc Length} = \frac{240^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 7 \] \[ \text{Arc Length} = \frac{2}{3} \times 44 = \frac{88}{3}\text{ cm} \approx 29.33\text{ cm} \] - Calculate the total perimeter of the sector (Arc Length + 2 radii):
\[ \text{Perimeter} = \frac{88}{3} + 2(7) = \frac{88}{3} + 14 = \frac{88 + 42}{3} = \frac{130}{3}\text{ cm} \approx 43.33\text{ cm} \]

• Part (ii): Area of Shaded Segment:
- Calculate the area of the minor sector $OAB$:
\[ \text{Area of Minor Sector} = \frac{120^\circ}{360^\circ} \times \pi r^2 \] \[ \text{Area of Minor Sector} = \frac{1}{3} \times \frac{22}{7} \times 7^2 \] \[ \text{Area of Minor Sector} = \frac{1}{3} \times 154 = \frac{154}{3}\text{ cm}^2 \approx 51.33\text{ cm}^2 \] - The area of $\Delta OAB$ is given as $21.2\text{ cm}^2$.
- Subtract the area of the triangle from the area of the minor sector to find the segment area:
\[ \text{Area of Shaded Segment} = \frac{154}{3} - 21.2 \] \[ \text{Area of Shaded Segment} \approx 51.33 - 21.2 = 30.13\text{ cm}^2 \]

Step 4: Final Answer:
(i) The perimeter of the major sector is approximately $43.33\text{ cm}$ (or $\frac{130}{3}\text{ cm}$).
(ii) The area of the shaded segment is approximately $30.13\text{ cm}^2$.