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: Find the full circumference and full area of the circle first, before looking at any sector.
Radius $r = 7$ cm.
Full circumference:
\[ 2\pi r = 2 \times \frac{22}{7} \times 7 = 44 \text{ cm} \]
Full area:
\[ \pi r^2 = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 154 \text{ cm}^2 \]

Step 2: Find the minor arc length, and get the major arc length by SUBTRACTING it from the full circumference, instead of applying the 240/360 fraction directly.
The minor sector's angle is $120^\circ$, so the minor arc is $\frac{120^\circ}{360^\circ} = \frac{1}{3}$ of the full circumference:
\[ \text{Minor arc} = \frac{1}{3} \times 44 = \frac{44}{3} \text{ cm} \]
The major arc is simply what remains of the full circumference after removing the minor arc:
\[ \text{Major arc} = 44 - \frac{44}{3} = \frac{132 - 44}{3} = \frac{88}{3} \text{ cm} \approx 29.33 \text{ cm} \]

Step 3: Add the two bounding radii to get the perimeter of the major sector.
\[ \text{Perimeter of major sector} = \text{Major arc} + 2r = \frac{88}{3} + 14 = \frac{88 + 42}{3} = \frac{130}{3} \text{ cm} \approx 43.33 \text{ cm} \]

Step 4: Find the minor sector's area by SUBTRACTING the major sector's area from the full circle's area, instead of applying the fraction directly to the minor sector.
The major sector's angle is $240^\circ$, which is $\frac{2}{3}$ of the full circle, so:
\[ \text{Major sector area} = \frac{2}{3} \times 154 = \frac{308}{3} \text{ cm}^2 \]
The minor sector's area is what remains:
\[ \text{Minor sector area} = 154 - \frac{308}{3} = \frac{462 - 308}{3} = \frac{154}{3} \text{ cm}^2 \approx 51.33 \text{ cm}^2 \]

Step 5: Subtract the given triangle area to get the shaded segment area.
We are told the area of $\Delta OAB$ is $21.2 \text{ cm}^2$.
\[ \text{Area of shaded segment} = \text{Minor sector area} - \text{Area of } \Delta OAB = \frac{154}{3} - 21.2 \approx 51.33 - 21.2 = 30.13 \text{ cm}^2 \]

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