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: Convert the angle to radians and use the radian versions of the sector formulas.
The chord subtends $\theta = 120^{\circ}$ at the centre. In radians:
\[ \theta = 120^{\circ} \times \frac{\pi}{180^{\circ}} = \frac{2\pi}{3} \]
The radian form of the arc length and sector area formulas are $L = r\theta$ and $A = \dfrac{1}{2}r^{2}\theta$, which avoid writing a "$\dfrac{\theta}{360^{\circ}}$" fraction at all.

Step 2: Find the perimeter of the major sector.
The major sector's angle in radians is:
\[ \theta_{\text{major}} = 2\pi - \frac{2\pi}{3} = \frac{4\pi}{3} \]
Its arc length is:
\[ L = r\,\theta_{\text{major}} = 7 \times \frac{4\pi}{3} = \frac{28\pi}{3} \]
Using $\pi = \dfrac{22}{7}$:
\[ L = \frac{28}{3}\times\frac{22}{7} = \frac{4\times 22}{3} = \frac{88}{3} \approx 29.33\text{ cm} \]
The perimeter of the major sector adds the two radii:
\[ P = 2r + L = 14 + 29.33 = 43.33\text{ cm} \]

Step 3: Find the area of the minor sector.
\[ A_{\text{sector}} = \frac{1}{2}r^{2}\theta = \frac{1}{2}(7)^{2}\left(\frac{2\pi}{3}\right) = \frac{49\pi}{3} \]
Using $\pi=\dfrac{22}{7}$:
\[ A_{\text{sector}} = \frac{49}{3}\times\frac{22}{7} = \frac{7\times 22}{3} = \frac{154}{3} \approx 51.33\text{ cm}^{2} \]

Step 4: Subtract the triangle's area to get the segment.
\[ A_{\text{segment}} = A_{\text{sector}} - A_{\Delta OAB} = 51.33 - 21.2 = 30.13\text{ cm}^{2} \]

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