Question

Find the area of the segment AYB shown in the figure, if the radius of the circle is 21 cm and \(\angle AOB = 120^\circ\). [Use \(\pi = \frac{22}{7}\)]

• A. \(\left(462 - \frac{441\sqrt{3}}{4}\right)\ \text{cm}^2\)
• B. \(\left(462 - \frac{441\sqrt{3}}{2}\right)\ \text{cm}^2\)
• C. \(\left(231 - \frac{441\sqrt{3}}{4}\right)\ \text{cm}^2\)
• D. \(\left(462 - 441\sqrt{3}\right)\ \text{cm}^2\)

Hint:

For any triangle \(\Delta OAB\) inside a circle of radius \(r\) with a central angle \(\theta\):
If \(\theta = 120^\circ\), drawing a perpendicular bisector from \(O\) to \(AB\) splits it into two \(30^\circ-60^\circ-90^\circ\) triangles.
The base of the triangle is \(2r \sin(60^\circ) = r\sqrt{3}\) and the height is \(r \cos(60^\circ) = \frac{r}{2}\).
The area is:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times r\sqrt{3} \times \frac{r}{2} = \frac{\sqrt{3}}{4} r^2 \]
This geometric approach is highly intuitive and leads directly to the same exact formula!

Answer 1

The Correct Option is A

Step 1: Draw the perpendicular from the centre to the chord, splitting the isosceles triangle into two right triangles, rather than using the general sine area formula in one shot.
Let $M$ be the foot of the perpendicular from $O$ to chord $AB$. Since $OA = OB = 21$ cm (radii), this perpendicular bisects both the chord and the angle at $O$.
\[ \angle AOM = \frac{120^\circ}{2} = 60^\circ \]

Step 2: Use right triangle $OAM$, right angled at $M$, to find the half chord and the perpendicular distance.
\[ AM = OA\sin 60^\circ = 21 \times \frac{\sqrt{3}}{2} = \frac{21\sqrt{3}}{2} \]
\[ OM = OA\cos 60^\circ = 21 \times \frac{1}{2} = \frac{21}{2} \]
So the full chord is $AB = 2 \times AM = 21\sqrt{3}$.

Step 3: Use the base times height formula for the area of triangle $OAB$, with $AB$ as base and $OM$ as height.
\[ A_{\triangle OAB} = \frac{1}{2} \times AB \times OM = \frac{1}{2} \times 21\sqrt{3} \times \frac{21}{2} = \frac{441\sqrt{3}}{4} \text{ cm}^2 \]

Step 4: Find the sector area and subtract to get the segment area.
\[ A_{\text{sector}} = \frac{120^\circ}{360^\circ} \times \frac{22}{7} \times 21^2 = \frac{1}{3} \times \frac{22}{7} \times 441 = \frac{1}{3} \times 22 \times 63 = 462 \text{ cm}^2 \]
\[ A_{\text{segment}} = A_{\text{sector}} - A_{\triangle OAB} = 462 - \frac{441\sqrt{3}}{4} \text{ cm}^2 \]

Final Answer:
The area of segment $AYB$ is $\left(462 - \frac{441\sqrt{3}}{4}\right)$ cm$^2$, which corresponds to option (A). \[ \boxed{\left(462 - \frac{441\sqrt{3}}{4}\right) \text{cm}^2} \]