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 \(\pi = \frac{22}{7}\)]
Show 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!
Step 1: Understanding the Question:
We are given a circle of radius \(r = 21\) cm. A chord \(AB\) subtends an angle of \(120^\circ\) at the center \(O\). We need to calculate the area of the minor segment \(AYB\) formed by this chord.
Step 2: Key Formula or Approach:
1. The area of a segment is calculated by subtracting the area of the corresponding triangle from the area of the sector:
\[ \text{Area of Segment } (A) = \text{Area of Sector } (A_{\text{sector}}) - \text{Area of Triangle } (A_{\text{triangle}}) \]
2. The area of the sector is given by:
\[ A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 \]
3. The area of triangle \(\Delta OAB\) with central angle \(\theta = 120^\circ\) can be found using the trigonometric formula:
\[ A_{\text{triangle}} = \frac{1}{2} r^2 \sin \theta \]
Step 3: Detailed Explanation:
1.
Calculate the area of the sector:
Given \(r = 21\ \text{cm}\), \(\theta = 120^\circ\), and \(\pi = \frac{22}{7}\).
\[ A_{\text{sector}} = \frac{120^\circ}{360^\circ} \times \frac{22}{7} \times 21 \times 21 \]
\[ A_{\text{sector}} = \frac{1}{3} \times 22 \times 3 \times 21 \]
Cancel the factor of 3:
\[ A_{\text{sector}} = 22 \times 21 = 462\ \text{cm}^2 \]
2.
Calculate the area of triangle \(\Delta OAB\):
We use the general trigonometric area formula for a triangle:
\[ A_{\text{triangle}} = \frac{1}{2} r^2 \sin(120^\circ) \]
Since \(\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\[ A_{\text{triangle}} = \frac{1}{2} \times 21^2 \times \frac{\sqrt{3}}{2} \]
\[ A_{\text{triangle}} = \frac{441\sqrt{3}}{4}\ \text{cm}^2 \]
3.
Calculate the area of the segment \(AYB\):
\[ A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} \]
\[ A_{\text{segment}} = \left( 462 - \frac{441\sqrt{3}}{4} \right)\ \text{cm}^2 \]
Step 4: Final Answer:
The area of the segment is \(\left(462 - \frac{441\sqrt{3}}{4}\right)\ \text{cm}^2\), which corresponds to option (A).
