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 \(\pi = 22/7\)]

Hint:

For \(\theta>90^\circ\), you can find the area of the triangle by using \(r^2 \sin(\theta/2) \cos(\theta/2)\) or simply \(\frac{1}{2}r^2 \sin \theta\).

Answer 1

Step 1: Understanding the Concept:
Area of segment = Area of sector - Area of triangle.
Step 2: Key Formula or Approach:
Area of sector \( = \frac{\theta}{360} \pi r^2 \)
Area of triangle \( = \frac{1}{2} r^2 \sin \theta \)
Step 3: Detailed Explanation:
Given: \(r = 21 \text{ cm}\), \(\theta = 120^\circ\).
1. Area of sector AOBY:
\[ = \frac{120}{360} \times \frac{22}{7} \times 21 \times 21 \]
\[ = \frac{1}{3} \times 22 \times 3 \times 21 = 22 \times 21 = 462 \text{ cm}^2 \]
2. Area of \(\Delta AOB\):
\[ = \frac{1}{2} \times r^2 \times \sin 120^\circ \]
We know \(\sin 120^\circ = \sin(180 - 60) = \sin 60^\circ = \frac{\sqrt{3}}{2}\).
\[ = \frac{1}{2} \times 21 \times 21 \times \frac{\sqrt{3}}{2} = \frac{441\sqrt{3}}{4} \text{ cm}^2 \]
3. Area of segment AYB:
\[ = 462 - \frac{441\sqrt{3}}{4} \text{ cm}^2 \]
Step 4: Final Answer:
The area is \((462 - 110.25\sqrt{3}) \text{ cm}^2\).