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: Concept of Area of Segment:
The area of a segment of a circle is given by the difference between the area of the sector and the area of the triangle formed by the radius lines and the chord.
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 Calculation:
Given values: \( r = 21 \text{ cm} \), \( \theta = 120^\circ \).
1. Area of Sector AOBY:
Area of sector: \[ \text{Area of sector} = \frac{120}{360} \times \frac{22}{7} \times 21 \times 21 \] \[ = \frac{1}{3} \times 22 \times 3 \times 21 = 22 \times 21 = 462 \text{ cm}^2 \] 2. Area of Triangle AOB:
\[ \text{Area of triangle} = \frac{1}{2} \times r^2 \times \sin 120^\circ \] We know \( \sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} \).
\[ = \frac{1}{2} \times 21 \times 21 \times \frac{\sqrt{3}}{2} = \frac{441\sqrt{3}}{4} \text{ cm}^2 \] 3. Area of Segment AYB:
\[ \text{Area of segment} = 462 - \frac{441\sqrt{3}}{4} \text{ cm}^2 \] Step 4: Final Answer:
The area of the segment is \( (462 - 110.25\sqrt{3}) \text{ cm}^2 \).