Question

Area of a segment of a circle of radius 'r' and central angle \(60^\circ\) is :

• A. \(\frac{\pi r^2}{2} - \frac{1}{2} r^2\)
• B. \(\frac{2 \pi r}{4} - \frac{\sqrt{3}}{4} r^2\)
• C. \(\frac{\pi r^2}{6} - \frac{\sqrt{3}}{4} r^2\)
• D. \(\frac{2 \pi r}{4} - r^2 \sin 60^\circ\)

Hint:

Remember that any circle triangle with a central angle of \(60^\circ\) is always an equilateral triangle.
Its area is always \(\frac{\sqrt{3}}{4} r^2\).
Combine this with the sector fraction \(\frac{60}{360} = \frac{1}{6}\) to instantly write down the segment area formula!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the algebraic formula representing the area of a segment of a circle with radius \(r\) and a central angle \(\theta = 60^\circ\).

Step 2: Key Formula or Approach:
The area of a segment of a circle is calculated by subtracting the area of the corresponding triangle from the area of the sector:
\[ \text{Area of Segment} = \text{Area of Sector} - \text{Area of Triangle} \]
The formulas for these areas are:
- Area of Sector:
\[ A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 \]
- Area of Triangle:
\[ A_{\text{triangle}} = \frac{1}{2} r^2 \sin \theta \]

Step 3: Detailed Explanation:
1. Calculate the area of the sector with \(\theta = 60^\circ\):
\[ A_{\text{sector}} = \frac{60^\circ}{360^\circ} \times \pi r^2 = \frac{1}{6} \pi r^2 = \frac{\pi r^2}{6} \]
2. Calculate the area of the corresponding triangle with \(\theta = 60^\circ\):
Since the central angle is \(60^\circ\) and the two sides are equal to the radius \(r\), the triangle is an equilateral triangle with side length \(r\).
The area of an equilateral triangle is:
\[ A_{\text{triangle}} = \frac{\sqrt{3}}{4} r^2 \]
Alternatively, using the general triangle formula:
\[ A_{\text{triangle}} = \frac{1}{2} r^2 \sin 60^\circ \]
Since \(\sin 60^\circ = \frac{\sqrt{3}}{2}\):
\[ A_{\text{triangle}} = \frac{1}{2} r^2 \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4} r^2 \]
3. Subtract the area of the triangle from the area of the sector to find the area of the segment:
\[ \text{Area of Segment} = A_{\text{sector}} - A_{\text{triangle}} = \frac{\pi r^2}{6} - \frac{\sqrt{3}}{4} r^2 \]
This matches option (C).

Step 4: Final Answer:
The area of the segment is \(\frac{\pi r^2}{6} - \frac{\sqrt{3}}{4} r^2\), which corresponds to option (C).