A chord PQ of a circle of radius 10 cm makes an angle of `60^(@)` at the centre of the circle. Find the area of the major and the minor segment

Answer

To solve the problem of finding the area of the major and minor segments of a circle with a radius of 10 cm and a chord PQ that makes an angle of 60° at the center, we can follow these steps:

### Step 1: Calculate the Area of the Sector
The area of the sector formed by the angle at the center can be calculated using the formula:
\[ \text{Area of Sector} = \frac{\theta}{360^\circ} \times \pi r^2 \]
where \( \theta \) is the angle in degrees and \( r \) is the radius of the circle.

Here, \( \theta = 60^\circ \) and \( r = 10 \, \text{cm} \).

\[ \text{Area of Sector} = \frac{60}{360} \times \pi \times (10)^2 \]
\[ = \frac{1}{6} \times \pi \times 100 \]
\[ = \frac{100\pi}{6} \, \text{cm}^2 \]
\[ = \frac{50\pi}{3} \, \text{cm}^2 \]

### Step 2: Calculate the Area of the Triangle
Next, we need to find the area of triangle OPQ, where O is the center of the circle. The formula for the area of a triangle when two sides and the included angle are known is:
\[ \text{Area of Triangle} = \frac{1}{2} \times a \times b \times \sin(C) \]
where \( a \) and \( b \) are the lengths of the sides and \( C \) is the included angle.

In this case, both sides (radii) are equal to 10 cm, and the angle \( C = 60^\circ \).

\[ \text{Area of Triangle} = \frac{1}{2} \times 10 \times 10 \times \sin(60^\circ) \]
\[ = \frac{1}{2} \times 100 \times \frac{\sqrt{3}}{2} \]
\[ = 25\sqrt{3} \, \text{cm}^2 \]

### Step 3: Calculate the Area of the Minor Segment
The area of the minor segment can be calculated by subtracting the area of the triangle from the area of the sector:
\[ \text{Area of Minor Segment} = \text{Area of Sector} - \text{Area of Triangle} \]
\[ = \frac{50\pi}{3} - 25\sqrt{3} \, \text{cm}^2 \]

### Step 4: Calculate the Area of the Major Segment
The area of the major segment can be found by subtracting the area of the minor segment from the total area of the circle:
\[ \text{Area of Circle} = \pi r^2 = \pi (10)^2 = 100\pi \, \text{cm}^2 \]
\[ \text{Area of Major Segment} = \text{Area of Circle} - \text{Area of Minor Segment} \]
\[ = 100\pi - \left(\frac{50\pi}{3} - 25\sqrt{3}\right) \]
\[ = 100\pi - \frac{50\pi}{3} + 25\sqrt{3} \]
\[ = \frac{300\pi}{3} - \frac{50\pi}{3} + 25\sqrt{3} \]
\[ = \frac{250\pi}{3} + 25\sqrt{3} \, \text{cm}^2 \]

### Summary of Results
- Area of Minor Segment: \( \frac{50\pi}{3} - 25\sqrt{3} \, \text{cm}^2 \)
- Area of Major Segment: \( \frac{250\pi}{3} + 25\sqrt{3} \, \text{cm}^2 \)