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 chord PQ that makes an angle of \(60^\circ\) at the center, follow these steps:

### Step 1: Understand the Circle and Given Data
- The radius \( r \) of the circle is given as \( 10 \, \text{cm} \).
- The angle \( \theta \) at the center of the circle is \( 60^\circ \).

### Step 2: Calculate the Area of the Sector
The area of the sector formed by the angle at the center can be calculated using the formula:
\[ \text{Area of Sector} = \frac{\theta}{360^\circ} \times \pi r^2 \]
Substituting the values:
\[ \text{Area of Sector} = \frac{60}{360} \times \pi \times (10)^2 = \frac{1}{6} \times \pi \times 100 = \frac{100\pi}{6} = \frac{50\pi}{3} \, \text{cm}^2 \]

### Step 3: Calculate the Area of Triangle OAB
Since the triangle OAB is an equilateral triangle (all angles are \(60^\circ\)):
- The formula for the area of an equilateral triangle is:
\[ \text{Area} = \frac{\sqrt{3}}{4} a^2 \]
where \( a \) is the length of a side. Here, \( a = r = 10 \, \text{cm} \).
\[ \text{Area of Triangle OAB} = \frac{\sqrt{3}}{4} \times (10)^2 = \frac{\sqrt{3}}{4} \times 100 = 25\sqrt{3} \, \text{cm}^2 \]

### Step 4: Calculate the Area of the Minor Segment
The area of the minor segment is given by:
\[ \text{Area of Minor Segment} = \text{Area of Sector} - \text{Area of Triangle OAB} \]
Substituting the values:
\[ \text{Area of Minor Segment} = \frac{50\pi}{3} - 25\sqrt{3} \, \text{cm}^2 \]

### Step 5: Calculate the Area of the Circle
The area of the entire circle is:
\[ \text{Area of Circle} = \pi r^2 = \pi \times (10)^2 = 100\pi \, \text{cm}^2 \]

### Step 6: Calculate the Area of the Major Segment
The area of the major segment is given by:
\[ \text{Area of Major Segment} = \text{Area of Circle} - \text{Area of Minor Segment} \]
Substituting the values:
\[ \text{Area of Major Segment} = 100\pi - \left(\frac{50\pi}{3} - 25\sqrt{3}\right) \]
\[ = 100\pi - \frac{50\pi}{3} + 25\sqrt{3} \]
To simplify, convert \( 100\pi \) to have a common denominator:
\[ = \frac{300\pi}{3} - \frac{50\pi}{3} + 25\sqrt{3} \]
\[ = \frac{250\pi}{3} + 25\sqrt{3} \, \text{cm}^2 \]

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