Chapter 11

Class 10 Mathematics - Chapter 11: Areas Related to Circles
Chapter 11: Areas Related to Circles
1. Perimeter and Area of a Circle
Perimeter (Circumference): The distance covered by traveling once around the circle. $$\text{Circumference} = 2\pi r = \pi d$$
Area of Circle: The measure of the region enclosed inside the boundary. $$\text{Area} = \pi r^2$$ (where $r$ is the radius and $d$ is the diameter of the circle)
2. Sector of a Circle
A sector is the portion of the circular region enclosed by two radii and the corresponding arc.

1. Area of a Sector of angle $\theta$: $$\text{Area of Sector} = \frac{\theta}{360^\circ} \times \pi r^2$$
2. Length of an Arc of a Sector of angle $\theta$: $$l = \frac{\theta}{360^\circ} \times 2\pi r$$
O A B θ r Arc (l)
Example 2.1: Finding Area and Arc Length of a Sector

Find the area of a sector of a circle with radius $6\text{ cm}$ if the angle of the sector is $60^\circ$. (Use $\pi = \frac{22}{7}$)

Solution:
Given $r = 6\text{ cm}$ and $\theta = 60^\circ$.
Using the sector area formula: $$\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$$ $$\text{Area} = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times (6)^2$$ $$\text{Area} = \frac{1}{6} \times \frac{22}{7} \times 36$$ $$\text{Area} = \frac{22 \times 6}{7} = \frac{132}{7}\text{ cm}^2$$ The area of the sector is $\frac{132}{7}\text{ cm}^2$ (or $\approx 18.86\text{ cm}^2$).

3. Segment of a Circle
A segment is the region bounded by a chord and the corresponding arc of the circle.

$$\text{Area of Minor Segment} = \text{Area of Sector } OAPB - \text{Area of } \triangle OAB$$
Where area of $\triangle OAB$ can be evaluated as: $$\text{Area of } \triangle OAB = \frac{1}{2} r^2 \sin\theta$$
Major vs Minor Regions:
  • Minor Segment: Region between chord and the smaller arc.
  • Major Segment: $\text{Total Circle Area} - \text{Minor Segment Area}$
  • Major Sector: $\text{Total Circle Area} - \text{Minor Sector Area} = \frac{360^\circ - \theta}{360^\circ} \times \pi r^2$
Example 3.1: Finding Area of a Minor Segment

A chord of a circle of radius $10\text{ cm}$ subtends a right angle at the centre. Find the area of the corresponding minor segment. (Use $\pi = 3.14$)

Solution:
Here $r = 10\text{ cm}$ and $\theta = 90^\circ$.

1. Area of Sector $OAPB$: $$\text{Area of Sector} = \frac{90^\circ}{360^\circ} \times 3.14 \times (10)^2$$ $$\text{Area of Sector} = \frac{1}{4} \times 3.14 \times 100$$ $$\text{Area of Sector} = \frac{314}{4} = 78.5\text{ cm}^2$$
2. Area of Right-Angled $\triangle OAB$: $$\text{Area of } \triangle OAB = \frac{1}{2} \times \text{base} \times \text{height}$$ $$\text{Area of } \triangle OAB = \frac{1}{2} \times 10 \times 10 = 50\text{ cm}^2$$
3. Area of Minor Segment: $$\text{Area of Minor Segment} = 78.5 - 50 = 28.5\text{ cm}^2$$ The area of the minor segment is $28.5\text{ cm}^2$.