Complete theory, formulas for area and circumference of circle, area of sector and segment, length of arc, areas of combinations of plane figures, and NCERT solved examples.
In this chapter, we will learn how to calculate the areas and perimeters (circumferences) of circles, as well as parts of circles like sectors and segments. These concepts are widely used in real life — from designing wheels and clocks to calculating the area of a pizza slice or a circular garden.
The distance around the circle.
C = 2πr or C = πd
where r = radius, d = diameter
The region enclosed by the circle.
A = πr²
d = 2r
r = √(A/π)
Find the circumference and area of a circle of radius 14 cm.
Solution: Circumference = 2 × 22/7 × 14 = 2 × 22 × 2 = 88 cm.
Area = πr² = 22/7 × 14 × 14 = 22 × 2 × 14 = 616 cm².
A sector is the region enclosed by an arc and the two radii joining the arc's endpoints to the centre. It looks like a slice of pizza or a cake.
Area = πr² - Area of minor sector
Area = (θ/360) × πr² (where θ < 180°)
Find the area of a sector of a circle with radius 6 cm and central angle 60°.
Solution: Area = (60/360) × π × 6² = (1/6) × 3.14 × 36 = (1/6) × 113.04 = 18.84 cm².
A segment is the region enclosed by an arc and a chord. There are two types: minor segment (smaller area) and major segment (larger area).
Find the area of the minor segment of a circle of radius 10 cm, when the central angle is 90°.
Solution: Area of sector = (90/360) × π × 100 = (1/4) × 314 = 78.5 cm².
Area of triangle = ½ × 10 × 10 = 50 cm².
Area of minor segment = 78.5 - 50 = 28.5 cm².
Find the length of an arc of a circle of radius 21 cm subtending an angle of 60° at the centre.
Solution: Arc length = (60/360) × 2 × 22/7 × 21 = (1/6) × 2 × 22 × 3 = (1/6) × 132 = 22 cm.
In many real-life problems, we encounter shapes that are combinations of circles, triangles, rectangles, etc. The key is to break down the complex figure into simpler shapes.
A square of side 14 cm has four circles of radius 7 cm drawn inside. Find the area of the shaded region (if circles are drawn with diameter as side of square).
Solution: Area of square = 14 × 14 = 196 cm². Area of one circle = π × 7² = 154 cm². Four circles → area = 616 cm² (but overlaps occur). Typically in board problems, you learn to handle combinations carefully.
Question: Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also find the area of the corresponding major sector (use π = 3.14).
Solution: Area of minor sector = (30/360) × 3.14 × 16 = (1/12) × 50.24 = 4.186 cm².
Area of major sector = πr² - 4.186 = 50.24 - 4.186 = 46.054 cm².
Question: A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding minor segment.
Solution: r = 10 cm, θ = 90°.
Area of sector = (90/360) × π × 100 = (1/4) × 314 = 78.5 cm².
Area of triangle = ½ × 10 × 10 = 50 cm².
Area of minor segment = 78.5 - 50 = 28.5 cm².
Question: The minute hand of a clock is 14 cm long. Find the area swept by the minute hand in 5 minutes.
Solution: In 60 minutes, minute hand sweeps 360°. In 5 minutes, angle = (5/60) × 360° = 30°.
Area swept = (30/360) × π × 14² = (1/12) × 22/7 × 196 = (1/12) × 616 = 51.33 cm².
Question: In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the length of the arc and the area of the sector.
Solution: Arc length = (60/360) × 2 × 22/7 × 21 = (1/6) × 132 = 22 cm.
Area of sector = (60/360) × 22/7 × 21 × 21 = (1/6) × 1386 = 231 cm².
| Concept | Formula |
|---|---|
| Circumference of circle | 2πr |
| Area of circle | πr² |
| Length of arc (angle θ in degrees) | (θ/360) × 2πr |
| Area of sector (angle θ in degrees) | (θ/360) × πr² |
| Area of sector (angle θ in radians) | ½r²θ |
| Area of minor segment | Sector area - Triangle area |
| Area of major segment | πr² - Minor segment area |
| Perimeter of sector | 2r + Arc length |