Introduction
You have known the formulae for the circumference and area of a circle since earlier classes. In this chapter we use them to measure parts of a circle, a sector (a slice like a piece of pizza) and a segment (the cap cut off by a chord), and then combine circles with other shapes such as squares, rectangles and triangles to find the area of more complicated figures.
All the formulae in this chapter come from one simple idea: if a full circle corresponds to an angle of 360° at the centre, then a part that subtends an angle θ at the centre is just the fraction θ/360 of the whole.
Perimeter and Area of a Circle
For a circle of radius r and diameter d = 2r:
Circumference (perimeter) of a circle:
C = 2πr = πd
Area of a circle:
A = πr2
Where π is taken as 22/7 unless stated otherwise, or as 3.14 if the question says so.
These two formulae are the starting point. A semicircle, for example,, has an arc length of πr and, if you include the diameter, a perimeter of πr + 2r. Its area is half the full area, ½πr². Similarly, a quarter-circle (quadrant) has area ¼πr².
NCERT also uses the idea of a circular path or ring. The area between two concentric circles (an annulus) with outer radius R and inner radius r is simply the difference of the two areas: πR² − πr² = π(R² − r²). The same subtraction idea is used throughout the chapter.
The radius of a circular park is 35 m. Find its circumference and area.
Circumference = 2πr = 2 × 22/7 × 35 = 220 m
Area = πr2 = 22/7 × 35 × 35 = 3850 m2
Areas of Sectors
A sector is the region enclosed by two radii and the arc between them. If you cut a circle along two radii, each piece you get is a sector. The angle between the two radii, measured at the centre, is called the angle of the sector (θ).
For a circle of radius r and a sector with central angle θ (in degrees):
Length of the arc = θ/360 × 2πr
Area of the sector = θ/360 × πr2
Also useful: Area of sector = ½ × (arc length) × r
The logic is straightforward: the whole circle is 360° with circumference 2πr and area πr². A sector that spans θ° is θ/360 of the full circle, so its arc and area are the same fraction of the circumference and area.
A sector with θ < 180° is called a minor sector, and the larger remaining part is the major sector. Their angles add to 360°, and their areas add to πr². So once you know one, the other is simply the total minus that value.
Semicircle (θ = 180°) and quadrant (θ = 90°) are just special cases: put θ = 180° or θ = 90° into the formula and you get ½πr² and ¼πr², as expected.
Find the area of a sector of a circle with radius 14 cm and central angle 60°.
Area = θ/360 × πr2 = 60/360 × 22/7 × 14 × 14
= 1/6 × 22/7 × 196 = 1/6 × 616 = 102.67 cm2
So the area is 308/3 cm² (about 102.67 cm²).
Find the length of the arc of the same sector (r = 14 cm, θ = 60°).
Arc length = θ/360 × 2πr = 60/360 × 2 × 22/7 × 14
= 1/6 × 88 = 44/3 cm ≈ 14.67 cm
Areas of Segments
A segment is the region between a chord and the corresponding arc. A chord divides the circle into two segments: the smaller one is the minor segment and the larger one is the major segment. The area of a segment is found by subtracting the area of a triangle from the area of its sector.
For a segment with central angle θ and radius r:
Area of minor segment = Area of sector OAB − Area of ΔOAB
Area of major segment = Area of circle − Area of minor segment
Where O is the centre and AB is the chord. For the triangle OAB with two sides r and included angle θ:
Area of ΔOAB = ½ × r2 × sin θ
In practice at Class 10 level, when θ is 60°, 90° or 120°, the triangle area is found using standard values: for θ = 60°, ΔOAB is equilateral with area (√3/4)r²; for θ = 90°, ΔOAB is right-angled isosceles with area ½r².
Think of it like this: the sector OAB is like a slice of pizza with a curved edge. If you cut off the triangular part OAB (the straight-sided slice), what remains, the curved cap, is exactly the segment.
NCERT expects you to handle a few standard angles. The sine values you need are:
sin 60° = √3/2 sin 90° = 1 sin 120° = √3/2
So for a 60° sector, area of ΔOAB = ½ r² × √3/2 = (√3/4)r², which matches the equilateral triangle formula. For 90°, it gives ½ r².
Find the area of the minor segment of a circle of radius 14 cm, when the central angle is 60°. (Use π = 22/7, √3 = 1.73.)
Area of sector = 60/360 × πr² = 1/6 × 22/7 × 14 × 14 = 102.67 cm² (from earlier).
Area of ΔOAB: since θ = 60° and OA = OB = r, the triangle is equilateral with side 14 cm.
Area of ΔOAB = (√3/4) r2 = (√3/4) × 196 = 49√3 ≈ 49 × 1.73 = 84.77 cm2
Area of minor segment = 102.67 − 84.77 = 17.90 cm2
Area of major segment = Area of circle − Minor segment = 616 − 17.90 = 598.10 cm².
Find the area of the segment when the central angle is 90° and r = 7 cm. (Use π = 22/7.)
Area of sector = 90/360 × πr2 = ¼ × 22/7 × 49 = 38.50 cm2
Area of ΔOAB = ½ × r2 × sin 90° = ½ × 49 × 1 = 24.50 cm2
Area of minor segment = 38.50 − 24.50 = 14 cm2
Areas of Combinations of Plane Figures
Many real-life figures are not a single shape but a combination, for example, a rectangular field with a semicircular flower bed, a window made of a rectangle and a semicircle, or a running track made of a rectangle and two semicircles. To find such areas, NCERT follows a single strategy: add or subtract the areas you already know.
To find the area of a combined figure:
1. Break the figure into known parts (circles, sectors, segments, triangles, rectangles, squares).
2. Calculate each part separately using the formulae above.
3. Add the parts if they together make the whole, or subtract if a part is a hole or overlap.
There is no new formula to memorise here, the skill is in seeing which parts to add and which to subtract. Always draw a rough diagram and label the radii and angles before calculating.
A square park of side 20 m has a semicircular flower bed on each side on the outside. Find the total area of the four flower beds.
Each flower bed is a semicircle with diameter equal to the side of the square, so radius r = 10 m.
Area of one semicircle = ½ πr2 = ½ × 22/7 × 10 × 10 = 1100/7 ≈ 157.14 m2
Area of four semicircles = 4 × 1100/7 = 4400/7 ≈ 628.57 m2
This is the same as the area of two full circles of radius 10 m: 2 × π × 100 = 200π ≈ 628.57 m².
A design is made of a circle of radius 7 cm with a sector of angle 90° removed (cut out). Find the area of the remaining design.
Area of circle = πr2 = 22/7 × 49 = 154 cm2
Area of removed sector = 90/360 × πr2 = ¼ × 154 = 38.50 cm2
Area of remaining design = 154 − 38.50 = 115.50 cm2
The remaining part is the major sector with angle 270°, and indeed 270/360 × 154 = 115.50 cm².
Another common pattern is the area swept by a pendulum or a wiper blade, this is simply a sector whose radius is the length of the blade and whose angle is the angle of the sweep. Similarly, the cost of fencing a circular path uses the circumference, while painting uses the area.
Summary
- Circumference of a circle of radius r: C = 2πr = πd; Area: A = πr².
- For a sector with central angle θ (in degrees): arc length = θ/360 × 2πr; sector area = θ/360 × πr². Also, sector area = ½ × arc length × r.
- Minor sector has θ < 180°, major sector has θ > 180°; their angles sum to 360° and areas sum to πr².
- Area of minor segment = area of sector OAB − area of ΔOAB, where area of ΔOAB = ½ r² sin θ (for 60°: √3/4 r²; for 90°: ½ r²).
- Area of major segment = area of circle − area of minor segment.
- For combinations of plane figures, break the figure into known shapes, calculate each area, and add or subtract as needed.
- Use π = 22/7 unless the question states otherwise.
Practice Questions
- Find the circumference and area of a circle of diameter 14 cm. (Use π = 22/7.)
- A sector of a circle of radius 21 cm subtends an angle of 120° at the centre. Find its arc length and area.
- Find the area of the minor segment of a circle of radius 10 cm when the central angle is 90°. (Use π = 3.14.)
- A chord of a circle of radius 14 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor segment. (Use π = 22/7, √3 = 1.73.)
- A square of side 14 cm has a quadrant of radius 7 cm cut from each corner. Find the area of the remaining part of the square.
- Two concentric circles have radii 14 cm and 7 cm. Find the area of the ring (annulus) between them.