Key Concepts & Formulae at a Glance
- Sector:The pie-slice region enclosed by two radii and the arc between them. The minor sector is the smaller one; the major sector's angle is 360°−θ.
- Segment:The region enclosed by a chord and its arc. The minor segment lies on the same side as the minor arc; the major segment is the rest of the circle.
- Area of a sector:(θ360) × πr², where θ is the sector's angle in degrees.
- Length of an arc:(θ360) × 2πr
- Area of a segment:Area of the sector − Area of the triangle formed by the two radii and the chord.
- Major from minor:Area of major sector = πr² − area of minor sector; Area of major segment = πr² − area of minor segment.
Exercise 11.1
14 Questions
1Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Area of sector = (θ360) × πr² = (60360) × (227) × 6² = (16) × (227) × 36
Area of the sector = 1327 cm² ≈ 18.86 cm².
2Find the area of a quadrant of a circle whose circumference is 22 cm.
Circumference = 2πr = 22 → r = 22/(2 × 227) = 22 × 744 = 72 = 3.5 cm.
A quadrant is a 90° sector: Area = (90360) × πr² = (14) × (227) × (72)² = (14) × (227) × (494)
A quadrant is a 90° sector: Area = (90360) × πr² = (14) × (227) × (72)² = (14) × (227) × (494)
Area of the quadrant = 778 cm² = 9.625 cm².
3The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
In 60 minutes the minute hand sweeps 360°, so in 5 minutes it sweeps (560) × 360° = 30°.
Area swept = (30360) × πr² = (112) × (227) × 14² = (112) × (227) × 196
Area swept = (30360) × πr² = (112) × (227) × 14² = (112) × (227) × 196
Area swept = 1543 cm² ≈ 51.33 cm².
4A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector. (Use π = 3.14)
(i) Minor segment
Area of sector (90°) = (90360) × 3.14 × 10² = (14) × 3.14 × 100 = 78.5 cm²
ΔOAB is right-angled at O with OA = OB = 10 cm (the radii), so its area = ½ × 10 × 10 = 50 cm²
Area of minor segment = sector − triangle = 78.5 − 50 = 28.5 cm²
ΔOAB is right-angled at O with OA = OB = 10 cm (the radii), so its area = ½ × 10 × 10 = 50 cm²
Area of minor segment = sector − triangle = 78.5 − 50 = 28.5 cm²
Minor segment = 28.5 cm².
(ii) Major sector
Area of major sector = πr² − area of minor sector = (3.14 × 100) − 78.5 = 314 − 78.5 = 235.5 cm²
Major sector = 235.5 cm².
5In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) the length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord
(i) Arc length = (60360) × 2πr = (16) × 2 × (227) × 21 = 22 cm
(ii) Sector area = (60360) × πr² = (16) × (227) × 441 = 231 cm²
(iii) With θ=60°, ΔOAB is equilateral (OA=OB=r=21, and the third angle also works out to 60°) (isosceles triangle with a 60° vertex angle is automatically equilateral), so its area = (√34) × 21² = 441√34.
Segment area = sector − triangle = 231 − 441√34
(ii) Sector area = (60360) × πr² = (16) × (227) × 441 = 231 cm²
(iii) With θ=60°, ΔOAB is equilateral (OA=OB=r=21, and the third angle also works out to 60°) (isosceles triangle with a 60° vertex angle is automatically equilateral), so its area = (√34) × 21² = 441√34.
Segment area = sector − triangle = 231 − 441√34
(i) 22 cm (ii) 231 cm² (iii) (231 − 441√34) cm² ≈ 40.04 cm².
6A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73)
Minor segment
Sector area = (60360) × 3.14 × 15² = (16) × 3.14 × 225 = 117.75 cm²
ΔOAB is equilateral (60° central angle, OA=OB=15), area = (√34) × 15² = (1.734) × 225 = 97.3125 cm²
Minor segment = 117.75 − 97.3125 = 20.4375 cm²
ΔOAB is equilateral (60° central angle, OA=OB=15), area = (√34) × 15² = (1.734) × 225 = 97.3125 cm²
Minor segment = 117.75 − 97.3125 = 20.4375 cm²
Minor segment ≈ 20.44 cm².
Major segment
Major segment = πr² − minor segment = (3.14 × 225) − 20.4375 = 706.5 − 20.4375
Major segment ≈ 686.06 cm².
7A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)
Sector area = (120360) × 3.14 × 12² = (13) × 3.14 × 144 = 150.72 cm²
ΔOAB has OA=OB=12 with included angle 120°, so area = ½ × 12 × 12 × sin120° = ½ × 144 × (√32) = 36√3 = 36 × 1.73 = 62.28 cm²
Segment area = 150.72 − 62.28
ΔOAB has OA=OB=12 with included angle 120°, so area = ½ × 12 × 12 × sin120° = ½ × 144 × (√32) = 36√3 = 36 × 1.73 = 62.28 cm²
Segment area = 150.72 − 62.28
Segment area = 88.44 cm².
8A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig. 11.8). Find:
(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)
The peg is at a corner of the square, so the interior angle available to the horse is 90° — the grazing area is a quadrant.
(i) With rope 5 m: Area = (90360) × 3.14 × 5² = (14) × 3.14 × 25 = 19.625 m²
(ii) With rope 10 m (still shorter than the 15 m side, so it stays a full quadrant): Area = (90360) × 3.14 × 10² = (14) × 3.14 × 100 = 78.5 m²
Increase = 78.5 − 19.625
(i) With rope 5 m: Area = (90360) × 3.14 × 5² = (14) × 3.14 × 25 = 19.625 m²
(ii) With rope 10 m (still shorter than the 15 m side, so it stays a full quadrant): Area = (90360) × 3.14 × 10² = (14) × 3.14 × 100 = 78.5 m²
Increase = 78.5 − 19.625
(i) 19.625 m² (ii) increase = 58.875 m².
9A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 11.9. Find:
(i) the total length of the silver wire required. (ii) the area of each sector of the brooch.
(i) Wire is used for the circumference plus 5 full diameters: Length = πd + 5d = d(π+5) = 35 × (227 + 5) = 35 × (577) = 5 × 57 = 285 mm
(ii) 5 diameters create 10 equal sectors, so each sector's angle = 360°10 = 36°, and radius = 352 = 17.5 mm.
Area of each sector = (36360) × πr² = (110) × (227) × 17.5² = (110) × (227) × 306.25
(ii) 5 diameters create 10 equal sectors, so each sector's angle = 360°10 = 36°, and radius = 352 = 17.5 mm.
Area of each sector = (36360) × πr² = (110) × (227) × 17.5² = (110) × (227) × 306.25
(i) 285 mm of wire (ii) each sector = 96.25 mm².
10An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
8 equally spaced ribs divide the circle into 8 equal sectors, each of angle 360°8 = 45°.
Area between consecutive ribs = (45360) × πr² = (18) × (227) × 45² = (18) × (227) × 2025
Area between consecutive ribs = (45360) × πr² = (18) × (227) × 45² = (18) × (227) × 2025
Area between two consecutive ribs = 2227528 cm² ≈ 795.54 cm².
11A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Each wiper cleans a sector of radius 25 cm and angle 115°: Area = (115360) × (227) × 25²
Since the wipers don't overlap, the total cleaned area is simply twice one wiper's sector area.
Since the wipers don't overlap, the total cleaned area is simply twice one wiper's sector area.
Total area cleaned ≈ 2 × 627.48 = 1254.96 cm².
12To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)
The warned region is a sector of radius 16.5 km and angle 80°:
Area = (80360) × 3.14 × 16.5² = (29) × 3.14 × 272.25
Area = (80360) × 3.14 × 16.5² = (29) × 3.14 × 272.25
Area of sea warned ≈ 189.97 km².
13A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹0.35 per cm². (Use √3 = 1.7)
The six designs are the six segments cut off by the sides of a regular hexagon inscribed in the circle — together, the 6 sectors make up the whole circle, and the 6 triangles make up the hexagon.
So total design area = Area of circle − Area of the hexagon.
Area of circle = πr² = (227) × 28² = (227) × 784 = 2464 cm²
The hexagon splits into 6 equilateral triangles of side r=28: Hexagon area = 6 × (√34) × 28² = (3√32) × 784 = (3 × 1.72) × 784 = 2.55 × 784 = 1999.2 cm²
Design area = 2464 − 1999.2 = 464.8 cm²
Cost = 464.8 × ₹0.35
So total design area = Area of circle − Area of the hexagon.
Area of circle = πr² = (227) × 28² = (227) × 784 = 2464 cm²
The hexagon splits into 6 equilateral triangles of side r=28: Hexagon area = 6 × (√34) × 28² = (3√32) × 784 = (3 × 1.72) × 784 = 2.55 × 784 = 1999.2 cm²
Design area = 2464 − 1999.2 = 464.8 cm²
Cost = 464.8 × ₹0.35
Cost of making the designs = ₹162.68.
14Tick the correct answer in the following: Area of a sector of angle p (in degrees) of a circle with radius R is
(A) (p180) × 2πR (B) (p180) × πR² (C) (p360) × 2πR (D) (p720) × 2πR²
The standard formula is Area = (p360) × πR². Rewriting option (D): (p720) × 2πR² = (2p720) × πR² = (p360) × πR² (exactly matches the standard formula).
Options (A) and (C) use the circumference formula 2πR instead of πR², so they don't give an area at all — those are dimensionally wrong. Option (B) uses the wrong denominator (180 instead of 360).
Options (A) and (C) use the circumference formula 2πR instead of πR², so they don't give an area at all — those are dimensionally wrong. Option (B) uses the wrong denominator (180 instead of 360).
(D) (p720) × 2πR²
Frequently Asked Questions
A sector is the pie-slice region enclosed by two radii and the arc between them. A segment is the region enclosed by a chord and the arc it cuts off — it does not include the centre unless the chord is a diameter. A segment's area is always found as sector area minus triangle area.
Area of a sector with angle θ (in degrees) and radius r is (θ360) × πr². This comes directly from the unitary method: a full circle is a 360° sector with area πr², so a θ° sector is that fraction of the whole circle's area.
Area of segment = Area of the corresponding sector − Area of the triangle formed by the two radii and the chord. The triangle is isosceles (two sides are radii), so its area can be found using ½ × r × r × sinθ, or by splitting it into two right triangles with trigonometry.
Length of an arc with angle θ (in degrees) and radius r is (θ360) × 2πr — the same unitary-method idea applied to the circle's circumference instead of its area.
Yes — a printable Formula Card covering all 7 units, including sector and segment area formulas, is available for ₹99.
Continue to the Next Chapter
Chapter 12 (Surface Areas and Volumes) moves from flat regions into 3D solids.