Areas Related to Circles

Why This Chapter Matters

This chapter combines circle geometry with area calculations — a favourite for board exam questions worth 4-6 marks. The formulae are straightforward, but combining them for complex figures requires careful reading. Questions often involve sectors, segments, and combination figures.

Prerequisites

Core Concepts

1. Basic Circle Formulas

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Value of π: Use 22/7 unless told to use 3.14

2. Sector and Arc

A sector is a "pie slice" of a circle — bounded by two radii and an arc.

The angle at the centre is called the angle of the sector (θ).

$$\text{Area of sector} = \frac{\theta}{360°} \times \pi r^2$$

$$\text{Length of arc} = \frac{\theta}{360°} \times 2\pi r$$

Special cases:

3. Segment

A segment is the region between a chord and its arc.

Minor segment = small piece (between chord and minor arc)

Major segment = large piece (between chord and major arc)

$$\text{Area of minor segment} = \text{Area of sector} - \text{Area of triangle}$$

$$\text{Area of segment (with central angle θ)} = \frac{\theta}{360°}\pi r^2 - \frac{1}{2}r^2 \sin\theta$$

Solved Examples

Example 1 — Sector Area

Q: Find area of sector of angle 45° in circle of radius 7 cm.

Area = (45/360) × π × 7² = (1/8) × (22/7) × 49 = 19.25 cm²

Example 2 — Combination Figure

Q: A brooch is made with silver wire in the form of a circle of radius 35mm. The wire is also used to make 5 diameters. Find total length of silver wire needed.

Circumference = 2π(35) = 2 × 22/7 × 35 = 220 mm

5 diameters = 5 × 70 = 350 mm

Total = 220 + 350 = 570 mm

Example 3 — Segment

Q: Find the area of a segment of a circle of radius 12 cm if the chord subtends 120° at the centre.

Area of sector (120°, r=12):

= (120/360) × π × 144 = (1/3) × (22/7) × 144 = 150.86 cm²

Area of triangle (isosceles, two sides = 12, angle = 120°):

= (1/2) × r² × sin120° = (1/2) × 144 × (√3/2) = 36√3 = 62.35 cm²

Area of segment = 150.86 − 62.35 = 88.44 cm² (approximately)

PYQs

2023

Q: The area of a sector with radius 6 cm and angle 60°:

= (60/360) × π × 36 = (1/6) × 22/7 × 36 = 18.86 cm²

2022

Q: Find area swept by minute hand of length 15 cm in 5 minutes.

In 60 min → 360°. In 5 min → 30°

Area = (30/360) × π × 15² = (1/12) × 22/7 × 225 = 58.93 cm²

2021

Q: In a circle of radius 21 cm, an arc subtends an angle of 60°. Find length of arc and area of sector.

Arc length = (60/360) × 2π × 21 = (1/6) × 2 × 22/7 × 21 = 22 cm

Area of sector = (60/360) × π × 441 = 231 cm²

2020

Q: Find area of shaded region: square of side 10 cm with 4 quarter circles at each corner (radius 5 cm each).

Area of 4 quarter circles = π × 5² = 78.57 cm²

Shaded area (remaining) = 100 − 78.57 = 21.43 cm²

MCQ Practice

Q1. Area of sector with radius r and angle 90°:

(A) πr²/2 (B) πr²/4 ✓ (C) 2πr (D) πr

Q2. If area of sector (radius 7, angle θ) = 77 cm², then θ =

(A) 120° (B) 270° ✓ (C) 180° (D) 90°

[θ/360 × 22/7 × 49 = 77 → θ = 270°]

Q3 (Hard). A horse is tied to a corner of a square plot of side 10m with a rope of 7m. Find the area the horse can graze.

Corner → 90° sector, radius = 7m (rope can't go past 10m side, so 7 < 10)

Area = (90/360) × π × 7² = (1/4) × 22/7 × 49 = 38.5 m²

Revision Notes

CIRCLE BASICS:
  Area = πr²
  Circumference = 2πr

SECTOR (angle θ at centre):
  Area of sector = (θ/360) × πr²
  Arc length    = (θ/360) × 2πr

SEGMENT:
  Area = Area of sector − Area of triangle

COMBINATION FIGURES STRATEGY:
  1. Break into known shapes (circle, semicircle, triangle, rectangle)
  2. Calculate each area separately
  3. Add or subtract as needed

Use π = 22/7 when radius is multiple of 7
Use π = 3.14 when the problem says so

Common Mistakes:

❌ Using circumference formula for area (πr² vs 2πr)

❌ In segment: subtracting circle area instead of just triangle area from sector

❌ Forgetting to convert θ correctly: (θ/360), not (θ/180)

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