Class 10 Maths — Chapter 11: Areas Related to Circles

By RAAM SETU | NEET, SSC, Railway & Police Bharti Focused Notes 📘

📘 Introduction

This chapter covers Areas Related to Circles — including circle, semicircle, quadrant, sector, and segment. Learn formulas, solved examples, and practice questions. Essential for NEET, SSC, Railway, Police Bharti and Class 10 exams.

🔹 Key Formulas

📊 Chapter Summary Table

Shape Formula Use
Circleπr²Basic area calculation
Semicircle½πr²Half circle area
Quadrant¼πr²Quarter circle area
Ringπ(R² − r²)Annulus area
Sector(θ/360) × πr²Part of circle
SegmentSector − TriangleCurved area portion

🧮 Practice Questions

  1. Find the area of a circle with radius 7 cm.
  2. Calculate the area of a semicircle with diameter 10 cm.
  3. Area of a quadrant with radius 8 cm?
  4. Find the area of a ring with R = 12 cm and r = 8 cm.
  5. Area of a sector with radius 14 cm and central angle 60°.
  6. Calculate the area of a segment of a circle with radius 10 cm and chord length 12 cm.

Areas Related to Circles — Class 10

Circle area • Sector & segment • Annulus • Shaded regions • Derivations • Practice
Jump to Practice

1. Area of a Circle

For a circle with radius r, area A is given by:

A = πr²

Use π ≈ 22/7 or 3.1416 depending on the exam instructions. For problems involving parts of circles, keep units consistent (e.g., m²).

2. Area of a Sector

A sector is the region bounded by two radii and the included arc. If central angle is θ (in degrees), sector area = fraction θ/360 of full circle:

Area (sector) = (θ/360) × πr²

If θ is in radians: Area = (1/2) r² θ.

3. Area of a Segment

A segment is part of a circle cut off by a chord. Area of a segment = area of sector − area of triangle formed by the two radii and the chord.

Area (segment) = (θ/360)πr² − (1/2) r² sin θ (θ in degrees)

For θ in radians: Area = (1/2)r²(θ − sin θ).

Important: θ is the central angle corresponding to the segment.

4. Annulus (Ring)

An annulus is region between two concentric circles of radii R and r (R > r). Area is difference of areas:

Area (annulus) = π(R² − r²)

Useful in problems with circular frames, washers, and rings.

5. Shaded Regions — Typical Patterns

Many exam problems ask the area of shaded regions — usually a combination of sectors, segments, triangles and rectangles. Strategy:

  1. Draw the figure cleanly and mark radii, angles and given lengths.
  2. Break shaded region into known shapes (sector, segment, triangle) or subtract unwanted parts from larger shapes.
  3. Use exact formulas and simplify before plugging numbers; cancel π where possible (e.g., when π used both numerator and denominator).

Common example: square of side 14 with four quarter-circles of radius 7 drawn at corners; shaded region = square area − 4×(quarter-circle area) = 14² − 4×(π×7²/4) = 196 − 49π.

6. Short Derivations (Why formulas look like they do)

Sector area as fraction of full circle: sector central angle θ out of 360 degrees ⇒ fraction θ/360 of area πr². Segment area = sector minus isosceles triangle area (triangle with two sides r and included angle θ has area 1/2 r² sin θ).

Annulus = area large circle − small circle by subtracting radii squares.

7. Solved Examples

Example 1. Find area of sector with r = 10 cm and θ = 60°.

Area = (60/360) × π × 10² = (1/6) × 100π = (50/3)π ≈ 52.36 cm² (if π = 3.1416).

Example 2. In a circle r = 14 cm, find area of segment with central angle 90°.

Sector area = (90/360)πr² = (1/4)π×196 = 49π.
Triangle area = (1/2) r² sin 90° = (1/2)×196×1 = 98.
Segment area = 49π − 98.

Example 3. A circular ring has outer radius 10 cm and inner radius 6 cm. Find area.

Area = π(10² − 6²) = π(100 − 36) = 64π ≈ 201.06 cm².

Example 4 (Shaded region). A semicircle of radius 7 is inscribed on top of a rectangle 14 by 7. Find shaded area (rectangle minus semicircle).

Rect area = 14×7 = 98. Semicircle area = (1/2)π×7² = 24.5π. Shaded = 98 − 24.5π.

8. Practice Questions

  1. Find the area of a circle whose circumference is 44 cm. (Use π = 22/7)
  2. Find area of segment in a circle radius 21 cm with central angle 120°.
  3. Square of side 20 cm has quarter-circles of radius 10 cm at each corner. Find area outside the quarter-circles but inside square.
  4. Find area of shaded region: sector of 120° in a circle radius 9 cm minus triangle formed by radii and chord.

9. Exam Tips & Shortcuts

  • Always note whether θ is in degrees or radians; formula differs for sector/segment.
  • Cancel π early when possible to simplify arithmetic, e.g., when comparing areas with same π factor.
  • Memorise triangle area formula in terms of r and θ: (1/2)r² sin θ (useful inside segment calculations).
  • Sketch and label clearly — most mistakes come from misidentifying sector angle or triangle boundaries.