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Areas Related To Circles

Class 10 . Mathematics . Mathematics . Areas Related to Circles

Chapter 11: Areas Related To Circles

Ch 11

Mathematics

Class 10

Lecture 1

Areas Of Sector And Segment Of A Circle

Reference video · Lecture 1Areas Of Sector And Segment Of A Circle

Sectors and Arc Length

Definition

Sector — A sector is the part of a circle enclosed by two radii and the arc between them.

Minor and major sectors of a circle.
Minor and major sectors of a circle.

Sector facts

  • Minor sector: The smaller sector; its central angle is usually denoted by \(\theta\).
  • Major sector: The remaining, larger sector; its central angle is \(360^\circ-\theta\).
  • Convention: Unless stated otherwise, “sector” means the minor sector.
Area and arc length of a sector

A sector with radius \(r\) and central angle \(\theta\) (in degrees) is the fraction \(\frac{\theta}{360^\circ}\) of the whole circle.

Sector area
$$A=\frac{\theta}{360^\circ}\pi r^2$$

Use when the radius and central angle in degrees are known.

Arc length
$$l=\frac{\theta}{360^\circ}\,2\pi r$$

The arc length is the same fraction of the circle’s circumference.

Major sector area
$$A_{\text{major}}=\pi r^2-A_{\text{minor}}=\frac{360^\circ-\theta}{360^\circ}\pi r^2$$

The major sector and minor sector together make one full circle.

Finding sector area and arc length

  1. Identify the radius and central angle in degrees.
  2. Substitute into the sector-area or arc-length formula.
    $$A=\frac{\theta}{360^\circ}\pi r^2,\qquad l=\frac{\theta}{360^\circ}\,2\pi r$$
  3. For a major sector, use the remaining angle or subtract the minor-sector area from the circle’s area.
Worked example

Find the areas of the minor and major sectors.

  1. Find the minor and major sector areas for radius \(4\) cm and angle \(30^\circ\), using \(\pi=3.14\).
  2. Calculate the minor-sector area.
    $$A_{\text{minor}}=\frac{30}{360}\times 3.14\times 4^2=\frac{12.56}{3}\approx 4.19\text{ cm}^2$$
  3. Subtract it from the area of the full circle to get the major-sector area.
    $$A_{\text{major}}=3.14\times 4^2-4.19=46.05\approx 46.1\text{ cm}^2$$
Answer: Minor sector: approximately \(4.19\text{ cm}^2\); major sector: approximately \(46.1\text{ cm}^2\).
⚠ Common mistake

Do not use the sector-area formula for arc length: area uses \(\pi r^2\), while arc length uses the circumference \(2\pi r\).

Segments of a Circle

Definition

Segment — A segment is the part of a circle enclosed by a chord and its corresponding arc.

A chord divides a circle into a minor segment and a major segment.
A chord divides a circle into a minor segment and a major segment.

Segment facts

  • Minor segment: The smaller region between a chord and its minor arc; by default, “segment” means the minor segment.
  • Major segment: The larger region on the other side of the chord.
  • Complement: Major segment area equals the circle’s area minus the minor segment area.
Area of a segment

The area of a segment equals the area of its corresponding sector minus the area of the triangle formed by the two radii and the chord.

$$A_{\text{minor segment}}=A_{\text{sector}}-A_{\triangle}$$
Major segment area
$$A_{\text{major segment}}=\pi r^2-A_{\text{minor segment}}$$

Use the full-circle area to find the complementary major segment.

Finding a segment area

  1. Find the corresponding sector area using its central angle.
    $$A_{\text{sector}}=\frac{\theta}{360^\circ}\pi r^2$$
  2. Find the area of the triangle inside that sector.
    $$A_{\triangle}=\frac{1}{2}\times\text{base}\times\text{height}$$
  3. Subtract the triangle area from the sector area.
    $$A_{\text{segment}}=A_{\text{sector}}-A_{\triangle}$$
The minor segment is the sector minus the triangle formed by the radii and chord.
The minor segment is the sector minus the triangle formed by the radii and chord.
Worked example

Find the area of the minor segment cut off by a chord subtending \(120^\circ\) at the centre of a circle of radius \(21\) cm.

  1. Find the minor segment area when \(r=21\) cm and \(\angle AOB=120^\circ\), using \(\pi=\frac{22}{7}\).
  2. Calculate the sector area.
    $$A_{\text{sector}}=\frac{120}{360}\times\frac{22}{7}\times 21^2=462\text{ cm}^2$$
  3. Drop a perpendicular from \(O\) to chord \(AB\) at \(M\). Since \(OA=OB\), \(M\) bisects \(AB\) and the central angle.
    $$\angle AOM=60^\circ,\qquad OM=21\cos 60^\circ=\frac{21}{2},\qquad AM=21\sin 60^\circ=\frac{21\sqrt{3}}{2}$$
  4. Find the triangle area using \(AB=2AM\).
    $$AB=21\sqrt{3},\qquad A_{\triangle OAB}=\frac{1}{2}\times 21\sqrt{3}\times\frac{21}{2}=\frac{441\sqrt{3}}{4}\text{ cm}^2$$
  5. Subtract the triangle area from the sector area.
    $$A_{\text{segment}}=462-\frac{441\sqrt{3}}{4}=\frac{21}{4}(88-21\sqrt{3})\text{ cm}^2$$
Answer: \(\frac{21}{4}(88-21\sqrt{3})\text{ cm}^2\).
⚠ Common mistake

A segment is not just the sector: subtract the triangle between the radii and the chord to get the minor-segment area.

Practice

A sector has radius \(10\) cm and angle \(72^\circ\). Find its area and the length of its arc.

Hint: The angle is one-fifth of a full turn.
Key step: Use \(\frac{72}{360}=\frac{1}{5}\) for both the circle’s area and circumference.
Solution: Area \(=\frac{1}{5}\pi(10)^2=20\pi\text{ cm}^2\); arc length \(=\frac{1}{5}(2\pi\times10)=4\pi\text{ cm}\).
Practice

Find the minor segment area for radius \(14\) cm and central angle \(90^\circ\). Use \(\pi=\frac{22}{7}\).

Hint: Subtract the right triangle’s area from the quarter-circle sector’s area.
Key step: The sector area is \(154\text{ cm}^2\) and the triangle area is \(\frac{1}{2}(14)(14)=98\text{ cm}^2\).
Solution: Segment area \(=154-98=56\text{ cm}^2\).

Quick recap

  • A sector is bounded by two radii and an arc; a segment is bounded by a chord and an arc.
  • Sector area: \(A=\frac{\theta}{360^\circ}\pi r^2\).
  • Arc length: \(l=\frac{\theta}{360^\circ}2\pi r\).
  • Major-sector area is the full-circle area minus the minor-sector area.
  • Minor-segment area = corresponding sector area − corresponding triangle area.
  • Major-segment area is the full-circle area minus the minor-segment area.

Formula sheet

Sector area
$$A=\frac{\theta}{360^\circ}\pi r^2$$
Arc length
$$l=\frac{\theta}{360^\circ}\,2\pi r$$
Major sector area
$$A_{\text{major}}=\pi r^2-A_{\text{minor}}=\frac{360^\circ-\theta}{360^\circ}\pi r^2$$
Major segment area
$$A_{\text{major segment}}=\pi r^2-A_{\text{minor segment}}$$

Mathematics — ALL CHAPTERS

1

Real Numbers

2

Polynomials

3

Pair Of Linear Equations In Two Variables

4

Quadratic Equations

5

Arithmetic Progressions

6

Triangles

7

Coordinate Geometry

8

Introduction To Trigonometry

9

Some Applications Of Trigonometry

10

Circles

11

Areas Related To Circles

12

Surface areas And Volumes

13

Statistics 

14

Probability