Areas Related to Circles notes: CBSE Class 10 Mathematics (Standard 041 / Basic 241)

1. Circle facts and the value of π

For a circle of radius $r$ : area $= \pi r^2$ , circumference $= 2\pi r$ .

Use the value of $\pi$ given in the question: $\dfrac{22}{7}$ (choose radii that cancel the $7$ ) or $3.14$ ; if none is given, leave $\pi$ in the answer.

A ring between concentric circles of radii $R > r$ has area $\pi(R^2 - r^2)$ — not $\pi(R - r)^2$ . A wheel covers one circumference per revolution.

Always use the value of $\pi$ the question gives.
Stem says	Use	Typical radius
$\pi = \dfrac{22}{7}$	cancel the $7$	$7, 14, 21, 35, 10.5$
$\pi = 3.14$	decimals	$10, 12, 20$
nothing	leave the answer in terms of $\pi$	any

Formulas

Circle
$A = \pi r^2, \quad C = 2\pi r$
Ring
$A_{\text{ring}} = \pi(R^2 - r^2)$

Key terms

Ring (annulus): The region between two concentric circles.

2. Sectors and arcs

Two radii divide a circle into a minor sector (angle $\theta$ ) and a major sector (angle $360^\circ - \theta$ ). A sector of angle $\theta$ is the fraction $\dfrac{\theta}{360}$ of the circle, so

\text{Area} = \frac{\theta}{360} \times \pi r^2, \qquad \text{Arc length } l = \frac{\theta}{360} \times 2\pi r.

Perimeter of a sector $= l + 2r$ . Also $\text{Area} = \dfrac12 lr$ .

Example: $r = 21$ cm, $\theta = 60^\circ$ : area $231$ cm², arc $22$ cm, perimeter $64$ cm.

Minor sector (shaded) and major sector: their angles add up to $360^\circ$.

A sector is the fraction $\dfrac{\theta}{360}$ of the circle.
Angle $\theta$	Fraction $\dfrac{\theta}{360}$	Sector area	Arc length
$360^\circ$	$1$	$\pi r^2$	$2\pi r$
$180^\circ$	$\dfrac12$	$\dfrac12\pi r^2$	$\pi r$
$90^\circ$	$\dfrac14$	$\dfrac14\pi r^2$	$\dfrac12\pi r$
$\theta$	$\dfrac{\theta}{360}$	$\dfrac{\theta}{360}\pi r^2$	$\dfrac{\theta}{360} \times 2\pi r$

Formulas

Area of a sector
$\frac{\theta}{360} \times \pi r^2$
Length of an arc
$\frac{\theta}{360} \times 2\pi r$
Area from the arc
$A = \frac12 l r$

Key terms

Sector: The region between two radii and the arc between them.
Quadrant: A sector of angle $90^\circ$ .

3. Finding the angle first; working backwards

Many problems hide the angle:

Clocks: minute hand $6^\circ$ per minute; hour hand $0.5^\circ$ per minute.
Equal slices: $n$ equal sectors have angle $\dfrac{360^\circ}{n}$ (a $24$ -spoke wheel: $15^\circ$ ).
Fractions: a sector that is $\dfrac15$ of the circle has angle $72^\circ$ .

Backwards: from an arc, $\dfrac{\theta}{360} = \dfrac{l}{2\pi r}$ (pendulum $35$ cm, arc $22$ cm gives $36^\circ$ ); from an area, $\dfrac{\theta}{360} = \dfrac{A}{\pi r^2}$ .

For the same angle, sector areas are in the ratio of $r^2$ and arcs in the ratio of $r$ .

Find the angle first, then use the sector formulas.
Hand	Full turn	Per minute	In 20 minutes
Minute hand	$360^\circ$ in $60$ min	$6^\circ$	$120^\circ$
Hour hand	$360^\circ$ in $12$ h	$0.5^\circ$	$10^\circ$
Seconds hand	$360^\circ$ in $1$ min	$360^\circ$	$20$ full turns

The bob moves along an arc of the circle whose radius is the string.

Formulas

Working backwards
$\frac{\theta}{360} = \frac{l}{2\pi r} = \frac{A}{\pi r^2}$

4. Segments

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

\text{Minor segment} = \text{sector } OAB - \triangle OAB, \qquad \text{Major segment} = \pi r^2 - \text{minor segment}.

The syllabus uses central angles of $60^\circ$ , $90^\circ$ and $120^\circ$ :

$90^\circ$ : $\triangle OAB = \dfrac12 r^2$ .
$60^\circ$ : $\triangle OAB$ is equilateral, area $\dfrac{\sqrt3}{4}r^2$ .
$120^\circ$ : drop $OM \perp AB$ ; $OM = \dfrac r2$ , $AB = \sqrt3\,r$ , area $\dfrac{\sqrt3}{4}r^2$ .

Perimeter of a segment $=$ arc $+$ chord ( $90^\circ$ : chord $= r\sqrt2$ ; $60^\circ$ : chord $= r$ ; $120^\circ$ : chord $= r\sqrt3$ ).

The syllabus uses only these three angles for segments.
Angle at $O$	$\triangle OAB$	Area of $\triangle OAB$	Minor segment
$60^\circ$	equilateral, side $r$	$\dfrac{\sqrt3}{4}r^2$	$\left(\dfrac{\pi}{6} - \dfrac{\sqrt3}{4}\right)r^2$
$90^\circ$	right isosceles	$\dfrac12 r^2$	$\left(\dfrac{\pi}{4} - \dfrac12\right)r^2$
$120^\circ$	height $\dfrac r2$ , base $\sqrt3\,r$	$\dfrac{\sqrt3}{4}r^2$	$\left(\dfrac{\pi}{3} - \dfrac{\sqrt3}{4}\right)r^2$

For $120^\circ$: $OM = \dfrac r2$, $AB = \sqrt3\,r$, so $\triangle OAB = \dfrac{\sqrt3}{4}r^2$.

Formulas

Minor segment, 90°
$\left(\frac{\pi}{4} - \frac12\right)r^2$
Minor segment, 60°
$\left(\frac{\pi}{6} - \frac{\sqrt3}{4}\right)r^2$
Minor segment, 120°
$\left(\frac{\pi}{3} - \frac{\sqrt3}{4}\right)r^2$

Key terms

Segment: The region between a chord and an arc of a circle.

5. Shaded regions

Plan: whole $\pm$ pieces, checking overlaps.

Useful facts:

Four equal quadrants at the corners of a square make one full circle.
The three corner sectors (equal radius) of any triangle make a semicircle, since the angles add to $180^\circ$ .
A leaf between two quadrants on a square's diagonal $=$ two $90^\circ$ segments $=$ quadrant $+$ quadrant $-$ square.
The region inside a circle but outside an inscribed square is four $90^\circ$ segments.

Examples: tile $28$ cm with corner quadrants and a central circle, radius $7$ : $784 - 154 - 154 = 476$ cm². Equilateral triangle side $20$ with corner sectors of radius $10$ : $173 - 157 = 16$ cm².

Every shaded-region question is "whole plus or minus pieces".

Shaded $=$ square $-$ four quadrants $-$ central circle.

Leaf $= 2 \times$ (quadrant $-$ right triangle).

Shaded $=$ triangle $-$ three $60^\circ$ sectors (together a semicircle).

6. Perimeters of curved figures

Walk round the boundary and add every piece:

sector: arc $+ 2r$ ;
segment: arc $+$ chord;
a quadrant-cornered park edge: straight parts $+$ arcs (four quadrant arcs $= 2\pi r$ );
a ring track: the outer edge exceeds the inner edge by $2\pi \times$ width.

Example: quadrant of radius $14$ cm: arc $22$ cm, perimeter $50$ cm; $90^\circ$ segment of radius $14$ cm: $22 + 14\sqrt2 \approx 41.80$ cm.

Perimeter of a segment $=$ arc $+$ chord.

Formulas

Perimeter of a sector
$P_{\text{sector}} = \frac{\theta}{360} \times 2\pi r + 2r$

Key terms

Perimeter: The total length of the boundary of a region.

Exam tips

How marks are awarded

Sector / arc (2 marks): formula with $\dfrac{\theta}{360}$ 1; answer with unit 1.
Segment (3 marks): sector 1; triangle 1; difference 1.
Shaded region (3–5 marks): plan 1; each component 1; final answer 1.

Common mistakes

$2\pi r$ in an area, or $\pi r^2$ in an arc.
Forgetting the two radii in the perimeter of a sector (or adding radii to a segment's perimeter).
$\dfrac12 r^2$ used for a $60^\circ$ or $120^\circ$ triangle.
$\pi(R - r)^2$ for a ring.
A different value of $\pi$ from the one given; rounding too early.
Missing units: cm for lengths, cm² for areas.

1. Circle facts and the value of π

Formulas

Key terms

2. Sectors and arcs

Formulas

Key terms

3. Finding the angle first; working backwards

Formulas

4. Segments

Formulas

Key terms

5. Shaded regions

6. Perimeters of curved figures

Formulas

Key terms

Exam tips

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