Areas Related to Circles Class 10: Concepts and Solved Questions

What Areas Related to Circles is about

Weightage: This chapter is part of Unit VI, Mensuration, which carries 10 of the 80 theory marks together with Surface Areas and Volumes.Source: CBSE Mathematics syllabus 2026-27, Unit VI Mensuration

In real board papers: Standard, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 4.9 marks per paper; Basic, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 4.7 marks per paper. See the questions

Key concepts

Circumference and area of a circle: the starting point

NCERT section 11.1 Areas of Sector and Segment of a Circle

The circumference is the length of the boundary: $C=2\pi r=\pi d$. The area is the space inside: $A=\pi r^2$. The number $\pi$ is the ratio of circumference to diameter, and it is the same for every circle. That is why one formula works for a coin and a racetrack.

Every result in this chapter is a fraction of these two quantities, so you must handle them fast and without slips. Two habits help:

• Always convert diameter to radius before using $\pi r^2$.
• With $\pi =\frac{22}{7}$, radii like 7, 14, 21, 35 give whole-number answers. If a question gives such numbers, expect the 7 to cancel.

For a semicircle, the perimeter is $\pi r+2r$, because the boundary is half the circle plus the diameter. Students often forget the straight edge.

Worked example

The circumference of a circle exceeds its diameter by 30 cm. Find the area of the circle. (Take $\pi =\frac{22}{7}$)

Try it first, then show the solutionSolution

1. Let the diameter be $d$ cm. Then circumference $=\pi d$.
2. Given: $\pi d-d=30$, so $d\left(\frac{22}{7}-1\right)=30$.
3. $d\times \frac{15}{7}=30$, so $d=14$ cm and radius $r=7$ cm.
4. Area $=\pi r^2=\frac{22}{7}\times 7\times 7=154$ cm${}^2$.
5. Check: circumference $=\frac{22}{7}\times 14=44$ cm, and $44-14=30$ cm. Correct.

Answer: Area of the circle $=154$ cm${}^2$.

Length of an arc

NCERT section 11.1 Areas of Sector and Segment of a Circle

An arc is a part of the circumference. If the arc subtends angle $\theta$ (in degrees) at the centre, its length is

$l=\frac{\theta }{360^{\circ }}\times 2\pi r$

Why it works: a full turn at the centre is $360^{\circ }$ and gives the whole circumference $2\pi r$. Arc length grows in direct proportion to the angle, so an angle of $\theta$ gives the fraction $\frac{\theta }{360^{\circ }}$ of the circumference. A $90^{\circ }$ arc is one quarter of the circle, a $60^{\circ }$ arc is one sixth.

When to use it: the distance moved by the tip of a clock hand or a pendulum, the curved edge of a sector or segment, and the perimeter of any shaded region with curved boundaries.

The formula also works backwards. Given the arc length and the angle, you can find the radius; given the arc and the radius, you can find the angle.

Worked example

An arc of a circle is 22 cm long and subtends an angle of $60^{\circ }$ at the centre. Find the radius of the circle and the perimeter of the sector formed. (Take $\pi =\frac{22}{7}$)

Try it first, then show the solutionSolution

1. Arc length $l=\frac{\theta }{360^{\circ }}\times 2\pi r$.
2. $22=\frac{60^{\circ }}{360^{\circ }}\times 2\times \frac{22}{7}\times r=\frac{1}{6}\times \frac{44}{7}\times r=\frac{22r}{21}$.
3. So $r=21$ cm.
4. Perimeter of sector $=r+r+l=21+21+22=64$ cm.
5. Check: full circumference $=2\times \frac{22}{7}\times 21=132$ cm, and $\frac{1}{6}$ of 132 is 22 cm. Correct.

Answer: Radius $=21$ cm; perimeter of the sector $=64$ cm.

Area of a sector

NCERT section 11.1 Areas of Sector and Segment of a Circle

A sector is the region enclosed by two radii and the arc between them. If the central angle is $\theta$, its area is

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

Why it works: the same proportion idea. The full circle ($360^{\circ }$) has area $\pi r^2$, so each degree of central angle carries $\frac{\pi r^2}{360}$ of area.

Notice the difference from arc length: arc uses $2\pi r$ (a length), sector uses $\pi r^2$ (an area). Mixing them is one of the commonest board errors.

A useful way to work: first find the full area, then take the fraction. For $r=14$ cm, the full area is 616 cm${}^2$, so a $90^{\circ }$ sector is $\frac{1}{4}\times 616=154$ cm${}^2$. Board MCQs often reverse this: given a sector area, find the angle or the radius.

Worked example

The area of a sector of a circle of radius 14 cm is $\frac{308}{3}$ cm${}^2$. The central angle of the sector is (Take $\pi =\frac{22}{7}$)
(a) $30^{\circ }$
(b) $45^{\circ }$
(c) $60^{\circ }$
(d) $90^{\circ }$

Try it first, then show the solutionSolution

1. Area of full circle $=\frac{22}{7}\times 14\times 14=616$ cm${}^2$.
2. $\frac{\theta }{360^{\circ }}\times 616=\frac{308}{3}$.
3. $\frac{\theta }{360^{\circ }}=\frac{308}{3\times 616}=\frac{1}{6}$.
4. $\theta =\frac{360^{\circ }}{6}=60^{\circ }$.

Answer: (c) $60^{\circ }$

Sector area from arc length: $A=\frac{1}{2}lr$

NCERT section 11.1 Areas of Sector and Segment of a Circle

Sometimes the question gives the arc length (or the perimeter) but not the angle. Then use

$A=\frac{1}{2}\times l\times r$

Why it works: write $A=\frac{\theta }{360^{\circ }}\times \pi r^2=\frac{1}{2}r\times \left(\frac{\theta }{360^{\circ }}\times 2\pi r\right)$. The bracket is exactly $l$. So a sector behaves like a triangle with base $l$ and height $r$.

When to use it: when the angle is not a neat number, or when the perimeter of the sector is given. From perimeter $P=2r+l$ you get $l=P-2r$, then the area directly, with no need to find $\theta$.

It also gives a quick check: for any sector, $\frac{\text{area}}{\text{arc}}=\frac{r}{2}$.

Worked example

The perimeter of a sector of a circle of radius 7 cm is 30 cm. Find the area of the sector.

Try it first, then show the solutionSolution

1. Perimeter of sector $=2r+l$, so $30=14+l$.
2. Arc length $l=16$ cm.
3. Area $=\frac{1}{2}\times l\times r=\frac{1}{2}\times 16\times 7=56$ cm${}^2$.
4. Check by proportion: circumference $=44$ cm, area of circle $=154$ cm${}^2$; area of sector $=\frac{16}{44}\times 154=16\times 3.5=56$ cm${}^2$. Correct.

Answer: Area of the sector $=56$ cm${}^2$.

Minor and major sectors and arcs

NCERT section 11.1 Areas of Sector and Segment of a Circle

Two radii split a circle into two sectors. The smaller one (angle $\theta$ less than $180^{\circ }$) is the minor sector; the rest of the circle is the major sector, with angle $360^{\circ }-\theta$. In the same way the boundary splits into a minor arc and a major arc.

You can find major quantities in two ways:

• Use the angle $360^{\circ }-\theta$ directly in the formula.
• Or subtract: major sector $=\pi r^2-$ minor sector, and major arc $=2\pi r-$ minor arc.

Doing both is a quick check. Read the question carefully: if it says the region 'not swept' or 'remaining part' of a circle, it usually wants the major sector. If no word is given, 'sector' and 'arc' mean the minor ones.

Worked example

In a circle of radius 21 cm, a minor arc subtends an angle of $120^{\circ }$ at the centre. Find (i) the area of the major sector and (ii) the length of the major arc. (Take $\pi =\frac{22}{7}$)

Try it first, then show the solutionSolution

1. Area of circle $=\frac{22}{7}\times 21\times 21=1386$ cm${}^2$.
2. Angle of major sector $=360^{\circ }-120^{\circ }=240^{\circ }$.
3. Area of major sector $=\frac{240^{\circ }}{360^{\circ }}\times 1386=\frac{2}{3}\times 1386=924$ cm${}^2$.
4. Check: minor sector $=\frac{1}{3}\times 1386=462$ cm${}^2$, and $1386-462=924$ cm${}^2$.
5. Circumference $=2\times \frac{22}{7}\times 21=132$ cm.
6. Major arc $=\frac{240^{\circ }}{360^{\circ }}\times 132=88$ cm.

Answer: (i) Major sector $=924$ cm${}^2$; (ii) major arc $=88$ cm.

Area of a minor segment (central angle $60^{\circ }$, $90^{\circ }$, $120^{\circ }$)

NCERT section 11.1 Areas of Sector and Segment of a Circle

A segment is the region between a chord $AB$ and its arc. The sector $OAB$ is made of the triangle $OAB$ plus the segment, so

**Area of minor segment $=$ area of sector $OAB-$ area of $△OAB$.**

The syllabus restricts you to three angles, and each has a neat triangle:

• $60^{\circ }$: $OA=OB$ and the angle between them is $60^{\circ }$, so $△OAB$ is equilateral with area $\frac{\sqrt{3}}{4}{r}^2$.
• $90^{\circ }$: a right triangle with legs $r$ and $r$, area $\frac{1}{2}{r}^2$.
• $120^{\circ }$: draw $OM\perp AB$. It bisects $\angle AOB$, so each half is $60^{\circ }$. Then $OM=r\cos 60^{\circ }=\frac{r}{2}$ and $AM=r\sin 60^{\circ }=\frac{\sqrt{3}}{2}r$, so $AB=\sqrt{3}r$ and area $=\frac{\sqrt{3}}{4}{r}^2$.

In a 3-mark answer, show why the triangle area is what it is; examiners give a mark for it.

Worked example

A chord $AB$ of a circle with centre $O$ and radius 12 cm subtends an angle of $120^{\circ }$ at the centre. Find the area of the minor segment. (Use $\pi =3.14$ and $\sqrt{3}=1.73$)

Try it first, then show the solutionSolution

1. Area of sector $OAB=\frac{120^{\circ }}{360^{\circ }}\times 3.14\times 12\times 12=\frac{1}{3}\times 452.16=150.72$ cm${}^2$.
2. Draw $OM\perp AB$. Since $OA=OB$, $OM$ bisects $\angle AOB$ and $AB$, so $\angle AOM=60^{\circ }$.
3. $OM=12\cos 60^{\circ }=6$ cm and $AM=12\sin 60^{\circ }=6\sqrt{3}$ cm, so $AB=12\sqrt{3}$ cm.
4. Area of $△OAB=\frac{1}{2}\times 12\sqrt{3}\times 6=36\sqrt{3}=36\times 1.73=62.28$ cm${}^2$.
5. Area of minor segment $=150.72-62.28=88.44$ cm${}^2$.

Answer: Area of the minor segment $=88.44$ cm${}^2$.

Area of a major segment

NCERT section 11.1 Areas of Sector and Segment of a Circle

The chord splits the circle into a minor segment and a major segment. The cleanest route is

**Major segment $=$ area of circle $-$ minor segment.**

There is a second route that checks the first: the major segment is the major sector plus triangle $OAB$ (the triangle lies on the major side of the chord, because the centre $O$ is in the major segment). So

Major segment $=$ major sector $+$ area of $△OAB$.

A common slip is writing 'major sector minus triangle'. Picture it: the major sector stops at the two radii, and the triangle fills the gap between those radii and the chord. You add it, not subtract it.

Questions on this often appear as 'find the areas of both segments' in a 3-mark question.

Worked example

A chord of a circle of radius 14 cm subtends a right angle at the centre. Find the areas of the minor segment and the major segment. (Take $\pi =\frac{22}{7}$)

Try it first, then show the solutionSolution

1. Area of circle $=\frac{22}{7}\times 14\times 14=616$ cm${}^2$.
2. Area of minor sector $=\frac{90^{\circ }}{360^{\circ }}\times 616=154$ cm${}^2$.
3. Area of $△OAB=\frac{1}{2}\times 14\times 14=98$ cm${}^2$ (right angle at $O$).
4. Minor segment $=154-98=56$ cm${}^2$.
5. Major segment $=616-56=560$ cm${}^2$.
6. Check: major sector $=616-154=462$ cm${}^2$; $462+98=560$ cm${}^2$. Correct.

Answer: Minor segment $=56$ cm${}^2$; major segment $=560$ cm${}^2$.

Shaded regions built from sectors and segments

NCERT section 11.1 Areas of Sector and Segment of a Circle

Most long questions show a figure with a shaded region made of sectors, segments, squares, rectangles and triangles. The method is always the same:

• Name every piece: which sector, centre, radius and angle; which square or triangle.
• Write the shaded area as a sum or difference of pieces, in words first. For example, 'square minus four quadrants'.
• Find angles from the figure: a square gives $90^{\circ }$ at each corner, an equilateral triangle gives $60^{\circ }$.
• Then calculate.

A classic trick: when two quarter circles of radius $a$ are drawn inside a square of side $a$ from opposite corners, they overlap in a 'leaf'. The leaf is two $90^{\circ }$ segments back to back, or equally, two quadrants minus the square.

For the perimeter of a shaded region, add only the boundary pieces: arcs plus any straight sides that actually form the edge.

Worked example

$ABCD$ is a square of side 14 cm. With centre $A$ and radius 14 cm, a quarter circle is drawn inside the square from $B$ to $D$. With centre $C$ and radius 14 cm, another quarter circle is drawn inside the square from $B$ to $D$. The two arcs enclose a leaf-shaped region between them. Find the area and the perimeter of the leaf. (Take $\pi =\frac{22}{7}$)

Try it first, then show the solutionSolution

1. Each arc joins $B$ and $D$, so $BD$ is a common chord and the leaf is made of two segments on either side of $BD$.
2. Area of one quadrant $=\frac{1}{4}\times \frac{22}{7}\times 14\times 14=154$ cm${}^2$.
3. Area of right $△ABD=\frac{1}{2}\times 14\times 14=98$ cm${}^2$.
4. One $90^{\circ }$ segment $=154-98=56$ cm${}^2$.
5. Area of leaf $=2\times 56=112$ cm${}^2$.
6. Check: two quadrants $-$ square $=308-196=112$ cm${}^2$.
7. Each arc $=\frac{1}{4}\times 2\times \frac{22}{7}\times 14=22$ cm, so perimeter of leaf $=2\times 22=44$ cm.

Answer: Area of the leaf $=112$ cm${}^2$; perimeter $=44$ cm.

Word problems: wheels, clocks, wipers and sprinklers

NCERT section 11.1 Areas of Sector and Segment of a Circle

Real-life questions hide a circle, sector or arc inside a story. Translate first:

• Wheel: one revolution covers one circumference. Distance $=$ number of revolutions $\times 2\pi r$. Convert all units (km to cm, hours to minutes) before dividing.
• Clock: the minute hand turns $360^{\circ }$ in 60 minutes, so $6^{\circ }$ per minute. The hour hand turns $30^{\circ }$ per hour, so ${\frac{1}{2}}^{\circ }$ per minute. Area swept by a hand is a sector; distance moved by its tip is an arc.
• Wiper, sprinkler, lighthouse beam, grazing animal tied by a rope: each sweeps a sector whose radius is the blade length, spray range, beam range or rope length.

Write one line saying what shape it is, for example 'area swept is a sector of radius 21 cm and angle $120^{\circ }$'. That line earns a mark and stops you using the wrong formula.

Worked example

Each wheel of a car has a diameter of 70 cm. The car is travelling at a steady speed of 66 km per hour. How many complete revolutions does each wheel make in one minute? (Take $\pi =\frac{22}{7}$)

Try it first, then show the solutionSolution

1. Radius $=35$ cm, so circumference $=2\times \frac{22}{7}\times 35=220$ cm.
2. Distance in 1 hour $=66$ km $=66\times 1,00,000=66,00,000$ cm.
3. Distance in 1 minute $=\frac{66,00,000}{60}=1,10,000$ cm.
4. Revolutions per minute $=\frac{1,10,000}{220}=500$.
5. Check: $500\times 220=1,10,000$ cm $=1.1$ km per minute, and $1.1\times 60=66$ km per hour. Correct.

Answer: Each wheel makes 500 revolutions per minute.

Maths in real life: Areas Related to Circles

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Length of an Arc of a Circle
• Area of a Sector of a Circle
• Area of a Segment of a Circle
• Shaded Region Problems with Sectors
• Word Problems on Circles: Wheels, Clocks, Fields

Areas Related to Circles formula sheet

Circumference of a circle

$C=2\pi r=\pi d$

Boundary length. Use for wheels (one revolution = one circumference) and as the base for arc length.

Area of a circle

$A=\pi r^2$

Base for every sector and segment. Halve the diameter first.

Semicircle: perimeter and area

$P=\pi r+2r,A=\frac{1}{2}\pi r^2$

The perimeter includes the diameter. Leave out $2r$ only if the straight edge is not part of the boundary.

Quadrant (quarter circle) area

$A=\frac{1}{4}\pi r^2$

A $90^{\circ }$ sector. Appears at the corners of squares and rectangles.

Length of an arc

$l=\frac{\theta }{360^{\circ }}\times 2\pi r$

Distance moved by a clock tip or pendulum; curved edges in perimeter questions.

Area of a sector

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

Area swept by a hand, wiper, beam or grazing rope.

Sector area from arc length

$A=\frac{1}{2}lr$

Use when the arc or perimeter is given and the angle is not.

Perimeter of a sector

$P=2r+l$

Two radii plus the arc.

Major sector and major arc

$\text{Major sector}=\pi r^2-\text{minor sector},\text{major arc}=2\pi r-l$

Or use the angle $360^{\circ }-\theta$ directly.

Area of a minor segment

$\text{Segment}=\text{sector }OAB-\text{area of }△OAB$

Only for central angles $60^{\circ }$, $90^{\circ }$, $120^{\circ }$ in this syllabus.

Area of triangle OAB for the allowed angles

$60^{\circ }:\frac{\sqrt{3}}{4}{r}^2,90^{\circ }:\frac{1}{2}{r}^2,120^{\circ }:\frac{\sqrt{3}}{4}{r}^2$

$60^{\circ }$ gives an equilateral triangle; $120^{\circ }$ needs the perpendicular from $O$ with $\sin 60^{\circ }$ and $\cos 60^{\circ }$.

Segment shortcut, 60 degrees

$\left(\frac{\pi }{6}-\frac{\sqrt{3}}{4}\right)r^2$

Sector is $\frac{1}{6}$ of the circle, triangle is equilateral.

Segment shortcut, 90 degrees

$\left(\frac{\pi }{4}-\frac{1}{2}\right)r^2$

With $\pi =\frac{22}{7}$ this equals $\frac{2}{7}{r}^2$, a fast check.

Segment shortcut, 120 degrees

$\left(\frac{\pi }{3}-\frac{\sqrt{3}}{4}\right)r^2$

Same triangle area as the $60^{\circ }$ case, but a sector twice as big.

Chord length for the allowed angles

$60^{\circ }:AB=r,90^{\circ }:AB=r\sqrt{2},120^{\circ }:AB=r\sqrt{3}$

Needed for the perimeter of a segment.

Area of a major segment

$\text{Major segment}=\pi r^2-\text{minor segment}=\text{major sector}+\text{area of }△OAB$

Use one route, check with the other.

Perimeter of a segment

$P=l+AB$

Arc plus chord.

Distance covered by a wheel

$\text{Distance}=n\times 2\pi r$

$n$ = number of revolutions. Match units before dividing.

Angles turned by clock hands

$\text{Minute hand}:6^{\circ }\text{ per minute},\text{hour hand}:{\frac{1}{2}}^{\circ }\text{ per minute}$

So 20 minutes = $120^{\circ }$ for the minute hand; 1 hour = $30^{\circ }$ for the hour hand.

How the board asks Areas Related to Circles

The question patterns to expect, and the method that works for each.

Tricks to remember Areas Related to Circles

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Angle as a fraction

  Learn the fractions: $60^{\circ }=\frac{1}{6}$, $90^{\circ }=\frac{1}{4}$, $120^{\circ }=\frac{1}{3}$, $30^{\circ }=\frac{1}{12}$, $45^{\circ }=\frac{1}{8}$ of the circle.

  Use it when: Find the full area or circumference once, then just divide. Fewer steps, fewer slips.

• Segment = Sector − Triangle

  Say 'SST': Segment equals Sector minus Triangle. For the major segment: 'Circle minus minor'.

  Use it when: Stops you subtracting the wrong way or forgetting the triangle.

• 60 and 120 share a triangle

  For both $60^{\circ }$ and $120^{\circ }$, triangle $OAB$ has area $\frac{\sqrt{3}}{4}{r}^2$. Only the sector changes. Chords go $r$, $r\sqrt{2}$, $r\sqrt{3}$ for $60^{\circ }$, $90^{\circ }$, $120^{\circ }$.

  Use it when: Quick triangle area and chord length in segment and perimeter questions.

• Length has 2πr, area has πr²

  Arc is a length, so it uses the length formula $2\pi r$. Sector is an area, so it uses $\pi r^2$. Units follow: cm for arcs, cm${}^2$ for sectors.

  Use it when: Check your formula by checking your unit before you write the answer.

• The r/2 check

  For any sector, area divided by arc length equals $\frac{r}{2}$.

  Use it when: After finding both the arc and the area, divide them. If you do not get half the radius, one of them is wrong.

• Clock speed: 6 and half

  Minute hand: $6^{\circ }$ every minute. Hour hand: half a degree every minute.

  Use it when: Turns any time interval into a central angle in one line.

Areas Related to Circles: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Areas Related to Circles

The slips examiners see most often, and what to write instead.

• ✗Mistake: Using the diameter as the radius

  The question gives the diameter of a wheel or a semicircle and the student plugs it straight into the formula.

  Wrong working

  Wheel diameter 42 cm: circumference $=2\times \frac{22}{7}\times 42=264$ cm

  Right working

  $r=21$ cm, circumference $=2\times \frac{22}{7}\times 21=132$ cm

  ✓Do this: Underline 'diameter' in the question and write $r=\dots$ as your first line.

  Usually costs: Usually all calculation marks after that line; method marks may survive.

• ✗Mistake: Mixing the arc and sector formulas

  Both formulas start with $\frac{\theta }{360^{\circ }}$, so students attach the wrong ending.

  Wrong working

  Area of $60^{\circ }$ sector, $r=7$: $\frac{60}{360}\times 2\times \frac{22}{7}\times 7=\frac{22}{3}$ cm${}^2$

  Right working

  $\frac{60}{360}\times \frac{22}{7}\times 7\times 7=\frac{77}{3}$ cm${}^2$

  ✓Do this: Ask: am I finding a length or an area? Length uses $2\pi r$, area uses $\pi r^2$.

  Usually costs: 1 mark in an MCQ; 1 to 2 marks in a short answer.

• ✗Mistake: Forgetting the two radii in the perimeter of a sector

  Students think 'perimeter' of a curved shape means only the curved part.

  Wrong working

  Perimeter of $60^{\circ }$ sector, $r=21$ cm: $=22$ cm

  Right working

  Perimeter $=2r+l=42+22=64$ cm

  ✓Do this: Trace the boundary with your finger and list every edge: radius, arc, radius.

  Usually costs: ½ to 1 mark.

• ✗Mistake: Wrong triangle area in a $120^{\circ }$ segment

  Students reuse the right-triangle formula $\frac{1}{2}{r}^2$ for every angle.

  Wrong working

  $r=14$, $\theta =120^{\circ }$: area of $△OAB=\frac{1}{2}\times 14\times 14=98$ cm${}^2$

  Right working

  Draw $OM\perp AB$: $OM=7$, $AB=14\sqrt{3}$; area $=\frac{1}{2}\times 14\sqrt{3}\times 7=49\sqrt{3}$ cm${}^2$

  ✓Do this: Use $\frac{1}{2}{r}^2$ only for $90^{\circ }$. For $120^{\circ }$, draw the perpendicular and show the working.

  Usually costs: 1 to 2 marks in a 3-mark question.

• ✗Mistake: Subtracting the triangle for a major segment

  Students copy the minor-segment rule without picturing where the triangle lies.

  Wrong working

  $r=14$, $\theta =90^{\circ }$: major segment $=462-98=364$ cm${}^2$

  Right working

  Major segment $=616-56=560$ cm${}^2$ (or $462+98=560$ cm${}^2$)

  ✓Do this: Use 'circle minus minor segment', and check with 'major sector plus triangle'.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Treating minutes as degrees in clock questions

  Students read '20 minutes' and write $\theta =20^{\circ }$.

  Wrong working

  Minute hand 21 cm, 20 minutes: area $=\frac{20}{360}\times 1386=77$ cm${}^2$

  Right working

  $\theta =20\times 6^{\circ }=120^{\circ }$; area $=\frac{1}{3}\times 1386=462$ cm${}^2$

  ✓Do this: Always convert time to angle in a separate line before using any formula.

  Usually costs: 1 mark or more; the whole answer is wrong.

• ✗Mistake: Wrong or missing units

  Rushing at the end, or writing cm for an area.

  Wrong working

  Area of sector $=154$ cm

  Right working

  Area of sector $=154$ cm${}^2$

  ✓Do this: Lengths in cm or m, areas in cm${}^2$ or m${}^2$. Write the unit in the final line every time.

  Usually costs: Often ½ mark.

NCERT exercise guide: Areas Related to Circles

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.