MathsCBSE Class 9 · Measuring Space: Perimeter and Area

Circumference, Arc, Sector and Segment of a Circle: Class 9

Updated 17 October 2026 · Fact-checked

For a circle of radius $r$: circumference $= 2\pi r$ and area $= \pi r^2$. For a sector of angle $\theta$ degrees: arc length $= \frac{\theta}{360} \times 2\pi r$, area $= \frac{\theta}{360} \times \pi r^2$ and perimeter $= \text{arc} + 2r$. The area of a segment is the area of the sector minus the area of the triangle formed by the two radii and the chord. A ring of radii $R$ and $r$ has area $\pi(R^2 - r^2)$.

Segment area = sector area minus triangle area: 154 − 98 = 56 cm², and the arc is a quarter of the circumference.

Illustration drawn for QuestionBuddy.

Understand Circumference, Arc Length, Sector and Segment

**Circumference and the number $π$ .** For every circle the ratio of the circumference $C$ to the diameter $d$ is the same number, called $π$ ( $\approx \frac{22}{7}$ or $3.14$ ). So $C = π d = 2 π r$ . If the circumference is $88$ cm, then $r = \frac{88}{2 π} = 14$ cm and the area is $π r^{2} = \frac{22}{7} \times 196 = 616$ cm $^{2}$ . A wheel of radius $21$ cm has circumference $132$ cm, so it turns $5000$ times to cover $6.6$ km.

Arc length and sector. A sector with central angle $θ$ is the fraction $\frac{θ}{360}$ of the whole circle, so
$ $arc length = \frac{θ}{360} \times 2 π r, sector area = \frac{θ}{360} \times π r^{2} .$ $
The perimeter of the sector adds the two radii: $arc + 2 r$ . For $r = 14$ cm and $θ = 9 0^{\circ}$ : arc $= \frac{1}{4} \times 88 = 22$ cm, area $= \frac{1}{4} \times 616 = 154$ cm $^{2}$ , perimeter $= 22 + 28 = 50$ cm. A minute hand of length $10.5$ cm sweeps $12 0^{\circ}$ in 20 minutes, so its tip travels $\frac{1}{3} \times 2 \times \frac{22}{7} \times 10.5 = 22$ cm and the area swept is $\frac{1}{3} \times \frac{22}{7} \times 110.25 = 115.5$ cm $^{2}$ .

Segment. A chord cuts the circle into a minor segment and a major segment. The area of the minor segment is the sector area minus the area of the triangle $O A B$ . With $r = 14$ and $θ = 9 0^{\circ}$ the triangle is right-angled with area $\frac{1}{2} \times 14 \times 14 = 98$ , so the segment is $154 - 98 = 56$ cm $^{2}$ . At this level the central angle is usually $6 0^{\circ}$ , $9 0^{\circ}$ or $12 0^{\circ}$ , so the triangle comes from simple geometry (an equilateral or right isosceles triangle, or $\frac{1}{2} \times base \times height$ ).

Ring (annulus). The region between two concentric circles of radii $R > r$ has area $π R^{2} - π r^{2} = π (R - r) (R + r)$ . A circular path of width $w$ around a circle of radius $r$ has area $π (R^{2} - r^{2})$ with $R = r + w$ .

Composite figures. A square of side $14$ cm with a quarter-circle of radius $7$ cm at each corner: the four quarter-circles together form one full circle of radius $7$ (area $154$ ), so the remaining area is $196 - 154 = 42$ cm $^{2}$ . Break figures into sectors, triangles and rectangles; add or subtract.

Why running tracks need a stagger. On a track with lanes of width $w$ , a runner in an outer lane runs a longer curve. For one full circle the outer lane is longer by $2 π w$ ; for one semicircular bend, by $π w$ . With $w = 1.4$ m the extra is $π \times 1.4 = 4.4$ m per bend. To make every runner cover the same distance, the outer lanes start further ahead; this is the stagger.

Method. Identify the fraction of the circle, write the whole-circle formula, multiply by the fraction, add radii for a perimeter, and subtract the triangle for a segment. Use $π = \frac{22}{7}$ when the radius contains 7 or 21; otherwise follow the question.

Circumference, Arc Length, Sector and Segment: formulas to remember

Circle: $C = 2 π r, A = π r^{2}$; $π \approx \frac{22}{7}$ or $3.14$ .
Arc length: $ℓ = \frac{θ}{360} \times 2 π r$; $θ$ in degrees.
Sector area: $A = \frac{θ}{360} \times π r^{2}$; Also $\frac{1}{2} ℓ r$ .
Perimeter of a sector: $P = ℓ + 2 r$; Arc plus two radii.
Segment: $A_{seg} = A_{sector} - A_{△ O A B}$; Minor segment.
Ring: $A = π (R^{2} - r^{2})$; Concentric circles.
Track stagger: $extra per bend = π w$; $w$ = lane width.

How to solve Circumference, Arc Length, Sector and Segment questions

Work with the fraction of the whole circle that the question describes.

1Find the radius (halve the diameter; derive $r$ from a given circumference or area).
2Write the fraction $\frac{θ}{360}$ for an arc or sector.
3Multiply the whole-circle formula by that fraction.
4For the perimeter of a sector add $2 r$ ; for a segment subtract the triangle area.
5For composite shapes, split into circles, sectors, triangles and rectangles; add or subtract areas.
6Give units: length in cm or m, area in square units.

Circumference, Arc Length, Sector and Segment in real life

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.

Sector

Suppose a circular rangoli of radius 21 cm has one sector of angle 60 degrees coloured red.

Find the length of the curved edge of the red part and its area. Use pi = 22/7.

Try it first, then show the solutionSolution

Arc length $= \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 132 = 22$ cm.
Area $= \frac{60}{360} \times \frac{22}{7} \times 441 = \frac{1}{6} \times 1386 = 231$ cm $^{2}$ .

Answer: The curved edge is 22 cm long and the red region has area 231 cm $^{2}$ .

Memory hook: Fraction of the circle times the whole.

Tricks to remember Circumference, Arc Length, Sector and Segment

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

Fraction first
Write $\frac{θ}{360}$ first; both arc and area use it.
Use it when: Arc and sector.
Sector minus triangle
A segment is a sector with the triangle cut away.
Use it when: Segment area.
Ring is outer minus inner
Subtract the inner circle from the outer.
Use it when: Paths and rings.

Circumference, Arc Length, Sector and Segment: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Wheel revolutions · 3 marks · Short answerBoard level

A wheel has a radius of 21 cm. How many complete revolutions does it make in covering 6.6 km? Use $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Circumference $= 2 π r = 2 \times \frac{22}{7} \times 21 = 132$ cm. (1)
Distance $= 6.6$ km $= 660000$ cm. (1)
Revolutions $= \frac{660000}{132} = 5000$ . (1)

Answer: The wheel makes 5000 revolutions.

Example 2: Circumference to area · 3 marks · Short answerBoard level

The circumference of a circle is 88 cm. Find its radius and area. Use $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

$2 π r = 88$ , so $r = \frac{88 \times 7}{2 \times 22} = 14$ cm. (1½)
Area $= π r^{2} = \frac{22}{7} \times 14 \times 14 = 616$ cm $^{2}$ . (1½)

Answer: $r = 14$ cm and area $= 616$ cm $^{2}$ .

Example 3: Sector, arc and segment · 5 marks · Long answerHarder board level

A chord $A B$ of a circle of radius 14 cm subtends a right angle at the centre $O$ . Find (i) the length of the minor arc, (ii) the perimeter of the sector, (iii) the area of the sector, (iv) the area of the minor segment. Use $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

(i) Arc $= \frac{90}{360} \times 2 \times \frac{22}{7} \times 14 = \frac{1}{4} \times 88 = 22$ cm. (1)
(ii) Perimeter $= 22 + 2 \times 14 = 50$ cm. (1)
(iii) Sector area $= \frac{1}{4} \times \frac{22}{7} \times 196 = 154$ cm $^{2}$ . (1)
(iv) $△ O A B$ is right-angled at $O$ : area $= \frac{1}{2} \times 14 \times 14 = 98$ cm $^{2}$ . (1)
Segment $= 154 - 98 = 56$ cm $^{2}$ . (1)

Answer: (i) 22 cm; (ii) 50 cm; (iii) 154 cm $^{2}$ ; (iv) 56 cm $^{2}$ .

Example 4: Square with quarter-circles · 3 marks · Short answerHarder board level

A square of side 14 cm has a quarter-circle of radius 7 cm cut from each of its four corners. Find the area of the remaining plate. Use $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Area of square $= 14 \times 14 = 196$ cm $^{2}$ . (½)
The four quarter-circles together make one full circle of radius 7 cm: $π r^{2} = \frac{22}{7} \times 49 = 154$ cm $^{2}$ . (1½)
Remaining area $= 196 - 154 = 42$ cm $^{2}$ . (1)

Answer: The remaining area is $42$ cm $^{2}$ .

Example 5: Lane stagger on a track · 4 marks · Case-basedHOTS / competency

Case: the straight parts of a running track are joined by semicircular bends, and the lanes are 1.4 m wide. (i) By how much is the outer edge of lane 2 longer than lane 1 for a full circle of bend? (ii) In a 200 m race with one bend, how much further does a runner in lane 2 go along the curve than one in lane 1? (iii) How much further does a runner in lane 3 go than lane 1 on one bend? (iv) Why are the starting positions staggered? Use $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

(i) Radii differ by $w = 1.4$ m, so full circles differ by $2 π w = 2 \times \frac{22}{7} \times 1.4 = 8.8$ m. (1)
(ii) One semicircular bend: $π w = 4.4$ m. (1)
(iii) Lane 3 has radius larger by $2 w = 2.8$ m, so the difference over one bend is $π \times 2.8 = 8.8$ m. (1)
(iv) Without a stagger, the outer-lane runners would run a longer distance; starting them further ahead makes all runners cover the same distance. (1)

Answer: (i) 8.8 m; (ii) 4.4 m; (iii) 8.8 m; (iv) the stagger equalises the distance.

Where students lose marks in Circumference, Arc Length, Sector and Segment

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

Mistake: Using the diameter as the radius
Not halving the given diameter.
Wrong working
Diameter 70 cm: area $= π (70)^{2}$ .
Right working
$r = 35$ cm.
Do this: Underline "diameter" and halve it.
Usually costs: The whole answer.
Mistake: Forgetting the radii in the perimeter of a sector
Giving only the arc.
Wrong working
Perimeter $= 22$ cm.
Right working
Perimeter $= 22 + 28 = 50$ cm.
Do this: Perimeter of a sector = arc + $2 r$ .
Usually costs: 1 mark.
Mistake: Not subtracting the triangle for a segment
Giving the sector area as the segment area.
Wrong working
Segment $= 154$ cm $^{2}$ .
Right working
Segment $= 154 - 98 = 56$ cm $^{2}$ .
Do this: Draw the triangle $O A B$ and subtract.
Usually costs: 2 marks.

Exam tips

Convert units first (km to cm, minutes to hours).
Write the fraction of the circle on its own line.
For segments, draw the triangle $O A B$ and label its sides.
Check whether the question says "area" or "perimeter"; they need different formulas.

Practice questions on Circumference, Arc Length, Sector and Segment

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The area of a circle is $154$ cm $^{2}$ . Its circumference is (use $π = \frac{22}{7}$ )
(a) 22 cm
(b) 44 cm
(c) 88 cm
(d) 14 cm
Hint
Find $r$ from $π r^{2} = 154$ .
Answer
(b) 44 cm. $r^{2} = 49$ , $r = 7$ , so $C = 2 \times \frac{22}{7} \times 7 = 44$ .
Q2 · 3 marks · Short answerBoard level
A sector of a circle of radius 12 cm has a central angle of $7 5^{\circ}$ . Find the length of its arc and its area. Use $π = 3.14$ .
Hint
Use the fraction $\frac{75}{360} = \frac{5}{24}$ .
Answer
Arc $= \frac{5}{24} \times 2 \times 3.14 \times 12 = \frac{5}{24} \times 75.36 = 15.7$ cm. Area $= \frac{5}{24} \times 3.14 \times 144 = \frac{5}{24} \times 452.16 = 94.2$ cm $^{2}$ .
Q3 · 3 marks · Short answerHarder board level
The minute hand of a clock is 10.5 cm long. Find the distance moved by its tip and the area swept by the hand in 20 minutes. Use $π = \frac{22}{7}$ .
Hint
In 20 minutes the hand turns through $\frac{20}{60} \times 36 0^{\circ}$ .
Answer
The angle is $12 0^{\circ}$ , a third of the circle. Distance $= \frac{1}{3} \times 2 \times \frac{22}{7} \times 10.5 = 22$ cm. Area $= \frac{1}{3} \times \frac{22}{7} \times 110.25 = 115.5$ cm $^{2}$ .
Q4 · 5 marks · Long answerHarder board level
A circular lawn of radius 21 m is surrounded by a path 3.5 m wide. Find the area of the path and the cost of gravelling it at 40 rupees per square metre. Use $π = \frac{22}{7}$ .
Hint
Outer radius $= 21 + 3.5$ ; subtract the inner circle.
Answer
$R = 24.5$ m, $r = 21$ m. $R^{2} = 600.25$ , $r^{2} = 441$ , difference $159.25$ . Area of path $= \frac{22}{7} \times 159.25 = 500.5$ m $^{2}$ . Cost $= 500.5 \times 40 = 20020$ rupees.
Q5 · 4 marks · Case-basedHOTS / competency
Case: a car wheel has a diameter of 70 cm. (i) Find its circumference. (ii) The car runs at 66 km per hour; find the number of revolutions per minute. (iii) How many revolutions in one hour? (iv) If the wheel were replaced by one of diameter 56 cm at the same speed, find the revolutions per minute. Use $π = \frac{22}{7}$ .
Hint
Convert speed to metres per minute.
Answer
(i) $\frac{22}{7} \times 70 = 220$ cm $= 2.2$ m. (ii) $66$ km/h $= 1100$ m/min, so $\frac{1100}{2.2} = 500$ revolutions per minute. (iii) $500 \times 60 = 30000$ . (iv) Circumference $= \frac{22}{7} \times 56 = 176$ cm $= 1.76$ m, so $\frac{1100}{1.76} = 625$ revolutions per minute.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Circumference of a circle?Show answer
$2 π r$ .
2.Arc length for angle $θ$ ?Show answer
$\frac{θ}{360} \times 2 π r$ .
3.Sector area for angle $θ$ ?Show answer
$\frac{θ}{360} \times π r^{2}$ .
4.Area of a minor segment?Show answer
Sector area minus triangle area.
5.Area of a ring?Show answer
$π (R^{2} - r^{2})$ .

Circumference, Arc Length, Sector and Segment: frequently asked questions

Which value of pi should I use?

Use \frac{22}{7} when the radius is a multiple of 7, otherwise use what the question gives; 3.14 is common.

Is the area of a sector the same as the area of a segment?

No. A sector is bounded by two radii and an arc; a segment is bounded by a chord and an arc.

Why does the outer lane need a head start?

Its curve is longer, so without a head start the runner would cover a different total distance.