MathsCBSE Class 10 · Areas Related to Circles

Length of an Arc of a Circle: Class 10 Formula and Questions

Updated 14 October 2026 · Fact-checked

The length of an arc is the part of a circle's circumference cut off by a central angle. It equals the angle divided by 360 degrees, times the full circumference. The board asks you to find arc lengths, perimeters of sectors and segments, a missing radius or angle, and distances moved by clock hands, pendulums and swinging gates.

The arc l is the fraction θ/360° of the whole circumference 2πr, so doubling the angle doubles the arc.

Illustration drawn for QuestionBuddy.

Understand Length of an Arc of a Circle

An arc is a piece of the boundary of a circle. Pick two points $A$ and $B$ on a circle with centre $O$ . The curved path from $A$ to $B$ is an arc, and $∠ A O B$ is the angle it subtends at the centre. The smaller path is the minor arc; the longer path the other way round is the major arc.

Why is there a simple formula? A full turn at the centre is $36 0^{\circ}$ and gives the whole circumference $2 π r$ . Double the angle and the arc doubles; halve it and the arc halves. So arc length is in direct proportion to the central angle, and an angle $θ$ gives the fraction $\frac{θ}{36 0 ^{\circ}}$ of the circumference: $l = \frac{θ}{36 0 ^{\circ}} \times 2 π r$ . A $9 0^{\circ}$ arc is a quarter of the circle, $6 0^{\circ}$ is a sixth, $12 0^{\circ}$ is a third, $18 0^{\circ}$ is a semicircle.

Arc length is also proportional to the radius when the angle is fixed. That is why the same $6 0^{\circ}$ turn gives a longer arc on a bigger circle. If two arcs have equal length, then $θ_{1} r_{1} = θ_{2} r_{2}$ , a fact behind many ratio questions.

The arc links to the rest of the chapter. The perimeter of a sector is $2 r + l$ . The perimeter of a segment is $l$ plus the chord. The perimeter of any shaded region with curved edges is a sum of arcs and straight sides. The sector area can be written as $\frac{1}{2} l r$ , so knowing the arc can give the area without the angle.

Conditions to watch: $θ$ must be in degrees here, $r$ must be the radius (halve a diameter), and all lengths must be in the same unit. The formula works for any angle from $0^{\circ}$ to $36 0^{\circ}$ , including major arcs, where you use $36 0^{\circ} - θ$ or subtract the minor arc from $2 π r$ . Take $π = \frac{22}{7}$ unless told otherwise; radii that are multiples of 7 give clean answers.

Length of an Arc of a Circle: formulas to remember

Length of an arc: $l = \frac{θ}{36 0 ^{\circ}} \times 2 π r$; $θ$ in degrees, $r$ is the radius. Find $2 π r$ first, then take the fraction.
Length of a major arc: $L = 2 π r - l = \frac{36 0 ^{\circ} - θ}{36 0 ^{\circ}} \times 2 π r$; Use either form; doing both is a quick check.
Finding the angle from the arc: $θ = \frac{l}{2 π r} \times 36 0^{\circ}$; Reverse use of the arc formula.
Finding the radius from the arc: $r = \frac{36 0 ^{\circ} \times l}{2 π θ}$; Given the arc length and the angle.
Perimeter of a sector: $P = 2 r + l$; Two radii plus the arc. Do not forget the radii.
Perimeter of a minor segment: $P = l + A B$; For $6 0^{\circ}$ : $A B = r$ ; $9 0^{\circ}$ : $A B = r 2$ ; $12 0^{\circ}$ : $A B = r 3$ .
Equal arcs in two circles: $θ_{1} r_{1} = θ_{2} r_{2}$; If arcs are equal, the radii are in the inverse ratio of the angles.
Semicircle arc: $l = π r$; Arc for $θ = 18 0^{\circ}$ . A semicircular region's perimeter is $π r + 2 r$ .
Angles turned by clock hands: $minute hand: 6^{\circ} per minute; hour hand: 0. 5^{\circ} per minute (3 0^{\circ} per hour)$; Turn the time into an angle first, then use the arc formula with the length of the hand as the radius.

How to solve Length of an Arc of a Circle questions

Every arc question, direct or reverse, is the same proportion. Use this order and you will not lose step marks.

1Identify the circle: write its centre and radius, halving a diameter if needed.
2Identify the central angle. In word problems convert time or position into an angle (minute hand $6^{\circ}$ per minute, hour hand $3 0^{\circ}$ per hour).
3Decide whether you need the minor arc ( $θ$ ) or the major arc ( $36 0^{\circ} - θ$ ).
4Write the formula $l = \frac{θ}{36 0 ^{\circ}} \times 2 π r$ before substituting; this line earns a mark.
5Compute the full circumference $2 π r$ first, then multiply by the fraction, letting the 7 cancel.
6For reverse problems, put the known values into the formula and solve the equation for $r$ or $θ$ .
7If asked for a perimeter, add the straight edges that bound the region: $2 r$ for a sector, the chord for a segment.
8Check: does the answer make sense against the full circumference, and does it carry the right unit?

Length of an Arc of a Circle 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.

Length of an arc

Imagine your school has a main gate made of two leaves, each 2.1 m wide, hinged at the two gateposts. Every morning the guard opens both leaves through $12 0^{\circ}$ and later closes them again. Each leaf sweeps a sector on the ground centred at its hinge, and the groundsman wants to paint the boundary of each swept sector so that nobody parks inside it.

(i) Find the total length of paint line needed for the boundaries of both swept sectors. (ii) Find the total distance moved by the outer corners of both leaves in one opening and one closing. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Each sector has radius $r = 2.1$ m (the leaf width) and central angle $12 0^{\circ}$ .
Full circumference $= 2 \times \frac{22}{7} \times 2.1 = 13.2$ m.
Arc for $12 0^{\circ}$ $= \frac{12 0 ^{\circ}}{36 0 ^{\circ}} \times 13.2 = \frac{1}{3} \times 13.2 = 4.4$ m.
Perimeter of one swept sector $= 2 r + l = 4.2 + 4.4 = 8.6$ m.
Paint for both sectors $= 2 \times 8.6 = 17.2$ m.
One corner in one opening and one closing moves $2 \times 4.4 = 8.8$ m.
Both corners: $2 \times 8.8 = 17.6$ m.

Answer: (i) 17.2 m of paint line; (ii) 17.6 m

Memory hook: The gate sweeps a sector: hinge is the centre, leaf width is the radius, paint is two radii plus the arc.

Length of an arc

Imagine a circular park with a walking track of radius 70 m around it, centred at a fountain O. Two friends start together from point P on the track and want to reach point Q on the track, where $∠ P O Q = 9 0^{\circ}$ . One friend walks along the track at 1.1 m per second; the other cuts straight across the grass along the chord $P Q$ at 0.94 m per second.

Who reaches $Q$ first, and by how many seconds? (Take $π = \frac{22}{7}$ and $2 = 1.41$ )

Try it first, then show the solutionSolution

Radius $r = 70$ m, central angle $θ = 9 0^{\circ}$ .
Full circumference $= 2 \times \frac{22}{7} \times 70 = 440$ m.
Arc $P Q = \frac{9 0 ^{\circ}}{36 0 ^{\circ}} \times 440 = 110$ m, so the track walker takes $\frac{110}{1.1} = 100$ s.
In $△ P O Q$ , $∠ O = 9 0^{\circ}$ , so $P Q = 7 0^{2} + 7 0^{2} = 702 = 70 \times 1.41 = 98.7$ m.
The friend crossing the grass takes $\frac{98.7}{0.94} = 105$ s.
Difference $= 105 - 100 = 5$ s.

Answer: The friend on the track reaches $Q$ first, by 5 seconds (100 s against 105 s).

Memory hook: Track or shortcut: the arc is longer than the chord, so speed decides who wins.

Tricks to remember Length of an Arc of a Circle

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

Fraction of the fence
Arc = the fraction $\frac{θ}{36 0 ^{\circ}}$ of the whole fence $2 π r$ . Find the whole fence first, then take the fraction.
Use it when: Stops you mixing up $2 π r$ (length) with $π r^{2}$ (area).
Easy fractions
$9 0^{\circ}$ is a quarter, $6 0^{\circ}$ a sixth, $12 0^{\circ}$ a third, $4 5^{\circ}$ an eighth, $3 6^{\circ}$ a tenth, $7 2^{\circ}$ a fifth.
Use it when: Turns most arc calculations into one division.
Two straight arms
A sector is a slice of pizza: crust (arc) plus two cut edges (radii). Perimeter $= l + 2 r$ .
Use it when: Perimeter of sector questions.
Same arc, swap the ratio
Equal arcs: bigger angle means smaller circle. $θ_{1} r_{1} = θ_{2} r_{2}$ .
Use it when: Ratio of radii or areas when arcs are equal.

Length of an Arc of a Circle: solved examples

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

Example 1: Perimeter of a quadrant-shaped sector · 1 mark · MCQBoard level

A sector of a circle of radius 14 cm has a central angle of $9 0^{\circ}$ . The perimeter of the sector is (Take $π = \frac{22}{7}$ )
(a) 22 cm
(b) 36 cm
(c) 50 cm
(d) 72 cm

Try it first, then show the solutionSolution

Circumference $= 2 \times \frac{22}{7} \times 14 = 88$ cm.
Arc $= \frac{9 0 ^{\circ}}{36 0 ^{\circ}} \times 88 = 22$ cm.
Perimeter $= 2 r + l = 28 + 22 = 50$ cm.

Answer: (c) 50 cm

Example 2: Path of a pendulum · 2 marks · Very short answerBoard level

The bob of a pendulum 21 cm long swings through an angle of $6 0^{\circ}$ from one end of its swing to the other. Find the length of the path traced by the bob in one swing. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

The bob moves along an arc of a circle of radius 21 cm with central angle $6 0^{\circ}$ . (½ mark)
Arc $= \frac{6 0 ^{\circ}}{36 0 ^{\circ}} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 132 = 22$ cm. (1½ marks)

Answer: 22 cm

Example 3: Equal arcs in two circles · 3 marks · Short answerBoard level

An arc of a circle $C_{1}$ subtends $6 0^{\circ}$ at its centre, and an arc of the same length in circle $C_{2}$ subtends $4 5^{\circ}$ at its centre. (i) Find the ratio of the radii of $C_{1}$ and $C_{2}$ . (ii) If the radius of $C_{1}$ is 21 cm, find the common arc length and verify it using $C_{2}$ . (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Equal arcs: $\frac{6 0 ^{\circ}}{36 0 ^{\circ}} \times 2 π r_{1} = \frac{4 5 ^{\circ}}{36 0 ^{\circ}} \times 2 π r_{2}$ , so $60 r_{1} = 45 r_{2}$ . (½ mark)
$r_{1} : r_{2} = 45 : 60 = 3 : 4$ . (½ mark)
Arc in $C_{1}$ : $\frac{1}{6} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 132 = 22$ cm. (1 mark)
$r_{2} = \frac{4}{3} \times 21 = 28$ cm. (½ mark)
Check in $C_{2}$ : $\frac{1}{8} \times 2 \times \frac{22}{7} \times 28 = \frac{1}{8} \times 176 = 22$ cm. Verified. (½ mark)

Answer: (i) $3 : 4$ ; (ii) arc length 22 cm, confirmed in $C_{2}$ with radius 28 cm.

Example 4: Perimeter of a 120 degree segment · 3 marks · Short answerHarder board level

A chord $A B$ of a circle with centre $O$ and radius 21 cm subtends $12 0^{\circ}$ at the centre. Find the perimeter of the minor segment cut off by $A B$ . (Take $π = \frac{22}{7}$ and $3 = 1.73$ )

Try it first, then show the solutionSolution

Minor arc $= \frac{12 0 ^{\circ}}{36 0 ^{\circ}} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{3} \times 132 = 44$ cm. (1 mark)
Draw $O M ⊥ A B$ ; it bisects $∠ A O B$ , so $∠ A O M = 6 0^{\circ}$ and $A M = A B /2$ . (½ mark)
$A M = 21 sin 6 0^{\circ} = \frac{21 3}{2}$ , so $A B = 213 = 21 \times 1.73 = 36.33$ cm. (1 mark)
Perimeter $= 44 + 36.33 = 80.33$ cm. (½ mark)

Answer: 80.33 cm

Example 5: Perimeter of a square with a quadrant cut off · 3 marks · Short answerHarder board level

$A B C D$ is a square of side 21 cm. A quadrant of a circle with centre $A$ and radius 14 cm is cut off from the square; its arc meets $A B$ at $P$ and $A D$ at $Q$ . (i) Find the perimeter of the remaining part of the square. (ii) Find the perimeter of the minor segment of this circle cut off by the chord $P Q$ . (Take $π = \frac{22}{7}$ and $2 = 1.41$ )

Try it first, then show the solutionSolution

The remaining part is bounded by $P B$ , $B C$ , $C D$ , $D Q$ and the arc $P Q$ . (½ mark)
Arc $P Q = \frac{9 0 ^{\circ}}{36 0 ^{\circ}} \times 2 \times \frac{22}{7} \times 14 = \frac{1}{4} \times 88 = 22$ cm. (½ mark)
$P B = D Q = 21 - 14 = 7$ cm, so the straight edges total $7 + 21 + 21 + 7 = 56$ cm. (½ mark)
Perimeter of the remaining part $= 56 + 22 = 78$ cm. (½ mark)
In $△ P A Q$ , $∠ A = 9 0^{\circ}$ , so by Pythagoras $P Q = 1 4^{2} + 1 4^{2} = 142 = 14 \times 1.41 = 19.74$ cm. (½ mark)
Perimeter of the minor segment $= 22 + 19.74 = 41.74$ cm. (½ mark)

Answer: (i) 78 cm; (ii) 41.74 cm

Example 6: From perimeter to angle, major arc and segment · 5 marks · Long answerHarder board level

The perimeter of a sector of a circle is 64 cm and the length of its arc is 22 cm. (i) Find the radius. (ii) Find the central angle. (iii) Find the length of the major arc. (iv) Find the perimeter of the minor segment formed by the chord joining the ends of the arc. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

$P = 2 r + l$ , so $64 = 2 r + 22$ . (½ mark)
$2 r = 42$ , $r = 21$ cm. (½ mark)
$22 = \frac{θ}{36 0 ^{\circ}} \times 2 \times \frac{22}{7} \times 21 = \frac{θ}{36 0 ^{\circ}} \times 132$ . (½ mark)
$θ = \frac{22}{132} \times 36 0^{\circ} = 6 0^{\circ}$ . (1 mark)
Major arc $= 132 - 22 = 110$ cm. (1 mark)
With $O A = O B = 21$ and $∠ A O B = 6 0^{\circ}$ , the base angles are equal and each is $6 0^{\circ}$ , so $△ O A B$ is equilateral and $A B = 21$ cm. (1 mark)
Perimeter of minor segment $= 22 + 21 = 43$ cm. (½ mark)

Answer: (i) 21 cm; (ii) $6 0^{\circ}$ ; (iii) 110 cm; (iv) 43 cm

Example 7: Case study: staggered start on a running track · 4 marks · Case-basedHOTS / competency

Suppose your school builds a running track whose two ends are semicircles. On each bend, the inner edge of lane 1 is a semicircle of radius 35 m, and the inner edge of lane 2 is a semicircle of radius 36.4 m, with the same centre. In a race that goes round both bends, the lane 2 runner is started further ahead so that both run the same distance.
(i) Find the length of one bend of lane 1. (1 mark)
(ii) Find the length of one bend of lane 2. (1 mark)
(iii) Find how far ahead the lane 2 runner must start, and explain why this distance depends only on the gap between the lanes, not on the radius. (2 marks)
(Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

(i) Semicircle arc $= π r = \frac{22}{7} \times 35 = 110$ m. (1 mark)
(ii) $π r = \frac{22}{7} \times 36.4 = 22 \times 5.2 = 114.4$ m. (1 mark)
(iii) Extra per bend $= 114.4 - 110 = 4.4$ m; two bends give $2 \times 4.4 = 8.8$ m. (1 mark)
Difference per bend $= π R - π r = π (R - r) = \frac{22}{7} \times 1.4 = 4.4$ m, which uses only $R - r$ , the lane gap. (1 mark)

Answer: (i) 110 m; (ii) 114.4 m; (iii) 8.8 m ahead, because the extra length per bend is $π (R - r)$ , depending only on the lane gap.

Where students lose marks in Length of an Arc of a Circle

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

Mistake: Using $π r^{2}$ instead of $2 π r$
The sector and arc formulas look alike and students mix them up.
Wrong working
Arc for $6 0^{\circ}$ , $r = 7$ : $\frac{1}{6} \times \frac{22}{7} \times 49 = 25.67$ cm
Right working
Arc $= \frac{1}{6} \times 2 \times \frac{22}{7} \times 7 = \frac{44}{6} = 7.33$ cm
Do this: Ask: am I measuring a length or an area? Length uses $2 π r$ .
Usually costs: All marks for the part.
Mistake: Leaving out the radii in the perimeter of a sector
Students give the curved edge only.
Wrong working
Perimeter of $9 0^{\circ}$ sector, $r = 14$ : 22 cm
Right working
$P = 2 r + l = 28 + 22 = 50$ cm
Do this: List every edge of the region before adding: two radii and the arc.
Usually costs: 1 mark in a 2- or 3-mark question.
Mistake: Using the diameter as the radius
Questions on wheels and pendulums often give a diameter.
Wrong working
Diameter 28 cm, $9 0^{\circ}$ : $\frac{1}{4} \times 2 \times \frac{22}{7} \times 28 = 44$ cm
Right working
$r = 14$ : $\frac{1}{4} \times 2 \times \frac{22}{7} \times 14 = 22$ cm
Do this: Write " $r = \dots$ " as your first line every time.
Usually costs: Usually 1 to 2 marks.
Mistake: Giving the hour hand $6^{\circ}$ per minute
The minute hand rate is remembered and applied to both hands.
Wrong working
Hour hand in 20 min: $20 \times 6^{\circ} = 12 0^{\circ}$
Right working
Hour hand: $\frac{1}{2}^{\circ}$ per minute, so $20 \times \frac{1}{2}^{\circ} = 1 0^{\circ}$
Do this: Minute hand: $36 0^{\circ}$ in 60 min. Hour hand: $36 0^{\circ}$ in 12 hours.
Usually costs: The whole angle and every step after it.

Exam tips

Write the formula $l = \frac{θ}{36 0 ^{\circ}} \times 2 π r$ before substituting; the method mark often sits on that line.
When a perimeter is asked, say in words which edges you are adding, for example "two radii and the arc".
In reverse questions, show the equation you are solving for $r$ or $θ$ ; do not jump to the answer.
For segment perimeters, justify the chord length (equilateral triangle for $6 0^{\circ}$ , perpendicular from the centre for $12 0^{\circ}$ ).
Check your arc is less than the full circumference; it catches the $π r^{2}$ slip instantly.

Practice questions on Length of an Arc of a Circle

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

Q1 · 5 marks · Long answerHarder board level
$O A B$ is a sector of a circle with centre $O$ and central angle $∠ A O B = 9 0^{\circ}$ . The perimeter of the sector is 50 cm. (i) Find the radius. (ii) Find the length of the major arc $A B$ . (iii) Find the perimeter of the minor segment cut off by the chord $A B$ . (iv) The square $O A C B$ is completed, with $O A$ and $O B$ as two of its sides and $C$ the vertex opposite $O$ . The region inside the square but outside the sector is shaded. Find the perimeter of the shaded region. (Take $π = \frac{22}{7}$ and $2 = 1.41$ )
Hint
Write $P = 2 r + \frac{1}{4} \times 2 π r$ and solve for $r$ . The chord of a $9 0^{\circ}$ angle is $r 2$ by Pythagoras.
Answer
(i) $2 r + \frac{1}{4} \times 2 \times \frac{22}{7} \times r = 2 r + \frac{11 r}{7} = \frac{25 r}{7} = 50$ , so $r = 14$ cm. (2 marks) (ii) Circumference $= 88$ cm, minor arc $= 22$ cm, major arc $= 88 - 22 = 66$ cm. (1 mark) (iii) $A B = 1 4^{2} + 1 4^{2} = 142 = 19.74$ cm, so the segment perimeter $= 22 + 19.74 = 41.74$ cm. (1 mark) (iv) The shaded region is bounded by $A C$ , $C B$ and the arc $A B$ , so its perimeter $= 14 + 14 + 22 = 50$ cm. (1 mark)
Q2 · 4 marks · Case-basedHOTS / competency
Suppose a park has a footpath in one corner shaped like a quarter of a ring. Its inner edge is an arc of radius 14 m and its outer edge is an arc of radius 21 m, both centred at the corner point $O$ of the park, and the path turns through $9 0^{\circ}$ . The two straight ends of the path lie along the two sides of the park that meet at $O$ .
(i) Find the length of the inner edge of the path. (1 mark)
(ii) Find the length of the outer edge of the path. (1 mark)
(iii) A metal edging strip is fixed along the entire boundary of the path, including both straight ends. Find the total length of strip needed and its cost at Rs 40 per metre. (2 marks)
(Take $π = \frac{22}{7}$ )
Hint
Both edges are $9 0^{\circ}$ arcs. Each straight end is the width of the path, $21 - 14$ .
Answer
(i) $\frac{1}{4} \times 2 \times \frac{22}{7} \times 14 = 22$ m. (ii) $\frac{1}{4} \times 2 \times \frac{22}{7} \times 21 = 33$ m. (iii) Each straight end $= 21 - 14 = 7$ m, so the boundary $= 22 + 33 + 7 + 7 = 69$ m; $69 \times 40 = 2760$ , so the cost is Rs 2760.
Q3 · 3 marks · Short answerBoard level
Two arcs of equal length subtend angles of $6 0^{\circ}$ and $7 5^{\circ}$ at the centres of two circles. Find the ratio of their radii and the ratio of their circumferences.
Hint
Equal arcs give $θ_{1} r_{1} = θ_{2} r_{2}$ ; circumference is proportional to radius.
Answer
$60 r_{1} = 75 r_{2}$ , so $r_{1} : r_{2} = 5 : 4$ . Circumferences $2 π r_{1} : 2 π r_{2} = 5 : 4$ .
Q4 · 3 marks · Short answerHarder board level
A wall clock has a minute hand 21 cm long and an hour hand 10.5 cm long. Find the distance travelled by the tip of each hand from 2:00 pm to 6:00 pm, and the ratio of these distances. (Take $π = \frac{22}{7}$ )
Hint
In 4 hours the minute hand makes 4 full turns; the hour hand turns $4 \times 3 0^{\circ}$ .
Answer
Minute tip: $4 \times 2 \times \frac{22}{7} \times 21 = 4 \times 132 = 528$ cm. Hour hand turns $12 0^{\circ}$ : $\frac{1}{3} \times 2 \times \frac{22}{7} \times 10.5 = \frac{1}{3} \times 66 = 22$ cm. Ratio $528 : 22 = 24 : 1$ .
Q5 · 3 marks · Short answerHOTS / competency
Riya writes: "Arc of a $12 0^{\circ}$ angle in a circle of radius 21 cm $= \frac{120}{360} \times \frac{22}{7} \times 2 1^{2} = 462$ cm." (i) Give a reason, without recalculating, why 462 cm cannot be correct. (ii) Find her error and the correct answer. (Take $π = \frac{22}{7}$ )
Hint
Compare her answer with the full circumference.
Answer
(i) The full circumference is $2 \times \frac{22}{7} \times 21 = 132$ cm, and a third of it cannot be 462 cm. (ii) She used the area $π r^{2}$ instead of $2 π r$ ; her 462 is a sector area in cm $^{2}$ . Correct arc $= \frac{1}{3} \times 132 = 44$ cm.
Q6 · 1 mark · Assertion-reasonBoard level
Assertion (A): An arc of a circle of radius $r$ that subtends $6 0^{\circ}$ at the centre has length $\frac{π r}{3}$ .
Reason (R): The length of an arc is proportional to the angle it subtends at the centre.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Use R to write the arc as a fraction of $2 π r$ , then check A.
Answer
(a). R is true: the arc is the fraction $\frac{θ}{36 0 ^{\circ}}$ of $2 π r$ . Using it, $\frac{6 0 ^{\circ}}{36 0 ^{\circ}} \times 2 π r = \frac{π r}{3}$ , so A is true and follows from R.

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.Formula for the length of an arc of angle $θ$ in a circle of radius $r$ ?Show answer
$l = \frac{θ}{36 0 ^{\circ}} \times 2 π r$
2.Arc of a $6 0^{\circ}$ angle in a circle of radius 21 cm?Show answer
$\frac{1}{6} \times 132 = 22$ cm
3.Perimeter of a sector?Show answer
$2 r + l$
4.Length of the major arc if the minor arc is $l$ ?Show answer
$2 π r - l$
5.Angle turned by the minute hand in 15 minutes?Show answer
$9 0^{\circ}$
6.Chord length for a $12 0^{\circ}$ central angle?Show answer
$r 3$

Length of an Arc of a Circle: frequently asked questions

What is the formula for the length of an arc in Class 10?

Arc length equals the central angle divided by 360 degrees, multiplied by the circumference 2πr. It works because arc length is in direct proportion to the angle at the centre. The angle must be in degrees.

What is the difference between arc length and sector area?

Arc length is a length, measured in cm or m, and uses 2πr. Sector area is an area, measured in square units, and uses πr squared. Both use the same fraction of angle over 360 degrees.

Do I need radians for arc length in Class 10?

No. The CBSE Class 10 syllabus uses degrees only, so always use the angle over 360 degrees form. Radian measure is not part of this chapter.

How do I find the radius if the arc length is given?

Put the arc length and the angle into the formula and solve for r. If the perimeter of a sector is given instead, subtract the arc from the perimeter and halve the result to get the radius.

Is the segment angle limit also applied to arc length?

The syllabus limits segment area questions to central angles of 60, 90 and 120 degrees. Arc length and sector questions can use other angles, but segment perimeters usually stick to these three because the chord has a neat length.