MathsCBSE Class 10 · Areas Related to Circles

Area of a Segment of a Circle for Class 10

Updated 13 October 2026 · Fact-checked

A segment is the region between a chord and its arc. Its area equals the area of the sector minus the area of the triangle formed by the two radii and the chord. The CBSE board asks this only for central angles of 60, 90 and 120 degrees, in short answers, shaded regions and case studies.

Minor segment = sector OAB − triangle OAB: cut the triangle out of the sector and only the curved cap beside the chord remains.

Illustration drawn for QuestionBuddy.

Understand Area of a Segment of a Circle

A chord $A B$ cuts a circle into two pieces. Each piece is a segment: the smaller one is the minor segment, the larger one is the major segment. A segment is bounded by the chord and an arc, not by radii.

The trick is to see the segment inside a sector you already know. Join the centre $O$ to $A$ and $B$ . The sector $O A B$ is now split into two parts: the triangle $O A B$ and the minor segment. So **minor segment $=$ sector $O A B -$ area of $△ O A B$ **. The sector formula $\frac{θ}{36 0 ^{\circ}} \times π r^{2}$ comes from the sector topic; the only new work is the triangle.

The syllabus restricts you to central angles of $6 0^{\circ}$ , $9 0^{\circ}$ and $12 0^{\circ}$ , because for these the triangle is easy. At $6 0^{\circ}$ , $O A = O B$ and the base angles are each $6 0^{\circ}$ , so $△ O A B$ is equilateral with area $\frac{3}{4} r^{2}$ and $A B = r$ . At $9 0^{\circ}$ it is a right isosceles triangle with area $\frac{1}{2} r^{2}$ and $A B = r 2$ .

At $12 0^{\circ}$ , draw $O M ⊥ A B$ . The perpendicular from the centre bisects the chord and, since $△ O A B$ is isosceles, also bisects $∠ A O B$ into two $6 0^{\circ}$ angles. Then $O M = r cos 6 0^{\circ} = \frac{r}{2}$ and $A M = r sin 6 0^{\circ} = \frac{3}{2} r$ , so $A B = 3 r$ and area $= \frac{1}{2} \times 3 r \times \frac{r}{2} = \frac{3}{4} r^{2}$ . Surprisingly, this is the same area as at $6 0^{\circ}$ .

The major segment is everything else: $π r^{2} -$ minor segment. A check: major segment $=$ major sector $+ △ O A B$ , because the centre lies on the major side. The perimeter of a segment is its arc plus its chord.

Segments appear inside larger questions: the leaf in a square, the regions between a triangle and its circumcircle, flower beds cut off by a path. Master the three triangle areas and these become routine.

Area of a Segment of a Circle: formulas to remember

Area of minor segment: $Segment = \frac{θ}{36 0 ^{\circ}} π r^{2} - ar (△ O A B)$; Only $θ = 6 0^{\circ}, 9 0^{\circ}, 12 0^{\circ}$ in the 2026-27 syllabus.
Triangle OAB, 60 degrees: $ar (△ O A B) = \frac{3}{4} r^{2}, A B = r$; Equilateral triangle.
Triangle OAB, 90 degrees: $ar (△ O A B) = \frac{1}{2} r^{2}, A B = r 2$; Right isosceles triangle.
Triangle OAB, 120 degrees: $ar (△ O A B) = \frac{1}{2} \cdot 3 r \cdot \frac{r}{2} = \frac{3}{4} r^{2}, A B = 3 r$; Drop $O M ⊥ A B$ ; use $sin 6 0^{\circ}$ and $cos 6 0^{\circ}$ .
Segment shortcut, 90 degrees: $(\frac{π}{4} - \frac{1}{2}) r^{2} = \frac{2}{7} r^{2} (π = \frac{22}{7})$; A fast check for right-angle chords.
Major segment: $π r^{2} - minor segment = major sector + ar (△ O A B)$; Use one, check with the other.
Perimeter of a segment: $P = \frac{θ}{36 0 ^{\circ}} \times 2 π r + A B$; Arc plus chord.

How to solve Area of a Segment of a Circle questions

Every segment question is the same three-piece calculation. Write it in words first, then fill in numbers.

1Identify the centre, radius $r$ and central angle $θ$ ; check it is $6 0^{\circ}$ , $9 0^{\circ}$ or $12 0^{\circ}$ (find it from the figure if not given, e.g. a chord equal to the radius means $6 0^{\circ}$ ).
2Write the plan: minor segment $=$ sector $O A B -$ $△ O A B$ .
3Find the sector area $\frac{θ}{36 0 ^{\circ}} π r^{2}$ , ideally as a fraction of the full circle area.
4Find the triangle area with a reason: equilateral for $6 0^{\circ}$ , right-angled for $9 0^{\circ}$ , perpendicular from $O$ for $12 0^{\circ}$ .
5Subtract to get the minor segment; keep $3$ until the end unless a decimal value is given.
6If asked, major segment $= π r^{2} -$ minor segment, and perimeter $=$ arc $+$ chord.
7Write the answer with square units (or units for perimeter) and check with a second route if time allows.

Area of a Segment 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.

Area of minor segment

Suppose you are painting a rangoli-style design on a circular steel plate (thali) of radius 14 cm for a Diwali stall. You draw a straight line across the plate as a chord whose ends make a right angle at the centre. The small region between the line and the rim is to be painted gold.

Find the area to be painted gold and the length of gold border paint needed if you paint along both the chord and the arc of that region. (Take $π = \frac{22}{7}$ , $2 = 1.41$ )

Try it first, then show the solutionSolution

Region is a minor segment with $r = 14$ cm, $θ = 9 0^{\circ}$ .
Sector $= \frac{1}{4} \times \frac{22}{7} \times 196 = 154$ cm $^{2}$ .
$△ O A B = \frac{1}{2} \times 14 \times 14 = 98$ cm $^{2}$ .
Segment $= 154 - 98 = 56$ cm $^{2}$ .
Arc $= \frac{1}{4} \times 2 \times \frac{22}{7} \times 14 = 22$ cm.
Chord $= 142 = 14 \times 1.41 = 19.74$ cm.
Border $= 22 + 19.74 = 41.74$ cm.

Answer: Gold area 56 cm², border length 41.74 cm

Memory hook: The thali line: slice of pizza minus the triangle leaves the crust strip.

Major segment

Imagine a circular water tank lid of radius 21 cm. A straight hinge is fixed along a chord $A B$ that subtends $6 0^{\circ}$ at the centre $O$ , and the small piece beyond the hinge flips open while the larger part stays fixed. A rubber edging is to be fitted all round the flap (along its curved rim and along the hinge) at a cost of Rs 4 per cm.

Find (i) the area of the flap, (ii) the area of the fixed part, and (iii) the length of edging needed and its cost. (Take $π = \frac{22}{7}$ , $3 = 1.73$ )

Try it first, then show the solutionSolution

$O A = O B = 21$ cm and $∠ A O B = 6 0^{\circ}$ , so the base angles are $6 0^{\circ}$ each and $△ O A B$ is equilateral: hinge $A B = 21$ cm.
Circle $= \frac{22}{7} \times 441 = 1386$ cm $^{2}$ .
Sector $O A B = \frac{1}{6} \times 1386 = 231$ cm $^{2}$ .
$△ O A B = \frac{1.73}{4} \times 441 = 190.7325$ cm $^{2}$ .
Flap (minor segment) $= 231 - 190.7325 = 40.2675 \approx 40.27$ cm $^{2}$ .
Fixed part (major segment) $= 1386 - 40.2675 = 1345.7325 \approx 1345.73$ cm $^{2}$ .
Arc $A B = \frac{1}{6} \times 2 \times \frac{22}{7} \times 21 = 22$ cm.
Edging $=$ arc $+$ hinge $= 22 + 21 = 43$ cm.
Cost $= 43 \times 4 = 172$ , so Rs 172.

Answer: Flap ≈ 40.27 cm², fixed part ≈ 1345.73 cm², edging 43 cm costing Rs 172

Memory hook: The tank lid: the flap is the minor segment, its edge is arc plus hinge, and the rest of the lid is the major segment.

Tricks to remember Area of a Segment of a Circle

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

Slice minus triangle
Segment = pizza slice minus the triangle you get by cutting straight across the crust ends.
Use it when: Writing the first line of any segment answer.
60 and 120 are twins
Same triangle area $\frac{3}{4} r^{2}$ for both; only the sector doubles.
Use it when: Saves recomputing the 120° triangle and catches errors.
Chord ladder r, r√2, r√3
Angles 60°, 90°, 120° give chords $r$ , $r 2$ , $r 3$ : root goes 1, 2, 3.
Use it when: Perimeter of a segment.
Two-sevenths rule
With $π = \frac{22}{7}$ , a 90° segment is $\frac{2}{7} r^{2}$ .
Use it when: Instant check for right-angle segments.

Area of a Segment of a Circle: solved examples

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

Example 1: Segment for a right-angle chord · 2 marks · Very short answerBoard level

A chord of a circle of radius 21 cm subtends a right angle at the centre. Find the area of the minor segment. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Sector $= \frac{90}{360} \times \frac{22}{7} \times 21 \times 21 = 346.5$ cm $^{2}$ . (½)
$△ O A B$ is right-angled at $O$ : area $= \frac{1}{2} \times 21 \times 21 = 220.5$ cm $^{2}$ . (½)
Segment $=$ sector $-$ triangle $= 346.5 - 220.5$ (½)
$= 126$ cm $^{2}$ . Check: $\frac{2}{7} \times 441 = 126$ . (½)

Answer: 126 cm²

Example 2: Both segments for a 120° chord · 3 marks · Short answerBoard level

A chord $A B$ of a circle with centre $O$ and radius 14 cm subtends $12 0^{\circ}$ at the centre. Find the areas of the minor and major segments. (Take $π = \frac{22}{7}$ , $3 = 1.73$ )

Try it first, then show the solutionSolution

Sector $= \frac{1}{3} \times \frac{22}{7} \times 196 = \frac{616}{3} = 205.33$ cm $^{2}$ . (½)
Draw $O M ⊥ A B$ ; it bisects $∠ A O B$ , so $∠ A O M = 6 0^{\circ}$ . (½)
$O M = 14 cos 6 0^{\circ} = 7$ , $A M = 14 sin 6 0^{\circ} = 73$ , so $A B = 143$ . (½)
$△ O A B = \frac{1}{2} \times 143 \times 7 = 493 = 84.77$ cm $^{2}$ . (½)
Minor segment $= 205.33 - 84.77 = 120.56$ cm $^{2}$ . (½)
Major segment $= 616 - 120.56 = 495.44$ cm $^{2}$ . (½)

Answer: Minor ≈ 120.56 cm², major ≈ 495.44 cm²

Example 3: Find the angle first from the chord · 3 marks · Short answerBoard level

In a circle of radius 6 cm, a chord $A B$ has length 6 cm. Find the area and the perimeter of the minor segment. (Use $π = 3.14$ , $3 = 1.73$ )

Try it first, then show the solutionSolution

$O A = O B = A B = 6$ cm, so $△ O A B$ is equilateral and $∠ A O B = 6 0^{\circ}$ . (½)
Sector $= \frac{1}{6} \times 3.14 \times 36 = 18.84$ cm $^{2}$ . (½)
$△ O A B = \frac{1.73}{4} \times 36 = 15.57$ cm $^{2}$ . (½)
Segment $= 18.84 - 15.57 = 3.27$ cm $^{2}$ . (½)
Arc $= \frac{1}{6} \times 2 \times 3.14 \times 6 = 6.28$ cm. (½)
Perimeter $= 6.28 + 6 = 12.28$ cm. (½)

Answer: Area 3.27 cm², perimeter 12.28 cm

Example 4: Square inscribed in a circle (5 marks) · 5 marks · Long answerHarder board level

A square $A B C D$ is inscribed in a circle with centre $O$ , so all four vertices lie on the circle. The diagonal of the square is 28 cm. The four regions inside the circle but outside the square are shaded. Find (i) the radius and the angle each side subtends at $O$ , (ii) the area of one shaded region using the segment method, (iii) the total shaded area, and verify it by a second method. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

The diagonal is a diameter, so $r = 14$ cm. (½)
Diagonals of a square are equal and bisect at right angles, so each side subtends $9 0^{\circ}$ at $O$ . (1)
Sector $= \frac{1}{4} \times \frac{22}{7} \times 196 = 154$ cm $^{2}$ . (½)
$△ O A B = \frac{1}{2} \times 14 \times 14 = 98$ cm $^{2}$ . (½)
One segment $= 154 - 98 = 56$ cm $^{2}$ . (½)
Total shaded $= 4 \times 56 = 224$ cm $^{2}$ . (1)
Check: circle $= 616$ ; square $= \frac{1}{2} \times 2 8^{2} = 392$ ; $616 - 392 = 224$ cm $^{2}$ . (1)

Answer: r = 14 cm, 90°; one region 56 cm²; total 224 cm²

Example 5: Case study: circular stage with a ramp line · 4 marks · Case-basedHOTS / competency

Imagine your school sets up a circular stage of radius 7 m with centre $O$ for the annual day. A straight carpet edge $A B$ is laid as a chord with $∠ A O B = 12 0^{\circ}$ ; the smaller region between the carpet edge and the rim is for the orchestra. A second chord $C D$ elsewhere has $∠ C O D = 6 0^{\circ}$ , and the minor segment it cuts off is for lights. (Take $π = \frac{22}{7}$ , $3 = 1.73$ )
(i) Find the length of $A B$ . (1)
(ii) Find the area of triangle $C O D$ . (1)
(iii) Find the total area of the orchestra and lights regions. (2)

Try it first, then show the solutionSolution

(i) $A B = 3 \times 7 = 12.11$ m. (1)
(ii) $△ C O D$ is equilateral: $\frac{1.73}{4} \times 49 = 21.1925$ m $^{2}$ . (1)
(iii) Circle $= 154$ m $^{2}$ . Orchestra $= \frac{154}{3} - 21.1925 = 51.3333 - 21.1925 = 30.1408$ m $^{2}$ . (½)
Lights $= \frac{154}{6} - 21.1925 = 25.6667 - 21.1925 = 4.4742$ m $^{2}$ . (½)
Total $= 30.1408 + 4.4742 = 34.615$ m $^{2}$ . (½)
Shortcut check: total $= \frac{154}{2} - 2 \times 21.1925 = 77 - 42.385 = 34.615$ m $^{2}$ . (½)

Answer: (i) 12.11 m (ii) 21.19 m² (iii) about 34.62 m²

Example 6: Equilateral triangle on a chord: area and perimeter (5 marks) · 5 marks · Long answerHarder board level

In a circle with centre $O$ and radius 14 cm, a chord $A B$ subtends $6 0^{\circ}$ at $O$ . An equilateral triangle $A B P$ is drawn on $A B$ on the side of $A B$ away from $O$ , so that $P$ lies outside the circle. The region inside $△ A B P$ but outside the circle is shaded; it is bounded by $A P$ , $P B$ and the arc $A B$ . Find (i) the length of $A B$ , (ii) the area of the minor segment on $A B$ , (iii) the area of the shaded region, and (iv) the perimeter of the shaded region. (Take $π = \frac{22}{7}$ , $3 = 1.73$ )

Try it first, then show the solutionSolution

(i) $O A = O B = 14$ cm and $∠ A O B = 6 0^{\circ}$ , so the base angles are $6 0^{\circ}$ each: $△ O A B$ is equilateral and $A B = 14$ cm. (½)
(ii) Sector $O A B = \frac{1}{6} \times \frac{22}{7} \times 196 = \frac{308}{3}$ cm $^{2}$ . (½)
$ar (△ O A B) = \frac{3}{4} \times 196 = 493$ cm $^{2}$ . (½)
Minor segment $= \frac{308}{3} - 493 = 102.67 - 84.77 = 17.90$ cm $^{2}$ . (½)
(iii) $△ A B P$ is equilateral with side 14 cm: area $= \frac{3}{4} \times 196 = 493$ cm $^{2}$ . (½)
The segment lies inside $△ A B P$ , so shaded $= ar (△ A B P) -$ segment $= 493 - \frac{308}{3} + 493 = 983 - \frac{308}{3}$ . (½)
$= 169.54 - 102.67 = 66.87$ cm $^{2}$ . (½)
(iv) Arc $A B = \frac{1}{6} \times 2 \times \frac{22}{7} \times 14 = \frac{44}{3} = 14.67$ cm. (½)
Perimeter $= A P + P B +$ arc $A B = 14 + 14 + 14.67 = 42.67$ cm. (1)

Answer: (i) 14 cm (ii) about 17.90 cm² (iii) about 66.87 cm² (iv) about 42.67 cm

Example 7: Two equal circles through each other's centres (5 marks) · 5 marks · Long answerHarder board level

Two circles with centres $P$ and $Q$ each have radius 14 cm. Each circle passes through the centre of the other, so $P Q = 14$ cm. The circles meet at $A$ and $B$ . The region common to both circles (the lens between the two arcs $A B$ ) is shaded. Find (i) $∠ A P B$ , (ii) the area of the shaded region, and (iii) its perimeter. (Take $π = \frac{22}{7}$ , $3 = 1.73$ )

Try it first, then show the solutionSolution

$P A = P Q = Q A = 14$ cm (radii of equal circles), so $△ P A Q$ is equilateral and $∠ A P Q = 6 0^{\circ}$ . (½)
In the same way $∠ B P Q = 6 0^{\circ}$ , so $∠ A P B = 12 0^{\circ}$ ; likewise $∠ A QB = 12 0^{\circ}$ . (½)
The lens is made of two minor segments on the common chord $A B$ : one of circle $P$ and one of circle $Q$ . (½)
Sector $P A B = \frac{1}{3} \times \frac{22}{7} \times 196 = \frac{616}{3}$ cm $^{2}$ . (½)
$ar (△ P A B) = \frac{3}{4} \times 196 = 493$ cm $^{2}$ (the perpendicular from $P$ halves the $12 0^{\circ}$ angle). (½)
One segment $= \frac{616}{3} - 493$ cm $^{2}$ . (½)
Lens $= 2 (\frac{616}{3} - 493) = \frac{1232}{3} - 983 = 410.67 - 169.54 = 241.13$ cm $^{2}$ . (1)
Perimeter $= 2 \times \frac{1}{3} \times 2 \times \frac{22}{7} \times 14 = 2 \times \frac{88}{3} = \frac{176}{3} \approx 58.67$ cm. (1)

Answer: (i) 120° (ii) about 241.13 cm² (iii) about 58.67 cm

Example 8: Case study: rectangular photo on a circular card · 4 marks · Case-basedHOTS / competency

Suppose an art teacher pastes a rectangular photo $A B C D$ on a circular card with centre $O$ so that all four corners of the photo lie on the edge of the card. The diagonal $A C$ of the photo is 14 cm and the side $A B$ is 7 cm. The four parts of the card outside the photo are coloured. (Take $π = \frac{22}{7}$ , $3 = 1.73$ )
(i) Find the radius of the card and $∠ A O B$ . (1)
(ii) Find $∠ B O C$ and the length $B C$ . (1)
(iii) Find the total coloured area using segment areas. (2)

Try it first, then show the solutionSolution

(i) The diagonals of the rectangle are diameters (each subtends a right angle at the corners), so $r = 7$ cm. Then $O A = O B = A B = 7$ cm, so $△ O A B$ is equilateral and $∠ A O B = 6 0^{\circ}$ . (1)
(ii) $A C$ is a straight line through $O$ , so $∠ B O C = 18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ , and $B C = 1 4^{2} - 7^{2} = 147 = 73$ cm. (1)
(iii) Segment on $A B$ ( $6 0^{\circ}$ ) $= \frac{1}{6} \times 154 - \frac{3}{4} \times 49 = \frac{77}{3} - \frac{49 3}{4}$ cm $^{2}$ . (½)
Segment on $B C$ ( $12 0^{\circ}$ ) $= \frac{1}{3} \times 154 - \frac{49 3}{4} = \frac{154}{3} - \frac{49 3}{4}$ cm $^{2}$ . (½)
There are two of each: total $= 2 (\frac{77}{3} + \frac{154}{3}) - 4 \times \frac{49 3}{4} = 154 - 493$ cm $^{2}$ . (½)
$= 154 - 84.77 = 69.23$ cm $^{2}$ . Check: circle $-$ rectangle $= 154 - 7 \times 73$ , the same. (½)

Answer: (i) 7 cm, 60° (ii) 120°, 7√3 cm (iii) about 69.23 cm²

Where students lose marks in Area of a Segment of a Circle

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

Mistake: Using $\frac{1}{2} r^{2}$ for every triangle
Only the 90° triangle is right-angled.
Wrong working
120°, r = 14: triangle $= \frac{1}{2} \times 14 \times 14 = 98$
Right working
Triangle $= \frac{3}{4} \times 196 = 493 \approx 84.77$ cm²
Do this: Match the triangle to the angle every time and write the reason.
Usually costs: 1 to 2 marks
Mistake: Major segment as major sector minus triangle
The triangle lies on the major side, so it must be added.
Wrong working
90°, r = 14: $462 - 98 = 364$
Right working
$462 + 98 = 560$ , or $616 - 56 = 560$ cm²
Do this: Use circle minus minor segment.
Usually costs: 1 to 2 marks
Mistake: Perimeter of segment taken as arc plus two radii
That is the sector perimeter; the segment edge is arc plus chord.
Wrong working
60°, r = 6: $6.28 + 12 = 18.28$ cm
Right working
$6.28 + 6 = 12.28$ cm
Do this: Trace the boundary of the shaded part before adding.
Usually costs: 1 mark
Mistake: No justification for triangle area
Examiners award a step mark for stating why the triangle is equilateral or for drawing the perpendicular.
Wrong working
Triangle $= \frac{3}{4} r^{2}$ written with no reason
Right working
$O M ⊥ A B$ bisects $∠ A O B$ ; $O M = \frac{r}{2}$ , $A B = 3 r$ ; area $= \frac{3}{4} r^{2}$
Do this: Add one line of reasoning.
Usually costs: ½ to 1 mark

Exam tips

Write 'segment = sector − triangle' as your first line; it signals the method and earns credit.
Use the value of π and √3 the question gives; switching values changes the answer.
Keep √3 in surd form until the last line to avoid rounding errors.
If the angle is not given, look for clues: chord equal to radius means 60°, a square's side means 90°, an equilateral triangle's side in its circumcircle means 120°.
Verify a total shaded area by a second route (circle minus polygon) when time permits.

Practice questions on Area of a Segment 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 · 2 marks · Very short answerBoard level
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the minor segment. (Use $π = 3.14$ )
Hint
Quadrant minus a right triangle with legs 10 cm.
Answer
Sector $= 78.5$ , triangle $= 50$ , segment $= 28.5$ cm².
Q2 · 3 marks · Short answerBoard level
A chord of a circle of radius 21 cm subtends $12 0^{\circ}$ at the centre. Find the area of the minor segment and the length of the chord. (Take $π = \frac{22}{7}$ , $3 = 1.73$ )
Hint
Drop a perpendicular from O; it splits the angle into two 60° angles.
Answer
Sector $= 462$ ; triangle $= \frac{3}{4} \times 441 = 190.7325$ ; segment ≈ 271.27 cm²; chord $= 213 = 36.33$ cm.
Q3 · 3 marks · Short answerHarder board level
The area of a sector of a circle with central angle $9 0^{\circ}$ is 154 cm². Find the area of the corresponding major segment. (Take $π = \frac{22}{7}$ )
Hint
Find r from the sector first.
Answer
$\frac{1}{4} π r^{2} = 154 \Rightarrow r = 14$ . Minor segment $= 154 - 98 = 56$ . Major $= 616 - 56 = 560$ cm².
Q4 · 5 marks · Long answerHarder board level
An equilateral triangle $A B C$ is inscribed in a circle of radius 7 cm with centre $O$ . Find the area of the part of the circle outside the triangle using segment areas, and verify by subtracting the triangle from the circle. (Take $π = \frac{22}{7}$ , $3 = 1.73$ )
Hint
Each side subtends 120° at O (central angle is twice the 60° inscribed angle, or 360° split equally).
Answer
One segment $= \frac{154}{3} - \frac{3}{4} \times 49 = 51.333 - 21.1925 = 30.141$ ; three segments ≈ 90.42 cm². Check: triangle $= 3 \times 21.1925 = 63.58$ ; $154 - 63.58 = 90.42$ cm².
Q5 · 3 marks · Short answerBoard level
Riya writes: in a circle of radius 14 cm with a $6 0^{\circ}$ chord, segment $= \frac{1}{6} \times \frac{22}{7} \times 1 4^{2} - \frac{1}{2} \times 14 \times 14 = 102.67 - 98 = 4.67$ cm². Find her error and give the correct area. (Take $3 = 1.73$ )
Hint
Is the triangle right-angled at 60°?
Answer
She used the 90° triangle. At 60° the triangle is equilateral: $\frac{1.73}{4} \times 196 = 84.77$ . Segment $= 102.67 - 84.77 = 17.90$ cm².
Q6 · 1 mark · MCQBoard level
A chord $A B$ of a circle of radius 12 cm subtends $6 0^{\circ}$ at the centre. The area of the minor segment is (Use $π = 3.14$ , $3 = 1.73$ )
(a) 75.36 cm²
(b) 62.28 cm²
(c) 13.08 cm²
(d) 26.16 cm²
Hint
Sector is one-sixth of the circle; the triangle is equilateral.
Answer
(c) 13.08 cm². Sector $= \frac{1}{6} \times 3.14 \times 144 = 75.36$ , triangle $= \frac{1.73}{4} \times 144 = 62.28$ , segment $= 13.08$ cm².
Q7 · 1 mark · MCQBoard level
A chord of length $103$ cm subtends $12 0^{\circ}$ at the centre of a circle. The radius of the circle is
(a) 5 cm
(b) $53$ cm
(c) 10 cm
(d) 20 cm
Hint
At 120° the chord is $3$ times the radius.
Answer
(c) 10 cm. Chord $= 3 r = 103$ , so $r = 10$ cm.
Q8 · 1 mark · Assertion-reasonBoard level
Assertion (A): In a circle of radius 7 cm, the triangle formed by the two radii and a chord that subtends $12 0^{\circ}$ at the centre has area $\frac{49 3}{4}$ cm².
Reason (R): The perpendicular from the centre to the chord splits the $12 0^{\circ}$ angle into two $6 0^{\circ}$ angles, so the height of the triangle is $\frac{r}{2}$ and the chord is $3 r$ .
(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 work out the area for r = 7 and compare with A.
Answer
(a). From R, area $= \frac{1}{2} \times 73 \times \frac{7}{2} = \frac{49 3}{4}$ cm², so A is true and R explains it.
Q9 · 3 marks · Short answerHarder board level
A chord of a circle of radius $r$ subtends $9 0^{\circ}$ at the centre. (i) Show that the minor and major segments are in the ratio $1 : 10$ . (ii) If the major segment has area 560 cm², find $r$ and the perimeter of the minor segment. (Take $π = \frac{22}{7}$ , $2 = 1.41$ )
Hint
Write both segments as fractions of $r^{2}$ first.
Answer
(i) Minor $= \frac{11}{14} r^{2} - \frac{1}{2} r^{2} = \frac{2}{7} r^{2}$ ; major $= \frac{22}{7} r^{2} - \frac{2}{7} r^{2} = \frac{20}{7} r^{2}$ ; ratio $1 : 10$ . (ii) $\frac{20}{7} r^{2} = 560 \Rightarrow r^{2} = 196$ , $r = 14$ cm. Perimeter $=$ arc $22 +$ chord $142 = 19.74$ , total 41.74 cm.
Q10 · 3 marks · Short answerHarder board level
A chord $A B$ of a circle with centre $O$ subtends $12 0^{\circ}$ at $O$ . The area of the minor segment is $(\frac{616}{3} - 493)$ cm². Find the radius of the circle and the length of $A B$ . (Take $π = \frac{22}{7}$ )
Hint
Write the segment as $r^{2} (\frac{22}{21} - \frac{3}{4})$ and take out a common factor from the given area.
Answer
Segment $= \frac{1}{3} \times \frac{22}{7} r^{2} - \frac{3}{4} r^{2} = r^{2} (\frac{22}{21} - \frac{3}{4})$ . Given area $= 196 (\frac{22}{21} - \frac{3}{4})$ , so $r^{2} = 196$ , $r = 14$ cm, and $A B = 143$ cm.
Q11 · 3 marks · Short answerHarder board level
$A B C D$ is a square of side 14 cm. With centre $B$ , an arc is drawn from $A$ to $C$ inside the square; with centre $D$ , another arc is drawn from $A$ to $C$ inside the square. The leaf-shaped region between the two arcs is shaded. Find (i) the area of the leaf, (ii) the area of the square outside the leaf, and (iii) the perimeter of the leaf. (Take $π = \frac{22}{7}$ )
Hint
Join $A C$ . Each half of the leaf is a segment with a $9 0^{\circ}$ central angle.
Answer
Each half $= 154 - 98 = 56$ cm² ( $∠ A B C = 9 0^{\circ}$ , $r = 14$ ). (i) Leaf $= 112$ cm². (ii) $196 - 112 = 84$ cm². (iii) Perimeter $= 2 \times 22 = 44$ cm.
Q12 · 3 marks · Short answerHOTS / competency
Aman claims that if the central angle of a chord is doubled from $6 0^{\circ}$ to $12 0^{\circ}$ in the same circle, the area of the minor segment also doubles. Test his claim for a circle of radius 21 cm and explain why it fails or holds. (Take $π = \frac{22}{7}$ , $3 = 1.73$ )
Hint
The sector doubles. What happens to the triangle?
Answer
Triangle $= \frac{1.73}{4} \times 441 = 190.7325$ cm² in both cases. At 60°: $231 - 190.73 = 40.27$ cm². At 120°: $462 - 190.73 = 271.27$ cm², nearly 6.7 times, not double. The claim fails because the sector doubles but the same triangle area is subtracted.
Q13 · 4 marks · Case-basedHOTS / competency
Suppose a wall clock has a circular face of radius 14 cm with centre $O$ ; the numerals 1 to 12 are equally spaced on the rim. A straight decorative strip joins the rim points at 12 and 4, and a second strip joins the rim points at 4 and 7. Each strip cuts off a small minor segment that is painted silver. (Take $π = \frac{22}{7}$ , $2 = 1.41$ , $3 = 1.73$ )
(i) Find the angle at $O$ between the rim points at 12 and 4. (1)
(ii) Find the length of the strip from 4 to 7. (1)
(iii) Find the total silver area. (2)
Hint
Adjacent numerals are $3 0^{\circ}$ apart at the centre.
Answer
(i) $4 \times 3 0^{\circ} = 12 0^{\circ}$ . (ii) 4 to 7 is $9 0^{\circ}$ , so strip $= 142 = 19.74$ cm. (iii) 120° segment $= \frac{616}{3} - 493 = 205.33 - 84.77 = 120.56$ cm²; 90° segment $= 154 - 98 = 56$ cm²; total ≈ 176.56 cm².
Q14 · 5 marks · Long answerHarder board level
$A B C D$ is a rectangle with $A B = 21$ cm and $A D = 14$ cm. Point $E$ lies on $A B$ with $A E = 14$ cm. A quadrant with centre $A$ is drawn through $D$ and $E$ . Find (i) the area of the minor segment between chord $D E$ and arc $D E$ , (ii) the area of the part of the rectangle outside the quadrant, using trapezium $E B C D$ and the segment, and verify it a second way, (iii) the perimeter of that part. (Take $π = \frac{22}{7}$ )
Hint
The part outside the quadrant is trapezium $E B C D$ with the segment taken away.
Answer
(i) Quadrant $= 154$ , $△ A D E = 98$ , segment $= 56$ cm². (ii) Trapezium $E B C D$ : parallel sides $E B = 7$ , $D C = 21$ , height 14, area $= \frac{1}{2} \times 28 \times 14 = 196$ ; outside part $= 196 - 56 = 140$ cm². Check: rectangle $-$ quadrant $= 294 - 154 = 140$ cm². (iii) Arc $D E = 22$ , so perimeter $= 22 + 7 + 14 + 21 = 64$ cm.

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.Minor segment area?Show answer
Sector OAB minus triangle OAB.
2.Triangle area at 60°?Show answer
$\frac{3}{4} r^{2}$ (equilateral).
3.Triangle area at 120°?Show answer
$\frac{3}{4} r^{2}$ , same as 60°.
4.Chord at 90°?Show answer
$r 2$ .
5.90° segment with π = 22/7?Show answer
$\frac{2}{7} r^{2}$ .
6.Major segment?Show answer
$π r^{2} -$ minor segment.

Area of a Segment of a Circle: frequently asked questions

Which central angles can be asked for segment area?

Only 60, 90 and 120 degrees in the 2026-27 syllabus. For these the triangle has a neat area, so no other angles are tested.

What is the difference between a sector and a segment?

A sector is bounded by two radii and an arc. A segment is bounded by a chord and an arc. The segment is the sector with the triangle removed.

Why does the 120 degree triangle have the same area as the 60 degree one?

Drop OM perpendicular to AB. For 120 degrees it splits the angle into two 60 degree angles, so OM = r/2 and AB = √3 r. The area is ½ × √3r × r/2 = (√3/4)r², the same as for the equilateral triangle at 60 degrees.

Do I need to draw a figure in the answer?

A quick labelled sketch with the centre, chord and perpendicular helps the examiner follow your steps. It is especially useful for the 120 degree case.