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Areas Related to Circles — CBSE Class 10 Maths Important Questions

14 CBSE Class 10 Maths practice questions on Areas Related to Circles, each with a full model answer, covering 5 topics from the chapter in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 30 September 2026

Last updated 25 October 2026

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Key concepts
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Areas Related to Circles — CBSE Class 10 Maths Important Questions

Sectors and Segments, Nailed

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Quick answer

For a sector of angle θ\thetaθ and radius rrr: length of arc =θ360×2πr=\dfrac{\theta}{360}\times 2\pi r=θ/360× 2π r, area of sector =θ360×πr2=\dfrac{\theta}{360}\times\pi r^2=θ/360× π r². Area of a segment = area of sector −-- area of the triangle formed by the two radii and the chord. Use π=227\pi=\dfrac{22}{7}π=22/7 unless stated otherwise.

Syllabus reference (2026–27): NCERT (National Council of Educational Research and Training) · CBSE (Central Board of Secondary Education)

About Areas Related to Circles

This chapter deals with the perimeter and area of a circle and, more importantly, with sectors and segments cut out of a circle. Questions are often set in real-life contexts such as clock hands, wheels, brooches, table-cover designs and grazing animals. The two formulas you must never confuse are the sector-area formula and the segment-area formula.

Circumference and area of a circleLength of an arc and area of a sectorArea of a segment (minor and major)Areas of combinations of plane figures (shaded regions)Real-life applications: clocks, wheels, designs, grazing

Key concepts & formulas

Sector of a circle

A region bounded by two radii and the corresponding arc. For central angle θ\thetaθ: Area =θ360×πr2=\dfrac{\theta}{360}\times\pi r^2=θ/360× π r² and arc length =θ360×2πr=\dfrac{\theta}{360}\times 2\pi r=θ/360× 2π r. Perimeter of a sector === arc length + 2r+\,2r+ 2r.

Segment of a circle

A region bounded by a chord and its arc. Area of minor segment === area of sector −-- area of △OAB\triangle OAB△ OAB. For θ=60∘\theta=60^\circθ=60° the triangle is equilateral (area 34r2\dfrac{\sqrt3}{4}r^2√3/4r²); for θ=90∘\theta=90^\circθ=90° its area is 12r2\dfrac12 r^21/2 r²; for θ=120∘\theta=120^\circθ=120°, drop a perpendicular from OOO to the chord. Major segment =πr2−=\pi r^2-=π r²- minor segment.

Angle swept by clock hands

In 606060 minutes the minute hand sweeps 360∘360^\circ360°, so it turns 6∘6^\circ6° per minute. In 121212 hours the hour hand sweeps 360∘360^\circ360°, i.e. 0.5∘0.5^\circ0.5° per minute. Convert the time to an angle, then use the sector-area formula.

Shaded (combination) regions

Split the figure into standard shapes (circles, sectors, squares, triangles). Add or subtract their areas carefully. A quadrant is a sector of 90∘90^\circ90°; a semicircle is a sector of 180∘180^\circ180°.

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Important questions with answers

Try each on paper first, then reveal the model answer to check your method.

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The area of a sector of a circle of radius 21 cm21\ \text{cm}21 cm with central angle 60∘60^\circ60° is:

  1. (a)

    231 cm2231\ \text{cm}^2231 cm²

  2. (b)

    462 cm2462\ \text{cm}^2462 cm²

  3. (c)

    154 cm2154\ \text{cm}^2154 cm²

  4. (d)

    115.5 cm2115.5\ \text{cm}^2115.5 cm²

Show model answer

Answer: (a) 231 cm2231\ \text{cm}^2231 cm²

Area =θ360×πr2=60360×227×21×21=16×22×3×21=231 cm2.=\dfrac{\theta}{360}\times\pi r^2=\dfrac{60}{360}\times\dfrac{22}{7}\times 21\times 21=\dfrac{1}{6}\times 22\times 3\times 21=231\ \text{cm}^2.=θ/360× π r²=60/360× 22/7× 21× 21=1/6× 22× 3× 21=231 cm².

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Q2MCQEasy1 mark

If the circumference and the area of a circle are numerically equal, then the radius of the circle is:

  1. (a)

    222 units

  2. (b)

    444 units

  3. (c)

    111 unit

  4. (d)

    777 units

Show model answer

Answer: (a) 222 units

2πr=πr2⇒r=2.2\pi r=\pi r^2\Rightarrow r=2.2π r=π r²⇒ r=2.

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Q3MCQModerate1 mark

The radii of two circles are 8 cm8\ \text{cm}8 cm and 6 cm6\ \text{cm}6 cm. The radius of the circle whose area is equal to the sum of the areas of these two circles is:

  1. (a)

    10 cm10\ \text{cm}10 cm

  2. (b)

    14 cm14\ \text{cm}14 cm

  3. (c)

    12 cm12\ \text{cm}12 cm

  4. (d)

    7 cm7\ \text{cm}7 cm

Show model answer

Answer: (a) 10 cm10\ \text{cm}10 cm

πR2=π(8)2+π(6)2⇒R2=64+36=100⇒R=10 cm.\pi R^2=\pi(8)^2+\pi(6)^2\Rightarrow R^2=64+36=100\Rightarrow R=10\ \text{cm}.π R²=π(8)²+π(6)²⇒ R²=64+36=100⇒ R=10 cm.

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Q4MCQHOTS1 mark

The area of the largest circle that can be drawn inside a rectangle of sides 7 cm7\ \text{cm}7 cm and 3.5 cm3.5\ \text{cm}3.5 cm is:

  1. (a)

    9.625 cm29.625\ \text{cm}^29.625 cm²

  2. (b)

    38.5 cm238.5\ \text{cm}^238.5 cm²

  3. (c)

    19.25 cm219.25\ \text{cm}^219.25 cm²

  4. (d)

    22 cm222\ \text{cm}^222 cm²

Show model answer

Answer: (a) 9.625 cm29.625\ \text{cm}^29.625 cm²

The diameter cannot exceed the shorter side, so diameter =3.5 cm=3.5\ \text{cm}=3.5 cm, radius =1.75 cm=1.75\ \text{cm}=1.75 cm. Area =227×(1.75)2=227×3.0625=9.625 cm2.=\dfrac{22}{7}\times(1.75)^2=\dfrac{22}{7}\times 3.0625=9.625\ \text{cm}^2.=22/7×(1.75)²=22/7× 3.0625=9.625 cm².

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): The area of the major sector of a circle of radius 7 cm7\ \text{cm}7 cm, whose minor sector has a central angle of 90∘90^\circ90°, is 115.5 cm2115.5\ \text{cm}^2115.5 cm².

Reason (R): Area of the major sector === area of the circle −-- area of the minor sector.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

Show model answer

Answer: (a) Both A and R are true and R is the correct explanation of A.

Area of circle =227×7×7=154 cm2=\dfrac{22}{7}\times7\times7=154\ \text{cm}^2=22/7×7×7=154 cm²; area of minor sector =90360×154=38.5 cm2=\dfrac{90}{360}\times154=38.5\ \text{cm}^2=90/360×154=38.5 cm².

Major sector =154−38.5=115.5 cm2=154-38.5=115.5\ \text{cm}^2=154-38.5=115.5 cm², exactly as R gives.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

The circumference of a circle is 22 cm22\ \text{cm}22 cm. Find the area of its quadrant.

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2πr=22⇒r=22×72×22=3.5 cm.2\pi r=22\Rightarrow r=\dfrac{22\times 7}{2\times 22}=3.5\ \text{cm}.2π r=22⇒ r=(22× 7)/(2× 22)=3.5 cm.

Area of quadrant =14πr2=14×227×3.5×3.5=14×38.5=9.625 cm2.=\dfrac{1}{4}\pi r^2=\dfrac{1}{4}\times\dfrac{22}{7}\times 3.5\times 3.5=\dfrac{1}{4}\times 38.5=9.625\ \text{cm}^2.=1/4π r²=1/4× 22/7× 3.5× 3.5=1/4× 38.5=9.625 cm².

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Q7Very ShortModerate2 marks

The minute hand of a clock is 12 cm12\ \text{cm}12 cm long. Find the area of the face of the clock described by the minute hand in 353535 minutes.

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In 606060 min the hand turns 360∘360^\circ360°, so in 353535 min it turns 3560×360∘=210∘.\dfrac{35}{60}\times 360^\circ=210^\circ.35/60× 360°=210°.

Area =210360×227×12×12=712×227×144=22×12=264 cm2.=\dfrac{210}{360}\times\dfrac{22}{7}\times 12\times 12=\dfrac{7}{12}\times\dfrac{22}{7}\times 144=22\times 12=264\ \text{cm}^2.=210/360× 22/7× 12× 12=7/12× 22/7× 144=22× 12=264 cm².

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

A chord of a circle of radius 12 cm12\ \text{cm}12 cm subtends an angle of 120∘120^\circ120° at the centre. Find the area of the corresponding minor segment (use π=3.14\pi=3.14π=3.14 and 3=1.73\sqrt{3}=1.73√3=1.73).

CBSE Class 10 Maths — Areas Related to Circles: A chord of a circle of radius 12\ \text{cm} subtends an angle of 120^\circ at the centre. Find the area of the corresponding minor s
Show model answer

Area of sector =120360×3.14×122=13×3.14×144=150.72 cm2.=\dfrac{120}{360}\times 3.14\times 12^2=\dfrac{1}{3}\times 3.14\times 144=150.72\ \text{cm}^2.=120/360× 3.14× 12²=1/3× 3.14× 144=150.72 cm².

Area of triangle OABOABOAB: draw OM⊥ABOM\perp ABOM⊥ AB. In right △OMA\triangle OMA△ OMA, ∠AOM=120∘2=60∘\angle AOM=\dfrac{120^\circ}{2}=60^\circ∠ AOM=120°/2=60°, so
OM=12cos⁡60∘=12×12=6 cm,AM=12sin⁡60∘=12×32=63 cm.OM=12\cos60^\circ=12\times\dfrac12=6\ \text{cm},\qquad AM=12\sin60^\circ=12\times\dfrac{\sqrt3}{2}=6\sqrt3\ \text{cm}.OM=12 cos 60°=12× 1/2=6 cm, AM=12 sin 60°=12× √3/2=6√3 cm.

Hence AB=2AM=123 cmAB=2AM=12\sqrt3\ \text{cm}AB=2AM=12√3 cm and
area of △OAB=12×AB×OM=12×123×6=363=36×1.73=62.28 cm2.\text{area of }\triangle OAB=\dfrac12\times AB\times OM=\dfrac12\times12\sqrt3\times6=36\sqrt3=36\times1.73=62.28\ \text{cm}^2.area of △ OAB=1/2× AB× OM=1/2×12√3×6=36√3=36×1.73=62.28 cm².

Area of minor segment =150.72−62.28=88.44 cm2.=150.72-62.28=88.44\ \text{cm}^2.=150.72-62.28=88.44 cm².

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Q9Short AnswerModerate3 marks

A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm25\ \text{cm}25 cm sweeping through an angle of 115∘115^\circ115°. Find the total area cleaned at each sweep of the two blades.

Show model answer

Each blade sweeps a sector of radius 25 cm25\ \text{cm}25 cm and angle 115∘115^\circ115°.

Area of one sector =115360×227×25×25=115×22×625360×7=15812502520=627.48 cm2.=\dfrac{115}{360}\times\dfrac{22}{7}\times 25\times 25=\dfrac{115\times 22\times 625}{360\times 7}=\dfrac{1581250}{2520}=627.48\ \text{cm}^2.=115/360× 22/7× 25× 25=(115× 22× 625)/(360× 7)=1581250/2520=627.48 cm².

Total area for two wipers =2×627.48=158125126≈1254.96 cm2.=2\times 627.48=\dfrac{158125}{126}\approx 1254.96\ \text{cm}^2.=2× 627.48=158125/126≈ 1254.96 cm².

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Q10Short AnswerHOTS3 marks

Three circles, each of radius 3.5 cm3.5\ \text{cm}3.5 cm, are drawn so that each touches the other two externally. Find the area enclosed between the three circles (take 3=1.732\sqrt3=1.732√3=1.732).

Show model answer

The centres form an equilateral triangle of side =3.5+3.5=7 cm.=3.5+3.5=7\ \text{cm}.=3.5+3.5=7 cm.

Area of triangle =34×72=1.7324×49=21.22 cm2.=\dfrac{\sqrt3}{4}\times 7^2=\dfrac{1.732}{4}\times 49=21.22\ \text{cm}^2.=√3/4× 7²=1.732/4× 49=21.22 cm².

At each vertex the angle is 60∘60^\circ60°, so the three sectors together make an angle of 180∘180^\circ180°; total sector area =180360×227×(3.5)2=12×38.5=19.25 cm2.=\dfrac{180}{360}\times\dfrac{22}{7}\times(3.5)^2=\dfrac{1}{2}\times 38.5=19.25\ \text{cm}^2.=180/360× 22/7×(3.5)²=1/2× 38.5=19.25 cm².

Enclosed area =21.22−19.25=1.97 cm2.=21.22-19.25=1.97\ \text{cm}^2.=21.22-19.25=1.97 cm².

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Q11Short AnswerModerate3 marks

A chord ABABAB of a circle of radius 14 cm14\ \text{cm}14 cm subtends a right angle at the centre OOO. Find the area of (i) the minor segment and (ii) the major segment. (Use π=227\pi=\dfrac{22}{7}π=22/7.)

CBSE Class 10 Maths — Areas Related to Circles: A chord AB of a circle of radius 14\ \text{cm} subtends a right angle at the centre O. Find the area of (i) the minor segment and (i
Show model answer

Area of sector OAB=90360×227×14×14=14×616=154 cm2.OAB=\dfrac{90}{360}\times\dfrac{22}{7}\times14\times14=\dfrac14\times616=154\ \text{cm}^2.OAB=90/360× 22/7×14×14=1/4×616=154 cm².

Area of △OAB=12×14×14=98 cm2\triangle OAB=\dfrac12\times14\times14=98\ \text{cm}^2△ OAB=1/2×14×14=98 cm² (the angle at OOO is 90∘90^\circ90°).

(i) Minor segment =154−98=56 cm2.=154-98=56\ \text{cm}^2.=154-98=56 cm².

(ii) Area of circle =227×14×14=616 cm2=\dfrac{22}{7}\times14\times14=616\ \text{cm}^2=22/7×14×14=616 cm², so the major segment =616−56=560 cm2.=616-56=560\ \text{cm}^2.=616-56=560 cm².

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Long answer questions (5 marks)

Q12Long AnswerModerate5 marks

A round table cover has six equal designs, each design being a segment cut off by a chord subtending 60∘60^\circ60° at the centre. If the radius of the cover is 28 cm28\ \text{cm}28 cm, find the total area of the designs and the cost of making them at the rate of Rs 0.35\text{Rs }0.35Rs 0.35 per cm2\text{cm}^2cm² (use 3=1.7\sqrt3=1.7√3=1.7).

Show model answer

Each design is a segment with θ=60∘, r=28 cm.\theta=60^\circ,\ r=28\ \text{cm}.θ=60°, r=28 cm.

Area of one sector =60360×227×28×28=16×2464=410.67 cm2.=\dfrac{60}{360}\times\dfrac{22}{7}\times 28\times 28=\dfrac{1}{6}\times 2464=410.67\ \text{cm}^2.=60/360× 22/7× 28× 28=1/6× 2464=410.67 cm².

Area of triangle =34×282=1.74×784=333.2 cm2.=\dfrac{\sqrt3}{4}\times 28^2=\dfrac{1.7}{4}\times 784=333.2\ \text{cm}^2.=√3/4× 28²=1.7/4× 784=333.2 cm².

Area of one segment =410.67−333.2=77.47 cm2.=410.67-333.2=77.47\ \text{cm}^2.=410.67-333.2=77.47 cm².

Six designs =6×77.47=464.8 cm2.=6\times 77.47=464.8\ \text{cm}^2.=6× 77.47=464.8 cm².

Cost =464.8×0.35=Rs 162.68.=464.8\times 0.35=\text{Rs }162.68.=464.8× 0.35=Rs 162.68.

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Q13Long AnswerHOTS5 marks

A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m15\ \text{m}15 m by means of a 5 m5\ \text{m}5 m long rope (take π=3.14\pi=3.14π=3.14). Find (i) the area of that part of the field in which the horse can graze, and (ii) the increase in the grazing area if the rope were 10 m10\ \text{m}10 m long instead of 5 m5\ \text{m}5 m.

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Tied at a corner of a square, the horse grazes a quarter circle (angle 90∘90^\circ90°).

(i) With rope =5 m=5\ \text{m}=5 m: area =90360×3.14×52=14×3.14×25=19.625 m2.=\dfrac{90}{360}\times 3.14\times 5^2=\dfrac{1}{4}\times 3.14\times 25=19.625\ \text{m}^2.=90/360× 3.14× 5²=1/4× 3.14× 25=19.625 m².

(ii) With rope =10 m=10\ \text{m}=10 m (still less than the side 15 m15\ \text{m}15 m, so it stays a quarter circle): area =14×3.14×100=78.5 m2.=\dfrac{1}{4}\times 3.14\times 100=78.5\ \text{m}^2.=1/4× 3.14× 100=78.5 m².

Increase =78.5−19.625=58.875 m2.=78.5-19.625=58.875\ \text{m}^2.=78.5-19.625=58.875 m².

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Case-based questions (4 marks)

Q14Case-basedModerate4 marks

A circular pizza of radius 14 cm14\ \text{cm}14 cm is cut from the centre into 666 equal slices, as shown. The soft crust runs along the curved edge (arc) of each slice.

CBSE Class 10 Maths — Areas Related to Circles: A circular pizza of radius 14\ \text{cm} is cut from the centre into 6 equal slices, as shown. The soft crust runs along the curved

(i) What is the central angle of each slice?

(ii) Find the area of one slice.

(iii) Find the length of crust (arc) along one slice, and hence the perimeter of one slice.

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(i) Angle of each slice =360∘6=60∘.=\dfrac{360^\circ}{6}=60^\circ.=360°/6=60°.

(ii) Area of one slice (sector) =60360×227×14×14=16×616=102.67 cm2.=\dfrac{60}{360}\times\dfrac{22}{7}\times 14\times 14=\dfrac{1}{6}\times 616=102.67\ \text{cm}^2.=60/360× 22/7× 14× 14=1/6× 616=102.67 cm².

(iii) Arc (crust) length =60360×2×227×14=16×88=14.67 cm.=\dfrac{60}{360}\times 2\times\dfrac{22}{7}\times 14=\dfrac{1}{6}\times 88=14.67\ \text{cm}.=60/360× 2× 22/7× 14=1/6× 88=14.67 cm.

Perimeter of one slice =arc+2r=14.67+28=42.67 cm.=\text{arc}+2r=14.67+28=42.67\ \text{cm}.=arc+2r=14.67+28=42.67 cm.

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All CBSE Class 10 Maths Chapters

Real NumbersPolynomialsPair of Linear Equations in Two VariablesQuadratic EquationsArithmetic ProgressionsTrianglesCoordinate GeometryIntroduction to TrigonometrySome Applications of TrigonometryCirclesAreas Related to CirclesSurface Areas and VolumesStatisticsProbability

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Sin, cos, tan, identities — all the Class 10 trigonometry formulas explained the way a real tutor would teach them, not memorised.

Frequently asked questions

  • What are the important questions in Areas Related to Circles Class 10?
    Sector and arc problems, segment areas for 60∘60^\circ60°, 90∘90^\circ90° and 120∘120^\circ120° angles, clock hands and car wipers, and shaded regions built from circles and squares. This page has 14 questions with answers, including a horse grazing in a square field and a pizza-slice case study.
  • How do you find the area of a segment of a circle?
    Subtract the triangle from the sector: segment =θ360πr2−area of △OAB=\dfrac{\theta}{360}\pi r^2-\text{area of }\triangle OAB=θ/360π r²-area of △ OAB. For θ=90∘\theta=90^\circθ=90° the triangle is 12r2\tfrac12r^21/2r²; for 60∘60^\circ60° it is equilateral, 34r2\tfrac{\sqrt3}{4}r^2√3/4r². The major segment is the whole circle minus the minor one.
  • How do you find the area swept by a clock's minute hand?
    The minute hand turns 6∘6^\circ6° each minute, so convert the time to an angle and use the sector formula. In 555 minutes it sweeps 30∘30^\circ30°; with a 141414 cm hand that is 30360×227×142≈51.3 cm2\dfrac{30}{360}\times\dfrac{22}{7}\times14^2\approx51.3\ \text{cm}^230/360× 22/7×14²≈51.3 cm².
  • If the circumference and area of a circle are numerically equal, what is the radius?
    Set 2πr=πr22\pi r=\pi r^22π r=π r² and divide by πr\pi rπ r: r=2r=2r=2 units. Questions like this, and finding one circle whose area equals the sum of two others (888 cm and 666 cm give 101010 cm), test the formulas rather than heavy arithmetic.
  • Where can I find Areas Related to Circles Class 10 notes?
    This page's Key concepts & formulas section is a ready revision note: the sector, the segment, the angle a clock hand sweeps and shaded regions built from combinations of shapes. The NCERT Solutions page for Chapter 11 solves all of Exercise 11.1.
  • Where can I get Areas Related to Circles extra questions?
    In addition to NCERT, this page has three touching circles of radius 3.53.53.5 cm, a chord subtending 90∘90^\circ90° in a 141414 cm circle, a round table cover with six segment designs, a horse grazing in a 151515 m square field, and a pizza-slice case study, each with a full answer.
  • Where can I practise Areas Related to Circles MCQs?
    Four MCQs and an assertion–reason item begin this page: a 60∘60^\circ60° sector of radius 212121 cm, the circle whose circumference and area are equal, one circle equal in area to two others, and the largest circle inside a 777 by 3.53.53.5 cm rectangle. Timed sets on the chapter come from Create Test after a free sign-up.

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