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Areas Related to Circles — CBSE Class 10 Maths Important Questions

13 hand-picked CBSE Class 10 Maths important questions for Areas Related to Circles, each with a full model answer — the formats and topics most likely to appear in your board exam.

13
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6
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32
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With answers

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\theta and radius rrr: length of arc =θ360×2πr=\dfrac{\theta}{360}\times 2\pi r=/360× 2π r, area of sector =θ360×πr2=\dfrac{\theta}{360}\times\pi r^2=/360×π r^2. 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. In CBSE board papers you can expect a short-answer and often a case-study or long-answer question (about 4-6 marks) from here, frequently 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\theta: Area =θ360×πr2=\dfrac{\theta}{360}\times\pi r^2=/360×π r^2 and arc length =θ360×2πr=\dfrac{\theta}{360}\times 2\pi r=/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 =θ360×πr2−12r2sin⁡θ=\dfrac{\theta}{360}\times\pi r^2-\dfrac{1}{2}r^2\sin\theta=/360×π r^2-1/2r^2, i.e. area of sector −-- area of triangle. Major segment =πr2−=\pi r^2-=π r^2- 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.

Question typeCountMarks
MCQ41
Assertion–Reason11
Very Short22
Short Answer33
Long Answer25
Case-based14

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 cm^2

  2. (b)

    462 cm2462\ \text{cm}^2462 cm^2

  3. (c)

    154 cm2154\ \text{cm}^2154 cm^2

  4. (d)

    115.5 cm2115.5\ \text{cm}^2115.5 cm^2

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Answer: (a) 231 cm2231\ \text{cm}^2231 cm^2

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.=/360×π r^2=60/360×22/7× 21× 21=1/6× 22× 3× 21=231 cm^2.

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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

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Answer: (a) 222 units

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

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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 R^2=π(8)^2+π(6)^2 R^2=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 cm^2

  2. (b)

    38.5 cm238.5\ \text{cm}^238.5 cm^2

  3. (c)

    19.25 cm219.25\ \text{cm}^219.25 cm^2

  4. (d)

    22 cm222\ \text{cm}^222 cm^2

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Answer: (a) 9.625 cm29.625\ \text{cm}^29.625 cm^2

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 cm^2.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonEasy1 mark

Assertion (A): The area of a sector of angle 90∘90^\circ90^ of a circle of radius 10 cm10\ \text{cm}10 cm is 78.5 cm278.5\ \text{cm}^278.5 cm^2 (take π=3.14\pi=3.14π=3.14).

Reason (R): The area of a sector of angle θ\theta is θ360×πr2\dfrac{\theta}{360}\times\pi r^2/360×π r^2.

  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

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

Area =90360×3.14×102=14×314=78.5 cm2=\dfrac{90}{360}\times 3.14\times 10^2=\dfrac{1}{4}\times 314=78.5\ \text{cm}^2=90/360× 3.14× 10^2=1/4× 314=78.5 cm^2, exactly what the formula in 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 r^2=1/4×22/7× 3.5× 3.5=1/4× 38.5=9.625 cm^2.

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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 cm^2.

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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
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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× 12^2=1/3× 3.14× 144=150.72 cm^2.

Area of triangle OAB=12r2sin⁡120∘=12×144×32=363=36×1.73=62.28 cm2.OAB=\dfrac{1}{2}r^2\sin 120^\circ=\dfrac{1}{2}\times 144\times\dfrac{\sqrt3}{2}=36\sqrt3=36\times 1.73=62.28\ \text{cm}^2.OAB=1/2r^2 120^=1/2× 144×3/2=363=36× 1.73=62.28 cm^2.

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 cm^2.

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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.

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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 cm^2.

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 cm^2.

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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=1.7323=1.732).

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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× 7^2=1.732/4× 49=21.22 cm^2.

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 cm^2.

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

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Long answer questions (5 marks)

Q11Long 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}^2cm^2 (use 3=1.7\sqrt3=1.73=1.7).

Show model answer

Each design is a segment with θ=60∘, r=28 cm.\theta=60^\circ,\ r=28\ \text{cm}.=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 cm^2.

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× 28^2=1.7/4× 784=333.2 cm^2.

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 cm^2.

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

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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Q12Long 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× 5^2=1/4× 3.14× 25=19.625 m^2.

(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 m^2.

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

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Case-based questions (4 marks)

Q13Case-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 cm^2.

(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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Frequently asked questions

  • Are these Areas Related to Circles important questions free?
    Yes. All 13 CBSE Class 10 Maths important questions for Areas Related to Circles are free, with full model answers and no login required.
  • Do these Areas Related to Circles questions follow the latest CBSE syllabus?
    Yes — they are aligned to the NCERT 2026–27 syllabus for CBSE Class 10 Maths, so nothing here is outside the current course.
  • How should I practise the Areas Related to Circles important questions?
    Attempt each question on paper first, then reveal the model answer to check your method — not just the final result. Re-do anything you got wrong the same day.
  • What types of questions are covered for Areas Related to Circles?
    A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the CBSE paper is covered.

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