Class 10 Maths Areas Related to Circles MCQ

Class 10 Mathematics — Areas Related to Circles

MCQs, Fill in the Blanks & Short Questions

A. MCQs

  1. The region bounded by two radii and their corresponding arc is called a:
    (a) chord (b) sector (c) segment (d) semicircle
  2. A sector has radius rr and central angle θ\theta^\circ. Its area is:
    (a) θ180πr2\frac{\theta}{180}\pi r^2 (b) θ360πr2\frac{\theta}{360}\pi r^2 (c) 360θπr2\frac{360}{\theta}\pi r^2 (d) 2πrθ2\pi r\theta
  3. The length of an arc subtending θ\theta^\circ at the centre is:
    (a) θ3602πr\frac{\theta}{360}2\pi r (b) θ180πr2\frac{\theta}{180}\pi r^2 (c) 360θ2πr\frac{360}{\theta}2\pi r (d) θ360πr2\frac{\theta}{360}\pi r^2
  4. The area of a minor segment is obtained by:
    (a) sector + triangle (b) circle − sector (c) sector − triangle (d) circle − triangle
  5. If a minor sector has angle 7070^\circ, the corresponding major sector has angle:
    (a) 290290^\circ (b) 110110^\circ (c) 360360^\circ (d) 7070^\circ
  6. A quadrant corresponds to a central angle of:
    (a) 4545^\circ (b) 9090^\circ (c) 180180^\circ (d) 270270^\circ
  7. The area of the major sector can be found by:
    (a) minor sector − circle (b) circle − minor sector (c) circle + minor sector (d) sector − triangle
  8. A chord divides the circular region into two:
    (a) sectors (b) arcs (c) segments (d) radii

B. Fill in the Blanks

  1. A sector is bounded by two ______ and the corresponding arc.
  2. A segment is bounded by a chord and its corresponding ______.
  3. The angle of a complete circular region at the centre is ______.
  4. Area of a sector of angle θ\theta^\circ is θ360×\frac{\theta}{360}\times ______.
  5. Arc length of a sector of angle θ\theta^\circ is θ360×\frac{\theta}{360}\times ______.
  6. Area of a minor segment = area of the corresponding ______ − area of the corresponding triangle.
  7. Major sector area = area of the complete circle − area of the ______ sector.
  8. Major segment area = area of the complete circle − area of the ______ segment.

C. True / False

  1. A sector is formed by a chord and an arc.
  2. The major sector corresponding to angle θ\theta^\circ has angle 360θ360^\circ-\theta^\circ.
  3. The area of a segment is always equal to the area of its sector.
  4. A quadrant is a sector of angle 9090^\circ.
  5. Arc length depends on both the radius and the central angle.
  6. The area of a major segment can be obtained by subtracting the minor segment from the area of the circle.

D. Very Short Answer

  1. What is meant by a minor sector?
  2. What is meant by a major segment?
  3. Write the formula for the area of a sector.
  4. Write the formula for the length of an arc.
  5. State the formula used to find the area of a minor segment.
  6. What angle does a semicircle subtend at the centre?

E. Application-Based Questions

  1. A sector has radius 1414 cm and angle 9090^\circ. Which formula would you use to find its area?
  2. A circular region has radius 2121 cm. An arc subtends 6060^\circ at the centre. Name the three quantities that can naturally be found from these data: arc length, sector area, and ______.
  3. A chord subtends 6060^\circ at the centre. To find the corresponding minor segment, which two areas must be subtracted?
  4. A horse tied at a corner of a square field moves while restrained by a rope. What part of a circle represents its possible grazing region, provided the rope does not reach another side of the field?

ANSWERS

A. MCQs

  1. (b) Sector
  2. (b) θ360πr2\frac{\theta}{360}\pi r^2
  3. (a) θ3602πr\frac{\theta}{360}2\pi r
  4. (c) Sector − triangle
  5. (a) 290290^\circ
  6. (b) 9090^\circ
  7. (b) Circle − minor sector
  8. (c) Segments

B. Fill in the Blanks

  1. radii
  2. arc
  3. 360∘360^\circ
  4. πr2\pi r^2
  5. 2πr2\pi r
  6. sector
  7. minor
  8. minor

C. True / False

  1. False
  2. True
  3. False
  4. True
  5. True
  6. True

D. Very Short Answer

  1. The smaller region formed by two radii and their corresponding arc.
  2. The larger region formed by a chord and its corresponding major arc.
  3. θ360πr2\boxed{\frac{\theta}{360}\pi r^2}
  4. θ360(2πr)\boxed{\frac{\theta}{360}(2\pi r)}
  5. Sector areaTriangle area\boxed{\text{Sector area} – \text{Triangle area}}
  6. 180∘180^\circ

E. Application-Based

  1. Area=θ360πr2\boxed{\text{Area}=\frac{\theta}{360}\pi r^2}
  2. area of the corresponding segment
  3. Area of the sector and area of the triangle formed by the two radii and chord
  4. A sector of a circle

Class 10 Mathematics

Chapter 11 — Areas Related to Circles

Board-Exam Style Question Bank


A. Assertion–Reason

Choose: (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.

  1. Assertion: The area of a sector with angle 9090^\circ is one-fourth of the area of the circle.
    Reason: A complete circle subtends 360360^\circ at its centre.
  2. Assertion: The area of a minor segment is less than the area of its corresponding sector.
    Reason: The minor segment is obtained by removing the corresponding triangle from the sector.
  3. Assertion: If the radius of a circle is doubled while the sector angle remains unchanged, its sector area becomes four times.
    Reason: Sector area is proportional to r2r^2.
  4. Assertion: Two sectors having equal central angles must have equal areas.
    Reason: Sector area depends only on the central angle.
  5. Assertion: The area of a major sector can be obtained by subtracting the minor sector from the complete circular area.
    Reason: Minor sector and major sector together form the complete circle.

B. Case-Based Questions

Case Study 1 — Clock Hand

The minute hand of a clock is 1414 cm long. During a particular interval, it turns through 6060^\circ.

  1. What type of region is swept by the minute hand?
  2. Find the length of the arc traced by its tip.
  3. Find the area swept by the minute hand.
  4. If the hand instead turns through 120120^\circ, how will the swept area change?

Case Study 2 — Circular Design

A circular design has radius 2121 cm. A chord divides the circle so that the smaller sector subtends 120120^\circ at the centre.

  1. Find the area of the 120120^\circ sector.
  2. To obtain the area of the corresponding minor segment, what area must be subtracted from the sector?
  3. If the major sector is required, which angle should be used?
  4. Write an expression for its area.

Case Study 3 — Lighthouse

A lighthouse illuminates a sector of the sea through an angle of 8080^\circ, up to a distance of 16.516.5 km.

  1. Identify the radius of the illuminated region.
  2. Which formula gives the illuminated area?
  3. Calculate the area using π=3.14\pi=3.14.
  4. If the angle were increased while the radius remained unchanged, would the illuminated area increase or decrease? Give a reason.

C. Competency-Based Questions

  1. A student calculates the area of a 6060^\circ sector of radius 1212 cm using 60180πr2\frac{60}{180}\pi r^2. Identify the error and write the correct expression.
  2. Two sectors have the same radius. Their central angles are 4040^\circ and 100100^\circ. Without calculating their actual areas, determine the ratio of their areas.
  3. A sector and its corresponding arc are given. A student uses the sector-area formula to find the arc length. Explain why this method is incorrect and state the correct formula.
  4. A chord subtends 9090^\circ at the centre of a circle. A student claims that the minor segment area is equal to the sector area. Do you agree? Justify mathematically.
  5. A circular sheet is divided into equal sectors by diameters. Explain how the number of equal sectors determines the central angle of each sector.
  6. A problem asks for the area of a major segment. A student first calculates the area of the major sector and then subtracts the triangle formed by the radii. Explain why this approach may not directly give the required major segment.

D. 2-Mark Questions

  1. Find the area of a sector of radius 1414 cm and angle 4545^\circ. Use π=227\pi=\frac{22}{7}.
  2. Find the length of an arc of radius 2121 cm subtending 6060^\circ at the centre.
  3. A sector has angle 120120^\circ and radius 77 cm. Find its area.
  4. Find the angle of a sector whose area is one-fifth of the area of the complete circle.
  5. A circular region has radius 1010 cm. Find the area of a quadrant.
  6. A sector has radius 1212 cm and angle 150150^\circ. Write the formula for its arc length and substitute the values.

E. 3-Mark Questions

  1. A sector of a circle has radius 2121 cm and central angle 6060^\circ. Find:
    (i) the arc length
    (ii) the area of the sector.
  2. A chord of a circle of radius 1010 cm subtends 9090^\circ at the centre. Find the area of the corresponding minor segment. Use π=3.14\pi=3.14.
  3. A circle of radius 1414 cm has a sector of angle 9090^\circ. Find the area of the corresponding major sector.
  4. The area of a sector is 154 cm2154\text{ cm}^2, its radius being 1414 cm. Find its central angle. Use π=227\pi=\frac{22}{7}.
  5. A chord subtends 120120^\circ at the centre of a circle of radius 2121 cm. Explain the steps required to calculate the area of the minor segment.

F. 5-Mark Questions

  1. A chord of a circle of radius 2121 cm subtends an angle of 120120^\circ at the centre. Find the area of the corresponding minor segment. Hence, determine the area of the major segment. Use π=227\pi=\frac{22}{7}.
  2. A horse is tied to a peg at one corner of a square field by a 55-m rope. Assuming the rope reaches only within the field, find the area over which the horse can graze. What happens to the grazing area if the rope length is increased to 1010 m? Use π=3.14\pi=3.14.
  3. A circular umbrella has radius 4545 cm and eight equally spaced ribs. Find the area of the region between two consecutive ribs.
  4. Two non-overlapping car wipers each have a blade length of 2525 cm and sweep through an angle of 115115^\circ. Find the total area cleaned in one complete sweep of both blades.
  5. A lighthouse spreads light through a sector of angle 8080^\circ to a distance of 16.516.5 km. Calculate the area of the sea covered by the light. Use π=3.14\pi=3.14.

G. HOTS / Challenge Questions

  1. Two sectors have equal areas but different radii. If their central angles are 6060^\circ and 120120^\circ, determine the ratio of their radii.
  2. A sector has radius rr and angle θ\theta^\circ. Another sector has radius 2r2r and angle θ/2\theta/2. Compare their areas without substituting numerical values.
  3. A minor segment is formed by a chord subtending 6060^\circ at the centre. Explain why calculating the segment area requires information about the triangle formed by the two radii and the chord.
  4. A circular region is divided into 1010 equal sectors by five diameters. Find the central angle of each sector and express the area of one sector in terms of the circle’s radius.

ANSWER KEY

A. Assertion–Reason

  1. A
  2. A
  3. A
  4. D
  5. A

B. Case-Based

1: (1) Sector (2) 60360(2π×14)\frac{60}{360}(2\pi\times14) (3) 60360π(14)2\frac{60}{360}\pi(14)^2 (4) It doubles, since the angle doubles.

2: (1) 120360π(21)2=462 cm2\frac{120}{360}\pi(21)^2=462\text{ cm}^2 (2) Area of AOB\triangle AOB (3) 240240^\circ (4) 240360π(21)2\frac{240}{360}\pi(21)^2

3: (1) 16.516.5 km (2) θ360πr2\frac{\theta}{360}\pi r^2 (3) 190.19 km2\approx190.19\text{ km}^2 (4) Increase, because sector area is proportional to θ\theta.

C. Competency-Based

  1. The denominator must be 360360, not 180180: 60360π(12)2\frac{60}{360}\pi(12)^2.
  2. 40:100=2:540:100=\boxed{2:5}.
  3. Arc length measures a length, so use θ360(2πr)\frac{\theta}{360}(2\pi r).
  4. No. Segment area = sector area − triangle area.
  5. For nn equal sectors, each central angle is 360/n\boxed{360^\circ/n}.
  6. Major segment = complete circle − minor segment; directly subtracting the triangle from the major sector does not represent the required standard construction.

D. 2-Mark

  1. 77 cm277\text{ cm}^2
  2. 22 cm22\text{ cm}
  3. 1543 cm2\frac{154}{3}\text{ cm}^2
  4. 7272^\circ
  5. 50π cm250\pi\text{ cm}^2
  6. 150360(2π×12)=10π cm\frac{150}{360}(2\pi\times12)=10\pi\text{ cm}

E. 3-Mark

  1. Arc =22=22 cm; sector area =231 cm2=231\text{ cm}^2.
  2. Sector area =78.5 cm2=78.5\text{ cm}^2; triangle area =50 cm2=50\text{ cm}^2; segment =28.5 cm2=28.5\text{ cm}^2.
  3. Minor sector =154 cm2=154\text{ cm}^2; major sector =462 cm2=\boxed{462\text{ cm}^2}.
  4. θ360×227×142=154θ=90\frac{\theta}{360}\times\frac{22}{7}\times14^2=154\Rightarrow\boxed{\theta=90^\circ}.
  5. Find sector area first, then find OAB\triangle OAB, and subtract: segment=sector\boxed{\text{segment}=\text{sector}-\triangle}.

F. 5-Mark

  1. Minor sector =462 cm2=462\text{ cm}^2; triangle =44134 cm2=\frac{441\sqrt3}{4}\text{ cm}^2.
    Minor segment =46244134 cm2=462-\frac{441\sqrt3}{4}\text{ cm}^2.
    Major segment =441πminor segment=441\pi-\text{minor segment}.
  2. For 55 m rope: 14π(5)2=6.25π m2\boxed{\frac14\pi(5)^2=6.25\pi\text{ m}^2}.
    For 1010 m rope: 14π(10)2=25π m2\boxed{\frac14\pi(10)^2=25\pi\text{ m}^2}.
    Increase =18.75π m2=\boxed{18.75\pi\text{ m}^2}.
  3. Each sector angle =360/8=45=360^\circ/8=45^\circ.
    Area =45360π(45)2=2025π8 cm2=\frac{45}{360}\pi(45)^2=\boxed{\frac{2025\pi}{8}\text{ cm}^2}.
  4. Area per wiper =115360π(25)2=\frac{115}{360}\pi(25)^2.
    For two wipers:2×115360π(25)2\boxed{2\times\frac{115}{360}\pi(25)^2}
  5. 80360×3.14×(16.5)2190.19 km2\frac{80}{360}\times3.14\times(16.5)^2 \approx\boxed{190.19\text{ km}^2}

G. HOTS

  1. Equal areas imply θ1r12=θ2r22\theta_1r_1^2=\theta_2r_2^2. Hence 60r12=120r2260r_1^2=120r_2^2, so r1:r2=2:1\boxed{r_1:r_2=\sqrt2:1}.
  2. First area =θ360πr2=\frac{\theta}{360}\pi r^2; second =θ/2360π(2r)2=\frac{\theta/2}{360}\pi(2r)^2. Ratio =1:2=\boxed{1:2}.
  3. Because segment area is obtained by subtracting the corresponding triangle from the sector.
  4. Each angle =36=\boxed{36^\circ}; area of one sector =110πr2=\boxed{\frac1{10}\pi r^2}.

CLASS 10 MATHEMATICS

CHAPTER 11 — AREAS RELATED TO CIRCLES

BOARD-STYLE TEST PAPER

Time: 1½ Hours  Maximum Marks: 40

General Instructions

  1. All questions are compulsory unless an internal choice is given.
  2. Use π=227\pi=\frac{22}{7} unless another value is specified.
  3. Show necessary steps in calculation-based questions.
  4. Figures, wherever required, are not necessarily drawn to scale.

SECTION A — Objective Questions

8 × 1 = 8 Marks

1. The area of a sector of radius rr and angle θ\theta^\circ is:
(a) θ180πr2\frac{\theta}{180}\pi r^2
(b) θ360πr2\frac{\theta}{360}\pi r^2
(c) 360θπr2\frac{360}{\theta}\pi r^2
(d) θ3602πr\frac{\theta}{360}2\pi r

2. An arc subtends 7272^\circ at the centre. Its length is what fraction of the circumference?
(a) 15\frac15 (b) 14\frac14 (c) 25\frac25 (d) 15πr\frac15\pi r

3. A sector has angle 9090^\circ. The corresponding major sector has angle:
(a) 180180^\circ (b) 270270^\circ (c) 300300^\circ (d) 360360^\circ

4. The area of a minor segment is equal to:
(a) sector + triangle
(b) sector − triangle
(c) circle − sector
(d) circle − triangle

5. A quadrant of a circle has radius 1414 cm. Its area is:
(a) 154 cm2154\text{ cm}^2
(b) 308 cm2308\text{ cm}^2
(c) 616 cm2616\text{ cm}^2
(d) 44 cm244\text{ cm}^2

6. Assertion (A): If the radius of a sector is doubled and its angle remains unchanged, its area becomes four times.
Reason (R): The area of a sector is proportional to the square of its radius.
Choose: (a) A and R true, R explains A (b) A and R true, R does not explain A (c) A true, R false (d) A false, R true

7. Fill in the blank: The area of a major segment is the area of the complete circle minus the area of the corresponding ______ segment.

8. State whether True or False: The length of an arc depends on the radius of the circle and the central angle.


SECTION B — Very Short Answer

4 × 2 = 8 Marks

9. Find the area of a sector of radius 1414 cm and angle 4545^\circ.

10. Find the length of the arc of a circle of radius 2121 cm subtending 6060^\circ at the centre.

11. A sector has an angle of 120120^\circ. What fraction of the complete circular region does it represent? Hence, write its area in terms of rr.

12. A student says: “To find the area of a minor segment, I only need the sector-area formula.” Is the statement correct? Give a reason.


SECTION C — Short Answer

3 × 3 = 9 Marks

13. A circle has radius 2121 cm. An arc subtends an angle of 6060^\circ at the centre. Find:
(i) the arc length
(ii) the area of the corresponding sector.

14. A chord of a circle of radius 1010 cm subtends 9090^\circ at the centre. Find the area of the corresponding minor segment. Use π=3.14\pi=3.14.

15. A sector of radius 1414 cm has an angle of 9090^\circ. Find the area of the corresponding major sector. Also state the angle of the major sector.


SECTION D — Case-Based / Competency-Based

1 × 5 = 5 Marks

16. Case Study — Circular Grazing Region

A horse is tied to a peg at one corner of a square field by a rope of length 55 m. Assume that the rope reaches only within the field. The horse can therefore graze over a quarter of a circular region.

Answer the following:

(a) What is the central angle of the grazing sector? (1)
(b) Write the formula for its area. (1)
(c) Calculate the grazing area using π=3.14\pi=3.14. (2)
(d) If the rope length is doubled, by what factor does the grazing area change? (1)


SECTION E — Long Answer

2 × 5 = 10 Marks

17. A chord of a circle of radius 2121 cm subtends 120120^\circ at the centre. Find the area of the corresponding minor segment.
Use π=227\pi=\frac{22}{7} and 3=1.73\sqrt3=1.73.

OR

A circular region has radius 2121 cm. A sector has central angle 120120^\circ. Find:
(i) area of the sector
(ii) area of the corresponding triangle OABOAB
(iii) area of the minor segment.

18. A lighthouse spreads light over a sector of angle 8080^\circ to a distance of 16.516.5 km. Find the area of the sea covered by the light. Use π=3.14\pi=3.14. Also explain briefly why increasing the angle while keeping the distance fixed increases the illuminated area.

OR

Two car wipers do not overlap. Each blade is 2525 cm long and sweeps through an angle of 115115^\circ. Find the total area cleaned by both blades in one sweep.

ANSWER KEY

Areas Related to Circles — 40 Marks

SECTION A

  1. (b) θ360πr2\frac{\theta}{360}\pi r^2
  2. (a) 15\frac15
  3. (b) 270270^\circ
  4. (b) Sector − triangle
  5. (a) 154 cm2154\text{ cm}^2
  6. (a) Both true, and R explains A
  7. minor
  8. True

SECTION B

45360×227×142=77 cm2\frac{45}{360}\times\frac{22}{7}\times14^2 =\boxed{77\text{ cm}^2}

60360×2×227×21=22 cm\frac{60}{360}\times2\times\frac{22}{7}\times21 =\boxed{22\text{ cm}}

120360=13\frac{120}{360}=\boxed{\frac13}

Area:13πr2\boxed{\frac13\pi r^2}

  1. No.
    Minor segment area is:

sector areaarea of corresponding triangle\boxed{\text{sector area}-\text{area of corresponding triangle}}

SECTION C

(i) Arc length:22 cm\boxed{22\text{ cm}}

(ii) Sector area:231 cm2\boxed{231\text{ cm}^2}

  1. Sector area:

90360×3.14×102=78.5\frac{90}{360}\times3.14\times10^2=78.5

Triangle area:12(10)(10)=50\frac12(10)(10)=50

Minor segment:78.550=28.5 cm278.5-50=\boxed{28.5\text{ cm}^2}

  1. Major-sector angle:

36090=270360^\circ-90^\circ=\boxed{270^\circ}

Minor sector area:14×227×142=154\frac14\times\frac{22}{7}\times14^2=154

Major sector:616154=462 cm2616-154=\boxed{462\text{ cm}^2}

SECTION D

(a) 90\boxed{90^\circ}
(b) θ360πr2\boxed{\frac{\theta}{360}\pi r^2}
(c)14(3.14)(5)2=19.625 m2\frac14(3.14)(5)^2 =\boxed{19.625\text{ m}^2}

(d) Radius doubles, so area becomes 4 times\boxed{4\text{ times}}.

SECTION E

  1. Sector area:

120360×227×212=462 cm2\frac{120}{360}\times\frac{22}{7}\times21^2 =462\text{ cm}^2

For OAB\triangle OAB:OM=21cos60=10.5OM=21\cos60^\circ=10.5AM=21sin60=21(1.732)AM=21\sin60^\circ =21\left(\frac{1.73}{2}\right)AB=2AM=36.33 cmAB=2AM=36.33\text{ cm}

Therefore,Area of OAB=12(36.33)(10.5)190.73 cm2\text{Area of }\triangle OAB =\frac12(36.33)(10.5) \approx190.73\text{ cm}^2

Hence,Minor segment271.27 cm2\boxed{\text{Minor segment}\approx271.27\text{ cm}^2}

OR: Same values obtained through the three requested parts.

Area=80360×3.14×(16.5)2\text{Area} =\frac{80}{360}\times3.14\times(16.5)^2=190.19 km2 (approximately)=\boxed{190.19\text{ km}^2\text{ (approximately)}}

The area increases because, for a fixed radius, sector area is directly proportional to the central angle.

OR

Area cleaned by one wiper:115360×3.14×252\frac{115}{360}\times3.14\times25^2

For two wipers:2×115360×3.14×2522\times\frac{115}{360}\times3.14\times25^21252.4 cm2\boxed{\approx1252.4\text{ cm}^2}