Chapter 11: Areas Related to Circles

Important questions — the ones the board asked most (20)

1. Q1Long answer5 marksAsked: 2024

   An arc of a circle of radius 21 cm subtends an angle of 60 degree at the centre. Find : (i) the length of the arc. (ii) the area of the minor segment of the circle made by the corresponding chord.

   

   Medha Ma’am

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   Model answer · arc length, area of segment

   1. 1.Given: a circle with centre O and radius r = 21 cm. The arc AB subtends ∠AOB = θ = 60° at the centre, and AB is the corresponding chord. Let L be any point on the minor arc AB, so OALB is the sector bounded by the radii OA, OB and the arc ALB. Use π = 22/7. (i) Length of an arc of a sector of angle θ = 2πr × θ/360 Length of the arc AB = 2 × 22/7 × 21 × 60/3601.5 marks
   2. 2.= 2 × 22 × 3 × 1/6 = 22 cm So, the length of the arc is 22 cm.0.5 marks
   3. 3.(ii) Area of the sector of angle θ = πr² × θ/360 Area of sector OALB = 22/7 × 21 × 21 × 60/360 = 22 × 3 × 21 × 1/6 = 231 cm²1 mark
   4. 4.In ΔOAB, OA = OB = 21 cm (radii) and ∠AOB = 60°, so ΔOAB is an equilateral triangle with side 21 cm. Area of ΔOAB = √3/4 × 21 × 21 = (441√3)/4 cm²1.5 marks
   5. 5.Area of segment of a circle = Area of the corresponding sector – Area of the corresponding triangle Area of minor segment ALB = Area of sector OALB – Area of ΔOAB = (231 − (441√3)/4) cm²0.5 marks

   • Write the substitution "2 × 22/7 × 21 × 60/360" before working out the arc length as 22 cm.
   • State that ΔOAB is equilateral because OA = OB = 21 cm and ∠AOB = 60°, then use √3/4 × 21 × 21.
   • Finish with "Area of minor segment ALB = Area of sector OALB – Area of ΔOAB" and write cm² with every area.

   Marks split as in the board’s published marking scheme

2. Q2Long answer5 marksAsked: 2024

   In the given figure, diameters AC and BD of the circle intersect at O. If angle AOB = 60 degree and OA = 10 cm, then : (i) find the length of the chord AB. (ii) find the area of shaded region. (Take pi = 3.14 and sqrt(3) = 1.73)

   

   Medha Ma’am

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   Hear it from Medha Ma’am · 165 s

   Model answer · Area of segment and sector

   1. 1.(i) Given: diameters AC and BD meet at the centre O, ∠AOB = 60° and OA = 10 cm. OA = OB = 10 cm (radii), so ∠OAB = ∠OBA. The three angles add to 180°, so ∠OAB = ∠OBA = (180° – 60°)/2 = 60°. So, Δ OAB is an equilateral triangle. Therefore, the chord AB = OA = 10 cm.1 mark
   2. 2.(ii) The shaded region is made of the segment APB on the chord AB and the sector OBC. Area of segment of a circle = Area of the corresponding sector – Area of the corresponding triangle. Area of the sector = (θ/360) × πr², and the area of an equilateral triangle of side a = (√3/4)a². Area of segment APB (A1) = 3.14 × 100 × 60/360 – (1.73/4) × 100 = 52.33 – 43.251 mark
   3. 3.A1 = 9.08 cm² (approx.)1 mark
   4. 4.AC is a diameter, so ∠AOB + ∠BOC = 180°. ∠BOC = 180° – 60° = 120°. Area of sector OBC (A2) = 3.14 × 100 × 120/3601 mark
   5. 5.A2 = 104.67 cm² (approx.) Area of shaded region = A1 + A2 = 9.08 + 104.67 = 113.75 cm² (approx.)1 mark

   • Write "Δ OAB is an equilateral triangle" and give the reason, OA = OB with ∠AOB = 60°, before you write AB = 10 cm.
   • Show that ∠BOC = 180° – 60° = 120° because AC is a diameter, then use it in the sector formula.
   • Write cm² after each area: A1, A2 and the final 113.75 cm².

   Marks split as in the board’s published marking scheme

3. Q3Long answer5 marksAsked: 2024

   The perimeter of a certain sector of a circle of radius 5.6 m is 20.0 m. Find the area of the sector.

   

   Medha Ma’am

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   Hear it from Medha Ma’am · 120 s

   Model answer · Perimeter and area of sector

   1. 1.The area of the sector is 24.64 m². It is worked out below. Given: radius of the circle, r = 5.6 m; perimeter of the sector = 20.0 m. Let θ be the angle of the sector in degrees. The perimeter of a sector is its two radii plus its arc. Length of an arc of a sector of angle θ = 2πr × θ/360 So, perimeter of the sector = 2r + 2πr × θ/360 2r + 2πr × θ/360 = 202 marks
   2. 2.Putting r = 5.6 m and π = 22/7: 2 × 5.6 + 2 × 22/7 × 5.6 × θ/360 = 20 ⇒ 11.2 + 2 × 22/7 × 5.6 × θ/360 = 201 mark
   3. 3.Solving: 35.2 × θ/360 = 20 − 11.2 = 8.8 θ/360 = 8.8/35.2 = 1/4 θ = 90° So the angle of the sector is 90°.1 mark
   4. 4.Area of the sector of angle θ = πr² × θ/360 ∴ Area of sector = 22/7 × 5.6 × 5.6 × 90/360 = 98.56 × 1/4 = 24.64 m² Hence, the area of the sector is 24.64 m².1 mark

   • Write the perimeter as "2r + 2πr × θ/360 = 20", which is the two radii plus the arc.
   • Find "θ = 90°" first, then put it into "πr² × θ/360".
   • End with the unit: "24.64 m²".

   Marks split as in the board’s published marking scheme

4. Q4Long answer5 marksAsked: 2023

   A chord of a circle of radius 14 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor segment of the circle. Also find the area of the major segment of the circle.

   How the model answer begins

   1. Given: a circle with centre O and radius r = 14 cm. A chord AB subtends ∠AOB = θ = 60° at the centre. We take π = 22/7 and √3 = 1.732. The area of the sector is found with (θ/360) × πr², and the area of the whole circle…

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5. Q5Long answer5 marksAsked: 2023

   33. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find the area of that part of the field in which the horse can graze. Also, find the increase in grazing area if length of rope is increased to 10 m. (Use pi = 3.14)

   How the model answer begins

   1. Given: a square grass field of side 15 m, with a horse tied to a peg at one corner by a rope 5 m long. Use π = 3.14. Each angle of a square is 90°. The rope (5 m) is shorter than the side (15 m), so the horse can graze…

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6. Q6Long answer5 marksAsked: 2024

   O is the centre of the circle. If AC = 28 cm, BC = 21 cm, angle BOD = 90 degree and angle BOC = 30 degree, then find the area of the shaded region given in the figure.

   How the model answer begins

   1. Given: O is the centre of the circle, AC = 28 cm, BC = 21 cm, ∠BOD = 90°, ∠BOC = 30°. AOB is taken to be a straight line through the centre O, so AB is the diameter of the circle. The angle in a semicircle is a right…

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7. Q7Long answer5 marksAsked: 2024

   A chord of a circle of radius 14 cm subtends an angle of 90 degree at the centre. Find the area of the corresponding minor and major segments of the circle.

   How the model answer begins

   1. Given: radius r = 14 cm, the chord AB subtends θ = 90° at the centre O. Use π = 22/7. Area of segment of a circle = Area of the corresponding sector – Area of the corresponding triangle Area of the sector = (θ/360) ×…

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8. Q8Long answer5 marksAsked: 2023

   (a) In the given figure, AB is a chord of a circle with centre O and radius 7 cm. If angle AOB = 90 degrees, find the area of the shaded region. Also find the length of minor arc AB. or (b) AB and CD are arcs of two concentric circles with centre O and radii 3.5 cm and 10.5 cm respectively. If angle AOB = 60 degrees, find the area of the shaded region and also find the length of arc CD.

   How the model answer begins

   1. Answering part (a). Given: a circle with centre O and radius r = 7 cm, and a chord AB with angle AOB = θ = 90°. The shaded region is the minor segment AB. Area of segment of a circle = Area of the corresponding sector –…

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9. Q9Long answer5 marksAsked: 2023

   (a) In the given figure, AB is a chord of a circle with centre O and radius 7 cm. If angle AOB = 90 degree, find the area of the shaded region. Also find the length of minor arc AB. OR (b) AB and CD are two arcs of two concentric circles with centre O and radii 3.5 cm and 10.5 cm respectively. If angle AOB = 60 degree, find the area of the shaded region and also find the length of arc CD.

   How the model answer begins

   1. Given: a circle with centre O, radius r = 7 cm, chord AB and ∠AOB = θ = 90°. The shaded region is the minor segment cut off by chord AB. Area of the segment = Area of the corresponding sector – Area of the corresponding…

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10. Q10Long answer5 marksAsked: 2023

    (a) In the given figure, AB is a chord of a circle of radius 7 cm and centred at O. Find the area of the shaded region if angle AOB = 90°. Also, find length of minor arc AB. OR (b) AB and CD are arcs of two concentric circles of radii 3.5 cm and 10.5 cm respectively and centred at O. Find the area of the shaded region if angle AOB = 60°. Also, find the length of arc CD.

    How the model answer begins

    1. Given: the circle has centre O and radius r = 7 cm. The chord AB makes ∠AOB = θ = 90° at the centre. The shaded region is the segment formed by the chord AB, so: Area of segment = Area of the corresponding sector – Area…

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11. Q11Case-based4 marksAsked: 2026

    A brooch is crafted from silver wire in the shape of a circle with a diameter of 35 cm. The wire is also used to create 5 diameters, dividing the circle into 10 equal sectors as shown in figure. Based on the above information, answer the following questions : (i) What is the radius of circle ? (ii) What is the circumference of the brooch ? (iii) (a) What is the total length of silver wire required ? OR (iii) (b) What is the area of each sector of the brooch ?

12. Q12Case-based4 marksAsked: 2025

    A farmer has put up a decorative windmill in his farm in which there are eight blades of equal width and equally placed in a circular arrangement. A circular wire goes through them. The diagram shows two blades OAB and OPQ in a quarter circle with centre O. angle AOB = angle POQ = 30°, OA = 28 cm, OC = 21 cm. O is the centre of both the circles. (i) Determine the measure of angle BOP. (ii) Find length of arc CD. (iii) (a) Find the area of region CABD. OR (iii) (b) Find perimeter of region CABD.

    How the model answer begins

    1. (i) The two blades OAB and OPQ lie in a quarter circle with centre O, so angle AOC = 90°. Angle AOB = angle POQ = 30°, so the two blades together take 60°. Because the blades are equally placed, the remaining angle is…

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13. Q13Case-based4 marksAsked: 2025

    38. A farmer installed a windmill in his field for decoration. It has eight blades of equal width which are equally arranged in a circle. A circular wire holds all of them together. The given diagram shows two blades OAB and OPQ in a quadrant of a circle. Both circles are concentric at point O. angleAOB = anglePOQ = 30°, OA = 28 cm and OC = 21 cm. (i) Find the value of angleBOP. (ii) Find the length of arc CD. (iii) (a) Find the area of region CABD. or (iii) (b) Find the perimeter of region CABD.

    How the model answer begins

    1. (i) The two blades OAB and OPQ lie in a quadrant of the circle, so angle AOC = 90°. The blades are equally placed, so the gap between the two blades is equal to the gap left at the end of the quadrant. The two blades…

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14. Q14Case-based4 marksAsked: 2024

    NSS (National Service Scheme) aims to connect the students to the community and to involve them in problem solving process. NSS symbol is based on the 'Rath' wheel of the Konark Sun Temple situated in Odisha. The wheel signifies the progress cycle of life. The diagramatic representation of the symbol is given below : Observe the given figure. The inner circle diameters are placed symmetrically. Given that OP = 21 cm and OS = 10 cm, then Based on the above information, answer the following questions : (i) Find m∠ROS. (ii) Find the perimeter of sector OPQ. (iii) (a) Find the area of shaded region PQRS. OR (iii) (b) Find the area of shaded region ACB (segment ACB).

    How the model answer begins

    1. (i) The diameters are placed symmetrically, so they divide the circle with centre O into 8 equal sectors. Together, the angles of these sectors make 360° at the centre O. m∠ROS = 360°/8 = 45°

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15. Q15Case-based4 marksAsked: 2024

    N.S.S. (National Service Scheme) aims to connect students with the community and involve them in the problem solving process. The symbol of N.S.S. is based on the chariot wheel of the Konark Sun Temple in Odisha. The wheel is a symbol of the cycle of progress of life. The schematic diagram of the symbol is given below : Observe the given figure. The diameters of the inner circle are placed evenly. Given that OP = 21 cm and OS = 10 cm, then Based on the above information, answer the following questions : (i) Find m angle ROS. (ii) Find the perimeter of sector OPQ. (iii) (a) Find the area of shaded region PQRS. OR (iii) (b) Find the area of shaded region ACB (segment ACB).

    How the model answer begins

    1. (i) Given: the wheel has centre O, outer radius OP = 21 cm and inner radius OS = 10 cm. Its 4 diameters are placed evenly, so they divide the full angle of 360° at O into 8 equal angles. ∠ROS = 360° / 8 ∠ROS = 45°

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16. Q16Case-based4 marksAsked: 2024

    NSS (National Service Scheme) aims to connect the students to the community and to involve them in problem solving process. NSS symbol is based on the ‘Rath’ wheel of the Konark Sun Temple situated in Odisha. The wheel signifies the progress cycle of life. The diagrammatic representation of the symbol is given below : Observe the figure given above. The diameters of inner circle are equally placed. Given that OP = 21 cm, OS = 10 cm. Based on the above information, answer the following questions : (i) Find m∠ROS. (ii) Find the perimeter of sector OPQ. (iii) (a) Find the area of shaded region PQRS. OR (iii) (b) Find the area of shaded region ACB i.e. the segment ACB.

    How the model answer begins

    1. (i) The diameters of the inner circle are equally placed. Four diameters divide the full angle of 360° at the centre O into 8 equal angles. So each of these angles, including ∠ROS, is one-eighth of 360°. m∠ROS = 360° /…

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17. Q17Case-based4 marksAsked: 2023

    Governing council of a local public development authority of Dehradun decided to build an adventurous playground on the top of a hill, which will have adequate space for parking. After survey, it was decided to build rectangular playground, with a semi-circular area allotted for parking at one end of the playground. The length and breadth of the rectangular playground are 14 units and 7 units, respectively. There are two quadrants of radius 2 units on one side for special seats. Based on the above information, answer the following questions : (i) What is the total perimeter of the parking area ? (ii) (a) What is the total area of parking and the two quadrants ? OR (b) What is the ratio of area of playground to the area of parking area ? (iii) Find the cost of fencing the playground and parking area at the rate of 2 per unit.

18. Q18Case-based4 marksAsked: 2023

    An inter-school rangoli competition was organised by one of the prestigious schools of Odisha. The theme of the rangoli competition was Diwali celebration where students had to make mathematical designs. Students from different schools participated in this competition and made beautiful rangoli designs. One such design is given below : Rangoli is in the shape of a square marked as ABCD. The side of the square is 40 cm. Quadrants of radius 10 cm are made at each corner of this square (in which diyas are placed). A circle of diameter 20 cm is also made in the middle of the square. Based on the above, answer the following questions : (i) What is the area of square ABCD ? (ii) What is the area of the circle ? (iii) If the circle and the four quadrants are cut out from the square ABCD, then find the area of the remaining square ABCD. OR (iii) Find the total area of the removed circle and four quadrants.

    How the model answer begins

    1. (i) Given: side of square ABCD = 40 cm. Area of square = side × side Area of square ABCD = (40)² cm² = 1600 cm²

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19. Q19Case-based4 marksAsked: 2023

    Case Study - 1 Flower beds in a garden look very beautiful. In one such circular park of radius 'r' m, there are flowers in two segments. One segment which subtends an angle of 90 degree at the centre has red roses, while the other segment, which subtends an angle of 60 degree at the centre, has yellow flowers. [See figure] It is given that the total area of the flower parts (both segments) is 256 2/3 m^2. Based on the above, answer the following questions : (i) Write the equation in terms of 'r' representing the sum of the areas of both segments. (1) (ii) Find the value of 'r'. (1) (iii) (a) Find the area of the segment with red roses. (2) OR (iii) (b) Find the area of the segment with yellow flowers. (2)

20. Q20Case-based4 marksAsked: 2023

    Flower beds look beautiful growing in gardens. One such circular park of radius 'r' m, has two segments with flowers. One segment which subtends an angle of 90 degree at the centre is full of red roses, while the other segment with central angle 60 degree is full of yellow coloured flowers. [See figure] It is given that the combined area of the two segments (of flowers) is 256 2/3 sq m. Based on the above, answer the following questions : (i) Write an equation representing the total area of the two segments in terms of 'r'. (ii) Find the value of 'r'. (iii) (a) Find the area of the segment with red roses. OR (iii) (b) Find the area of the segment with yellow flowers.

51 more questions of this chapter, each with its model answer, in the AskMedha app.

Objective questions with the board’s key (10)

1. 1. CBSE 2026

   The length of the arc of a sector of a circle with radius 21 cm and central angle 60 degrees is :

   • (A) 22 cm ✓
   • (B) 44 cm
   • (C) 88 cm
   • (D) 11 cm

   Answer: (A)

2. 2. CBSE 2026

   The minute hand of a clock is 7 cm long. The angle swept by it between 7:00 a.m. and 8:10 a.m. is :

   • (A) (35/4) degrees
   • (B) (35/2) degrees
   • (C) 35 degrees ✓
   • (D) 70 degrees

   Answer: (C)

3. 3. CBSE 2026

   The hour hand of a clock is 7 cm long. The angle swept by it between 7:00 a.m. and 8:10 a.m. is :

   • (A) (35/4)°
   • (B) (35/2)°
   • (C) 35° ✓
   • (D) 70°

   Answer: (C)

4. 4. CBSE 2026

   In the given figure, a circle with centre O is shown. The area of the minor sector is 7 cm^2. The area of the circle is :

   • (A) 84 pi cm^2
   • (B) 84/11 cm^2
   • (C) 84 cm^2 ✓
   • (D) sqrt(84)/sqrt(pi) cm^2

   Answer: (C)

5. 5. CBSE 2026

   In the given figure, O is the centre of the circle. XYZ is an arc of the circle, which subtends an angle of 45 degree at the centre. If the radius of the circle is 32 cm, then the length of arc XYZ is :

   • (A) 4 pi cm
   • (B) 8 pi cm ✓
   • (C) 64 pi cm
   • (D) 128 pi cm

   Answer: (B)

6. 6. CBSE 2026

   The area of a sector with radius 'r' and central angle 60 degree is :

   • (A) pi r^2 / 2 - 1/2 r^2
   • (B) 2 pi r / 4 - sqrt(3)/4 r^2
   • (C) pi r^2 / 6 - sqrt(3)/4 r^2 ✓
   • (D) 2 pi r / 4 - r^2 sin 60 degree

   Answer: (C)

7. 7. CBSE 2026

   Area of a segment of a circle of radius 'r' and central angle 60° is :

   • (A) pi*r^2/2 - 1/2*r^2
   • (B) 2*pi*r/4 - sqrt(3)/4*r^2
   • (C) pi*r^2/6 - sqrt(3)/4*r^2 ✓
   • (D) 2*pi*r/4 - r^2 sin 60°

   Answer: (C)

8. 8. CBSE 2026

   The area of a semicircle of diameter 'd' is :

   • (A) pi d^2 / 16
   • (B) pi d^2 / 4
   • (C) pi d^2 / 8 ✓
   • (D) pi d^2 / 2

   Answer: (C)

9. 9. CBSE 2026

   Chord AB subtends an angle of 40 degree at the centre of the circle of radius 9 cm. The length of arc AB is :

   • (A) 22/7 cm
   • (B) 198/7 cm
   • (C) 44/7 cm ✓
   • (D) 54/7 cm

   Answer: (C)

10. 10. CBSE 2026

    An arc of length 11 cm subtends an angle of 105 degrees at the centre of the circle. The radius of the circle is :

    • (A) 8 cm
    • (B) 4sqrt(3) cm
    • (C) 6 cm ✓
    • (D) 7 cm

    Answer: (C)

Questions parents and students ask

How many marks does "Areas Related to Circles" carry in CBSE Class 10 Maths?

In the CBSE papers of 2023–2026, this chapter carried about 5 marks and about 2 questions a year; by marks it ranks 13 in Maths.

Which are the most important (VVI) questions of "Areas Related to Circles"?

The 20 questions above are from the board’s real papers, each with the years it was asked; the most-asked come first with their full model answers. All 71 written questions of this chapter, with answers, are in the AskMedha app.

Do the answers follow the board’s marking scheme?

Yes. CBSE publishes its marking schemes; every model answer here is written point by point to the marks the board’s own scheme gives, and checked by a second, independent model.