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Areas Related to Circles: 4 marks Questions (CBSE Class 10)

13 different 4 marks questions on Areas Related to Circles from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (63)2 marks (12)3 marks (30)4 marks (13)5 marks (9)

Q384 marksCase StudyAreas Related to Circles CBSE 2026 · Standard 30/1/1

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 ? (1)
(ii) What is the circumference of the brooch ? (1)
(iii) (a) What is the total length of silver wire required ? (2)
OR
(iii) (b) What is the area of each sector of the brooch ? (2)

Diagram for CBSE 2026 Class 10 Maths question 38

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Answer: (i) 17.5 cm (ii) 110 cm (iii) (a) 285 cm OR (b) 96.25 cm

^{2}

(i) Radius $= \frac{35}{2} = 17.5$ cm.
(ii) Circumference $= π d = \frac{22}{7} \times 35 = 110$ cm.
(iii) (a) Wire = circumference + 5 diameters $= 110 + 5 \times 35 = 285$ cm.
(iii) (b) Each sector has central angle $3 6^{\circ}$ , i.e. $\frac{1}{10}$ of the circle.
Area $= \frac{1}{10} \times \frac{22}{7} \times 17.5 \times 17.5 = \frac{1}{10} \times 962.5 = 96.25$ cm $^{2}$ .

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q374 marksCase StudyAreas Related to Circles CBSE 2025 · Basic 430/5/1

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. $∠ A O B = ∠ P O Q = 3 0^{\circ}$ , OA = 28 cm, OC = 21 cm.
O is the centre of both the circles.
(i) Determine the measure of $∠ B O P$ . (1)
(ii) Find length of arc CD. (1)
(iii) (a) Find the area of region CABD. (2)
OR
(iii) (b) Find perimeter of region CABD. (2)

Diagram for CBSE 2025 Class 10 Maths question 37

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Answer: (i)

1 5^{\circ}

(ii) 11 cm (iii) (a)

\frac{539}{6}

^{2}

\approx 89.83

^{2}

OR (b)

\frac{119}{3}

\approx 39.67

(i) The quarter circle AOR is $9 0^{\circ}$ ; the two blades take $3 0^{\circ} + 3 0^{\circ} = 6 0^{\circ}$ . The blades are equally placed, so the remaining $3 0^{\circ}$ is shared equally by $∠ B O P$ and $∠ QO R$ : $∠ B O P = 1 5^{\circ}$ (check: 8 blades of $3 0^{\circ}$ and 8 equal gaps of $1 5^{\circ}$ make $36 0^{\circ}$ ).
(ii) Arc CD $= \frac{30}{360} \times 2 \times \frac{22}{7} \times 21 = 11$ cm.
(iii) (a) Area CABD = sector OAB – sector OCD $= \frac{30}{360} \times \frac{22}{7} \times (2 8^{2} - 2 1^{2}) = \frac{1}{12} \times \frac{22}{7} \times 343 = \frac{539}{6} \approx 89.83$ cm $^{2}$ .
(iii) (b) Arc AB $= \frac{30}{360} \times 2 \times \frac{22}{7} \times 28 = \frac{44}{3}$ cm; CA = BD = 28 – 21 = 7 cm.
Perimeter $= \frac{44}{3} + 11 + 7 + 7 = \frac{119}{3} \approx 39.67$ cm.

Also asked in: 2025 Basic 430/5/2, 2025 Basic 430/5/3

Q374 marksCase StudyAreas Related to Circles CBSE 2025 · Standard 30/1/1

A brooch is a decorative piece often worn on clothing like jackets, blouses or dresses to add elegance. Made from precious metals and decorated with gemstones, brooches come in many shapes and designs.
One such brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the figure.
Based on the above given information, answer the following questions :
(i) Find the central angle of each sector. (1)
(ii) Find the length of the arc ACB. (1)
(iii) (a) Find the area of each sector of the brooch. (2)
OR
(iii) (b) Find the total length of the silver wire used. (2)

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Answer: (i)

3 6^{\circ}

(ii)

11

mm (iii) (a)

96.25

^{2}

OR (iii) (b)

285

Radius $r = \frac{35}{2} = 17.5$ mm; circumference $= 2 \times \frac{22}{7} \times 17.5 = 110$ mm.
(i) Central angle of each sector $= \frac{36 0 ^{\circ}}{10} = 3 6^{\circ}$ .
(ii) Arc ACB is the arc of one sector: length $= \frac{36}{360} \times 110 = 11$ mm.
(iii) (a) Area of each sector $= \frac{36}{360} \times \frac{22}{7} \times (17.5)^{2} = \frac{1}{10} \times 962.5 = 96.25$ mm $^{2}$ .
(iii) (b) Wire $=$ circumference $+ 5$ diameters $= 110 + 5 \times 35 = 110 + 175 = 285$ mm.

Also asked in: 2025 Standard 30/1/2, 2025 Standard 30/1/3

Q364 marksCase StudyAreas Related to Circles CBSE 2025 · Standard 30/4/1

The Olympic symbol comprising five interlocking rings represents the union of the five continents of the world and the meeting of athletes from all over the world at the Olympic games. In order to spread awareness about Olympic games, students of Class-X took part in various activities organised by the school. One such group of students made 5 circular rings in the school lawn with the help of ropes. Each circular ring required 44 m of rope.
Also, in the shaded regions as shown in the figure, students made rangoli showcasing various sports and games. It is given that $△ O A B$ is an equilateral triangle and all unshaded regions are congruent.
Based on above information, answer the following questions :
(i) Find the radius of each circular ring. (1)
(ii) What is the measure of $∠ A O B$ ? (1)
(iii) (a) Find the area of shaded region $R_{1}$ . (2)
OR (iii) (b) Find the length of rope around the unshaded regions. (2)

Diagram for CBSE 2025 Class 10 Maths question 36

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Answer: (i) 7 m (ii)

6 0^{\circ}

(iii) (a)

(\frac{308}{3} + \frac{49 3}{2})

^{2} \approx 145.1

^{2}

OR (iii) (b)

\frac{176}{3}

\approx 58.67

(i) $2 π r = 44 \Rightarrow r = 44 \times \frac{7}{44} = 7$ m
(ii) $△ O A B$ is equilateral, so $∠ A O B = 6 0^{\circ}$
(iii)(a) Each unshaded region (lens) is made of two congruent segments with chord AB = 7 m subtending $6 0^{\circ}$ .
Segment area $= \frac{60}{360} \times \frac{22}{7} \times 49 - \frac{3}{4} \times 49 = \frac{77}{3} - \frac{49 3}{4}$
Lens area $= \frac{154}{3} - \frac{49 3}{2}$
$R_{1}$ = circle $-$ lens $= 154 - \frac{154}{3} + \frac{49 3}{2} = \frac{308}{3} + \frac{49 3}{2} \approx 102.67 + 42.44 = 145.1$ m $^{2}$
(iii)(b) Each lens is bounded by two arcs of $6 0^{\circ}$ : length $= 2 \times \frac{60}{360} \times 44 = \frac{44}{3}$ m
There are 4 unshaded regions: total $= 4 \times \frac{44}{3} = \frac{176}{3} \approx 58.67$ m

Also asked in: 2025 Standard 30/4/2, 2025 Standard 30/4/3

Q364 marksCase StudyAreas Related to Circles CBSE 2025 · Standard 30/5/1

A farmer has a circular piece of land. He wishes to construct his house in the form of largest possible square within the land as shown below.
The radius of circular piece of land is 35 m.
Based on given information, answer the following questions :
(i) Find the length of wire needed to fence the entire land. (1)
(ii) Find the length of each side of the square land on which house will be constructed. (1)
(iii) (a) The farmer wishes to grow grass on the shaded region around the house. Find the cost of growing the grass at the rate of ₹ 50 per square metre. (2)
OR (iii) (b) Find the ratio of area of land on which house is built to remaining area of circular piece of land. (2)

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Answer: (i) 220 m (ii)

352

m (iii)(a) ₹ 70000 OR (iii)(b) 7 : 4

(i) Length of wire $= 2 π r = 2 \times \frac{22}{7} \times 35 = 220$ m
(ii) Diagonal of square = diameter = 70 m, so side $= \frac{70}{2} = 352$ m
(iii)(a) Area of circle $= \frac{22}{7} \times 3 5^{2} = 3850$ m $^{2}$ ; area of square $= (352)^{2} = 2450$ m $^{2}$
Shaded area $= 3850 - 2450 = 1400$ m $^{2}$ ; cost $= 1400 \times 50 =$ ₹ 70000
(iii)(b) Ratio $= 2450 : 1400 = 7 : 4$

Also asked in: 2025 Standard 30/5/2, 2025 Standard 30/5/3

Q364 marksCase StudyAreas Related to Circles CBSE 2025 · Standard 30/6/1

Anurag purchased a farmhouse which is in the form of a semicircle of diameter 70 m. He divides it into three parts by taking a point P on the semicircle in such a way that $∠ P A B = 3 0^{\circ}$ as shown in the following figure, where O is the centre of semicircle.
In part I, he planted saplings of Mango tree, in part II, he grew tomatoes and in part III, he grew oranges. Based on given information, answer the following questions.
(i) What is the measure of $∠ P O A$ ? (1)
(ii) Find the length of wire needed to fence entire piece of land. (1)
(iii) (a) Find the area of region in which saplings of Mango tree are planted. (2)
OR (iii) (b) Find the length of wire needed to fence the region III. (2)

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Answer: (i)

12 0^{\circ}

(ii) 180 m (iii) (a)

(\frac{1925}{3} - \frac{1225 3}{4})

^{2} \approx 111.24

^{2}

OR (iii) (b)

(\frac{220}{3} + 353)

\approx 133.96

(i) $∠ P O B = 2∠ P A B = 6 0^{\circ}$ (angle at centre), so $∠ P O A = 18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ .
(ii) $r = 35$ m. Fence = semicircular arc + diameter $= \frac{22}{7} \times 35 + 70 = 110 + 70 = 180$ m.
(iii) (a) Part I is the segment cut off by chord PB, with $∠ P O B = 6 0^{\circ}$ .
Sector area $= \frac{60}{360} \times \frac{22}{7} \times 3 5^{2} = \frac{1925}{3} \approx 641.67$ m $^{2}$
$△ P O B$ is equilateral: area $= \frac{3}{4} \times 3 5^{2} = \frac{1225 3}{4} \approx 530.43$ m $^{2}$
Area of part I $\approx 641.67 - 530.43 = 111.24$ m $^{2}$
(iii) (b) Region III is bounded by chord AP and arc AP, with $∠ A O P = 12 0^{\circ}$ .
Arc AP $= \frac{120}{360} \times 2 \times \frac{22}{7} \times 35 = \frac{220}{3} \approx 73.33$ m
$∠ A P B = 9 0^{\circ}$ , so $A P = A B cos 3 0^{\circ} = 70 \times \frac{3}{2} = 353 \approx 60.62$ m
Wire needed $\approx 73.33 + 60.62 \approx 133.96$ m

Also asked in: 2025 Standard 30/6/2, 2025 Standard 30/6/3

Q374 marksCase StudyAreas Related to Circles CBSE 2024 · Basic 430/3/1

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 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 $m ∠ R O S$ . (1)
(ii) Find the perimeter of sector OPQ. (1)
(iii) Find the area of shaded region PQRS. (2)
OR Find the area of shaded region ACB i.e. the segment ACB. (2)

Diagram for CBSE 2024 Class 10 Maths question 37

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Answer: (i)

4 5^{\circ}

(ii) 58.5 cm (iii)

\frac{3751}{28} \approx 133.96

^{2}

; OR

\frac{200}{7} \approx 28.57

^{2}

(i) Four equally placed diameters make 8 equal angles: $∠ R O S = \frac{36 0 ^{\circ}}{8} = 4 5^{\circ}$ .
(ii) Arc PQ $= \frac{45}{360} \times 2 \times \frac{22}{7} \times 21 = 16.5$ cm; perimeter $= 21 + 21 + 16.5 = 58.5$ cm.
(iii) Area PQRS $= \frac{45}{360} \times \frac{22}{7} \times (2 1^{2} - 1 0^{2}) = \frac{1}{8} \times \frac{22}{7} \times 341 = \frac{3751}{28} \approx 133.96$ cm $^{2}$ .
OR: $∠ A O B = 2 \times 4 5^{\circ} = 9 0^{\circ}$ . Segment ACB = sector OACB $-$ $△ O A B$ .
$= \frac{90}{360} \times \frac{22}{7} \times 100 - \frac{1}{2} \times 10 \times 10 = \frac{550}{7} - 50 = \frac{200}{7} \approx 28.57$ cm $^{2}$ .

Also asked in: 2024 Basic 430/3/2, 2024 Basic 430/3/3

Q364 marksCase StudyAreas Related to Circles CBSE 2024 · Standard 30/3/1

A stable owner has four horses. He usually tie these horses with 7 m long rope to pegs at each corner of a square shaped grass field of 20 m length, to graze in his farm. But tying with rope sometimes results in injuries to his horses, so he decided to build fence around the area so that each horse can graze.
Based on the above, answer the following questions :
(i) Find the area of the square shaped grass field. (1)
(ii) Find the area of the total field in which these horses can graze. (2)
OR If the length of the rope of each horse is increased from 7 m to 10 m, find the area grazed by one horse. (Use $π$ = 3.14) (2)
(iii) What is area of the field that is left ungrazed, if the length of the rope of each horse is 7 cm ? (1)

Diagram for CBSE 2024 Class 10 Maths question 36

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Answer: (i) 400 m

^{2}

(ii) 154 m

^{2}

; OR 78.5 m

^{2}

(iii) 399.9846 m

^{2}

as printed (7 cm); 246 m

^{2}

if the rope is 7 m

(i) Area of the square = $20 \times 20 = 400$ m $^{2}$ .
(ii) Each horse grazes a quadrant of radius 7 m; four quadrants make one full circle: $\frac{22}{7} \times 7^{2} = 154$ m $^{2}$ .
OR: Area grazed by one horse = $\frac{1}{4} \times 3.14 \times 1 0^{2} = 78.5$ m $^{2}$ .
(iii) With a 7 cm = 0.07 m rope, grazed area = $\frac{22}{7} \times 0.0 7^{2} = 0.0154$ m $^{2}$ , so ungrazed area = $400 - 0.0154 = 399.9846$ m $^{2}$ .
If the rope is 7 m (as in the passage), ungrazed area = $400 - 154 = 246$ m $^{2}$ .

Also asked in: 2024 Standard 30/3/2, 2024 Standard 30/3/3

Q374 marksCase StudyAreas Related to Circles CBSE 2023 · Basic 430/1/1

Interschool Rangoli Competition was organized by one of the reputed schools of Odissa. The theme of the Rangoli Competition was Diwali celebrations where students were supposed to make mathematical designs. Students from various schools participated and made beautiful Rangoli designs. One such design is given below.
Rangoli is in the shape of square marked as ABCD, side of square being 40 cm. At each corner of a square, a quadrant of circle of radius 10 cm is drawn (in which diyas are kept). Also a circle of diameter 20 cm is drawn inside the square.
(i) What is the area of square ABCD ? (1)
(ii) Find the area of the circle. (1)
(iii) If the circle and the four quadrants are cut off from the square ABCD and removed, then find the area of remaining portion of square ABCD. (2)
OR (iii) Find the combined area of 4 quadrants and the circle, removed. (2)

Diagram for CBSE 2023 Class 10 Maths question 37

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Answer: (i) 1600 cm² (ii)

\frac{2200}{7}

cm² ≈ 314.29 cm² (iii)

\frac{6800}{7}

cm² ≈ 971.43 cm²; OR (iii)

\frac{4400}{7}

cm² ≈ 628.57 cm²

(i) Area of square $= 4 0^{2} = 1600$ cm²
(ii) Radius of circle = 10 cm; area $= \frac{22}{7} \times 1 0^{2} = \frac{2200}{7} \approx 314.29$ cm²
Four quadrants of radius 10 cm make one full circle: area $= \frac{2200}{7}$ cm².
Area removed $= \frac{2200}{7} + \frac{2200}{7} = \frac{4400}{7}$ cm²
(iii) Remaining area $= 1600 - \frac{4400}{7} = \frac{6800}{7} \approx 971.43$ cm²
OR (iii) Combined area removed $= \frac{4400}{7} \approx 628.57$ cm²

Also asked in: 2023 Basic 430/1/2, 2023 Basic 430/1/3

Q364 marksCase StudyAreas Related to Circles CBSE 2023 · Basic 430/2/1

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 $9 0^{\circ}$ at the centre is full of red roses, while the other segment with central angle $6 0^{\circ}$ is full of yellow coloured flowers. [See figure]
It is given that the combined area of the two segments (of flowers) is $256 \frac{2}{3}$ 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’. (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)

Diagram for CBSE 2023 Class 10 Maths question 36

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Answer: (i)

\frac{90}{360} π r^{2} + \frac{60}{360} π r^{2} = 256 \frac{2}{3}

(ii) r = 14 m (iii) (a) 154 m

^{2}

OR (iii) (b)

102 \frac{2}{3}

^{2}

The given total works out only if the flower regions are taken as the regions with central angles $9 0^{\circ}$ and $6 0^{\circ}$ (sector areas), which is the intended reading.
(i) $\frac{90}{360} π r^{2} + \frac{60}{360} π r^{2} = 256 \frac{2}{3}$ , i.e. $\frac{5}{12} π r^{2} = \frac{770}{3}$
(ii) $\frac{5}{12} \times \frac{22}{7} \times r^{2} = \frac{770}{3}$ , so $r^{2} = \frac{770 \times 84}{3 \times 110} = 196$ and r = 14 m
(iii) (a) Red roses: $\frac{1}{4} \times \frac{22}{7} \times 196 = 154$ m $^{2}$
(iii) (b) Yellow flowers: $\frac{1}{6} \times \frac{22}{7} \times 196 = \frac{308}{3} = 102 \frac{2}{3}$ m $^{2}$
(If true segments are used, red = $\frac{π r ^{2}}{4} - \frac{r ^{2}}{2} = \frac{2}{7} r^{2}$ and yellow = $\frac{π r ^{2}}{6} - \frac{3}{4} r^{2}$ , which gives r ≈ 26.1 m, not a neat value.)

Also asked in: 2023 Basic 430/2/2, 2023 Basic 430/2/3

Q364 marksCase StudyAreas Related to Circles CBSE 2023 · Basic 430/6/1

For the inauguration of ‘Earth day’ week in a school, badges were given to volunteers. Organisers purchased these badges from an NGO, who made these badges in the form of a circle inscribed in a square of side 8 cm.
O is the centre of the circle and $∠ A O B = 9 0^{\circ}$ :
Based on the above information, answer the following questions :
(i) What is the area of square ABCD ? (1)
(ii) What is the length of diagonal AC of square ABCD ? (1)
(iii) Find the area of sector OPRQO. (2)
OR (iii) Find the area of remaining part of square ABCD when area of circle is excluded. (2)

Show answer & solution

Answer: (i) 64 cm

^{2}

(ii)

82

cm (iii)

\frac{88}{7}

^{2}

\approx 12.57

^{2}

OR (iii)

\frac{96}{7}

^{2}

\approx 13.71

^{2}

(i) Area $= 8 \times 8 = 64$ cm $^{2}$ .
(ii) $A C = 8^{2} + 8^{2} = 82$ cm.
(iii) Radius of circle $= \frac{8}{2} = 4$ cm, angle $= 9 0^{\circ}$ . Area of sector $= \frac{90}{360} \times \frac{22}{7} \times 4^{2} = \frac{88}{7} \approx 12.57$ cm $^{2}$ .
OR (iii) Area of circle $= \frac{22}{7} \times 16 = \frac{352}{7}$ cm $^{2}$ . Remaining area $= 64 - \frac{352}{7} = \frac{96}{7} \approx 13.71$ cm $^{2}$ .

Also asked in: 2023 Basic 430/6/2, 2023 Basic 430/6/3

Q364 marksCase StudyAreas Related to Circles CBSE 2023 · Standard 30/2/1

In an annual day function of a school, the organizers wanted to give a cash prize along with a memento to their best students. Each memento is made as shown in the figure and its base ABCD is shown from the front side. The rate of silver plating is ₹ 20 per cm².
Based on the above, answer the following questions :
(i) What is the area of the quadrant ODCO ? (1)
(ii) Find the area of $△ A O B$ . (1)
(iii) (a) What is the total cost of silver plating the shaded part ABCD ? (2)
OR (iii) (b) What is the length of arc CD ? (2)

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Answer: (i) 38.5 cm² (ii) 50 cm² (iii) (a) ₹ 230 OR (b) 11 cm

(i) Area of quadrant $= \frac{1}{4} \times \frac{22}{7} \times 7^{2} = 38.5$ cm²
(ii) OA = OB = 7 + 3 = 10 cm and $∠ A O B = 9 0^{\circ}$ , so area $= \frac{1}{2} \times 10 \times 10 = 50$ cm²
(iii)(a) Shaded area $= 50 - 38.5 = 11.5$ cm²; cost $= 11.5 \times 20 = ₹ 230$
(iii)(b) Arc CD $= \frac{1}{4} \times 2 \times \frac{22}{7} \times 7 = 11$ cm

Also asked in: 2023 Standard 30/2/2, 2023 Standard 30/2/3

Q384 marksCase StudyAreas Related to Circles CBSE 2023 · Standard 30/4/1

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 ? (1)
(ii) (a) What is the total area of parking and the two quadrants ? (2)
OR (b) What is the ratio of area of playground to the area of parking area ? (2)
(iii) Find the cost of fencing the playground and parking area at the rate of ₹ 2 per unit. (1)

Diagram for CBSE 2023 Class 10 Maths question 38

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Answer: (i) 18 units (ii) (a)

\frac{715}{28} \approx 25.54

sq units OR (b) 56 : 11 (iii) ₹92

Parking area is a semicircle of diameter 7 units, radius $\frac{7}{2}$ units.
(i) Perimeter $= π r + 2 r = \frac{22}{7} \times \frac{7}{2} + 7 = 11 + 7 = 18$ units
(ii) (a) Parking area $= \frac{1}{2} \times \frac{22}{7} \times \frac{49}{4} = \frac{77}{4} = 19.25$ sq units
Two quadrants $= 2 \times \frac{1}{4} \times \frac{22}{7} \times 2^{2} = \frac{44}{7} \approx 6.29$ sq units
Total $= \frac{77}{4} + \frac{44}{7} = \frac{715}{28} \approx 25.54$ sq units
(ii) (b) Playground area $= 14 \times 7 = 98$ ; ratio $= 98 : \frac{77}{4} = 392 : 77 = 56 : 11$
(iii) Outer boundary $= 14 + 7 + 14 + 11 = 46$ units (the side shared with the parking area is not fenced)
Cost $= 46 \times 2 =$ ₹92

Also asked in: 2023 Standard 30/4/2, 2023 Standard 30/4/3

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