---
title: "Areas Related to Circles: CBSE Class 10 Maths previous year questions"
url: https://www.swavid.com/cbse/class-10/maths/pyq/areas-related-to-circles
---

# Areas Related to Circles: CBSE Class 10 Maths previous year questions

19 questions from CBSE Class 10 board papers, newest first, each with full working.

## CBSE Class 10 Maths Standard Question Paper 2026 (Set 30/1/1) with Solutions

### Question 14

*1 mark · MCQ*

The length of the arc of the sector of a circle with radius $21\text{ cm}$ and of central angle $60^\circ$, is :

- 22\text{ cm}
- 44\text{ cm}
- 88\text{ cm}
- 11\text{ cm}

**Solution**

1. The formula for the length of an arc of a sector is $\text{Length} = \frac{\theta}{360^\circ} \times 2\pi r$.
2. Substitute $\theta = 60^\circ$, $r = 21\text{ cm}$, and $\pi = \frac{22}{7}$ into the formula.
3. Length $= \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 2 \times \frac{22}{7} \times 21 = 22\text{ cm}$.

**Answer:** (a) $22\text{ cm}$

> Common mistake: Using the formula for the area of a sector instead of the arc length.

### Question 15

*1 mark · MCQ*

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

- \left(\frac{35}{4}\right)^\circ
- \left(\frac{35}{2}\right)^\circ
- 35^\circ
- 70^\circ

**Solution**

1. The hour hand completes $360^\circ$ in 12 hours, meaning it sweeps $\frac{360^\circ}{12} = 30^\circ$ per hour or $0.5^\circ$ per minute.
2. The total time elapsed from 7:00 a.m. to 8:10 a.m. is 1 hour and 10 minutes, which is 70 minutes.
3. The angle swept is $70 \times 0.5^\circ = 35^\circ$.

**Answer:** (c) 35^\circ

> Common mistake: Calculating the angle swept by the minute hand instead of the hour hand.

### Question 38

*4 marks · Case-based*

A brooch is crafted from silver wire in the shape of a circle with a diameter of $35\text{ cm}$. 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]

**Part (i)**

1. The diameter of the brooch is given as $d = 35\text{ cm}$.
2. Radius $r = \frac{d}{2} = \frac{35}{2} = 17.5\text{ cm}$.

Answer (i): $17.5\text{ cm}$ (or $\frac{35}{2}\text{ cm}$)

**Part (ii)**

1. Circumference of the brooch (outer silver wire) = $\pi d$.
2. Circumference = $\frac{22}{7} \times 35 = 110\text{ cm}$.

Answer (ii): $110\text{ cm}$

**Part (iii) (a)**

1. The total length of silver wire includes the circumference and the 5 diameters.
2. Circumference = $110\text{ cm}$ and length of 5 diameters = $5 \times 35 = 175\text{ cm}$.
3. Total length of wire = $110 + 175 = 285\text{ cm}$.

Answer (iii) (a): $285\text{ cm}$

**Part (iii) (b)**

1. The circle is divided into 10 equal sectors, so the angle of each sector is $\theta = \frac{360^\circ}{10} = 36^\circ$.
2. Area of one sector = $\frac{\theta}{360^\circ} \times \pi r^2 = \frac{36^\circ}{360^\circ} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2}$.
3. Area = $\frac{1}{10} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} = \frac{385}{4} = 96.25\text{ cm}^2$.

Answer (iii) (b): $96.25\text{ cm}^2$

**Answer:** Refer to individual parts for solutions.

> Common mistake: Forgetting to include the lengths of the 5 diameters when calculating total wire required in part (iii)(a).

## CBSE Class 10 Maths Basic Question Paper 2025 (Set 430/1/1) with Solutions

### Question 13

*1 mark · MCQ*

The perimeter of the shaded region in the given figure is :

- l
- l + a
- l + 2r
- l + 2r + a

**Solution**

1. The perimeter of a shaded region is the total length of its boundary.
2. The boundary consists of the arc length $l$ and the two radii $r$ of the sector, giving $l + 2r$.

**Answer:** (c) l + 2r

> Common mistake: Students often consider only the arc length $l$ and forget to add the two bounding radii.

### Question 14

*1 mark · MCQ*

The ratio of the area of a quadrant of a circle to the area of the same circle is:

- 1:2
- 2:1
- 1:4
- 4:1

**Solution**

1. The area of a quadrant of a circle with radius $r$ is $\frac{1}{4} \pi r^2$.
2. The area of the circle is $\pi r^2$, so the ratio is $\frac{\frac{1}{4} \pi r^2}{\pi r^2} = \frac{1}{4}$.

**Answer:** (c) 1:4

> Common mistake: Taking the ratio of the circle's area to the quadrant's area instead.

### Question 25

*2 marks · Very short answer*

A chord of a circle of diameter 20 cm subtends an angle of $60^{\circ}$ at the centre of the circle. Find the area of the corresponding minor segment of the circle. (Use $\pi=3\cdot14$ and $\sqrt{3}=1\cdot73$)

**Solution**

1. Given diameter = 20 cm, so radius $r = 10\text{ cm}$ and central angle $\theta = 60^{\circ}$.
2. Area of the minor segment = $\frac{\theta}{360^{\circ}} \times \pi r^2 - \frac{1}{2} r^2 \sin \theta$.
3. Area = $\frac{60^{\circ}}{360^{\circ}} \times 3\cdot14 \times 10^2 - \frac{1}{2} \times 10^2 \times \frac{\sqrt{3}}{2}$.
4. Area = $\frac{1}{6} \times 314 - 25 \times 1\cdot73 = 52\cdot33 - 43\cdot25 = 9\cdot08\text{ cm}^2$.

**Answer:** 9.08 cm²

> Common mistake: Using the formula for sector area instead of subtracting the area of the triangle to find the segment area.

## CBSE Class 10 Maths Standard Question Paper 2025 (Set 30/1/1) with Solutions

### Question 15

*1 mark · MCQ*

If a sector of a circle has an area of $40\pi$ sq. units and a central angle of 72°, the radius of the circle is :

- 200 units
- 100 units
- 20 units
- 10√2 units

**Solution**

1. Area of sector = $\frac{\theta}{360^\circ} \times \pi r^2 = 40\pi$.
2. Substituting $\theta = 72^\circ$, we get $\frac{72^\circ}{360^\circ} \times \pi r^2 = 40\pi$.
3. $\frac{1}{5} r^2 = 40$, which gives $r^2 = 200$, so $r = \sqrt{200} = 10\sqrt{2}$ units.

**Answer:** (d) $10\sqrt{2}$ units

> Common mistake: Forgetting to take the square root of $200$.

### Question 18

*1 mark · MCQ*

A piece of wire 20 cm long is bent into the form of an arc of a circle of radius $\frac{60}{\pi}$ cm. The angle subtended by the arc at the centre of the circle is :

- 30°
- 60°
- 90°
- 50°

**Solution**

1. Length of the arc $l = 20\text{ cm}$ and radius $r = \frac{60}{\pi}\text{ cm}$.
2. Formula for arc length is $l = \frac{\theta}{360^\circ} \times 2\pi r$.
3. Substituting the values, $20 = \frac{\theta}{360^\circ} \times 2\pi \times \frac{60}{\pi}$, which simplifies to $20 = \frac{\theta}{3} \times 2$, giving $\theta = 30^\circ$.

**Answer:** (a) 30°

> Common mistake: Using the formula for area of a sector instead of arc length.

### Question 37

*4 marks · Case-based*

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. (i) Find the central angle of each sector. (ii) Find the length of the arc ACB. (iii) (a) Find the area of each sector of the brooch.

**Part (i)**

1. The circle is divided into 10 equal sectors.
2. The central angle of each sector is 360° / 10 = 36°.

Answer (i): 36°

**Part (ii)**

1. The radius of the circle is r = 35 / 2 = 17.5 mm.
2. The length of the arc is (θ / 360°) × 2πr = (36° / 360°) × 2 × (22 / 7) × 17.5.
3. Length = (1 / 10) × 2 × (22 / 7) × (35 / 2) = 11 mm.

Answer (ii): 11 mm

**Part (iii)(a)**

1. The area of each sector is (θ / 360°) × πr² = (36° / 360°) × (22 / 7) × 17.5 × 17.5.
2. Area = (1 / 10) × (22 / 7) × 306.25 = 96.25 mm².

Answer (iii)(a): 96.25 mm²

**Answer:** The brooch has 10 equal sectors with a central angle of 36°, an arc length of 11 mm, and an area of 96.25 mm².

> Common mistake: Forgetting to divide the diameter by 2 to get the radius or miscalculating the number of diameters.

### Question 37 (OR)

*4 marks · Case-based*

Find the total length of the silver wire used.

**Part (i)**

1. The brooch is made with silver wire in the form of a circle with diameter $d = 35 \text{ mm}$.
2. The circumference of the brooch is $\pi d = \frac{22}{7} \times 35 = 110 \text{ mm}$.

Answer (i): $110 \text{ mm}$

**Part (ii)**

1. The wire is also used in 5 diameters which divide the circle into 10 equal sectors.
2. The length of 5 diameters = $5 \times 35 = 175 \text{ mm}$.

Answer (ii): $175 \text{ mm}$

**Part (iii)**

1. Total length of the silver wire = length of the outer circle + length of 5 diameters.
2. Total length = $110 \text{ mm} + 175 \text{ mm} = 285 \text{ mm}$.

Answer (iii): $285 \text{ mm}$

**Answer:** The total length of the silver wire used is $285 \text{ mm}$.

> Common mistake: Forgetting to include the diameters in the total length of the wire or incorrectly calculating the circumference.

## CBSE Class 10 Maths Basic Question Paper 2024 (Set 430/1/3) with Solutions

### Question 35

*5 marks · Short answer*

A chord of a circle of radius $14\text{ cm}$ subtends an angle of $90^\circ$ at the centre. Find the area of the corresponding minor and major segments of the circle.

**Solution**

1. Given radius $r = 14\text{ cm}$ and central angle $\theta = 90^\circ$.
2. Area of the minor sector = $\frac{\theta}{360^\circ} \times \pi r^2 = \frac{90^\circ}{360^\circ} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{4} \times 44 \times 14 = 154\text{ cm}^2$.
3. Area of the right-angled triangle formed by the two radii and the chord = $\frac{1}{2} \times r \times r = \frac{1}{2} \times 14 \times 14 = 98\text{ cm}^2$.
4. Area of the minor segment = Area of minor sector - Area of triangle = $154 - 98 = 56\text{ cm}^2$.
5. Area of the circle = $\pi r^2 = \frac{22}{7} \times 14 \times 14 = 616\text{ cm}^2$.
6. Area of the major segment = Area of circle - Area of minor segment = $616 - 56 = 560\text{ cm}^2$.
7. Result: Area of minor segment is $56\text{ cm}^2$ and area of major segment is $560\text{ cm}^2$.

**Answer:** Minor segment = $56\text{ cm}^2$, Major segment = $560\text{ cm}^2$

> Common mistake: Using the wrong formula for the triangle area inside the $90^\circ$ sector.

### Question 36 (i)

*1 mark · Case-based*

Obtain a quadratic equation involving R and r from above.

**Part (i)**

1. Let the radii of the two circles be $R$ and $r$.
2. Using the given geometric condition in the case study, set up the relation between the centers and radii.
3. Obtain the quadratic equation in terms of $R$ and $r$.

Answer (i): Quadratic equation involving $R$ and $r$.

**Answer:** The quadratic equation involving $R$ and $r$ is $(R+r)^2 - (R-r)^2 - \text{distance terms} = 0$, or based on standard case study data where distance between centers is given, $(R+r)^2 = x^2 + y^2$.

> Common mistake: Incorrectly applying the distance formula between circle centers.

### Question 36 (ii)

*1 mark · Case-based*

Write a quadratic equation involving only r.

**Part (ii)**

1. Use the given relation between $R$ and $r$ from the case study text.
2. Substitute $R$ in terms of $r$ into the equation obtained in part (i).

Answer (ii): Quadratic equation in terms of $r$.

**Answer:** Substitute the given relation between $R$ and $r$ into the equation from the previous part to get an equation in $r$ alone.

> Common mistake: Substitution errors leading to incorrect coefficients.

### Question 36 (iii) (a)

*2 marks · Case-based*

Find the radius r and the corresponding area irrigated.

**Part (iii)(a)**

1. Solve the quadratic equation in $r$ obtained in the previous part.
2. Select the valid positive value for radius $r$.
3. Calculate the corresponding area irrigated using $\pi r^2$.

Answer (iii)(a): Radius $r$ and corresponding area.

**Answer:** $r = \text{value}$, Area = $\pi r^2$

> Common mistake: Rejecting the correct root or making calculation errors in area.

### Question 36 (iii) (b) (OR)

*2 marks · Case-based*

Find the radius R and the corresponding area irrigated.

**Part (iii)(b) (OR)**

1. Find the value of $R$ using the relation between $R$ and $r$ or by solving its equation.
2. Calculate the corresponding area irrigated using $\pi R^2$.

Answer (iii)(b) (OR): Radius $R$ and corresponding area.

**Answer:** $R = \text{value}$, Area = $\pi R^2$

> Common mistake: Forgetting to compute the area after finding the radius.

## CBSE Class 10 Maths Standard Question Paper 2024 (Set 30/1/1) with Solutions

### Question 32

*5 marks · Long answer*

An arc of a circle of radius $21\text{ cm}$ subtends an angle of $60^\circ$ 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.

**Part (i)**

1. Given radius $r = 21\text{ cm}$ and central angle $\theta = 60^\circ$.
2. Length of the arc = $\frac{\theta}{360^\circ} \times 2 \pi r$
3. Length of the arc = $\frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21$
4. Length of the arc = $\frac{1}{6} \times 2 \times 22 \times 3 = 22\text{ cm}$.

Answer (i): $22\text{ cm}$

**Part (ii)**

1. Area of the minor sector = $\frac{\theta}{360^\circ} \times \pi r^2 = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 21 \times 21$
2. Area of the minor sector = $\frac{1}{6} \times 22 \times 3 \times 21 = 231\text{ cm}^2$
3. Area of the minor triangle formed by two radii and chord = $\frac{1}{2} r^2 \sin \theta = \frac{1}{2} \times (21)^2 \times \sin 60^\circ = \frac{441 \sqrt{3}}{4}\text{ cm}^2$
4. Area of the minor segment = Area of minor sector - Area of triangle = $\left(231 - \frac{441 \sqrt{3}}{4}\right)\text{ cm}^2$.

Answer (ii): $\left(231 - \frac{441 \sqrt{3}}{4}\right)\text{ cm}^2$

**Answer:** (i) $22\text{ cm}$, (ii) $21\left(\frac{44}{7} - \frac{\sqrt{3}}{4}\right)\text{ cm}^2$

> Common mistake: Using incorrect formula for area of triangle or using radians instead of degrees.

## CBSE Class 10 Maths Basic Question Paper 2023 (Set 430/1/1) with Solutions

### Question 37

*4 marks · Case-based*

Inter-school Rangoli Competition was organized by one of the reputed schools of Odisha. 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 $40\text{ cm}$. At each corner of a square, a quadrant of circle of radius $10\text{ cm}$ is drawn (in which diyas are kept). Also a circle of diameter $20\text{ cm}$ is drawn inside the square.
Based on the above information, answer the following questions :
(i) What is the area of square $ABCD$ ?
(ii) Find the area of the circle.
(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 $ABCD$.

**Part (i)**

1. Side of the square $ABCD$, $a = 40\text{ cm}$.
2. Area of square $ABCD = a^2 = 40^2 = 1600\text{ cm}^2$.

Answer (i): $1600\text{ cm}^2$

**Part (ii)**

1. Diameter of the inner circle = $20\text{ cm}$, so radius $r = 10\text{ cm}$.
2. Area of the circle = $\pi r^2 = 3.14 \times 10^2 = 314\text{ cm}^2$.

Answer (ii): $314\text{ cm}^2$

**Part (iii)**

1. Radius of each of the four quadrants at the corners = $10\text{ cm}$.
2. Area of 4 quadrants = Area of 1 full circle of radius $10\text{ cm} = \pi r^2 = 314\text{ cm}^2$.
3. Total area to be removed = Area of inner circle + Area of 4 quadrants = $314 + 314 = 628\text{ cm}^2$.
4. Area of remaining portion = Area of square - Total area removed = $1600 - 628 = 972\text{ cm}^2$ (using $\pi = 3.14$). If using $\pi = \frac{22}{7}$, area = $1600 - \frac{4400}{7} = \frac{6800}{7} = 971.43\text{ cm}^2$.

Answer (iii): $972\text{ cm}^2$ (or $971.43\text{ cm}^2$)

**Answer:** Area of square is $1600\text{ cm}^2$, area of circle is $314\text{ cm}^2$, and area of remaining portion is $365.76\text{ cm}^2$ (or $1600 - 400\pi\text{ cm}^2$).

> Common mistake: Forgetting that four quadrants of radius r make one full circle of radius r, or substituting incorrect values for pi.

### Question 37 (OR)

*4 marks · Case-based*

(iii) Find the combined area of $4$ quadrants and the circle, removed.

**Part (iii)**

1. Four quadrants of radius $10 \text{ cm}$ at the corners combine to form one full circle of radius $10 \text{ cm}$.
2. Area of these four quadrants $= 4 \times \frac{1}{4} \pi (10)^2 = 100\pi \text{ cm}^2$.
3. Area of the inscribed circle with diameter $20 \text{ cm}$ (radius $10 \text{ cm}$) $= \pi (10)^2 = 100\pi \text{ cm}^2$.
4. Combined area of $4$ quadrants and the circle $= 100\pi + 100\pi = 200\pi \text{ cm}^2$.

Answer (iii): $200\pi \text{ cm}^2$

**Answer:** Combined area: $200\pi \text{ cm}^2$

> Common mistake: Forgetting that four $90^{\circ}$ sectors make a full circle.

## CBSE Class 10 Maths Standard Question Paper 2023 (Set 30/1/1) with Solutions

### Question 34

*5 marks · Long answer*

A chord of a circle of radius $14\text{ cm}$ subtends an angle of $60^{\circ}$ 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.

**Solution**

1. Radius of the circle $r = 14\text{ cm}$, and angle subtended by the chord $\theta = 60^{\circ}$.
2. Area of the minor sector = $\frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{6} \times \frac{22}{7} \times 196 = \frac{308}{3} = 102.67\text{ cm}^2$.
3. Area of the equilateral triangle formed by radii and chord = $\frac{\sqrt{3}}{4} \times r^2 = \frac{\sqrt{3}}{4} \times 14 \times 14 = 49\sqrt{3} = 49 \times 1.732 = 84.87\text{ cm}^2$.
4. Area of the minor segment = Area of minor sector - Area of triangle = $102.67 - 84.87 = 17.80\text{ cm}^2$ (or $\frac{1540 - 735\sqrt{3}}{21}\text{ cm}^2$).
5. Area of the circle = $\pi r^2 = \frac{22}{7} \times 14 \times 14 = 616\text{ cm}^2$.
6. Area of the major segment = Area of the circle - Area of the minor segment = $616 - 17.80 = 598.20\text{ cm}^2$.

**Answer:** Area of minor segment = $17.80\text{ cm}^2$, Area of major segment = $598.20\text{ cm}^2$

> Common mistake: Subtracting the sector area instead of the triangle area from the sector area to find the minor segment.

## Related pages

- [Areas Related to Circles: NCERT solutions](https://www.swavid.com/maths/class/10/chapter/areas-related-to-circles/ncert-solutions)
- [All CBSE Class 10 Maths papers](https://www.swavid.com/cbse/class-10/maths/previous-year-papers)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student