---
title: "NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.3"
url: https://www.swavid.com/maths/class/9/chapter/measuring-space-perimeter-and-area/ncert-solutions/exercise-6-3
dateModified: 2026-10-07T15:48:27+00:00
---

# NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.3

Chapter 6: Measuring Space: Perimeter and Area. Every question from Exercise 6.3, with full working and the final answer.

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## Exercise Set 6.3

### Question 1

*3 marks · Short answer*

Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.

**Solution**

1. Given: Radius $r = 7 \text{ cm}$, central angle $\theta = 60^\circ$.
2. Formula: $\text{Area of sector} = \pi r^2 \times \frac{\theta^\circ}{360^\circ}$.
3. Substitution: $\text{Area} = \frac{22}{7} \times 7^2 \times \frac{60^\circ}{360^\circ}$.
4. Result: $\frac{22}{7} \times 49 \times \frac{1}{6} = \frac{77}{3} \text{ cm}^2$.

**Answer:** $\frac{77}{3} \text{ cm}^2$

> Common mistake: Using circumference formula instead of area formula for the circle.

### Question 2

*3 marks · Short answer*

Find the area of a quadrant of a circle whose circumference is 44 cm.

**Solution**

1. Given: Circumference $C = 44 \text{ cm}$.
2. Formula: $C = 2\pi r$, so $2 \times \frac{22}{7} \times r = 44$, giving $r = 7 \text{ cm}$.
3. Formula: $\text{Area of quadrant} = \frac{1}{4} \pi r^2$.
4. Result: $\frac{1}{4} \times \frac{22}{7} \times 7^2 = \frac{77}{2} \text{ cm}^2$.

**Answer:** $\frac{77}{2} \text{ cm}^2$

> Common mistake: Forgetting to divide the area of the circle by 4 for the quadrant.

### Question 3

*3 marks · Short answer*

The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.

**Solution**

1. Given: Radius $r = 7 \text{ cm}$. In 60 minutes, the minute hand sweeps $360^\circ$.
2. In 10 minutes, the angle swept is $\theta = \frac{10}{60} \times 360^\circ = 60^\circ$.
3. Formula: $\text{Area} = \pi r^2 \times \frac{\theta^\circ}{360^\circ}$.
4. Substitution and Result: $\frac{22}{7} \times 7^2 \times \frac{60^\circ}{360^\circ} = \frac{77}{3} \text{ cm}^2$.

**Answer:** $\frac{77}{3} \text{ cm}^2$

> Common mistake: Incorrectly calculating the angle swept in 10 minutes.

### Question 4

*3 marks · Short answer*

A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding: (i) minor sector (that subtends 90° at the centre), and (ii) major sector (that subtends 270° at the centre). (Use $\pi \approx 3.14$.)

**Part (i)**

1. Given: $r = 10 \text{ cm}$, $\theta = 90^\circ$, $\pi \approx 3.14$.
2. Formula: $\text{Area} = \pi r^2 \times \frac{90^\circ}{360^\circ}$.
3. Result: $3.14 \times 10^2 \times \frac{1}{4} = 78.5 \text{ cm}^2$.

Answer (i): $78.5 \text{ cm}^2$

**Part (ii)**

1. Given: $r = 10 \text{ cm}$, $\theta = 270^\circ$, $\pi \approx 3.14$.
2. Formula: $\text{Area} = \pi r^2 \times \frac{270^\circ}{360^\circ}$.
3. Result: $3.14 \times 10^2 \times \frac{3}{4} = 235.5 \text{ cm}^2$.

Answer (ii): $235.5 \text{ cm}^2$

**Answer:** (i) $78.5 \text{ cm}^2$, (ii) $235.5 \text{ cm}^2$

> Common mistake: Using wrong central angles for minor and major sectors.

### Question 5

*3 marks · Short answer*

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi \approx 3.14$ and $\sqrt{3} \approx 1.73$.)

**Part (i)**

1. Given: $r = 15 \text{ cm}$, $\theta = 60^\circ$. Area of sector $= \pi r^2 \times \frac{60}{360} = 3.14 \times 225 \times \frac{1}{6} = 117.75 \text{ cm}^2$.
2. Area of triangle $= \frac{\sqrt{3}}{4} r^2 = \frac{1.73}{4} \times 225 = 97.3125 \text{ cm}^2$.
3. Area of minor segment $= 117.75 - 97.3125 = 20.4375 \text{ cm}^2$.

Answer (i): $20.4375 \text{ cm}^2$

**Part (ii)**

1. Area of circle $= \pi r^2 = 3.14 \times 225 = 706.5 \text{ cm}^2$.
2. Area of major segment $= \text{Area of circle} - \text{Area of minor segment}$.
3. Result: $706.5 - 20.4375 = 686.0625 \text{ cm}^2$.

Answer (ii): $686.0625 \text{ cm}^2$

**Answer:** Minor segment $= 20.4375 \text{ cm}^2$, Major segment $= 686.0625 \text{ cm}^2$

> Common mistake: Subtracting the triangle area from the wrong sector area.

### Question 6

*3 marks · Short answer*

A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm and sweeps through an angle of 120°. Find the total area cleaned at each sweep of the blades.

**Solution**

1. Given that each wiper has a blade length (radius $r$) of $28~\text{cm}$ and sweeps through an angle ($\theta^\circ$) of $120^\circ$.
2. The area cleaned by one wiper in a single sweep is equal to the area of a sector with radius $r = 28~\text{cm}$ and central angle $\theta = 120^\circ$, given by $\pi r^2 \times \frac{\theta^\circ}{360^\circ}$.
3. Area of one sector = $\frac{22}{7} \times 28 \times 28 \times \frac{120^\circ}{360^\circ} = \frac{22}{7} \times 28 \times 28 \times \frac{1}{3} = \frac{24640}{3}~\text{cm}^2$.
4. Since the car has two non-overlapping wipers, the total area cleaned at each sweep is $2 \times \frac{24640}{3} = \frac{49280}{3}~\text{cm}^2 = 16426.67~\text{cm}^2$.

**Answer:** $16426.67~\text{cm}^2$ (or $16426\frac{2}{3}~\text{cm}^2$)

> Common mistake: Forgetting to multiply the area by 2 for the two wipers.

### Question 7

*4 marks · Proof*

A chord of a circle of radius $r$ subtends an angle of 60° at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to $\pi r^2 \left(\frac{1}{6} - \frac{\sqrt{3}}{4}\right)$.

**Solution**

1. Given: A circle of radius $r$ with a chord subtending an angle of $60^\circ$ at the centre.
2. To prove: Area of the corresponding minor segment = $\pi r^2 \left(\frac{1}{6} - \frac{\sqrt{3}}{4}\right)$.
3. The area of the minor sector corresponding to the $60^\circ$ angle is given by $\text{Area of sector} = \pi r^2 \times \frac{60^\circ}{360^\circ} = \frac{1}{6} \pi r^2$.
4. The triangle formed by the two radii and the chord is an isosceles triangle with an angle of $60^\circ$ between the equal sides of length $r$, making it an equilateral triangle.
5. The area of this equilateral triangle of side $r$ is $\frac{\sqrt{3}}{4} r^2$.
6. The area of the minor segment is the difference between the area of the minor sector and the area of this triangle.
7. Thus, $\text{Area of minor segment} = \frac{1}{6} \pi r^2 - \frac{\sqrt{3}}{4} r^2 = \pi r^2 \left(\frac{1}{6} - \frac{\sqrt{3}}{4}\right)$.
8. Hence proved.

**Answer:** Hence proved.

> Common mistake: Subtracting the sector area from the triangle area instead of the other way around.

### Question 8

*4 marks · Proof*

An equilateral triangle is inscribed in a circle of radius $r$. Show that the ratio of the area of the triangle to the area of the circle is equal to $\frac{3\sqrt{3}}{4\pi} \approx 0.413$.

**Solution**

1. Given: An equilateral triangle inscribed in a circle of radius $r$.
2. To prove: Ratio of the area of the triangle to the area of the circle is $\frac{3\sqrt{3}}{4\pi} \approx 0.413$.
3. The area of the circle of radius $r$ is $\pi r^2$.
4. An equilateral triangle inscribed in a circle of radius $r$ can be divided into three congruent isosceles triangles meeting at the centre, each with two sides equal to $r$ and the included angle being $120^\circ$.
5. The height of each such triangle from the centre to the side is $r \cos(60^\circ) = \frac{r}{2}$, and the base length is $2 \times r \sin(60^\circ) = \sqrt{3}r$.
6. The area of each small triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \sqrt{3}r \times \frac{r}{2} = \frac{\sqrt{3}}{4} r^2$.
7. The total area of the equilateral triangle is $3 \times \frac{\sqrt{3}}{4} r^2 = \frac{3\sqrt{3}}{4} r^2$.
8. The ratio of the area of the triangle to the area of the circle is $\frac{\frac{3\sqrt{3}}{4} r^2}{\pi r^2} = \frac{3\sqrt{3}}{4\pi} \approx 0.413$.
9. Hence proved.

**Answer:** Hence proved.

> Common mistake: Incorrectly finding the side length or height of the inscribed equilateral triangle.

### Question 9

*4 marks · Proof*

A square is inscribed in a circle of radius $r$. Show that the ratio of the area of the square to the area of the circle is equal to $\frac{2}{\pi} \approx 0.637$.

**Solution**

1. Given: A square inscribed in a circle of radius $r$.
2. To prove: Ratio of the area of the square to the area of the circle is $\frac{2}{\pi} \approx 0.637$.
3. The area of the circle of radius $r$ is $\pi r^2$.
4. The diagonal of the square inscribed in the circle is equal to the diameter of the circle, which is $2r$.
5. The area of a square in terms of its diagonal $d$ is given by $\frac{1}{2} d^2$.
6. Substituting $d = 2r$, the area of the square is $\frac{1}{2} (2r)^2 = \frac{1}{2} \times 4r^2 = 2r^2$.
7. The ratio of the area of the square to the area of the circle is $\frac{2r^2}{\pi r^2} = \frac{2}{\pi} \approx 0.637$.
8. Hence proved.

**Answer:** Hence proved.

> Common mistake: Taking the side of the square as $r$ instead of $r\sqrt{2}$.

### Question 10

*4 marks · Proof*

A hexagon is inscribed in a circle of radius $r$. Show that the ratio of the area of the hexagon to the area of the circle is equal to $\frac{3\sqrt{3}}{2\pi} \approx 0.827$. Can you see why the answer is exactly twice the answer to Question 8?

**Solution**

1. Given: A regular hexagon inscribed in a circle of radius $r$.
2. To prove: Ratio of the area of the hexagon to the area of the circle is $\frac{3\sqrt{3}}{2\pi} \approx 0.827$, and it is twice the answer to Question 8.
3. The area of the circle of radius $r$ is $\pi r^2$.
4. A regular inscribed hexagon is composed of 6 congruent equilateral triangles, each with side length equal to the radius $r$ of the circle.
5. The area of one such equilateral triangle of side $r$ is $\frac{\sqrt{3}}{4} r^2$.
6. The total area of the hexagon is $6 \times \frac{\sqrt{3}}{4} r^2 = \frac{3\sqrt{3}}{2} r^2$.
7. The ratio of the area of the hexagon to the area of the circle is $\frac{\frac{3\sqrt{3}}{2} r^2}{\pi r^2} = \frac{3\sqrt{3}}{2\pi} \approx 0.827$.
8. Comparing this with the ratio for the equilateral triangle in Question 8 (which is $\frac{3\sqrt{3}}{4\pi}$), we see that $\frac{3\sqrt{3}}{2\pi} = 2 \times \frac{3\sqrt{3}}{4\pi}$, so the hexagon's ratio is exactly twice the equilateral triangle's ratio because a hexagon consists of 6 equilateral triangles while the inscribed triangle consists of 3.
9. Hence proved.

**Answer:** Hence proved.

> Common mistake: Forgetting to multiply the area of one constituent equilateral triangle by 6.

## Related pages

- [All Chapter 6 NCERT solutions](https://www.swavid.com/maths/class/9/chapter/measuring-space-perimeter-and-area/ncert-solutions)
- [Exercise 6.1](https://www.swavid.com/maths/class/9/chapter/measuring-space-perimeter-and-area/ncert-solutions/exercise-6-1)
- [Exercise 6.2](https://www.swavid.com/maths/class/9/chapter/measuring-space-perimeter-and-area/ncert-solutions/exercise-6-2)

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