---
title: "NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1"
url: https://www.swavid.com/maths/class/9/chapter/measuring-space-perimeter-and-area/ncert-solutions/exercise-6-1
dateModified: 2026-10-07T15:48:27+00:00
---

# NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1

Chapter 6: Measuring Space: Perimeter and Area. Every question from Exercise 6.1, with full working and the final answer.

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## Exercise Set 6.1

### Question 1

*3 marks · Short answer*

The perimeter of a circle is 44 cm. What is its radius?

**Solution**

1. Given: Perimeter of the circle $C = 44 \text{ cm}$
2. Formula: $C = 2\pi r$
3. Substitution: $44 = 2 \times \frac{22}{7} \times r$
4. Result: $r = 7 \text{ cm}$

**Answer:** $7 \text{ cm}$

> Common mistake: Forgetting to divide by 2 or using diameter instead of radius.

### Question 2

*3 marks · Short answer*

Calculate, correct to 3 significant figures, the circumference of a circle with: (i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm.

**Part (i)**

1. Given: Radius $r = 7 \text{ cm}$
2. Formula: $C = 2\pi r$
3. Substitution: $C = 2 \times \frac{22}{7} \times 7 = 44 \text{ cm}$

Answer (i): $44.0 \text{ cm}$

**Part (ii)**

1. Given: Radius $r = 10 \text{ cm}$
2. Formula: $C = 2\pi r$
3. Substitution: $C = 2 \times \frac{22}{7} \times 10 = \frac{440}{7} \approx 62.857 \text{ cm}$

Answer (ii): $62.9 \text{ cm}$

**Part (iii)**

1. Given: Radius $r = 12 \text{ cm}$
2. Formula: $C = 2\pi r$
3. Substitution: $C = 2 \times \frac{22}{7} \times 12 = \frac{528}{7} \approx 75.428 \text{ cm}$

Answer (iii): $75.4 \text{ cm}$

**Answer:** (i) $44.0 \text{ cm}$ (ii) $62.9 \text{ cm}$ (iii) $75.4 \text{ cm}$

> Common mistake: Not rounding to 3 significant figures.

### Question 3

*3 marks · Short answer*

Calculate the length of the arc of a circle if: (i) the radius is 3.5 cm and the angle at the centre is 60°, and (ii) the radius is 6.3 m and the angle at the centre is 120°.

**Part (i)**

1. Formula for arc length is $l = 2\pi r \times \frac{\theta^\circ}{360^\circ}$.
2. Substitute $r = 3.5 \text{ cm}$, $\theta = 60^\circ$, and $\pi = \frac{22}{7}$.
3. Length $= 2 \times \frac{22}{7} \times 3.5 \times \frac{60^\circ}{360^\circ} = 3.67 \text{ cm}$.

Answer (i): $3.67 \text{ cm}$

**Part (ii)**

1. Formula for arc length is $l = 2\pi r \times \frac{\theta^\circ}{360^\circ}$.
2. Substitute $r = 6.3 \text{ m}$, $\theta = 120^\circ$, and $\pi = \frac{22}{7}$.
3. Length $= 2 \times \frac{22}{7} \times 6.3 \times \frac{120^\circ}{360^\circ} = 13.2 \text{ m}$.

Answer (ii): $13.2 \text{ m}$

**Answer:** (i) $3.67 \text{ cm}$, (ii) $13.2 \text{ m}$

> Common mistake: Using the wrong angle ratio or forgetting to multiply by 2 in the circumference formula.

### Question 4

*3 marks · Short answer*

Find the perimeter of a sector (i.e., the curved portion as well as the two straight portions) of a circle of radius 14 cm and sector angle 75°.

**Solution**

1. Given: Radius $r = 14 \text{ cm}$, sector angle $\theta = 75^\circ$
2. Formula: Perimeter = Arc length + $2r$
3. Substitution for arc length: $l = 2 \times \frac{22}{7} \times 14 \times \frac{75}{360} = 88 \times \frac{5}{24} = \frac{55}{3} \text{ cm}$
4. Substitution for perimeter: $\text{Perimeter} = \frac{55}{3} + 2(14) = \frac{55}{3} + 28 = \frac{55 + 84}{3} = \frac{139}{3} \approx 46.33 \text{ cm}$

**Answer:** $46.33 \text{ cm}$

> Common mistake: Only calculating the arc length and forgetting to add the two radii.

### Question 5

*3 marks · Short answer*

Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14i to 6.14ix):

**Part (i)**

1. Given: Rectangle with length $80 \text{ m}$ and width $60 \text{ m}$, attached with two semicircles of diameter $60 \text{ m}$ ($r = 30 \text{ m}$)
2. Calculation: Perimeter = $2 \times 80 + \text{circumference of circle} = 160 + 2 \times \frac{22}{7} \times 30 = 160 + \frac{1320}{7} \approx 160 + 188.57 = 348.57 \text{ m}$

Answer (i): $348.57 \text{ m}$

**Part (ii)**

1. Given: Two semicircles of diameters $8 \text{ cm}$ and $12 \text{ cm}$ forming an enclosed shape
2. Calculation: Perimeter = $\pi r_1 + \pi r_2 + \text{straight parts} = \frac{22}{7} \times 4 + \frac{22}{7} \times 6 + 4 + 4 = \frac{220}{7} + 8 \approx 31.43 + 8 = 39.43 \text{ cm}$

Answer (ii): $39.43 \text{ cm}$

**Part (iii)**

1. Given: Four quarter circles forming a shape with inner/outer boundaries of diameter $10 \text{ cm}$
2. Calculation: Perimeter = circumference of one full circle of diameter $10 \text{ cm}$ = $\pi d = \frac{22}{7} \times 10 \approx 31.43 \text{ cm}$

Answer (iii): $31.43 \text{ cm}$

**Part (iv)**

1. Given: Three-quarter circle with radius $12 \text{ cm}$ plus two radii
2. Calculation: Perimeter = $\frac{3}{4} \times 2 \pi r + 2r = \frac{3}{4} \times 2 \times \frac{22}{7} \times 12 + 2(12) = \frac{396}{7} + 24 \approx 56.57 + 24 = 80.57 \text{ cm}$

Answer (iv): $80.57 \text{ cm}$

**Part (v)**

1. Given: Circle of diameter $14 \text{ cm}$ with internal quarter-circle arcs
2. Calculation: Perimeter = $\pi d = \frac{22}{7} \times 14 = 44 \text{ cm}$

Answer (v): $44 \text{ cm}$

**Part (vi)**

1. Given: Shape with base $28 \text{ cm}$ and semicircular top
2. Calculation: Perimeter = $\pi r + 28 = \frac{22}{7} \times 14 + 28 = 44 + 28 = 72 \text{ cm}$

Answer (vi): $72 \text{ cm}$

**Answer:** (i) $285.71 \text{ m}$ (ii) $50.28 \text{ cm}$ (iii) $31.4 \text{ cm}$ (iv) $30.85 \text{ cm}$ (v) $88 \text{ cm}$ (vi) $72 \text{ cm}$ (vii) $28.56 \text{ cm}$ (viii) $18.28 \text{ cm}$ (ix) $51.4 \text{ cm}$

> Common mistake: Including internal boundaries that are not part of the outer perimeter.

### Question 6

*3 marks · Short answer*

If the diameter of a car tyre is 56 cm, then: (i) How far does the car need to travel for the tyre to complete one revolution? (ii) How many revolutions does the tyre make if the car travels 10 km?

**Part (i)**

1. Given: Diameter $d = 56 \text{ cm}$
2. Formula: Distance in one revolution = Circumference $C = \pi d$
3. Substitution: $C = \frac{22}{7} \times 56 = 176 \text{ cm} = 1.76 \text{ m}$

Answer (i): $1.76 \text{ m}$

**Part (ii)**

1. Given: Total distance = $10 \text{ km} = 10,00,000 \text{ cm}$
2. Formula: Number of revolutions = $\frac{\text{Total Distance}}{\text{Circumference}}$
3. Substitution: $\text{Revolutions} = \frac{10,00,000}{176} = \frac{125,000}{22} = \frac{62,500}{11} \approx 5681.82$

Answer (ii): $5681.82 \text{ revolutions}$

**Answer:** (i) $1.76 \text{ m}$ (ii) $5681.82 \text{ revolutions}$

> Common mistake: Mixing units between centimeters and kilometers.

### Question 7

*3 marks · Short answer*

Find the total perimeter of all the petals in each of the given flowers.

**Part (i)**

1. In Fig. 6.15A, the flower consists of 4 semicircular petals whose centres are the midpoints of the sides of a square of side $14 \text{ cm}$.
2. The diameter of each semicircle is equal to the side of the square, which is $14 \text{ cm}$, so the radius is $7 \text{ cm}$.
3. Each petal has a perimeter equal to the length of a semicircle, which is $\pi r$. Total perimeter of 4 petals $= 4 \times \pi r = 4 \times \frac{22}{7} \times 7 = 176 \text{ cm}$.

Answer (i): $176 \text{ cm}$

**Part (ii)**

1. In Fig. 6.15B, the flower consists of 6 petals formed by arcs whose centres are the vertices of a hexagon of side $42 \text{ cm}$.
2. The radius of each circular arc is equal to the side of the hexagon, which is $42 \text{ cm}$, and each arc is a quarter circle (sector angle $90^\circ$).
3. Total perimeter $= 6 \times \left(\frac{1}{4} \times 2\pi r\right) = 6 \times \frac{1}{4} \times 2 \times \frac{22}{7} \times 42 = 264 \text{ cm}$.

Answer (ii): $264 \text{ cm}$

**Answer:** (i) $176 \text{ cm}$, (ii) $264 \text{ cm}$

> Common mistake: Confusing the radius of the arcs with the side length or miscounting the number of petal arcs.

### Question 8

*3 marks · Short answer*

The ratio of the perimeters of two circles is 5:4. What is the ratio of their radii?

**Solution**

1. Let the radii of the two circles be $r_1$ and $r_2$.
2. Their perimeters (circumferences) are $C_1 = 2\pi r_1$ and $C_2 = 2\pi r_2$.
3. Given that $\frac{C_1}{C_2} = \frac{5}{4}$, we have $\frac{2\pi r_1}{2\pi r_2} = \frac{5}{4}$.
4. Canceling $2\pi$ from numerator and denominator gives $\frac{r_1}{r_2} = \frac{5}{4}$.
5. Therefore, the ratio of their radii is $5:4$.

**Answer:** $5:4$

> Common mistake: Assuming the ratio of areas is the same as the ratio of perimeters.

## Related pages

- [All Chapter 6 NCERT solutions](https://www.swavid.com/maths/class/9/chapter/measuring-space-perimeter-and-area/ncert-solutions)
- [Exercise 6.2](https://www.swavid.com/maths/class/9/chapter/measuring-space-perimeter-and-area/ncert-solutions/exercise-6-2)
- [Exercise 6.3](https://www.swavid.com/maths/class/9/chapter/measuring-space-perimeter-and-area/ncert-solutions/exercise-6-3)

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