Our ncert solutions for Exercise 13.7 Class 9 maths 13. Surface Areas and Volumes - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Exercise 13.7 Class 9 maths 13. Surface Areas and Volumes - ncert solutions - Toppers Study

Topper Study classes prepares ncert solutions on practical base problems and comes out with the best result that helps the students and teachers as well as tutors and so many ecademic coaching classes that they need in practical life. Our ncert solutions for Exercise 13.7 Class 9 maths 13. Surface Areas and Volumes - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

## Exercise 13.7 Class 9 maths 13. Surface Areas and Volumes - ncert solutions - Toppers Study

CBSE board students who preparing for **class 9 ncert solutions maths and Mathematics** solved exercise **chapter 13. Surface Areas and Volumes** available and this helps in upcoming exams
2023-2024.

### You can Find Mathematics solution Class 9 Chapter 13. Surface Areas and Volumes

- All Chapter review quick revision notes for chapter 13. Surface Areas and Volumes Class 9
- NCERT Solutions And Textual questions Answers Class 9 Mathematics
- Extra NCERT Book questions Answers Class 9 Mathematics
- Importatnt key points with additional Assignment and questions bank solved.

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### Exercise 13.7 class 9 Mathematics Chapter 13. Surface Areas and Volumes

Sure! The following topics will be covered in this article

- Exercise 13.7 Class 9 Maths 13. Surface Areas And Volumes - Ncert Solutions - Toppers Study
- Class 9 Ncert Solutions
- Solution Chapter 13. Surface Areas And Volumes Class 9
- Solutions Class 9
- Chapter 13. Surface Areas And Volumes Exercise 13.7 Class 9

## 13. Surface Areas and Volumes

### | Exercise 13.7 |

## Exercise 13.7 Class 9 maths 13. Surface Areas and Volumes - ncert solutions - Toppers Study

**EXERCISE 13.7**

**Assume π** **= 22/7****, unless stated otherwise.**

**1. Find the volume of the right circular cone with**

**(i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm**

**Solution:**

**2. Find the capacity in litres of a conical vessel with**

**(i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm**

**Solution:**

(i) h^{2} = 25^{2} - 7^{2}

**3. The height of a cone is 15 cm. If its volume is 1570 cm ^{3}, find the radius of the base.(Use π**

**= 3.14)**

**Solution:**

Height = 15 cm, volume = 1570 cm^{3}

^{}

**4. If the volume of a right circular cone of height 9 cm is 48π** **cm ^{3}, find the diameter of its base.**

**Solution:**

Height = 9 cm, volume = 48π cm^{3}

^{}

**5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?**

**Solution:**

Diameter = 3.5 cm , height = 12 cm , radius = 1.75

**6. The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find**

**(i) height of the cone (ii) slant height of the cone**

**(iii) curved surface area of the cone**

**Solution:**

Volume = 9856 cm^{3} , diameter = 28 cm , radius = 14cm

**7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.**

**Solution:**

Height = 12 cm , radius = 5 cm

**8. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.**

**Solution:**

Height = 5 m , radius = 12 m

**9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.**

**Solution:**

Diameter = 10.5 m, height = 3 m , radius = 5.25

##### Other Pages of this Chapter: 13. Surface Areas and Volumes

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