NCERT Solutions for Class 9 Maths Exercise 13.7 Chapter 13 Surface Areas and Volumes

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 11.

We aim to bring you a detailed collection of questions and solutions from the exercises with relevant answers, as per the NCERT syllabus. NCERT Solutions for Class 9 Maths are meticulously created by our subject-matter experts and experienced teaching faculty. NCERT solutions help students to score high in class, board and competitive examinations. Students can practise and enhance their Math skills by solving the NCERT solutions chapter-wise for Class 9 Maths that is provided here. These solutions include questions from the exercises given in the NCERT Textbooks as per the syllabus guidelines.

The main aim of creating these questions is to enable the students to score well in Class 9 CBSE exams. NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.7 helps students to face the board exams with confidence and also score marks as they gain a hold on the chapter by solving these exercises.

 

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Access Other Exercise Solutions of Class 9 Maths Chapter 13 Surface Areas and Volumes

Exercise 13.1 solution (9 questions)

Exercise 13.2 solution (8 questions)

Exercise 13.3 solution (9 questions)

Exercise 13.4 solution (5 questions)

Exercise 13.5 solution (5 questions)

Exercise 13.6 solution (8 questions)

Exercise 13.8 solution (10 questions)

Exercise 13.9 solution (3 questions)

Access Answers to NCERT Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.7

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 (Assume π = 22/7)

Solution:

Volume of cone = (1/3) πr2h cube units

Where r be radius and h be the height of the cone

(i) Radius of the cone, r = 6 cm

Height of cone, h = 7cm

Let V be the volume of the cone; we have

V = (1/3)×(22/7)×36×7

= (12×22)

= 264

The volume of the cone is 264 cm3.

(ii) Radius of the cone, r = 3.5cm

Height of the cone, h = 12cm

Volume of the cone = (1/3)×(22/7)×3.52×7 = 154

Hence,

The volume of the cone is 154 cm3.

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

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

(Assume π = 22/7)

Solution:

(i) Radius of the cone, r =7 cm

Slant height of the cone, l = 25 cm

Ncert solutions class 9 chapter 13-12

or h = 24

Height of the cone is 24 cm

Now,

Volume of cone, V = (1/3) πr2h (formula)

V = (1/3)×(22/7) ×72×24

= (154×8)

= 1232

So, the volume of the vessel is 1232 cm3

Therefore, capacity of the conical vessel = (1232/1000) liters (because 1L = 1000 cm3)

= 1.232 Liters.

(ii) Height of the cone, h = 12 cm

Slant height of the cone, l = 13 cm

Ncert solutions class 9 chapter 13-13

r = 5

Hence, the radius of the cone is 5 cm.

Now, Volume of cone, V = (1/3)Ï€r2h

V = (1/3)×(22/7)×52×12 cm3

= 2200/7

Volume of the cone is 2200/7 cm3

Now, Capacity of the conical vessel= 2200/7000 litres (1L = 1000 cm3)

= 11/35 litres

3. The height of a cone is 15cm. If its volume is 1570cm3, find the diameter of its base. (Use π = 3.14)

Solution:

Height of the cone, h = 15 cm

Volume of the cone =1570 cm3

Let r be the radius of the cone

As we know: Volume of the cone, V = (1/3) πr2h

So, (1/3) πr2h = 1570

(1/3)×3.14×r2 ×15 = 1570

r2 = 100

r = 10

Radius of the base of the cone is 10 cm.

4. If the volume of a right circular cone of height 9cm is 48Ï€cm3, find the diameter of its base.

Solution:

Height of cone, h = 9cm

Volume of cone =48Ï€ cm3

Let r be the radius of the cone.

As we know: Volume of the cone, V = (1/3) πr2h

So, 1/3 π r2(9) = 48 π

r2 = 16

r = 4

Radius of the cone is 4 cm.

So, diameter = 2×Radius = 8

Thus, diameter of the base is 8 cm.

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

(Assume π = 22/7)

Solution:

Diameter of the conical pit = 3.5 m

Radius of the conical pit, r = diameter/ 2 = (3.5/2)m = 1.75m

Height of the pit, h = Depth of the pit = 12m

Volume of the cone, V = (1/3) πr2h

V = (1/3)×(22/7) ×(1.75)2×12 = 38.5

Volume of the cone is 38.5 m3

Hence, capacity of the pit = (38.5×1) kiloliters = 38.5 kiloliters.

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

(i) height of the cone

(ii) slant height of the cone

(iii) curved surface area of the cone

(Assume π = 22/7)

Solution:

Volume of a right circular cone = 9856 cm3

Diameter of the base = 28 cm

(i) Radius of cone, r = (28/2) cm = 14 cm

Let the height of the cone be h

Volume of the cone, V = (1/3) πr2h

(1/3) πr2h = 9856

(1/3)×(22/7) ×14×14×h = 9856

h = 48

The height of the cone is 48 cm.

Ncert solutions class 9 chapter 13-14

Slant height of the cone is 50 cm.

(iii) curved surface area of the cone = πrl

= (22/7)×14×50

= 2200

The curved surface area of the cone is 2200 cm2.

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

Solution:

Height (h)= 12 cm

Radius (r) = 5 cm, and

Slant height (l) = 13 cm

Ncert solutions class 9 chapter 13-15

Volume of cone, V = (1/3) πr2h

V = (1/3)×π×52×12

= 100Ï€

The volume of the cone so formed is 100Ï€ cm3.

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

Solution:

Ncert solutions class 9 chapter 13-16

A right-angled ΔABC is revolved about its side 5cm, a cone will be formed of radius 12 cm, height 5 cm, and slant height 13 cm.

Volume of cone = (1/3) πr2h; where r is the radius and h is the height of cone

= (1/3)×π×12×12×5

= 240 π

The volume of the cones formed is 240Ï€ cm3.

So, the required ratio = (result of question 7) / (result of question 8) = (100Ï€)/(240Ï€) = 5/12 = 5:12.

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.

(Assume π = 22/7)

Solution:

Radius (r) of heap = (10.5/2) m = 5.25

Height (h) of heap = 3m

Volume of heap = (1/3)Ï€r2h

= (1/3)×(22/7)×5.25×5.25×3

= 86.625

The volume of the heap of wheat is 86.625 m3.

Again,

Ncert solutions class 9 chapter 13-17

= (22/7)×5.25×6.05

= 99.825

Therefore, the area of the canvas is 99.825 m2.


This exercise of Chapter 13 of NCERT Solutions for Class 9 Maths helps students to learn how to find the surface area and volume of various geometrical objects in an easy and smart way. It helps students to tackle the questions in the best possible way. For detailed questions and explanations, students can refer to the exercises in the NCERT chapter-wise solutions.

Key Features of NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volume Exercise 13.7

Solving these solutions helps students in the following ways:

  • Improves their speed in answering problems.
  • Helps students to find the volume of the right circular cone with given specifications.
  • Aids students in finding the capacity of a conical vessel in litres.
  • Pupils can self-assess their knowledge about the chapter easily.

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  1. Best and useful for class9

  2. Yeah it solved all my queries

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