NCERT Solutions for Class 9 Maths Exercise 13.3 Chapter 13 Surface Areas and Volumes

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

Students can practise and enhance their Math skills by solving the chapter-wise NCERT Solutions for Class 9 Maths which 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 final exams. NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.3 helps students to score well and also to face the exams more confidently as they gain practice solving these exercises.

We bring you a detailed collection of questions and solutions from the exercises with relevant answers created by our subject-matter experts and experienced teaching faculty. NCERT solutions aim to help students to score high in the CBSE Board exams. We provide proper illustrations and explanations so that students can understand the concepts in a better way.

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

Exercise 13.1 Solutions (9 questions)

Exercise 13.2 Solutions (8 questions)

Exercise 13.4 Solutions (5 questions)

Exercise 13.5 Solutions (5 questions)

Exercise 13.6 Solutions (8 questions)

Exercise 13.7 Solutions (9 questions)

Exercise 13.8 Solutions (10 questions)

Exercise 13.9 Solutions (3 questions)

Access Answers to Maths NCERT Class 9 Chapter 13 Surface Areas and Volumes Exercise 13.3

1. Diameter of the base of a cone is 10.5 cm, and its slant height is 10 cm. Find its curved surface area (Assume π=22/7)

Solution:

Radius of the base of cone = diameter/2 = (10.5/2)cm = 5.25cm

The slant height of the cone, say l = 10 cm

CSA of the cone is = πrl

= (22/7)×5.25×10 = 165 cm2

Therefore, the curved surface area of the cone is 165 cm2.

2. Find the total surface area of a cone, if its slant height is 21 m and the diameter of its base is 24 m. (Assume π = 22/7)

Solution:

Radius of cone, r = 24/2 m = 12m

Slant height, l = 21 m

Formula: Total Surface area of the cone = πr(l+r)

Total Surface area of the cone = (22/7)×12×(21+12) m2

= 1244.57m2

3. Curved surface area of a cone is 308 cm2, and its slant height is 14 cm. Find

(i) the radius of the base and (ii) the total surface area of the cone.

(Assume π = 22/7)

Solution:

The slant height of the cone, l = 14 cm

Let the radius of the cone be r.

(i) We know that the CSA of cone = πrl

Given: Curved surface area of a cone is 308 cm2

(308 ) = (22/7)×r×14

308 = 44 r

r = 308/44 = 7 cm

Therefore, the radius of the cone base is 7 cm.

(ii) Total surface area of cone = CSA of cone + Area of base (Ï€r2)

Total surface area of cone = 308+(22/7)×72 = 308+154 = 462 cm2

Therefore, the total surface area of the cone is 462 cm2.

4. A conical tent is 10 m high, and the radius of its base is 24 m. Find

(i) slant height of the tent.

(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs 70.

(Assume π=22/7)

Solution:

Ncert solutions class 9 chapter 13-5

Let ABC be a conical tent

Height of conical tent, h = 10 m

The radius of the conical tent, r = 24m

Let the slant height of the tent be l.

(i) In the right triangle ABO, we have

AB2 = AO2+BO2(using Pythagoras theorem)

l2 = h2+r2

= (10)2+(24)2

= 676

l = 26 m

Therefore, the slant height of the tent is 26 m.

(ii) CSA of tent = πrl

= (22/7)×24×26 m2

Cost of 1 m2 canvas = Rs 70

Cost of (13728/7)m2 canvas is equal to Rs (13728/7)×70 = Rs 137280

Therefore, the cost of the canvas required to make such a tent is Rs 137280.

5. What length of tarpaulin 3 m wide will be required to make a conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. [Use π=3.14]

Solution:

Height of conical tent, h = 8m

The radius of the base of the tent, r = 6m

The slant height of the tent, l2 = (r2+h2)

l2 = (62+82) = (36+64) = (100)

or l = 10 m

Again, CSA of conical tent = πrl

= (3.14×6×10) m2

= 188.4m2

Let the length of the tarpaulin sheet required be L

As 20 cm will be wasted,

The effective length will be (L-0.2m).

The breadth of tarpaulin = 3m (given)

Area of sheet = CSA of the tent

[(L–0.2)×3] = 188.4

L-0.2 = 62.8

L = 63 m

Therefore, the length of the required tarpaulin sheet will be 63 m.

6. The slant height and base diameter of a conical tomb are 25m and 14 m, respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2. (Assume π = 22/7)

Solution:

The slant height of the conical tomb, l = 25m

Base radius, r = diameter/2 = 14/2 m = 7m

CSA of conical tomb = πrl

= (22/7)×7×25 = 550

CSA of conical tomb= 550m2

Cost of white-washing 550 m2 area, which is Rs (210×550)/100

= Rs. 1155

Therefore, the cost will be Rs. 1155 while white-washing the tomb.

7. A joker’s cap is in the form of a right circular cone with a base radius of 7 cm and a height of 24cm. Find the area of the sheet required to make 10 such caps. (Assume π =22/7)

Solution:

The radius of the conical cap, r = 7 cm

Height of conical cap, h = 24cm

Slant height, l2 = (r2+h2)

= (72+242)

= (49+576)

= (625)

Or l = 25 cm

CSA of 1 conical cap = πrl

= (22/7)×7×25

= 550 cm2

CSA of 10 caps = (10×550) cm2 = 5500 cm2

Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.

8. A bus stop is barricaded from the remaining part of the road by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and a height of 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m2, what will be the cost of painting all these cones? (Use π = 3.14 and take √(1.04) =1.02)

Solution:

Given:

Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m

Height of cone, h = 1m

The slant height of the cone is l, and l2 = (r2+h2)

Using given values, l2 = (0.22+12)

= (1.04)

Or l = 1.02 m

The slant height of the cone is 1.02 m

Now,

CSA of each cone = πrl

= (3.14×0.2×1.02)

= 0.64056 m

CSA of 50 such cones = (50×0.64056) = 32.028

CSA of 50 such cones = 32.028 m2

Again,

Cost of painting 1 m2 area = Rs 12 (given)

Cost of painting 32.028 m2 area = Rs (32.028×12)

= Rs.384.336

= Rs.384.34 (approximately)

Therefore, the cost of painting all the cones is Rs. 384.34.


From this exercise of Chapter 13 of NCERT Solutions for Class 9 Maths, students will learn how to find the surface area and volume of various geometrical objects in a simplified 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 13.3

Solving these solutions help students to:

  • Self-assess their knowledge of the concept
  • Improve efficiency in solving the problems
  • Know the type of questions that appear in the exams
  • Remember the formulas easily

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  1. Very good

  2. Nice

  3. very helpful for quick revise thank you

  4. IT IS VERY USEFUL FOR WRITING NOTES THANKS

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