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 in creating these questions is to enable the students to score well in Class 9 first and second term 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 experts and experienced teaching faculty. NCERT solutions aim to help students to score high in first and second term exams. We provide proper illustrations and explanations, so that students can understand the concepts in a better way.
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 13.4 solution (5 questions)
Exercise 13.5 solution (5 questions)
Exercise 13.6 solution (8 questions)
Exercise 13.7 solution (9 questions)
Exercise 13.8 solution (10 questions)
Exercise 13.9 solution (3 questions)
Access Answers of 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)
Radius of the base of cone = diameter/ 2 = (10.5/2)cm = 5.25cm
Slant height of cone, say l = 10 cm
CSA of cone is = πrl
= (22/7)×5.25×10 = 165
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 diameter of its base is 24 m. (Assume π = 22/7)
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
3. Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find
(i) radius of the base and (ii) total surface area of the cone.
(Assume π = 22/7)
Slant height of cone, l = 14 cm
Let the radius of the cone be r.
(i) We know, 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
Radius of a cone base is 7 cm.
(ii) Total surface area of cone = CSA of cone + Area of base (πr2)
Total surface area of cone = 308+(22/7)×72 = 308+154
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.
Let ABC be a conical tent
Height of conical tent, h = 10 m
Radius of conical tent, r = 24m
Let the slant height of the tent be l.
(i) In right triangle ABO, we have
AB2 = AO2+BO2(using Pythagoras theorem)
l2 = h2+r2
l = 26
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 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]
Height of conical tent, h = 8m
Radius of base of tent, r = 6m
Slant height of tent, l2 = (r2+h2)
l2 = (62+82) = (36+64) = (100)
or l = 10
Again, CSA of conical tent = πrl
= (3.14×6×10) m2
Let the length of tarpaulin sheet required be L
As 20 cm will be wasted, therefore,
Effective length will be (L-0.2m).
Breadth of tarpaulin = 3m (given)
Area of sheet = CSA of tent
[(L–0.2)×3] = 188.4
L-0.2 = 62.8
L = 63
Therefore, the length of the required tarpaulin sheet will be 63 m.
6. The slant height and base diameter of 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)
Slant height of 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, cost will be Rs. 1155 while white-washing tomb.
7. A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. (Assume π =22/7)
Radius of conical cap, r = 7 cm
Height of conical cap, h = 24cm
Slant height, l2 = (r2+h2)
Or l = 25 cm
CSA of 1 conical cap = πrl
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 height 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)
Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m
Height of cone, h = 1m
Slant height of cone is l, and l2 = (r2+h2)
Using given values, l2 = (0.22+12)
Or l = 1.02
Slant height of the cone is 1.02 m
CSA of each cone = πrl
CSA of 50 such cones = (50×0.64056) = 32.028
CSA of 50 such cones = 32.028 m2
Cost of painting 1 m2 area = Rs 12 (given)
Cost of painting 32.028 m2 area = Rs (32.028×12)
= Rs.384.34 (approximately)
Therefore, the cost of painting all these 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 about the concept
- Improve the efficiency in solving the problems
- Can know the type of questions that appear for the exams
- Remember the formulas easily