Class 9 Maths Chapter 13 Exercise 13.3 - Surface Areas and Volumes NCERT Solutions
NCERT Solutions for Class 9 Maths Exercise 13.3: Surface Areas and Volumes will help clarify your understanding of the surface area of cones, which are common three-dimensional objects found in daily life, such as a joker’s cap, conical tombs, and other conical structures. These solutions are crucial study material for preparing for the Term 2 CBSE Board exam, covering all questions related to the topic. Created by an experienced CBSE teacher, these NCERT Solutions for Class 9 Maths Exercise 13.3 are designed using a step-by-step approach to aid Class 9 students in mastering the concepts.
Q1. The 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: The diameter of base of the cone 10.5 cm
Radius of the base,r =10.5/2 = 5.25 cm
The slant height of the cone is,l = 10 cm
Curved surface area of the cone,CSA = ?
CSA = πrl
= (22/7)×5.25×10
= 22 ×0.75 ×10
=22× 7.5
=165 cm²
Hence Curved surface area of the cone is 165 cm²
Q2.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)
Solution: The slant height of the cone is,l = 21 cm
Diameter of the base of the cone= 24 cm
Radius of the base,r =24/2 = 12cm
Total surface area of the cone,TSA =πr(r +l)
TSA = (22/7)×12(12 +21)
= (22/7)×12×33
= 1244.57 cm²
Hence total surface area of the cone is 1244.57 cm²
Q3.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)
Solution: Curved surface area of a cone .CSA= 308 cm²
The slant height of the cone is,l = 14 cm
(i) CSA of the cone = πrl
(22/7)×r×14 = 308
r = (308×7)/(22×14)
= (14×7)/14
= 7 cm
Hence radius of the base is 7 cm
Q4. 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: The height(h) of the conical tent = 10 m
Radius of its base,r = 24 m
(i)Let slant height of the conical tent is =l
Applying Pythogorus theorem
l = r² + h²
l = √(r² + h²)
= √(24² + 10²)
=√(576 + 100)
=√676 = 26
Hence slant height of the conical tent is 26 m
(ii) Cost of 1 m2 canvas is Rs 70
The canvas required to make the tent = curve surface area of the tent
∴The canvas required to make the tent
=πrl
=(22/7)×24×26
=13728/7
Therefore canvas required to make the tent is 13728/7 m²
Cost of the canvas required to make the tent = cost of 1m² canvas ×canvas required to make the tent
= 70 ×13728/7
=10 ×13728 = 137280
Hence the cost of the canvas required to make the tent is Rs 137280
Q5.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]
Solution: Let the length of tarpaulin required to make conical tent is x
Breadh of tarpaulin is given 3 m
Height(h) of the conical tent is given 8 m and radius(r) 6 m
Slant height(l) = √(8² +6²) = √(64 +36)=√100 = 10 m
The extra length of material that will be required for stitching margins and wastage in cutting is given 20 cm=0.20 m
Curved surface area of the conical tent is = πrl
Area of the tarpaoline = Length × Breadh
Since 0.20 m of the length is used for stitching margins and wastage
∴ (Length +0.20)× Breadh = πrl
(x -0.20)3 = (22/7) ×6×10
7(3x – 0.60 )= 22×6×10
21x – 4.20 = 1320
21x = 1320 +4.20
21x = 1320
x = 1320/21 =62.89≈63
Hence the length of the tarpauline required to make conical tent is 63 m
Q6.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)
Solution: The given dimensions of conical tomb are given as following
The slant height of the conical tomb is,l = 25 m
Diameter of the base of the conical tomb= 14 m
Radius of the base,r =14/2 = 7m
Curved surface area of the conical tomb
=πrl
=(22/7)×7×25
=22×25 =550
Hence the curved surface area of the conical tent is 550 m²
The cost of whitewashing its curved surface area is Rs 210 per 100 m²
The cost of whitewashing in 1 m² = 210 /100
The cost of whitewashing the curved surface area of the tomb
= 550 ×(210 /100)
=55×21 =1155
Hence the cost of white-washing curved surface area of the tomb is Rs 1155
Q7.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)
Solution: The given dimensions of joker’s cap are given as following
The height of the joker’s cap is,h = 24cm
Radius of the base,r = 7cm
The height of the joker’s cap is,h = 24cm
Slant height(l) = √(24² +7²) = √(576 +49)=√625 = 25 cm
Curved surface area of the joker’s cap
=πrl
=(22/7)×7×25 = 550
Hence the curved surface area of the joker’s cap is 550 cm²
The area of the sheet required to make 10 such caps = 10× 550 =5500 cm²
Q8.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)
Solution: The base diaeter of the hollow cone is 40 cm,therefore radius(r) is 20 cm=0.20m
Height of the hollow cone is given 1m
Slant height of the hollow cone , l = √(0.20² +1²) = √(0.04 +1)=√1.04= 1.02 m
Curved surface area of the hollow cone
=πrl
=3.14×0.20×1.02 = 0.64056
The curved surface area of single hollow cone is 0.64056 m²
The curved surface area of 50 hollow cone is 0.64056 ×50=32.028 m²
Cost of painting is Rs. 12 per m²
The cost of painting all these cones is = 12×32.028 =384.336≈Rs 384.34
Understanding Class 9 Maths Chapter 13 Exercise 13.3 is important for getting the hang of Surface Areas and Volumes. Our simple, step-by-step solutions for Class 9 Maths Chapter 13 Exercise 13.3 will help you work through each problem easily. With these solutions, Surface Areas and Volumes will feel much easier to learn. Be sure to download the PDF for Class 9 Chapter 13 Maths Exercise 13.3, so you can practice anytime and keep these key concepts handy. Keep practicing, and you’ll do great in your exams!
NCERT Solutions of class 9 maths
| Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |
| Chapter 2-Polynomial | Chapter 10-Circles |
| Chapter 3- Coordinate Geometry | Chapter 11-Construction |
| Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |
| Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |
| Chapter 6-Lines and Angles | Chapter 14-Statistics |
| Chapter 7-Triangles | Chapter 15-Probability |
| Chapter 8- Quadrilateral |