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NCERT Solutions for Class 9 Maths Exercise 13.2 -Chapter 13 Surface Areas and Volumes
NCERT Solutions for Class 9 Maths Exercise 13.2 -Chapter 13 Surface Areas and Volumes are created here for students of class 9 for the preparation of term 2 CBSE Board exams. Chapter 13-Surface Areas and Volumes is the most important chapter in the Class 9 term 2 maths exam, this chapter is based on surface areas and volumes of three-dimensional figures like cubes, cuboids, cylinders, cones, and the sphere. Here in this chapter, you will study the problems on surface areas and volumes one by one from exercises 13.1 to 13.9.NCERT Solutions for Class 9 Maths Exercise 13.2 -Chapter 13 Surface Areas and Volumes are the most important inputs to correct the maths concepts. These questions are also useful for the preparation of competitive entrance exams like SSC, CPO, Railway, Bank clerk, PO, etc.
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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Term 2 )CBSE Board exam
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NCERT Solutions for Class 9 Maths Exercise 13.2 -Chapter 13 Surface Areas and Volumes
Q1.The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.(Assume π = 22/7).
Ans. The curved surface area of a right circular cylinder = 2πrh
Height,h given =14 cm
CSA is given = 88 cm²
2πrh = 88
2×(22/7) ×r×14 = 88
r = 88×7/(2×22×14) = 14/14 =1
The radius of the right circular cylinder is 1 cm
Hence the diameter of the cylinder is 2(1) = 2cm
Q2.It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same?
Ans. The height(h) of closed cylindrical tank is = 1m
Base diameter of the cylinder is =140 cm
Radius of the base is = 140/2 = 70 cm = 0.70 m
Total surface area of the cylinder =2πr(r +h)
TSA of cylinder = 2(22/7) 0.70(0.70 + 1)
= 2×22×0.10×1.7
= 7.48 cm²
Therefore , required metallic sheet to make a closed cylindrical tank is 7.48 cm²
Q3.A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4 cm. Find the following:
(i) Inner curved surface area
(ii)Outer curved surface area
(iii)Total surface area
Ans. The length of metal pipe = 77 cm
Inner diameter of the pipe = 4 cm,corresponding radius = 4/2 = 2 cm
Outer diameter of the pipe = 4.4 cm,corresponding radius =4.4/2 = 2.2 cm
(i)Inner curve surface area of the pipe =2πrh =2×(22/7)×2×77 = 2×22×2×11 = 44×22 =968 cm²
(ii)Outer curve surface area of the pipe =2πrh =2×(22/7)×2.2×77 = 2×22×2.2×11 = 44×24.2 =1064.8 cm²
(iii) Total surface area =Inner surface area + Outer surface area + Area of both cross-sections
Area of both cross-sections =2π(outer radius² – Inner radius²)
= 2×(22/7) [2.2² -2²] =(44/7) [4.84 – 4] =(44/7)×.84 =44×.12 =5.28 cm²
Total surface area = 968 + 1064.8 +5.28 =2038.08 cm²
Q4.The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2? (Assume π = 22/7)
Ans. The diameter of the roller is given 84 cm,therefore radius r = 84/2=42 cm=0.42 m
Length,h is 120 cm =1.20 m
Area of the playground = No.of revolution × Area covered in a single revolution
Area covered in a single revolution = curve surface area of the cylinder
Curve surface area of the cylinder = 2πrh = 2×(22/7)× 0.42 ×1.20 =2 ×22×0.06×1.20 =3.168 m²
Area of the playground =500 × 3.168 =1584 m²
Q5.A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2.
Ans. Diameter of cylindrical pillar =50 cm ,therefore radius r =50/2=0.25 m
Height (h) of cylindrical pillar is =3.5 m
Curve surface area of cylindrical pillar =2πrh = 2×(22/7)× 0.25×3.5 = 2×22×0.25×0.5=5.5 m²
Cost of painting the curved surface of the pillar=5.5 ×12.50=68.75
Hence the cost of painting the curved surface of Piller is Rs 68.75
Q6.Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height. (Assume π = 22/7)
Ans.Curved surface area of a right circular cylinder = 4.4 m²
Radius(r) of the base of the cylinder =0.7 m
Let the height of circular cylinder =h
Curve surface area of the cylinder =2πrh =2×(22/7)×0.7 h
∴2×(22/7)×0.7 h = 4.4
2×22 ×0.1h =4.4
h = 44/44 =1
Hence height of the cylinder is 1 m
Q7.The inner diameter of a circular well is 3.5m. It is 10m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2.
(Assume π = 22/7)
Ans. The inner diameter of a circular well is 3.5m, therefore radius r =3.5/2 =1.75 m
The depth (h) of the well is = 10 m
(i) Inner curved surface area =2πrh
= 2×(22/7)×1.75 ×10
=2×22×0.25×10=44×2.5=110 m²
(ii) The rate of plastering inner curve surface area of the well =Rs. 40 per m²
Cost of plastering inner curved surface area of the well = 40×110 =4400
Hence the cost of plastering inner curved surface area of the well is Rs 4400
Q8.In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find
the total radiating surface in the system. (Assume π = 22/7)
Ans. The length, h of the cylindrical pipe = 28 m
Diameter of the pipe is 5 cm,therefore radius ,r of the pipe= 5/2 =2.5 cm =0.025 m
Total radiating surface of the pipe = Curve surface area of the pipe
Total radiating surface of the pipe
= 2πrh
= 2×(22/7)×0.025 ×28
=2×22×0.025×4
=44×0.1
=4.4
Hence the total radiating surface of the pipe is 4.4 m
Q9.Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in
diameter and 4.5m high.
(ii) How much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank. (Assume π = 22/7)
Ans. The diameter of the closed cylindrical petrol storage tank is 4.2 m,therefore its radius(r) is 4.2/2 =2.1 m
Height(h) of the cylindrical petrol storage tank is =4.5 m
Curved surface area of a closed cylindrical tank
=2πrh
= 2×(22/7)×2.1 ×4.5
= 2×22×0.3×4.5
=59.4
Hence the lateral or curved surface area of a closed cylindrical petrol storage tank is 59.4 m²
(ii) Let the total steel actually used in making the tank is x m²
The steel wasted in making the tank is (1/12) of x =x/12
Total surface area of the tank=Total steel used – wasted steel =x -x/12 =11x/12
Total surface area of the tank
=2πr(r +h)
=2×(22/7)×2.1(2.1+4.5)
=44×0.3×6.6
=87.12
11x/12 = 87.12
x = (87.12×12)/11 = 95.04
Total steel used in making the tank is 95.04 m²
Q10.In fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth.The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. (Assume π = 22/7)
Ans. The diameter of the lampshade frame is 20 cm, therefore radius r of the lampshade frame is =20/2 =10 cm
Height h of the lampshade frame is 30 cm
The cloth required for covering the lampshade
=2πrH ,where H is length of the cover
= 2×(22/7)×10[ h + length of margin at the top and base)]
= 2×(22/7)×10[ 30+ 2×2.5)]
=(440/7)[35]
= 440×5
=2200
Hence cloth is required for covering the lampshade is 2200 cm²
Q11.The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? (Assume π =22/7)
Ans. Rdius r of the penholder is 3 cm and its height h is 10.5 cm
Cardboard required for a single penholder =Curved surface area of penholder +area of its circular base
Cardboard required for a single penholder
=2πrh +πr²
=πr(2h +r)
=(22/7) ×3(2×10.5 +3)
=(66/7)(21 +3)
=66×24/7
=1584/7
Cardboard required for 35 competitors is
=(35 ×1584)/7
=5 ×1584
=7920
Hence cardboard was required to be bought for the competition is 7920 cm²
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NCERT Solutions of Science and Maths for Class 9,10,11 and 12
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 |
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NCERT solutions of class 11 maths
| Chapter 1-Sets | Chapter 9-Sequences and Series |
| Chapter 2- Relations and functions | Chapter 10- Straight Lines |
| Chapter 3- Trigonometry | Chapter 11-Conic Sections |
| Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |
| Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |
| Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |
| Chapter 7- Permutations and Combinations | Chapter 15- Statistics |
| Chapter 8- Binomial Theorem | Chapter 16- Probability |
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NCERT solutions of class 12 maths
| Chapter 1-Relations and Functions | Chapter 9-Differential Equations |
| Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |
| Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |
| Chapter 4-Determinants | Chapter 12-Linear Programming |
| Chapter 5- Continuity and Differentiability | Chapter 13-Probability |
| Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |
| Chapter 7- Integrals | |
| Chapter 8-Application of Integrals |
