Class 10 Maths NCERT Solutions of Chapter 13 Exercise 13.1 -Surface Areas and Volume - Future Study Point

Class 10 Maths NCERT Solutions of Chapter 13 Exercise 13.1 -Surface Areas and Volume

chapter 13 class 10 math

Class 10 Maths NCERT Solutions of Chapter 13 exercise 13.1

Class 10 maths NCERT solutions of chapter 13-Surace areas and volumes are presented here for helping the class 10 students in their preparation of maths paper in forthcoming assessments and CBSE board exam 2020-21. The NCERTquestions of the chapter 13-Surface areas and volume will help the students of 10 class in complete comprehension of the areas and volumes of three-dimensional figures like 3D shapes, cubes, cuboids, cylinders, cones, spheres and three-dimensional composite shapes made of at least two or more than two figures.All questions of the chapter 13 exercise 13.1 are solved by a specialist of maths subject according to the CBSE standards. The NCERT solutions published here by Future Study Point will benifit all categories of students in light of the fact that each question is explained by step by step strategy so every students could comprehend the concept unmistakably with no assistance of teacher or tutor.

chapter 13 class 10 math

 

You can also study

Exercise 13.2

Exercise 13.3

Exercise -13.4

Class 10 Maths NCERT Solutions of Chapter 13 exercise 13.1

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

Q1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Ans.

The volume of cubes = Side³ = 64

Side = 4 cm

2 cubes adjacant

 

Length of resulting cuboid  = 4 + 4 = 8 cm

Breadth of cuboid = 4 cm

Height of resulting cuboid = 4 cm

Surface area of resulting cuboid = 2 ( lb + bh + hl)

l = 8 cm, b = 4 cm, h = 4 cm

Surface area of resulting cuboid = 2 ( 8 ×4 + 4×4 + 4×8) = 2(32 + 16 + 32) = 2× 80 = 160

Hence the required surface area of resulting cuboid = 160 cm²

Q2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:

Ans.Height of vessel is=13cm

Diameter of hemisphere = 14 cm, radius of hemishere = 14/2 = 7 cm

Exercise 13.1 q2 class 10 maths

 

Height of cylinder = 13 – 7 = 6 cm

The inner surface area of the vessel = CSA of hemisphere + CSA of cylinder

CSA of hemisphere = 2πr²

The curved surface area of the hemisphere is = 308 cm²

CSA of cylinder = 2πrh

The curved surface area of cylinder = 264 cm²

Hence surface area of the vessel = 308 + 264 = 572 cm²

Q3.A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Ans.

Height of cone = 13.5 – 3.5= 12 cm, radius of hemishere = 3.5 cm

The total surface area of the toy = CSA of hemisphere + CSA of cone

q3 ex 13.1 maths class 10

 

CSA of hemisphere = 2πr²

CSA of cone = πrl, where l is slant height

CSA  of cone

The total surface area of the toy = 77 + 137.5 = 214.5

Hence the total surface area of the toy is 214.5 sq.cm

Q4.A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Ans. The greatest diameter of the hemisphere surmounted on a cube will be side of cube i.e 7 cm

Q4 EX.13.1 class 10 maths

The surface area of so formed solid = Total surface area of cube + curved surface area of the hemisphere – area of the base of the hemisphere

Total surface area of cube = 6×side² = 6 × 7 × 7 = 294

The total surface area of cube = 294 sq.cm

The curved surface area of hemisphere = 2πr²

Area of base (i.e circle) = πr²

Surface area of given solid = 294 + 77 – 38.5 = 332.5 sq.cm

Hence the surface area of the solid is 332.5 sq.cm

Q5.A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter d of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:

Ans. Let the side of the cubical wooden block is = a unit

Diameter d of the hemisphere = a unit

The radius of hemisphere depression = a/2

Q5.EX 13.1 Class 10

 

 

The remaining surface area of the solid = Total surface area of cube + curved surface area of the hemisphere – area of the base of the hemisphere

The total surface area of cube = 6a²

The curved surface area of the hemisphere = 2πr²

Area of the circular base of hemisphere =πr²

The remaining surface area of the solid

Q6.A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Ans. The length of the capsule = 14 mm, length of the cylindrical portion of capsule = Length of the capsule – 2×2.5 = 14 – 5 = 9 mm

 

Q6 Chapter 13.1 maths 10

The surface area of the capsule = CSA of the cylindrical portion of the capsule + CSA of two hemispheres

CSA of the cylindrical portion of the capsule = 2πrh

CSA of two hemispheres  = 2× 2πr² = 4πr²

r = 2.5 mm

CSA of two hemispheres

The surface area of the capsule

The surface area of the capsule is =220 mm²

Q7.A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not be covered with canvas.)

Ans.q7 ch.13.1 maths class 10

 

Area of the canvas used for making the tent = Curved surface area of the cylindrical part of tent + Curved surface area of the conical part of the tent

The curved surface area of the cylindrical part of tent = 2πrh

Radius r of the cylindrical shape = 4/2 = 2 m

Height h of the cylindrical part = 2.1 m

The curved surface area of the cylindrical part of the tent

The curved surface area of cylindrical shape = 26.4 m²

The curved surface area of the conical part of the tent = πrl

Radius r of the conical shape = 4/2 = 2 m

Slant height l of the conical part = 2.8 m

The curved surface area of the conical part of the tent

The curved surface area of conical shape = 17.6 m²

Area of the canvas used for making the tent = 26.4 + 17.6 = 44 m²

The rate of  canvas = Rs 500 per m²

∴The cost of the canvas  =  500 × 44 =  Rs 22000

Q8.From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.

Ans. Height of the given cylinder = 2.4 cm

The diameter of the given cylinder = 1.4 cm

Q8 13-1 ex maths class 10

 

Radius of the given cylinder = 1.4/2 = 0.7 cm

The height of the hollowed-out conical cavity = 2.4 cm

The remaining surface area of the cylinder = Curved surface area of cylinder + area of the circular base of cylinder + Curved surface area of the conical cavity

= 2πrh + πr² + πrl

= πr( 2h + r + l)

Therefore the remaining surface area of the cylinder = 17.6 ≈ 18 cm²

Q9.A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Ans.

Q9 ex 13.1 class 10 maths

The total surface area of the article = Curved surface area of the cylinder + Curved surface area of scooped out hemispheres

The radius of cylinder and hemispheres = 3.5 cm

Height of the cylinder = 10 cm

The curved surface area of the cylinder = 2πrh

The curved surface area of scooped out hemispheres = 2× 2πr² = 4πr²

The total surface area of the article = 220 + 154 = 374 cm²

You can compensate us by donating any amount of money for our survival

Our Paytm NO 9891436286

NCERT Solutions of  Science and Maths for Class 9,10,11 and 12

NCERT Solutions of class 9 maths

Chapter 1- Number SystemChapter 9-Areas of parallelogram and triangles
Chapter 2-PolynomialChapter 10-Circles
Chapter 3- Coordinate GeometryChapter 11-Construction
Chapter 4- Linear equations in two variablesChapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s GeometryChapter 13-Surface Areas and Volumes
Chapter 6-Lines and AnglesChapter 14-Statistics
Chapter 7-TrianglesChapter 15-Probability
Chapter 8- Quadrilateral

CBSE Maths Class IX  solutions of important questions of last years question papers

CBSE Class 9 maths assignment for SA-1

CBSE class 9 Notes on Lines, angles, and triangles

Addition, subtraction, and division of polynomials

Solutions of maths specific questions of mensuration

NCERT Solutions of class 9 science 

Chapter 1-Matter in our surroundingsChapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?Chapter 10- Gravitation
Chapter3- Atoms and MoleculesChapter 11- Work and Energy
Chapter 4-Structure of the AtomChapter 12- Sound
Chapter 5-Fundamental unit of lifeChapter 13-Why do we fall ill ?
Chapter 6- TissuesChapter 14- Natural Resources
Chapter 7- Diversity in living organismChapter 15-Improvement in food resources
Chapter 8- MotionLast years question papers & sample papers

CBSE Class 9-Question paper of science 2020 with solutions

CBSE Class 9-Sample paper of science

CBSE Class 9-Unsolved question paper of science 2019

NCERT Solutions of class 10 maths

Chapter 1-Real numberChapter 9-Some application of Trigonometry
Chapter 2-PolynomialChapter 10-Circles
Chapter 3-Linear equationsChapter 11- Construction
Chapter 4- Quadratic equationsChapter 12-Area related to circle
Chapter 5-Arithmetic ProgressionChapter 13-Surface areas and Volume
Chapter 6-TriangleChapter 14-Statistics
Chapter 7- Co-ordinate geometryChapter 15-Probability
Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

NCERT solutions of class 10 science

Chapter 1- Chemical reactions and equationsChapter 9- Heredity and Evolution
Chapter 2- Acid, Base and SaltChapter 10- Light reflection and refraction
Chapter 3- Metals and Non-MetalsChapter 11- Human eye and colorful world
Chapter 4- Carbon and its CompoundsChapter 12- Electricity
Chapter 5-Periodic classification of elementsChapter 13-Magnetic effect of electric current
Chapter 6- Life ProcessChapter 14-Sources of Energy
Chapter 7-Control and CoordinationChapter 15-Environment
Chapter 8- How do organisms reproduce?Chapter 16-Management of Natural Resources

Solutions of class 10 last years Science question papers

CBSE Class 10 – Question paper of science 2020 with solutions

CBSE class 10 -Latest sample paper of science

NCERT solutions of class 11 maths

Chapter 1-SetsChapter 9-Sequences and Series
Chapter 2- Relations and functionsChapter 10- Straight Lines
Chapter 3- TrigonometryChapter 11-Conic Sections
Chapter 4-Principle of mathematical inductionChapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbersChapter 13- Limits and Derivatives
Chapter 6- Linear InequalitiesChapter 14-Mathematical Reasoning
Chapter 7- Permutations and CombinationsChapter 15- Statistics
Chapter 8- Binomial Theorem Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

NCERT solutions of class 12 maths

Chapter 1-Relations and FunctionsChapter 9-Differential Equations
Chapter 2-Inverse Trigonometric FunctionsChapter 10-Vector Algebra
Chapter 3-MatricesChapter 11 – Three Dimensional Geometry
Chapter 4-DeterminantsChapter 12-Linear Programming
Chapter 5- Continuity and DifferentiabilityChapter 13-Probability
Chapter 6- Application of DerivationCBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals

 

 

 

 

 

 

Scroll to Top