This chapter is an important part of the mathematics syllabus and deals with the calculation of surface areas and volumes of various geometric shapes such as cubes, cylinders, cones, spheres, etc.
The chapter provides students with the fundamental concepts of surface areas and volumes, such as the formulae for calculating the surface area and volume of different objects. Additionally, the chapter also covers various practical applications of surface areas and volumes, making it an essential topic for students to learn.
The NCERT Solutions provided here have been designed to help students understand the concepts covered in this chapter in a clear and concise manner. These solutions have been prepared by experienced teachers and subject matter experts, keeping in mind the latest CBSE guidelines and syllabus.
The solutions provide step-by-step explanations to all the questions in the NCERT textbook, ensuring that students understand the concepts thoroughly. Additionally, the solutions also provide tips and tricks to solve problems quickly and accurately.
We hope that these NCERT Solutions will help students in their preparation for the Class 10 board exams and also help them build a strong foundation for higher studies in mathematics.
Download PDF of NCERT Solutions for Class 10th Maths Chapter 13 Surface Areas And Volumes
Answers of Maths NCERT Solutions for Class 10th Maths Chapter 13 Surface Areas And Volumes
Unit 13
Surface Areas and Volumes
Exercise 13.1
Question 1:
2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboids.
Answer:
Given that,
Volume of cubes = 64 cm3 (Edge)3 = 64
Edge = 4 cm
If cubes are joined end to end, the dimensions of the resulting cuboid will be 4 cm, 4 cm, 8 cm.
*Question 2:
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.
Answer:
It can be observed that radius (r) of the cylindrical part and the hemispherical part is the same (i.e., 7 cm).
Height of hemispherical part = Radius = 7 cm Height of cylindrical part
Inner surface area of the vessel = CSA of cylindrical part + CSA of hemispherical part
*Question 3:
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Answer:
It can be observed that the radius of the conical part and the hemispherical part is same (i.e., 3.5 cm).
Height of hemispherical part = Radius (r) = 3.5 = cm
Height of conical part (h) = 15.5 −3.5 = 12 cm
Total surface area of toy = CSA of conical part + CSA of hemispherical part
*Question 4:
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.
Answer:
From the figure, it can be observed that the greatest diameter possible for such hemisphere is equal to the cube’s edge, i.e., 7cm.
Radius (r) of hemispherical part = = 3.5cm
Total surface area of solid = Surface area of cubical part + CSA of hemispherical part
− Area of base of hemispherical part
Question 5:
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Answer:
Diameter of hemisphere = Edge of cube = l
Radius of hemisphere =
Total surface area of solid = Surface area of cubical part + CSA of hemispherical part
− Area of base of hemispherical part
Question 6:
A medicine capsule is in the shape of cylinder with two hemispheres stuck to each of its ends (see the given figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Answer:
It can be observed that
Radius (r) of cylindrical part = Radius (r) of hemispherical part
Length of cylindrical part (h) = Length of the entire capsule − 2 × r
= 14 − 5 = 9 cm
Surface area of capsule = 2×CSA of hemispherical part + CSA of cylindrical part
Question 7:
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.)
Answer:
Given that,
Height (h) of the cylindrical part = 2.1 m Diameter of the cylindrical part = 4 m
Radius of the cylindrical part = 2 m Slant height (l) of conical part = 2.8 m
Area of canvas used = CSA of conical part + CSA of cylindrical part
Cost of 1 m2 canvas = Rs 500
Cost of 44 m2 canvas = 44 × 500 = 22000
Therefore, it will cost Rs 22000 for making such a tent.
Question 8:
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.
Answer:
Given that,
Height (h) of the conical part = Height (h) of the cylindrical part = 2.4 cm Diameter of the cylindrical part = 1.4 cm
Therefore, radius (r) of the cylindrical part = 0.7 cm
Total surface area of the remaining solid will be
= CSA of cylindrical part + CSA of conical part + Area of cylindrical base
The total surface area of the remaining solid to the nearest cm2 is 18 cm2.
*Question 9:
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in given 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.
Answer:
Given that,
Radius (r) of cylindrical part = Radius (r) of hemispherical part = 3.5 cm Height of cylindrical part (h) = 10 cm
Surface area of article = CSA of cylindrical part + 2 × CSA of hemispherical part
Exercise 13.2
Question 1:
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
Answer:
Given that,
Height (h) of conical part = Radius(r) of conical part = 1 cm Radius(r) of hemispherical part = Radius of conical part (r) = 1 cm
Volume of solid = Volume of conical part + Volume of hemispherical part
=13r2h+23r3
=13121+2313
=3+2π3=π
Question 2:
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminum sheet. The diameter of the model is 3 cm and its length is 12 cm. if each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Answer:
From the figure, it can be observed that Height (h1) of each conical part = 2 cm
Height (h2) of cylindrical part = 12 − 2 × Height of conical part
= 12 − 2 ×2 = 8 cm
Radius (r) of cylindrical part = Radius of conical part
Volume of air present in the model = Volume of cylinder + 2 × Volume of cones
*Question 3:
A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see the given figure).
Answer:
It can be observed that
Radius (r) of cylindrical part = Radius (r) of hemispherical part
Length of each hemispherical part = Radius of hemispherical part = 1.4 cm Length (h) of cylindrical part = 5 − 2 × Length of hemispherical part
= 5 − 2 × 1.4 = 2.2 cm
Volume of one gulab jamun = Vol. of cylindrical part + 2 × Vol. of hemispherical part
Volume of 45 gulab jamuns
Volume of sugar syrup = 30% of volume
Question 4:
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboids are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in
the entire stand (see the following figure).
Answer:
Depth (h) of each conical depression = 1.4 cm
Radius (r) of each conical depression = 0.5 cm
Volume of wood = Volume of cuboid − 4 × Volume of cones
*Question 5:
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Answer:
Height (h) of conical vessel = 8 cm Radius (r1) of conical vessel = 5 cm
Radius (r2) of lead shots = 0.5 cm
Let n number of lead shots were dropped in the vessel.
Volume of water spilled = Volume of dropped lead shots
Hence, the number of lead shots dropped in the vessel is 100.
*Question 6:
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. [Use π = 3.14]
Answer:
From the figure, it can be observed that Height (h1) of larger cylinder = 220 cm
Radius (r1) of larger cylinder
Height (h2) of smaller cylinder = 60 cm Radius (r2) of smaller cylinder = 8 cm
Mass of iron= 8 g
Mass of 111532.8iron = 111532.8 × 8 = 892262.4 g = 892.262 kg
Question 7:
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Answer:
Radius (r) of hemispherical part = Radius (r) of conical part = 60 cm Height (h2) of conical part of solid = 120 cm
Height (h1) of cylinder = 180 cm
Radius (r) of cylinder = 60 cm
Volume of water left = Volume of cylinder − Volume of solid
Question 8:
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Answer:
Height (h) of cylindrical part = 8 cm
Radius (r2) of cylindrical part
Radius (r1) spherical part
Volume of vessel = Volume of sphere + Volume of cylinder
Hence, she is wrong.
Exercise 13.3
Question 1:
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Answer:
Radius (r1) of hemisphere = 4.2 cm Radius (r2) of cylinder = 6 cm
Let the height of the cylinder be h.
The object formed by recasting the hemisphere will be the same in volume.
Volume of sphere = Volume of cylinder
Hence, the height of the cylinder so formed will be 2.74 cm.
Question 2:
Metallic spheres of radii 6 cm, 8 cm, and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Answer:
Radius (r1) of 1st sphere = 6 cm Radius (r2) of 2nd sphere = 8 cm Radius (r3) of 3rd sphere = 10 cm
Let the radius of the resulting sphere be r.
The object formed by recasting these spheres will be same in volume as the sum of the volumes of these spheres.
Volume of 3 spheres = Volume of resulting sphere
Therefore, the radius of the sphere so formed will be 12 cm.
Question 3:
A 20 m deep well with a diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Answer:
The shape of the well will be cylindrical. Depth (h) of well = 20 m
Radius (r) of circular end of well
Area of platform = Length × Breadth = 22 × 14 m2 Let height of the platform = H
Volume of soil dug from the well will be equal to the volume of soil scattered on the platform.
Volume of soil from well = Volume of soil used to make such platform
Therefore, the height of such platform will be 2.5 m.
*Question 4:
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Answer:
The shape of the well will be cylindrical. Depth (h1) of well = 14 m
Radius (r1) of the circular end of well =
Width of embankment = 4 m
From the figure, it can be observed that our embankment
will be in a cylindrical shape having outer radius (r2) as
and inner radius (r1) as
Let the height of embankment be h2.
Volume of soil dug from well = Volume of earth used to form embankment
Therefore, the height of the embankment will be 1.125 m.
Question 5:
A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Answer:
Height (h1) of cylindrical container = 15 cm
Radius (r1) of circular end of container =
Radius (r2) of circular end of ice-cream cone
Height (h2) of conical part of ice-cream cone = 12 cm
Let n ice-cream cones be filled with ice-cream of the container.
Volume of ice-cream in cylinder = n × (Volume of 1 ice-cream cone + Volume of hemispherical shape on the top)
Therefore, 10 ice-cream cones can be filled with the ice-cream in the container.
Question 6:
How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted
to form a cuboid of dimensions?
Answer:
Coins are cylindrical in shape.
Height (h1) of cylindrical coins = 2 mm = 0.2 cm
Radius (r) of circular end of coins
Let n coins be melted to form the required cuboids. Volume of n coins = Volume of cuboids
Therefore, the number of coins melted to form such a cuboid is 400.
Question 7:
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm. Find the radius and slant height of the heap.
Answer:
Height (h1) of cylindrical bucket = 32 cm Radius (r1) of circular end of bucket = 18 cm Height (h2) of conical heap = 24 cm
Let the radius of the circular end of conical heap be r2.
The volume of sand in the cylindrical bucket will be equal to the volume of sand in the conical heap.
Volume of sand in the cylindrical bucket = Volume of sand in conical heap
Slant height
Therefore, the radius and slant height of the conical heap are 36 cm and respectively.
*Question 8:
Water in canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. how much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Answer:
Consider an area of cross-section of canal as ABCD. Area of cross-section = 6 × 1.5 = 9 m2
Speed of water
Volume of water that flows in 1 minute from canal
Volume of water that flows in 30 minutes from canal = 30 × 1500 = 45000 m3
Let the irrigated area be A. Volume of water irrigating the required area will be equal to the volume of water that flowed in 30 minutes from the canal.
Vol. of water flowing in 30 minutes from canal = Vol. of water irrigating the reqd. area
A = 562500 m2
Therefore, area irrigated in 30 minutes is 562500 m2.
Question 9:
A farmer connects a pipe of internal diameter 20 cm form a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Answer:
Consider an area of cross-section of pipe as shown in the figure.
Radius (r1) of circular end of pipe
Area of cross-section
Speed of water
Volume of water that flows in 1 minute from pipe
Volume of water that flows in t minutes from pipe = t × 0.5π m3
Radius (r2) of circular end of cylindrical tank
Depth (h2) of cylindrical tank = 2 m
Let the tank be filled completely in t minutes.
Volume of water filled in tank in t minutes is equal to the volume of water flowed in t minutes from the pipe.
Volume of water that flows in t minutes from pipe = Volume of water in tank t × 0.5π = π ×(r2)2 ×h2
t × 0.5 = 52 ×2
t = 100
Therefore, the cylindrical tank will be filled in 100 minutes.
Exercise 13.4
Question 1:
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Answer:
Radius (r1) of upper base of glass
Radius (r2) of lower base of glass
Capacity of glass = Volume of frustum of cone
Therefore, the capacity of the glass is.
Question 2:
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. find the curved surface area of the frustum.
Answer:
Perimeter of upper circular end of frustum = 2πr1=18
Perimeter of lower end of frustum = 2πr2= 6
Slant height (l) of frustum = 4 cm CSA of frustum = π(r1 + r2) l
Therefore, the curved surface area of the frustum is 48 cm2.
*Question 3:
A fez, the cap used by the Turks, is shaped like the frustum of a cone (see the figure given below). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material use for making it.
Answer:
Radius (r2) at upper circular end = 4 cm
Radius (r1) at lower circular end = 10 cm Slant height (l) of frustum = 15 cm
Area of material used for making the fez = CSA of frustum + Area of upper circular end
= π (10 + 4) 15 + π (4)2
= π (14) 15 + 16 π
Therefore, the area of material used for making it is.
*Question 4:
A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs.20 per litre. Also, find the cost of metal sheet used to make the container, if it costs Rs.8 per 100 cm2. [Take π = 3.14]
Answer:
Radius (r1) of upper end of container = 20 cm
Radius (r2) of lower end of container = 8 cm Height (h) of container = 16 cm
Slant height (l) of frustum
Capacity of container = Volume of frustum
Cost of 1 litre milk = Rs 20
Cost of 10.45 litre milk = 10.45 × 20
= Rs 209
Area of metal sheet used to make the container
= π (20 + 8) 20 + π (8)2
= 560 π + 64 π = 624 π cm2
Cost of 100 cm2 metal sheet = Rs 8
Therefore, the cost of the milk which can completely fill the container is
Rs 209 and the cost of metal sheet used to make the container is Rs 156.75.
Question 5:
A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained is drawn into a wire of diameter cm, find the length of the wire.
Answer:
In ΔAEG,
In ΔABD,
Radius (r1) of upper end of frustum
Radius (r2) of lower end of container
Height (h) of container = 10 cm
Volume of frustum
Radius (r) of wire
Let the length of wire be l.
Volume of wire = Area of cross-section × Length
= (πr2) (l)
Volume of frustum = Volume of wire
220009=π1322×l
220009×32×32=l
220009×32×32×722=l
l=7964.44m
Exercise 13.5
*Question 1:
A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.
Answer:
It can be observed that 1 round of wire will cover 3 mm height of cylinder.
Length of wire required in 1 round = Circumference of base of cylinder
= 2πr = 2π × 5 = 10π
Length of wire in 40 rounds = 40 × 10π
= 1257.14 cm = 12.57 m
Radius of wire (Equation alignment needed)
Volume of wire = Area of cross-section of wire × Length of wire
= π(0.15)2 × 1257.14
= 88.898 cm3
Mass = Volume × Density
= 88.898 × 8.88= 789.41 gm
Question 2:
A right triangle whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate.)
Answer:
The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.
Hypotenuse
=5 cm
Area of ΔABC
Volume of double cone = Volume of cone 1 + Volume of cone 2
= 30.14 cm3
Surface area of double cone = Surface area of cone 1 + Surface area of cone 2
= πrl1 + πrl2
= 52.75 cm2
Question 3:
A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?
Answer:
Volume of cistern = 150 × 120 × 110
= 1980000 cm3
Volume to be filled in cistern = 1980000 – 129600
= 1850400 cm3
Let n numbers of porous bricks were placed in the cistern. Volume of n bricks = n × 22.5 × 7.5 × 6.5
= 1096.875n
As each brick absorbs one-seventeenth of its volume, therefore, volume absorbed by these bricks
n = 1792.41
Therefore, 1792 bricks were placed in the cistern.
Question 4:
In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.
Answer:
Area of the valley = 7280 km2
If there was a rainfall of 10 cm in the valley then amount of rainfall in the valley = Area of the valley × 10 cm
Amount of rainfall in the valley = 7280 km2 × 10 cm
Length of each river,
Breadth of each river, b = 75 m Depth of each river, h = 3 m Volume of each river = l × b × h
= 1072000 × 75 × 3 m3
= 2.412 × 108 m3
Volume of three such rivers = 3 × Volume of each river
= 3 × 2.412 × 108 m3
= 7.236 × 108 m3
Thus, the total rainfall is approximately same as the volume of the three rivers.
Question 5:
An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see the given figure).
Answer:
Radius (r1) of upper circular end of frustum part
Radius (r2) of lower circular end of frustum part = Radius of circular end of cylindrical part
Height (h1) of frustum part = 22 − 10 = 12 cm Height (h2) of cylindrical part = 10 cm
Slant height (l) of frustum part
Area of tin sheet required = CSA of frustum part + CSA of cylindrical part
Question 6:
Derive the formula for the curved surface area and total surface area of the frustum of cone.
Answer:
Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base. Let r1 and r2 be the radii of the ends of the frustum of the cone and h be the height of the frustum of the cone.
In ΔABG and ΔADF, DF||BG
∴ ΔABG ∼ ΔADF
CSA of frustum DECB = CSA of cone ABC − CSA cone ADE
CSA of frustum
Question 7:
Derive the formula for the volume of the frustum of a cone.
Answer:
Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base.
Let r1 and r2 be the radii of the ends of the frustum of the cone and h be the height of the frustum of the cone.
In ΔABG and ΔADF, DF||BG
∴ ΔABG ∼ ΔADF
Volume of frustum of cone = Volume of cone ABC − Volume of cone ADE
Chapter 13, Surface Areas and Volumes, of class 10 Maths, is one of the most important chapters. The weightage of this chapter in the final exam is around 12 to 13 marks. On average, there are 4 questions asked from this chapter, based on surface areas and volumes. The distribution of marks with respect to questions are 3+3+3+4, where marks could vary depending upon the question. Topics covered in Chapter 13, Surface Areas and Volumes are;
- The surface area of a combination of Solids
- The volume of a combination of solids
- Conversion of solid from one shape to another
- Frustum of a cone
List of Exercises in class 10 Maths Chapter 13:
Exercise 13.1 Solutions 9 Question ( 7 long, 2 short)
Exercise 13.2 Solutions 8 Question ( 7 long, 1 short)
Exercise 13.3 Solutions 9 Question ( 9 long)
Exercise 13.4 Solutions 5 Question ( 5 long)
Exercise 13.5 Solutions 7 Question ( 7 long)
NCERT Solutions for Class 10 Maths Chapter 13- Surface Areas and Volumes. These NCERT solutions of chapter Surface Areas and Volumes are made available for the students who want to excel in the board exam. Let us discuss here, the importance of chapter in your academic and real life. You will face many real-life scenarios where the fundamentals of surface areas and volumes of real objects have to be configured. Some of the examples are rectangular boxes, gas cylinders, footballs, etc. These have 3D shapes which have both surface area and volume. And to calculate these quantities, we have to learn the formulas based on the dimensions of the object. Therefore, this solution of class 10, will help you learn about the formulas of surface areas and volumes of such 3D shapes such as Sphere, Cylinder, Cone, Cuboid, and combination of any two solids. With the help of these solved questions, you can not only solve the questions present in your academics but also real-life ones.
Key Features of NCERT Solutions for Class 10 Maths Chapter 13 – Surface Areas and Volumes
- The solutions are the great source for the students who want to get the unsolved questions present in each exercise of 10th Standard Maths book for the 13th chapter.
- These works as reference material for the student.
- It serves a great help for the revision of chapter 13, of all types of questions, asked in the exam.
- Student’s can attain really good marks from surface area and volume chapter, with the help of solutions.
- It is completely based on the syllabus prescribed by Central Board of Secondary Education(CBSE) guidelines.
Conclusion
The NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes are an essential resource for students looking to develop a strong understanding of calculating surface areas and volumes of different geometric shapes. The chapter covers various topics, such as the formulae for calculating the surface area and volume of objects such as cubes, cylinders, cones, and spheres.
Experienced teachers and subject matter experts have carefully designed these solutions to provide step-by-step explanations for all the NCERT textbook questions. The solutions also offer helpful tips and tricks to help students solve problems quickly and accurately.
Using these solutions, students can strengthen their understanding of the subject, improve their problem-solving skills, and prepare effectively for the Class 10 board exams. These solutions are an excellent resource for students who wish to excel in mathematics and build a strong foundation for higher studies.
Overall, the NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes are indispensable for students looking to score well in mathematics and enhance their analytical and mathematical abilities.
How can we score full marks in Class test of NCERT Solutions for Class 10 Maths Chapter 13?
By using the NCERT Solutions for Class 10 Maths Chapter 13 provided in SWC’s website makes you reach full marks in class tests as well as in board exams. These solutions are highly imperative for easy and quick revision during the class tests and exams. Meanwhile, this serves as the best study material for learners.
How NCERT Solutions for Class 10 Maths Chapter 13 helpful in board exams?
NCERT Solutions for Class 10 Maths Chapter 13 implements answers with detailed descriptions as per term limit specified by the SWC for self-evaluation. Practising these questions will ensure that students have a good preparation for all sorts of questions that can be devised in the finals.
What are the main concepts covered in NCERT Solutions for Class 10 Maths Chapter 13?
The main topics of NCERT Solutions for Class 10 Maths Chapter 13 are the surface area of a combination of solids, the volume of a combination of solids, conversion of solid from one shape to another and frustum of a cone.
