NCERT Solutions for Class 10 Maths Ch 13 Surface Areas and Volumes
In this page, you will find Class 10 Maths NCERT Solutions Chapter 13 Surface Areas and Volumes which are very helpful in knowing the important topics in the chapter and completing your homework in no time. These NCERT Solutions for Class 10 Maths will give you step by step solutions of every questions. Our experts have updated these solutions according to the latest pattern of CBSE.
Study Materials for Class 10 Maths Chapter 13 Surface Areas and Volumes 


Exercise 13.1
Unless stated otherwise, take Ï€ = 22/7.
1. 2 cubes each of volume 64 cm^{3} are joined end to end. Find the surface area of the resulting cuboid.
Answer
⇒ a^{3} = 64 cm^{3} ^{}
⇒ a = 4 cm
Side of the cube = 4 cm
Length of the resulting cuboid = 4 cm
Breadth of the resulting cuboid = 4 cm
Height of the resulting cuboid = 8 cm
∴ Surface area of the cuboid = 2(lb + bh + lh)
= 2(8×4 + 4×4 + 4×8) cm^{2}^{}
= 2(32 + 16 + 32) cm^{2}^{}
= (2 × 80) cm^{2} = 160 cm^{2}^{}
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
Radius of the hemisphere(r) = 7 cm
Height of the cylinder(h) = 13  7 = 6 cm
Also, radius of the hollow hemisphere = 7 cm
Inner surface area of the vessel = CSA of the cylindrical part + CSA of hemispherical part
= (2Ï€rh+2Ï€r^{2}) cm^{2}^{ }= 2Ï€r(h+r) cm^{2}
= 2 × 22/7 × 7 (6+7) cm^{2}^{ }= 572 cm^{2}
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
Total height of the toy = 15.5 cm
Height of the cone(h) = 15.5  3.5 = 12 cm
= 275/2 cm^{2}
Curved Surface Area of hemisphere = 2Ï€r^{2}
= 2 × 22/7 × (7/2)^{2}
= 77 cm^{2}
Total surface area of the toy = CSA of cone + CSA of hemisphere
= (275/2 + 77) cm^{2}
= (275+154)/2 cm^{2 }
= 429/2 cm^{2} = 214.5^{ }cm^{2}
The total surface area of the toy is 214.5^{ }cm^{2}
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
∴ Radius of the hemisphere = 7/2 cm
Total surface area of solid = Surface area of cubical block + CSA of hemisphere  Area of base of hemisphere
TSA of solid = 6×(side)^{2 }+ 2Ï€r^{2 }^{ }Ï€r^{2}
= 6×(side)^{2 }+ Ï€r^{2}
= 6×(7)^{2 }+ (22/7 × 7/2 × 7/2)
= 6×(side)^{2 }+ Ï€r^{2}
= 6×(7)^{2 }+ (22/7 × 7/2 × 7/2)
= (6×49) + (77/2)
= 294 + 38.5 = 332.5 cm^{2}
= 294 + 38.5 = 332.5 cm^{2}
The surface area of the solid is 332.5 cm^{2}
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 = l/2
Total surface area of solid = Surface area of cube + CSA of hemisphere  Area of base of hemisphere
TSA of remaining solid = 6 (edge)^{2 }+ 2Ï€r^{2 } Ï€r^{2}
= 6l^{2 }+ Ï€r^{2}
= 6l^{2 }+ Ï€(l/2)^{2}
= 6l^{2 }+ Ï€l^{2}/4^{ }
= l^{2}/4^{ }(24 + Ï€) sq units
6. 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.
Surface area of the cylinder = 2Ï€rh
= 2 × 22/7 × 2.5 x 9
= 22/7 × 45
990/7 mm^{2}
∴ Required surface area of medicine capsule
= 2 × Surface area of hemisphere + Surface area of cylinder
= (2 × 275/7) × 990/7
= 550/7 + 990/7
= 1540/7 = 220 mm^{2}
Page No. 245
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 m^{2}. (Note that the base of the tent will not be covered with canvas.)
Answer
Tent is combination of cylinder and cone.
Diameter = 4 m
Slant height of the cone (l) = 2.8 m
Radius of the cone (r) = Radius of cylinder = 4/2 = 2 m
Height of the cylinder (h) = 2.1 m
∴ Required surface area of tent = Surface area of cone+Surface area of cylinder
= Ï€rl + 2Ï€rh
= Ï€r(l+2h)
= 22/7 × 2 (2.8 + 2×2.1)
= 44/7 (2.8 + 4.2)
= 44/7 × 7 = 44 m^{2}
Cost of the canvas of the tent at the rate of ₹500 per m^{2}
= Surface area × cost per m^{2}
= 44 × 500 = ₹22000
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 cm^{2}.
Answer
Go Back To NCERT Solutions for Class 10 Maths
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
Radius of hemisphere = l/2
Total surface area of solid = Surface area of cube + CSA of hemisphere  Area of base of hemisphere
TSA of remaining solid = 6 (edge)^{2 }+ 2Ï€r^{2 } Ï€r^{2}
= 6l^{2 }+ Ï€r^{2}
= 6l^{2 }+ Ï€(l/2)^{2}
= 6l^{2 }+ Ï€l^{2}/4^{ }
= l^{2}/4^{ }(24 + Ï€) sq units
6. 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.
Answer
Two hemisphere and one cylinder are given in the figure.
Diameter of the capsule = 5 mm
∴ Radius = 5/2 = 2.5 mm
Length of the capsule = 14 mm
∴ Length of the cylinder = 14  (2.5 + 2.5) = 9mm
Surface area of a hemisphere = 2Ï€r^{2} = 2 × 22/7 × 2.5 × 2.5
= 275/7 mm^{2}
Surface area of the cylinder = 2Ï€rh
= 2 × 22/7 × 2.5 x 9
= 22/7 × 45
990/7 mm^{2}
∴ Required surface area of medicine capsule
= 2 × Surface area of hemisphere + Surface area of cylinder
= (2 × 275/7) × 990/7
= 550/7 + 990/7
= 1540/7 = 220 mm^{2}
Page No. 245
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 m^{2}. (Note that the base of the tent will not be covered with canvas.)
Answer
Tent is combination of cylinder and cone.
Slant height of the cone (l) = 2.8 m
Radius of the cone (r) = Radius of cylinder = 4/2 = 2 m
Height of the cylinder (h) = 2.1 m
∴ Required surface area of tent = Surface area of cone+Surface area of cylinder
= Ï€rl + 2Ï€rh
= Ï€r(l+2h)
= 22/7 × 2 (2.8 + 2×2.1)
= 44/7 (2.8 + 4.2)
= 44/7 × 7 = 44 m^{2}
Cost of the canvas of the tent at the rate of ₹500 per m^{2}
= Surface area × cost per m^{2}
= 44 × 500 = ₹22000
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 cm^{2}.
Answer
Diameter of cylinder = diameter of conical cavity = 1.4 cm
∴ Radius of cylinder = Radius of conical cavity = 1.4/2 = 0.7
Height of cylinder = Height of conical cavity = 2.4 cm
TSA of remaining solid = Surface area of conical cavity+TSA of cylinder
= Ï€rl + (2Ï€rh + Ï€r^{2})
= Ï€r (l + 2h + r)
= 22/7 × 0.7 (2.5 + 4.8 + 0.7)
= 2.2 × 8 = 17.6 cm^{2}
9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in 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,
Height of the cylinder, h = 10cm and radius of base of cylinder = Radius of hemisphere (r) = 3.5 cm
Now, required total surface area of the article = 2 × Surface area of hemisphere + Lateral surface area of cylinder
Exercise 13.2
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 Ï€.
Answer
Given, solid is a combination of a cone and a hemisphere.
Also, we have radius of the cone (r) = Radius of the hemisphere = 1cm and height of the cone (h) = 1cm
∴ Required volume of the solid = Volume of the cone + Volume of the hemisphere
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 aluminium 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 mode. (Assume the outer and inner dimensions of the model to be nearly the same.)
Answer
Given, model is a combination of a cylinder and two cones. Also, we have, diameter of the model,BC = ED = 3 cm.
3. A gulab jamun, contains sugar syrup upto 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 figure).
Answer
Let r be the radius of the hemisphere and cylinder both. h1 be the height of the hemisphere which is equal to its radius and h2 be the height of the cylinder.
Given, length = 5 cm, diameter = 2.8 cm
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 cuboid 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 figure).
Answer
Given, length of cuboid (l) = 15 cm
Breadth of cuboid (b) = 10cm
and height of cuboid (h) = 3.5 cm
∴ Volume of cuboid = l×b×h = 15 ×10 × 3.5 = 525^{3}
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 upto the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel,onefourth of the water flows out. Find the number of lead shots dropped in the vessel.
Answer
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 cm^{3} of iron has approximately 8 g mass. (Use Ï€ = 3.14)
Answer
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
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 cm^{3}. Check whether she is correct, taking the above as the inside measurements and Ï€ = 3.14.
Answer
She is not correct.
Exercise 13.3
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
Given, the radius of the sphere (r) = 4.2 cm
Radius of the cylinder (r1) = 6 cm
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
Let r1, r2 and r3 be the radius of given three spheres and R be the radius of a single solid sphere.
Given, r1 = 6 cm , r2 = 8 cm and r3 =10 cm
3. A 20 m deep well with 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
Given, the height of deep well which form a cylinder (h1) = 20 m
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 4m to form an embankment. Find the height of the embankment.
Answer
Given, the height of deep well which form a cylinder (h1) = 14 m
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
Let the height and radius of ice cream container (cylinder) be h1 and r1.
6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 10 3 5 . . cm cm cm × ?
Answer
We know that, every coin has a shape of cylinder. Let radius and height of the coin are r1 and h1 respectively.
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
Let the radius and slant height of the heap of sand are r and l.
Given, the height of the heap of sand h = 24 cm.
8. Water in a canal, 6 m wide and 1.5 m deep is flowing with a speed of 10 kmh^{−1}. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Answer
Given, speed of flow of water (l) = 10 kmh1 = 10 × 1000 mh^{−1}
Area of canal = 6 × 1.5 = 9m^{2}
9. A farmer connects a pipe of internal diameter 20 cm from 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 kmh^{−1}, in how much time will the tank be filled?
Answer
Given, speed of flow of water = 3 kmh^{−1} = 3×1000 mh^{−1}
∴ Length of water in 1 h = 3000 m
Exercise 13.4
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
Let the height of the frustum of a cone be h.
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
Let the slant height of the frustum be l and radius of the both ends of the frustum be r1 and r2.
3. A fez, the cap used by the turks, is shaped like the frustum of a cone (see figure). 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 used for making it.
Answer
Let the slant height of fez be l and the radius of upper end which is closed be r1 and the other end which is open be r2.
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 ` 20 per litre. Also find the cost of metal sheet used to make the container. If it costs ₹8 per 100 cm^{2}. (Take Ï€ = 3.14)
Answer
Let h be the height of the container, which is in the form of a frustum of a cone whose lower end is closed and upper end is opened. Also, let the radius of its lower end be r1 and upper end be r2.
5. A metallic right circular cone 20 cm high and whose vertical angle is 60Âº is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.
Answer
Let r1 and r2 be the radii of the frustum of upper and lower ends cut by a plane. Given, height of the cone = 20 cm.
∴ Height of the frustum = 10 cm
Exercise 13.5 (Optional)
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 cm^{3}.
Answer
Since, the diameter of the wire is 3 mm.
When a wire is one round wound about a cylinder, it covers a 3 mm of length of the cylinder.
Given, length of the cylinder = 12 cm = 120 mm
∴ Number of rounds to cover 120 mm = 120/3 = 40
Given, diameter of a cylinder is d = 10 cm
∴ Radius, r = 10/2 = 5 cm
∴ Length of wire required to complete one round = 2Ï€r = 2Ï€(5) = 10Ï€ cm
∴ Length of the wire in covering the whole surface
= Length of the wire in completing 40 rounds
= 10Ï€ × 40 = 400Ï€ cm
= 400 × 3.14 = 1256 cm
Now, radius of copper wire = 3/2 mm = 3/20 cm
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 Ï€ as found appropriate).
Answer
Here, ABC is a right angled triangle at A and BC is the hypotenuse.
3. A cistern, internally measuring 150 cm × 120 cm× 110 cm, has 129600 cm^{3} of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs oneseventeenth 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
Given, internally dimensions of cistern = 150 cm×120 cm×110 cm
∴ Volume of cistern = 150×120×110
= 1980000 cm^{3}
Volume of water = 129600 cm^{3}
∴ Volume of cistern to be filled = 1980000  129600 = 1850400 cm^{3}
Let required number of bricks = n
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 97280 km^{2}, 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
Given, area of the valley = 97280 km^{2}
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.
Answer
Given, oil funnel is a combination of a cylinder and a frustum of a cone.
Also, given height of cylindrical portion h = 10 cm
6. Derive the formula for the curved surface area and total surface area of the frustum of a cone. Using the symbols as explained.
Answer
Leth be the height, l be the slant height and r_{1} and r_{2} be the radii of the bases (r_{1}>r_{2}) of the frustum of a cone. We complete the conical part OCD.
The frustum of the right circular cone can be viewed as the difference of the two right circular cones OAB and OCD. Let slant height of the cone OAB be l_{1} and its height be h_{1} i.e.,OB = OA = l_{1} and OP = h_{1}
The frustum of the right circular cone can be viewed as the difference of the two right circular cones OAB and OCD. Let slant height of the cone OAB be l_{1} and its height be h_{1} i.e., OB = OA = l_{1} and OP = h_{1}
Then in ∆ACE,
7. Derive the formula for the volume of the frustum of a cone given to you in the section 13.5 using the symbols as explained.
Answer
Leth be the height, l be the slant height and r_{1} and r_{2} be the radii of the bases (r_{1}>r_{2}) of the frustum of a cone. We complete the conical part OCD.
The frustum of the right circular cone can be viewed as the difference of the two right circular cones OAB and OCD. Let the height of the cone OAB be h_{1} and its slant height be l_{1}.
i.e., OP = h_{1} and OA = OB = l_{1}
Then, height of the cone OCD = h_{1} −1
∆OQD ~ ∆OPB (AA similarity criterion)
NCERT Solutions for Class 10 Maths Ch 13 Surface Areas and Volumes
Chapter 13 Class 10 Maths NCERT Solutions plays very important role during the preparation of board exams as a lot of questions from this topic can be asked. There are total 5 topics in this chapter which will guide students in a better way.
• Introduction: We came across various types of solids made up of combinations of two or more of the basic solids such as water tank in our daily lives. We are going to learn about how to find surface areas and volumes of these types of objects.
• Surface Area of a Combination of Solids: We can breakdown complex figure into simpler one and can find their areas. The total surface area of the new solid is the sum of the curved surface areas of each of the individual parts. TSA of new solid = CSA of one hemisphere + CSA of cylinder + CSA of other hemisphere where TSA, CSA stand for ‘Total Surface Area’ and ‘Curved Surface Area’ respectively.
• Volume of a Combination of Solids: We will discuss how to calculate volumes of solids. The volume of the solid formed by joining two basic solids will be the sum of the volumes of the constituents.
• Conversion of Solid from One Shape to Another: We will learn to find the radius and volume when one shape got converted from one to another, or when a liquid which originally filled one container of a particular shape is poured into another container of a different shape or size.
• Frustum of a Cone: When we cut through a cone with a plane parallel to its base and remove the cone that is formed on one side of that plane, the part that is now left over on the other side of the plane is called a frustum of the cone. In this topics, we will dealing with the questions based on finding the volume of a frustum of a cone.
If you want to find exercisewise NCERT Solutions of Chapter 13 Surface Areas and Volumes the you can find them below.
NCERT Solutions for Class 10 Maths Chapters:
FAQ on Chapter 13 Surface Areas and Volumes
How many exercises in Chapter 13 Surface Areas and Volumes
There are total 5 exercise in the Chapter 13 Surface Areas and Volumes which will make students while solving any questions. These are also useful for competitive exams and higher grades.
What do you mean by Cube?
Cube is a special type of cuboid in which length = breadth = height which is also called an edge of a cube.
What is lateral surface area of cuboid?
Lateral surface area of cuboid = 2 (1 + b)h
If two identical solid cubes of side ‘x’ are joined end to end, then the total surface area of the resulting cuboid is 12x2. Is it true?
∵ The total surface area of a cube of side x is 6x^{2}
When they are joined end to end, the length bcomes 2x
∴Total surface area
= 2[lh + bh + hl]
= 2 [(2x · x) + (x · x) + (2x · x)]
= 2 [2x^{2} + x^{2} + 2x^{2}]
= 2 [5x^{2}] = 10x^{2} ≠ 12x^{2}
∴ False