NCERT Solutions for Class 10 Maths Chapter 13 : Surface Areas and Volumes

Q: Question 1. How is the surface area of a solid which is a combination of solid shapes calculated?

Answer. In order to find out the surface area of a solid which is a combination of solid shapes, firstly, the surface area of individual solid shapes needs to be calculated separately.

Q: Question 2. How is the volume of a combination of solids calculated?

Answer. The volume of a combination of solids is calculated by adding the volumes of individual solids. For example, a figure is made up of two solids, i.e. a cuboid and a pyramid. So the total volume of the solid is obtained by adding up the volume of the cuboid and the volume of the pyramid.

Q: Question 3. Does the volume of a solid change when we convert it into another solid by the method of melting or remoulding?

Answer. No, the volume of a solid does not change when we convert it into another solid by the method of melting or remoulding, despite the change in its shape.

Q: Question 4. What do you understand by the Frustum of a cone?

Answer. ‘Frustum’ is a Latin word that means ‘piece cut off’. The plural of the frustum is ‘frusta’. When we slice or cut a given cone through it 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 leftover on the other side of the plane is called a frustum of the cone. The frustum of the cone has two circular ends with different radii.

Question 1

Two cubes each of volume 64

c m^{3}

are joined end to end. Find the surface area of the resulting cuboid.

Medium

Solution

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Answer

The volume of cube

64 c m^{3}

Side of cube

364 = 4 c m

Length of resulting cuboid

4 + 4 = 8

Surface area

= 2 (l b + h l + b h)

= 2 (4 (4) + 4 (8) + 8 (4)) = 2 (16 + 32 + 32) = 2 (80) = 160 c m^{2}

Answer

The volume of cube

64 c m^{3}

Side of cube

364 = 4 c m

Length of resulting cuboid

4 + 4 = 8

Surface area

= 2 (l b + h l + b h)

= 2 (4 (4) + 4 (8) + 8 (4)) = 2 (16 + 32 + 32) = 2 (80) = 160 c m^{2}

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.

A

557 c m^{2}

B

517 c m^{2}

C

537 c m^{2}

D

572 c m^{2}

Medium

Solution

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Answer

Radius=

7

cm

Height of the cylindrical portion

13 - 7 = 6

cm.

Area of a Curved surface of cylindrical portion is,

=

2 π r h

=

2 \times \frac{2 2}{7} \times 7 \times 6

=

264 c m^{2}

Area of a curved of hemispherical portion is

=

2 π r^{2}

=

2 \times \frac{2 2}{7} \times 7 \times 7

=

308 c m^{2}

∴

total surface area is

= 308 + 264 = 572 c m^{2}

.

Answer

Radius=

7

cm

Height of the cylindrical portion

13 - 7 = 6

cm.

Area of a Curved surface of cylindrical portion is,

=

2 π r h

=

2 \times \frac{2 2}{7} \times 7 \times 6

=

264 c m^{2}

Area of a curved of hemispherical portion is

=

2 π r^{2}

=

2 \times \frac{2 2}{7} \times 7 \times 7

=

308 c m^{2}

∴

total surface area is

= 308 + 264 = 572 c m^{2}

.

Question 3

A toy in the form of 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 toy.

Medium

Solution

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Answer

Total surface area of toy

=

Curved surface area of cone

+

surface area of hemisphere

Curved surface area of cone

=

π r l

Where

r = 3.5 c m

, Height

= 15.5 - 3.5 = 12 c m

And hence,

l = 12.5 c m

(by using formula

l^{2} = h^{2} + b^{2}

)

Therefore C.S.A. of cone

= π \times 3.5 \times 12.5

= 137.5 c m^{2}

Surface area of hemisphere

= 2 π r^{2}

= 2 \times π \times (3.5)^{2}

= 77 c m^{2}

Hence T.S.A of toy

= 77 + 137.5 = 214.5 c m^{2}

Answer

Total surface area of toy

=

Curved surface area of cone

+

surface area of hemisphere

Curved surface area of cone

=

π r l

Where

r = 3.5 c m

, Height

= 15.5 - 3.5 = 12 c m

And hence,

l = 12.5 c m

(by using formula

l^{2} = h^{2} + b^{2}

)

Therefore C.S.A. of cone

= π \times 3.5 \times 12.5

= 137.5 c m^{2}

Surface area of hemisphere

= 2 π r^{2}

= 2 \times π \times (3.5)^{2}

= 77 c m^{2}

Hence T.S.A of toy

= 77 + 137.5 = 214.5 c m^{2}

Question 4

A hemispherical depression is cut out from one face of the 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.

Medium

Solution

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Answer

Consider the diagram shown below.

It is given that a hemisphere of radius

\frac{l}{2}

is cut out from the top face of the cuboidal wooden block.

Therefore, surface area of the remaining solid

=

surface area of the cuboidal box whose each edge is of length

l -

Area of the top of the hemispherical part

+

curved surface area of the hemispherical part

= 6 l^{2} - π r^{2} + 2 π r^{2}

= 6 l^{2} - π (\frac{l}{2})^{2} + 2 π (\frac{l}{2})^{2}

= 6 l^{2} - \frac{π l ^{2}}{4} + \frac{π l ^{2}}{2}

= \frac{l ^{2}}{4} (24 + π) s q . u n i t s

Answer

Consider the diagram shown below.

It is given that a hemisphere of radius

\frac{l}{2}

is cut out from the top face of the cuboidal wooden block.

Therefore, surface area of the remaining solid

=

surface area of the cuboidal box whose each edge is of length

l -

Area of the top of the hemispherical part

+

curved surface area of the hemispherical part

= 6 l^{2} - π r^{2} + 2 π r^{2}

= 6 l^{2} - π (\frac{l}{2})^{2} + 2 π (\frac{l}{2})^{2}

= 6 l^{2} - \frac{π l ^{2}}{4} + \frac{π l ^{2}}{2}

= \frac{l ^{2}}{4} (24 + π) s q . u n i t s

Question 5

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 m m

and the diameter of the capsule is

5 m m

. Find its surface area. (Use

π = \frac{2 2}{7}

).

Medium

Solution

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Answer

Given:
Length of Capsule (l) =

14

mm

Diameter of Capsule = Diameter of Cylinder

= 5

mm

Radius =

\frac{D i a m e t e r}{2}

Therefore, Radius of each Hemisphere = Radius of Cylinder = r =

\frac{5}{2}

= 2.5 mm

Length of Cylinder = AB = Total length of Capsule - Radius of left Hemisphere - Radius of Right Hemisphere

= 14 - 2.5 - 2.5 = 9

mm

Surface Area of Capsule = Curved Surface Area of Cylinder + Surface Area of Left Hemisphere + Surface Area of Right Hemisphere

= 2 π r l + 2 π r^{2} + 2 π r^{2}

= 2 π r l + 4 π r^{2}

= 2 \times \frac{2 2}{7} \times 2.5 \times 9 + 4 \times \frac{2 2}{7} \times 2 . 5^{2}

= \frac{2 2}{7} [45 + 25] = \frac{2 2}{7} \times 70 = 220

mm

^{2}

Answer

Given:
Length of Capsule (l) =

14

mm

Diameter of Capsule = Diameter of Cylinder

= 5

mm

Radius =

\frac{D i a m e t e r}{2}

Therefore, Radius of each Hemisphere = Radius of Cylinder = r =

\frac{5}{2}

= 2.5 mm

Length of Cylinder = AB = Total length of Capsule - Radius of left Hemisphere - Radius of Right Hemisphere

= 14 - 2.5 - 2.5 = 9

mm

Surface Area of Capsule = Curved Surface Area of Cylinder + Surface Area of Left Hemisphere + Surface Area of Right Hemisphere

= 2 π r l + 2 π r^{2} + 2 π r^{2}

= 2 π r l + 4 π r^{2}

= 2 \times \frac{2 2}{7} \times 2.5 \times 9 + 4 \times \frac{2 2}{7} \times 2 . 5^{2}

= \frac{2 2}{7} [45 + 25] = \frac{2 2}{7} \times 70 = 220

mm

^{2}

Question 6

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.) (Use

π = \frac{2 2}{7}

).

Hard

Solution

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Answer

Given:
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

= π r l + 2 π r h

= π \times 2 \times 2.8 + 2 π \times 2 \times 2.1

= 2 π [2.8 + 4.2]

= 2 \times \frac{2 2}{7} \times 7

= 44 m^{2}

Cost of

1

m

^{2}

canvas

= 500

rupees
Cost of

44

m

^{2}

canvas

= 44 \times 500 = 22000

rupees.

Therefore, it will cost

22000

rupees for making such a tent.

Answer

Given:
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

= π r l + 2 π r h

= π \times 2 \times 2.8 + 2 π \times 2 \times 2.1

= 2 π [2.8 + 4.2]

= 2 \times \frac{2 2}{7} \times 7

= 44 m^{2}

Cost of

1

m

^{2}

canvas

= 500

rupees
Cost of

44

m

^{2}

canvas

= 44 \times 500 = 22000

rupees.

Therefore, it will cost

22000

rupees for making such a tent.

Question 7

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

c m^{2}

. (Use

π = \frac{2 2}{7}

).

Hard

Solution

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Answer

Given:
Height (h) of the conical part = Height (h) of the cylindrical part

= 2.4

cm
Diameter of the cylindrical part

= 1.4

cm

Radius =

\frac{D i a m e t e r}{2}

Radius(r) of the cylindrical part

= 0.7

cm

Slant height (l) of conical part =

r^{2} + h^{2}

= 0 . 7^{2} + 2 . 4^{2} = 0.49 + 5.76 = 6.25 = 2.5

Total surface area of the remaining solid = CSA of cylindrical part + CSA of conical part + Area of cylindrical base

= 2 π r h + π r l + π r^{2}

= 2 \times \frac{2 2}{7} \times 0.7 \times 2.4 + \frac{2 2}{7} \times 0.7 \times 2.5 + \frac{2 2}{7} \times 0.7 \times 0.7

= 4.4 \times 2.4 + 2.2 \times 2 . + 2.2 \times 0.7

= 10.56 + 5.50 + 1.54 = 17.60

cm

^{2}

The total surface area of the remaining solid to the nearest

c m^{2}

is

18

c m^{2}

Answer

Given:
Height (h) of the conical part = Height (h) of the cylindrical part

= 2.4

cm
Diameter of the cylindrical part

= 1.4

cm

Radius =

\frac{D i a m e t e r}{2}

Radius(r) of the cylindrical part

= 0.7

cm

Slant height (l) of conical part =

r^{2} + h^{2}

= 0 . 7^{2} + 2 . 4^{2} = 0.49 + 5.76 = 6.25 = 2.5

Total surface area of the remaining solid = CSA of cylindrical part + CSA of conical part + Area of cylindrical base

= 2 π r h + π r l + π r^{2}

= 2 \times \frac{2 2}{7} \times 0.7 \times 2.4 + \frac{2 2}{7} \times 0.7 \times 2.5 + \frac{2 2}{7} \times 0.7 \times 0.7

= 4.4 \times 2.4 + 2.2 \times 2 . + 2.2 \times 0.7

= 10.56 + 5.50 + 1.54 = 17.60

cm

^{2}

The total surface area of the remaining solid to the nearest

c m^{2}

is

18

c m^{2}

Question 8

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to

1 c m

and the height of the cone is equal to its radius. Find the volume of the solid in terms of

π

.

Medium

Solution

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Answer

We have
Radius of the cone

= r_{1} = 1 c m

Height of the cone

= h_{1} = 1 c m

Radius of the hemisphere

= r_{2} = 1 c m

Volume of the solid = volume of the cone +volume of the hemisphere

= \frac{1}{3} π r_{1}^{2} h_{1}^{2} + \frac{2}{3} π r_{2}^{3}

= (\frac{1}{3} π \times 1^{2} \times 1 + \frac{2}{3} π \times 1^{3}) c m^{3}

= π c m^{3}

Answer

We have
Radius of the cone

= r_{1} = 1 c m

Height of the cone

= h_{1} = 1 c m

Radius of the hemisphere

= r_{2} = 1 c m

Volume of the solid = volume of the cone +volume of the hemisphere

= \frac{1}{3} π r_{1}^{2} h_{1}^{2} + \frac{2}{3} π r_{2}^{3}

= (\frac{1}{3} π \times 1^{2} \times 1 + \frac{2}{3} π \times 1^{3}) c m^{3}

= π c m^{3}

Question 9

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 c m

and its length is

12 c m

. If each cone has a height of

2 c m

, 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.)

Hard

Solution

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Answer

Step 1: Finding the radius and height of cone and cylinder

Radius of the conical and cylindrical part (r) = 1.5 c m

Height of the cylindrical part (H) = 8 c m

Height of the conical part (h) = 2 c m

Step 2: Finding the volume of air in the model

Volume of air in the model=Volume of cylinder + 2 \times Volume of cone

Volume of cylinder = π r^{2} H

= \frac{2 2}{7} \times (1.5)^{2} \times 8

Volume of cone = \frac{1}{3} π r^{2} h

= \frac{1}{3} \times \frac{2 2}{7} \times (1.5)^{2} \times 2

Volume of air = \frac{2 2}{7} \times (1.5)^{2} \times 8 + 2 \times \frac{1}{3} \times \frac{2 2}{7} \times (1.5)^{2} \times 2 \approx 66 c m^{3}

Hence , the volume of air in the model is 66 c m^{3}

Answer

Step 1: Finding the radius and height of cone and cylinder

Radius of the conical and cylindrical part (r) = 1.5 c m

Height of the cylindrical part (H) = 8 c m

Height of the conical part (h) = 2 c m

Step 2: Finding the volume of air in the model

Volume of air in the model=Volume of cylinder + 2 \times Volume of cone

Volume of cylinder = π r^{2} H

= \frac{2 2}{7} \times (1.5)^{2} \times 8

Volume of cone = \frac{1}{3} π r^{2} h

= \frac{1}{3} \times \frac{2 2}{7} \times (1.5)^{2} \times 2

Volume of air = \frac{2 2}{7} \times (1.5)^{2} \times 8 + 2 \times \frac{1}{3} \times \frac{2 2}{7} \times (1.5)^{2} \times 2 \approx 66 c m^{3}

Hence , the volume of air in the model is 66 c m^{3}

Question 10

A gulab jamun, contains sugar syrup up to about 30% of its volume. Find the 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 .

Medium

Solution

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Answer

Step 1: Find out the volume .

Given h = 5 c m

Length of cylinder h = 5 - 2.8 = 2.2

Volume of gulab jamun = volume of cylinder + volume of 2 hemisphere

= 3 π r^{2} h + \frac{4}{3} π r^{3}

= \frac{2 2}{7} \times 1.4 \times 1.4 \times \frac{( 3 \times 2 . 2 + 4 \times 1 . 4 )}{3}

= \frac{6 . 1 6 ( 1 2 . 2 )}{3}

= \frac{7 5 . 1 5 2}{3}

= 25.05 c m

There are 45 gulab jamuns .

Step 2: Find the volume of 45 gulab jamuns .

Volume = 45 \times 25 .05

= 1127.2830 percentage of syrup

= 1127.28 \times \frac{3 0}{1 0 0}

= 338.184 c m

Hence, there are 338.184 cm of syrup be found in 45 gulab jamuns .

Answer

Step 1: Find out the volume .

Given h = 5 c m

Length of cylinder h = 5 - 2.8 = 2.2

Volume of gulab jamun = volume of cylinder + volume of 2 hemisphere

= 3 π r^{2} h + \frac{4}{3} π r^{3}

= \frac{2 2}{7} \times 1.4 \times 1.4 \times \frac{( 3 \times 2 . 2 + 4 \times 1 . 4 )}{3}

= \frac{6 . 1 6 ( 1 2 . 2 )}{3}

= \frac{7 5 . 1 5 2}{3}

= 25.05 c m

There are 45 gulab jamuns .

Step 2: Find the volume of 45 gulab jamuns .

Volume = 45 \times 25 .05

= 1127.2830 percentage of syrup

= 1127.28 \times \frac{3 0}{1 0 0}

= 338.184 c m

Hence, there are 338.184 cm of syrup be found in 45 gulab jamuns .

NCERT Solutions for Class 10 Maths Chapter 13 : Surface Areas and Volumes

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