# SELINA Solutions for Class 9 Maths Chapter 21 - Solids [Surface Area and Volume of 3-D Solids]

Learn to find the volume and the surface area of a given cube with our Selina Solutions for ICSE Class 9 Maths Chapter 21 Solids - Surface Area and Volume of 3-D Solids. TopperLearning’s model answers by experts can help you to revise the textbook exercises with ease. Understanding the structure of solids such as cube, cuboid etc. is easy when you have accurate answers.

Our ICSE Class 9 Mathematics Selina solutions can assist you with revision of topics like volume of a cuboid, area of a cross-section etc. For more clarity, go through the online doubts and solutions available for Class 9 round the clock at our learning portal.

## Chapter 21 - Solids [Surface Area and Volume of 3-D Solids] Exercise Ex. 21(C)

Each face of a cube has perimeter equal to 32 cm. Find its surface area and its volume.

The perimeter of a cube formula is, Perimeter = 4a where (a= length)

A school auditorium is 40 m long, 30 m broad and 12 m high. If each student requires 1.2 m^{2} of the floor area; find the maximum number of students that can be accommodated in this auditorium. Also, find the volume of air available in the auditorium, for each student.

The internal dimensions of a rectangular box are 12 cm x cm 9 cm. If the length of the longest rod that can be placed in this box is 17 cm; find x.

Length of longest rod=Length of the diagonal of the box

The internal length, breadth and height of a box are 30 cm, 24 cm, and 15 cm. Find the largest number of cubes which can be placed inside this box if the edge of each cube is

(i) 3 cm(ii) 4 cm(iii) 5 cm

(i)

No. of cube which can be placed along length.

No. of cube along the breadth

No. of cubes along the height.

The total no. of cubes placed

(ii)

Cubes along length

Cubes along widthand cubes along height

The total no. of cubes placed

(iii)

Cubes along length

Cubes along widthand cubes along height

The total no. of cubes placed

A rectangular field is 112 m long and 62 m broad. A cubical tank of edge 6 m is dug at each of the four corners of the field and the earth so removed is evenly spread on the remaining field. Find the rise in level.

Vol. of the tank= vol. of earth spread

When length of each side of a cube is increased by 3 cm, its volume is increased by 2457 cm^{3}. Find its side. How much will its volume decrease, if length of each side of it is reduced by 20%?

A rectangular tank 30 cm × 20 cm × 12 cm contains water to a depth of 6 cm. A metal cube of side 10 cm is placed in the tank with its one face resting on the bottom of the tank. Find the volume of water, in litres, that must be poured in the tank so that the metal cube is just submerged in the water.

The dimensions of a solid metallic cuboid are 72 cm × 30 cm × 75 cm. It is melted and recast into identical solid metal cubes with each of edge 6 cm. Find the number of cubes formed.

Also, find the cost of polishing the surfaces of all the cubes formed at the rate Rs. 150 per sq. m.

The dimensions of a car petrol tank are 50 cm × 32 cm × 24 cm, which is full of petrol. If car's average consumption is 15 km per litre, find the maximum distance that can be covered by the car.

The dimensions of a rectangular box are in the ratio
4 : 2 : 3. The difference between cost of covering
it with paper at Rs. 12 per m^{2} and with paper at the rate of 13.50
per m^{2} is Rs. 1,248. Find the dimensions of the box.

## Chapter 21 - Solids [Surface Area and Volume of 3-D Solids] Exercise Ex. 21(A)

Four identical cubes are joined end to end to form a cuboid. If the total surface area of the resulting cuoid as 648 m^{2}; find the length of edge of each cube.

Also, find the ratio between the surface area of resulting cuboid and the surface area of a cube.

Let *l* be the length of the edge of each cube.

The length of the resulting cuboid=

Let width (b)* *= *l* cm and its height (h)= *l* cm

The total surface area of the resulting cuboid

Therefore, the length of each cube is 6 cm.

The volume of a cube is 729 cm^{3}. Find its total surface area.

Let *a* be the one edge of a cube.

Volume

Total surface area=6

The dimensions of a Cinema Hall are 100 m, 60 m and 15 m. How many persons can sit in the hall, if each requires 150 m^{3} of air?

Volume of cinema hall

150 requires= 1 person

90000requires=persons

Therefore, 600 persons can sit in the hall.

75 persons can sleep in a room 25 m by 9.6 m. If each persons requires 16 m^{3} of air; find the height of the room.

Let *h* be height of the room.

1 person requires 16

75 person requires

Volume of room is 1200

The edges of three cubes of metal are 3 cm, 4 cm and 5 cm. They are melted and formed into a single cube. Find the edge of the new cube.

Volume of melted single cube

Let *a* be the edge of the new cube.

Volume

Therefore, 6 cm is the edge of cube.

Three cubes, whose edges are x cm, 8 cm and 10 cm respectively, are melted and recasted into a single cube of edge 12 cm. Find 'x'.

Three equal cubes are placed adjacently in a row. Find the ratio of the total surfaced area of the resulting cuboid to that of the sum of the total surface areas of the three cubes.

Let the side of a cube be 'a' units.

Total surface area of one cube

Total surface area of 3 cubes

After joining 3 cubes in a row, length of Cuboid =3a

Breadth and height of cuboid = a

Total surface area of cuboid

Ratio of total surface area of cuboid to the total surface area of 3 cubes

The cost of papering the four walls of a room at 75 paisa per square meter Rs. 240. The height of the room is 5 metres. Find the length and the breadth of the room, if they are in the ratio 5 : 3.

The area of a playground is 3650 m^{2}. Find the cost of covering it with gravel 1.2 cm deep, if the gravel costs Rs. 6.40 per cubic metre.

The area of the playground is 3650 m^{2} and the gravels are 1.2 cm deep. Therefore the total volume to be covered will be:

3650 x 0.012 =43.8 m^{3}.

Since the cost of per cubic meter is Rs. 6.40, therefore the total cost will be:

43.8 x Rs.6.40 = Rs.280.32

A square plate of side 'x' cm is 8 mm thick. If its volume is 2880 cm^{3}; find the value of x.

The external dimensions of a closed wooden box are 27 cm, 19 cm and 11 cm. If the thickness of the wood in the box is 1.5 cm; find:

(i) Volume of the wood in the box;

(ii) The cost of the box, if wood costs Rs. 1.20 per cm^{3};

(iii) Number of 4 cm cubes that could be placed into the box.

External volume of the box=

Since, external dimensions are 27 cm, 19 cm, 11 cm; thickness of the wood is 1.5 cm.

Internal dimensions

Hence, internal volume of box=

(i)

Volume of wood in the box=

(ii)

Cost of wood

(iii)

Vol. of 4 cm cube=

Number of 4 cm cubes that could be placed into the box

A tank 20 m long, 12 m wide and 8 m deep is to be made of iron sheet. If it is open at the top. Determine the cost of iron-sheet, at the rate of Rs. 12.50 per metre, if the sheet is 2.5 m wide.

Area of sheet= Surface area of the tank

Length of the sheetits width=Area of 4 walls of the tank +Area of its base

Length of the sheet 2.5 m=

Length of the sheet= 300.8 m

Cost of the sheet = 300.8 Rs 12.50 = Rs 3760

A closed rectangular box is made of wood of 1.5 cm thickness. The exterior length and breadth are respectively 78 cm and 19 cm, and the capacity of the box is 15 cubic decimeters. Calculate the exterior height of the box.

Let exterior height is h cm. Then interior dimensions are 78-3=75, 19-3=16 and h-3 (subtract two thicknesses of wood). Interior volume = 75 x 16 x (h-3) which must = 15 cu dm

= 15000 cm^3

(1 dm = 10cm, 1 cu dm = 10^3 cm^3).

15000= 75 x 16 x (h-3)

h-3 = 15000/(75x16) = 12.5 cm h = 15.5 cm.

The square on the diagonal of a cube has an area of 1875 sq. cm. Calculate:

(i) The side of the cube.

(ii) The total surface area of the cube.

(i)

If the side of the cube= *a* cm

The length of its diagonal= cm

And,

(ii)

Total surface area of the cube=

=

A hollow square-shaped tube open at both ends is made of iron. The internal square is of 5 cm side and the length of the tube is 8 cm. There are 192 cm^{3} of iron in this tube. Find its thickness.

Given that the volume of the iron in the tube 192 cm^{3}

Let the thickness of the tube

Side of the external square=

Ext. vol. of the tube its internal vol.= volume of iron in the tube, we have,

Therefore, thickness is 1 cm.

Four identical cubes are joined end to end to form a cuboid. If the total surface area of the resulting cuoid as 648 m^{2}; find the length of edge of each cube.

Also, find the ratio between the surface area of resulting cuboid and the surface area of a cube.

Let *l* be the length of the edge of each cube.

The length of the resulting cuboid=

Let width (b)* *= *l* cm and its height (h)= *l* cm

The total surface area of the resulting cuboid

Therefore, the length of each cube is 6 cm.

## Chapter 21 - Solids [Surface Area and Volume of 3-D Solids] Exercise Ex. 21(B)

The following figure shows a solid of uniform cross-section. Find the volume of the solid. All measurements are in centimetres.

Assume that all angles in the figures are right angles.

The given figure can be divided into two cuboids of dimensions 6 cm, 4 cm, 3 cm, and 9 cm respectively. Hence, volume of solid

A swimming pool is 40 m long and 15 m wide. Its shallow and deep ends are 1.5 m and 3 m deep respectively. If the bottom of the pool slopes uniformly, find the amount of water in litres required to fill the pool.

Area of cross section of the solid

Volume of solid

The cross-section of a tunnel perpendicular to its length is a trapezium ABCD as shown in the following figure; also given that:

AM = BN; AB = 7 m; CD = 5 m. The height of the tunnel is 2.4 m. The tunnel is 40 m long. Calculate:

(i) The cost of painting the internal surface of the tunnel (excluding the floor) at the rate of Rs. 5 per m^{2} (sq. metre).

(ii) The cost of paving the floor at the rate of Rs. 18 per m^{2}.

The cross section of a tunnel is of the trapezium shaped ABCD in which and AM = BN. The height is 2.4 m and its length is 40m.

(i)

Perimeter of the cross- section of the tunnel=

Length=40 m

Internal surface area of the tunnel(except floor)

Rate of painting=Rs 5 per

Hence, total cost of painting=Rs 5408=Rs 2040

(ii)

Area of floor of tunnel

Rate of cost of paving

Total cost=

Water is discharged from a pipe of cross-section area 3.2 cm^{2} at the speed of 5m/s. Calculate the volume of water discharged:

(i) In cm^{3} per sec.

(ii) In litres per minute.

(i)

The rate of speed

Volume of water flowing per sec

(ii)

Vol. of water flowing per min

Since 1000= 1 lt

Therefore, Vol. of water flowing per min=

A hose-pipe of cross-section area 2 cm^{2} delivers 1500 litres of water in 5 minutes. What is the speed of water in m/s through the pipe?

Vol. of water flowing in 1 sec=

Vol. of water flowing =area of cross sectionspeed of water

The cross-section of a piece of metal 4 m in length is shown below. Calculate:

(i) The area of the cross-section;

(ii) The volume of the piece of metal in cubic centimetres.

If 1 cubic centimetre of the metal weighs 6.6 g, calculate the weight of the piece of metal to the nearest kg.

(i)

Area of total cross section= Area of rectangle abce+ area of

=

(ii)

The volume of the piece of metal in cubic centimeters= Area of total cross section

=

1 cubic centimetre of the metal weighs 6.6 g

of the metal weighs

The weight of the piece of metal to the nearest Kg is 352 Kg.

A rectangular water-tank measuring 80 cm 60 cm 60 cm is filled form a pipe of cross-sectional area 1.5 cm^{2}, the water emerging at 3.2 m/s. How long does it take to fill the tank?

Vol. of rectangular tank

One liter= 1000

Vol. of water flowing in per sec=

Vol. of water flowing in 1 min=

Hence,

can be filled = 1 min

can be filled

A rectangular card-board sheet has length 32 cm and breadth 26 cm.Squares each of side 3 cm, are cut from the corners of the sheet and the sides are folded to make a rectangular container. Find the capacity of the container formed.

Length of sheet=32 cm

Breadth of sheet=26 cm

Side of each square=3cm

Inner length=32-23=32-6=26 cm

Inner breadth=cm

By folding the sheet, the length of the container=26 cm

Breadth of the container= 20 cm and height of the container= 3 cm

Vol. of the container=

=

A swimming pool is 18 m long and 8 m wide. Its deep and shallow ends are 2 m and 1.2 m respectively. Find the capacity of the pool, assuming that the bottom of the pool slopes uniformly.

Length of pool= 18 m

Breadth of pool= 8 m

Height of one side= 2m

Height on second side=1.2 m

Volume of pool=

The following figure shows a closed victory-stand whose dimensions are given in cm.

Find the volume and the surface are of the victory stand.

Consider the box 1

Thus, the dimensions of box 1 are: 60 cm, 40 cm and 30 cm.

Consider the box 2

Thus, the dimensions of box 2 are: 40 cm, 30 cm and 30 cm.

Consider the box 3

Thus, the dimensions of box 2 are: 40 cm, 30 cm and 20 cm.

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