Surface Areas and Volumes, Class 10, Maths Detailed Chapter Notes PDF Download

INTRODUCTION

In this chapter, we shall discuss problems on conversion of one of the solids like cuboid, cube, right circular cylinder, right circular cone and sphere in another.

In our day-to-day life we come across various solids which are combinations of two or more such solids. For example, a conical circus tent with cylindrical base is a combination of a right circular cylinder and a right circular cone, also an ice-cream cone is a combination of a cone and a hemi-sphere. We shall discuss problems on finding surface areas and volumes of such solids. We also come across solids which are a part of a cone. For example, a bucket,a glass tumbler, a friction clutch etc. These solids are known as frustums of a cone. In the end of the chapter, we shall discuss problems on surface area and volume of frustum of a cone.

UNITS OF MEASUREMENT OF AREA AND VOLUME
The inter-relationships between various units of measurement of length, area and volume are listed below for ready reference:

LENGTH
1 Centimetre (cm) = 10 milimetre (mm)
1 Decimetre (dm) = 10 centimetre
1 Metre(m) = 10 dm = 100 cm = 1000mm
1 Decametre (dam)= 10 m = 1000 cm
1 Hectometre (hm) = 10 dam = 100 m
1 Kilometre (km) = 1000 m = 100 dam = 10 hm
1 Myriametre = 10 kilometre
 

AREA

1 cm² = 1 cm × 1 cm = 10 mm × 10 mm = 100 mm²
1 dm² =1 dm × 1 dm =10 cm × 10 cm =100 cm²
1 m2 = 1 m × 1 m =10 dm × 10 dm =100 dm²
1 dam² =1 dam × 1 dam = 10 m × 10 m = 100 m²
1 hm² = 1 hectare = 1 hm × 1 hm = 100 m × 100 m = 10000 m² = 100 dm²
1 km² = 1 km × 1 km = 10 hm × 10 hm = 100 hm² or 100 hectare

VOLUME
1 cm³ = 1 ml = 1 cm × 1 cm × 1 cm = 10 mm × 10 mm × 10 mm = 1000 mm³
1 litre = 1000 ml = 1000 cm³
1 m³ = 1 m × 1 m × 1 m = 100 cm × 100 cm × 100 cm = 106 cm³ = 1000 litre = 1 kilolitre
1 dm³ = 1000 cm³
1 m³ = 1000 dm³
1 km³ = 109 m³

CUBOID
A rectangular solid bounded by six rectangular plane faces is called a cuboid. A match box, a teapacket, a brick, a book, etc., are all examples of a cuboid.

A cuboid has 6 rectangular faces, 12 edges and 8 vertices.

The following are some definitions of terms related to a cuboid:

(i) The space enclosed by a cuboid is called its volume.

(ii) The line joining opposite corners of a cuboid is called its diagonal.
A cuboid has four diagonals.
A diagonal of a cuboid is the length of the longest rod that can be placed in the cuboid.
 

(iii) The sum of areas of all the six faces of a cuboid is known as its total surface area.

(iv) The four faces which meet the base of a cuboid are called the lateral faces of the cuboid.

(v) The sum of areas of the four walls of a cuboid is called its lateral surface area.

Formulae
For a cuboid of length = units, breadth = b units and height = h units, we have :

REMARK : For the calculation of surface area, volume etc. of a cuboid, the length, breadth and
height must be expressed in the same units.
 

CUBE
A cuboid whose length, breadth and height are all equal is called a cube .

Ice-cubes, Sugar cubes, Dice, etc. are all examples of a cube.

Each edge of a cube is called its side.
 

Formulae

For a cube of edge = a units, we have;

CROSS SECTION

A cut which is made through a solid perpendicular to its length is called its cross section. If the cut has the same shape and size at every point of its length, then it is called uniform cross-section.

Volume of a solid with uniform cross section = (Area of its cross section) × (length).

Lateral Surface Area of a solid with uniform cross section = (Perimeter of cross section) × (length).

Ex.1 The length, breadth and height of a rectangular solid are in the ratio 6 : 5 : 4. If the total
surface area is 5328 cm2, find the length, breadth and height of the solid.

Sol. Let length = (6x) cm, breadth = (5x) cm and height = (4x) cm.
Then, total surface area = [2(6x × 5x + 5x × 4x + 4x × 6x)] cm² = [2(30x2 + 20x2 + 24x2)] cm²
= (148x2) cm².

Hence, length = 36 cm, breadth = 30 cm, height = 24 cm.

Ex.2 An open rectangular cistern is made of iron 2·5 cm thick. When measured from outside, it is
1 m 25 cm long, 1 m 5 cm broad and 90 cm deep.

Find:

(i) the capacity of the cistern in litres;
(ii) the volume of iron used;
(iii) the total surface area of the cistern.

Sol. External dimensions of the cistern are :
Length = 125 cm, Breadth = 105 cm and Depth = 90 cm.
Internal dimensions of the cistern are :
Length = 120 cm, Breadth = 100 cm and Depth = 87·5 cm.
(i) Capacity = Internal volume = (120 × 100 × 87.5) cm³ =  litres = 1050 litres.
The vertex of a right circular cone is farthest from its base.

(ii) Volume of iron = (External volume) – (Internal volume) = [(125 × 105 × 90) – (120 × 100 × 87·5)] cm³
= (1181250 – 1050000) cm³ = 131250 cm³.

(iii) External area = (Area of 4 faces) + (Area of the base) = ([2(125 + 105) × 90] + (125 × 105)} cm²
= (41400 + 13125) cm2 = 54525 cm².

Internal area = ([2(120 + 100) × 87·5] + (120 × 100)} cm² = (38500 + 12000) cm2 = 50500 cm².
Area at the top = Area between outer and inner rectangles = [(125 × 105) – (120 × 100)] cm²
= (13125 - 12000) cm² = 1125 cm².
Total surface area = (54525 + 50500 + 1125) cm2 = 106150 cm².

Ex.3.A field is 80 m long and 50 m broad. In one corner of the field, a pit which is 10 m long, 7.5 m broad and 8 m deep has been dug out. The earth taken out of it is evenly spread over the remaining part of the field. Find the rise in the level of the field.

Sol.

Area of the field = (80 × 50) m2 = 4000 m²
Area of the pit = (10 × 7.5) m² = 75 m²
Area over which the earth is spread out = (4000 – 75) m² = 3925 m²
Volume of earth dug out = (10 × 7.5 × 8) m³ = 600 m³.
 Rise in level   = 15.3 cm

Ex.4 A room is half as long again as it is broad. The cost of carpeting the room at Rs 18 per m² is Rs 972 and the cost of white-washing the four walls at Rs 6 per m² is Rs 1080. Find the dimensions of the room.

Sol. Let breadth = (x) m. Then, length = 
Let height of the room = y m.
Area of the floor = 

x = 6. 
So, breadth = 6 m and length
Now, area of four walls

Hence, length = 9 m, breadth = 6 m, height = 6 m.

Ex.5 The water in a rectangular reservoir having a base 80 m × 60 m, is 6·5 m deep. In what time can the water be emptied by a pipe of which the cross section is a square of side 20 cm, if water runs through the pipe at the rate of 15 km/ hr ?

Sol. Volume of water in the reservoir = (80 × 60 × 6.5) m3 = 31200 m³.
Area of cross section of the pipe
Volume of water emptied in 1 hr
Time taken to empty the reservoir

RIGHT CIRCULAR CYLINDER
Solids like circular pillars, circular pipes, circular pencils, measuring jars, road rollers and gas cylinders, etc., are said to be in cylindrical shape.

In mathematical terms, a right circular cylinder is a solid generated by the revolution of a rectangle about its sides.

Let the rectangle ABCD revolve about its side AB, so as to describe a right circular cylinder as shown in the figure.
You must have observed that the cross-sections of a right circular cylinder are circles congruent and parallel to
each other.
 

Cylinders Not Right Circular
There are two cases when the cylinder is not a right circular cylinder.

Case-I : In the following figure, we see a cylinder, which is certainly circular, but is not at right angles to the
base. So we cannot say it is a right circular cylinder.

Case-II : In the following figure, we see a cylinder with a non-circular base as the base is not circular. So we
cannot call it a right circular cylinder.

REMARK : Unless stated otherwise, here in this chapter the word cylinder would mean a right circular cylinder.
The following are definitions of some terms related to a right circular cylinder :

(i) The radius of any circular end is called the radius of the right circular cylinder.
Thus, in the above figure, AD as well as BC is a radius of the cylinder.

(ii) The line joining the centres of circular ends of the cylinder, is called the axis of the right circular cylinder.
In the above figure, the line AB is the axis of the cylinder. Clearly, the axis is perpendicular to the circular ends.

REMARK : If the line joining the centres of circular ends of a cylinder is not perpendicular to
the circular ends, then the cylinder is not a right circular cylinder.

(iii) The length of the axis of the cylinder is called the height or length of the cylinder.

(iv) The curved surface joining the two bases of a right circular cylinder is called its lateral surface.

Formulae
For a right circular cylinder of radius = r units & height = h units, we have :

The above formulae are applicable to solid cylinders only.
 

Hollow Right Circular Cylinders

Solids like iron pipes, rubber tubes, etc., are in the shape of hollow cylinders.

A solid bounded by two coaxial cylinders of the same height and different radii is called a hollow
cylinder.

Formulae
For a hollow cylinder of height h and with external and internal radii R and r respectively, we have :

Ex.6 2.2 cu dm of brass is to be drawn into a cylindrical wire of diameter 0.50 cm. Find the length
of the wire.
Sol. Volume of brass = 2.2 cu dm = (2.2 × 10 × 10 × 10) cm³ = 2200 cm³. Let the required length of wire
be x cm.
Then, its volume  

= 11200 cm = 112 m.

Hence, the length of wire is 112 m.

Ex.7 A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread
all around it to a width of 21 m to form an embankment. Find the height of the embankment.

Sol. Volume of earth dug out from the well
Area of the embankment
Height of the embankment = = 53.3 cm.

Ex.8 The difference between the outside and inside surface of a cylindrical metallic pipe 14 cm long
is 44 cm². If the pipe is made of 99 cu cm of metal, find the outer and inner radii of the pipe.
 

Sol. Let, external radius = R cm and internal radius = r cm.

Then, outside surface
Inside surface

External volume
Internal volume

On dividing (ii) by (i), we get:
Solving (i) and (iii), we get, R = 2.5 and r = 2.
Hence, outer radius = 2.5 cm and inner radius = 2 cm.

Ex.9 A solid iron rectangular block of dimensions 4.4 m, 2.6 m and 1 m is cast into a hollow cylindrical
pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.
 

Sol. Volume of iron = (440 x 260 x 100) cm³.

Internal radius of the pipe = 30 cm.

External radius of the pipe = (30 + 5) cm = 35 cm.

Let the length of the pipe be h cm.

Volume of iron in the pipe = (External volume) - (Internal volume)

Hence, the length of the pipe is 112 m.

Ex.10 A cylindrical pipe has inner diameter of 7 cm and water flows through it at 192.5 litres per
minute. Find the rate of flow in kilometres per hour.

Sol. Volume of water that flows per hour = (192.50 × 60) litres = (192.5 × 60 × 1000) cm³.

Inner radius of the pipe = 3.5 cm.

Let the length of column of water that flows in 1 hour be h cm.