Cube and Cuboids - Surface Areas and Volumes, CBSE, Class 10, Mathematics 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 cm^₂, 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