ICSE Class 9 Maths Chapter 21 Solids Surface Area and Volume of 3D Solids

Chapter 21: Solids

Surface Area And Volume Of 3-D Solids

21.1 Introduction

1. Solid: Anything which occupies space and has a definite shape is called a solid.

2. Volume: The space occupied by a solid is called its volume.

(a) The capacity of a container = Its internal volume.

(b) The volume of material in a hollow body = Its external volume - Its internal volume.

3. Surface area of a solid: The sum of the areas of all the surfaces of a solid is called its surface area or its total surface area.

Since, a solid has three dimensions i.e. length, breadth and height, a solid is a three-dimensional (3-D) figure.

In the same way:

(i) area is a 2-dimensional (2-D) figure as it has two dimensions i.e. length and breadth.

(ii) perimeter is a single dimensional (1-D) figure as it has only one dimension i.e. length.

21.2 Cuboid

A rectangular solid which has six faces, each of which is a rectangle, is called a cuboid.

For a cuboid:

1. Volume = length \(\times\) breadth \(\times\) height

= \(l \times b \times h\)

2. Total surface area = \(2 (l \times b + b \times h + h \times l)\)

3. Lateral surface area of a cuboid = The sum of areas of the four walls (vertical faces) of the cuboid.

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

4. Length of diagonal = \(\sqrt{l^2 + b^2 + h^2}\)

The length of the longest rod that can be placed in a rectangular box or in a room = its diagonal.

Every cuboid has four diagonals.

In the figure, given above, diagonals are: AG, BH, CE and DF such that, AG = BH = CE = DF

In right triangle ABD, \(BD^2 = l^2 + b^2\) implies \(BD = \sqrt{l^2 + b^2}\)

In right triangle DBH, \(BH^2 = \left(\sqrt{l^2 + b^2}\right)^2 + h^2 = l^2 + b^2 + h^2\) implies \(BH = \sqrt{l^2 + b^2 + h^2}\)

Hence, the diagonal of a cuboid = \(\sqrt{l^2 + b^2 + h^2}\)

21.3 Cube

A rectangular solid, in which each face is a square, is called a cube.

For a square:

If length of each edge = \(a\) unit.

Its length = breadth = height = edge

implies \(l = b = h = a\) (edge)

1. Volume = \(l \times b \times h = a \times a \times a = a^3\) = (edge)\(^3\)

2. Total surface area = \(6a^2\)

3. Lateral surface area = \(4a^2\)

4. Length of diagonal = \(\sqrt{a^2 + a^2 + a^2} = a\sqrt{3}\)

21.4 Cost Of An Article

Multiply known quantity of the article with its rate to get the cost of the article.

(a) We buy petrol by its volume:

Cost of petrol bought = The quantity of petrol bought \(\times\) Rate of petrol

(b) We buy land by its area:

Cost of land bought = The area of land bought \(\times\) Rate of land

(c) We buy cloth by its length:

Cost of cloth bought = Length of cloth bought \(\times\) Rate of cloth

Cost = Rate \(\times\) Quantity

Problem 1

The outer dimensions of a closed wooden box are 22 cm, 15 cm and 10 cm. Thickness of the wood is 1 cm. Find the cost of wood required to make the box, if 1cm\(^3\) of wood costs \(\text{₹}\) 7.50.

Solution:

External volume of box = 22 \(\times\) 15 \(\times\) 10 cm\(^3\) = 3300 cm\(^3\)

Since, external dimensions are 22 cm, 15 cm and 10 cm; and thickness of the wood is 1 cm

Internal dimensions = (22 - 2 \(\times\) 1) cm, (15 - 2 \(\times\) 1) cm and (10 - 2 \(\times\) 1) cm = 20 cm, 13 cm and 8 cm.

Hence, internal volume of the box = 20 \(\times\) 13 \(\times\) 8 cm\(^3\) = 2080 cm\(^3\)

and volume of wood in the box = 3300 cm\(^3\) - 2080 cm\(^3\) = 1220 cm\(^3\)

Cost of wood = \(\text{₹}\) 7.50 \(\times\) 1220 = \(\text{₹}\) 9,150

Teacher's Note

When buying materials like wood or metal, we calculate volume to determine cost. This applies to real-world scenarios like purchasing bricks for construction or metal for manufacturing.

Problem 2

A cube of a metal of 5 cm edge is melted and casted into a cuboid whose base is 2.50 cm \(\times\) 0.50 cm. Find the height of the cuboid. Also, find the surface areas of cube and cuboid.

Solution:

Here, volume of cuboid formed = volume of cube melted

i.e. 2.50 \(\times\) 0.50 \(\times\) h = (5)\(^3\)

i.e. h = \(\frac{5 \times 5 \times 5}{2.50 \times 0.50}\) cm = 100 cm

Surface area of cube = 6a\(^2\) = 6(5)\(^2\) cm\(^2\) = 150 cm\(^2\)

Surface area of cuboid = 2(lb + bh + hl) = 2 (2.50 \(\times\) 0.50 + 0.50 \(\times\) 100 + 100 \(\times\) 2.50) cm\(^2\) = 602.50 cm\(^2\)

Teacher's Note

Conservation of volume is crucial in manufacturing. When melting and recasting materials, volume remains constant even though shape changes dramatically.

Problem 3

A small indoor green house (herbarium) is made entirely of glass panes (including base) held together with tape. The dimensions of the green house are 40 cm \(\times\) 30 cm \(\times\) 25 cm. Find:

(i) the area of the glass used.

(ii) the length of the tape required.

Solution:

(i) Since, the green house is in the shape of a cuboid and its length (l) = 40 cm, breadth (b) = 30 cm and height (h) = 25 cm

Area of the glass used = 2(lb + bh + hl) = 2(40 \(\times\) 30 + 30 \(\times\) 25 + 25 \(\times\) 40) cm\(^2\) = 5900 cm\(^2\)

(ii) Length of the tape used = Perimeter of the top + Perimeter of the bottom + Four vertical edges = 2(l + b) + 2(l + b) + 4h = 4(40 + 30) cm + 4 \(\times\) 25 cm = 380 cm\(^2\)

Teacher's Note

Greenhouses require careful measurement to calculate glass panels and support tape. Gardeners and manufacturers use these calculations for efficient construction planning.

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