ICSE Class 8 Maths Chapter 29 Volume and Surface Area of Cuboid

Chapter 29: Volume and Surface Area of Cuboid

Cuboid

The following diagram shows a cuboid. It has 6 rectangular faces, 8 vertices and 12 edges.

Let its length, breadth and height be denoted by \(l\), \(b\) and \(h\) respectively.

Volume

Volume of a cuboid = \(l \times b \times h\)

Surface Area

Surface area of a cuboid = \(2(lb + lh + bh)\)

Lateral Surface Area

Lateral surface of a cuboid (area of four walls) = \(2(l + b) \times h\)

Length of Diagonal

A cuboid has four diagonals. In the above cuboid, the four diagonals are AE, BF, CG and DH. All these four diagonals have equal length.

Consider the diagonal AE.

In rectangle ABCD, the length of its diagonal \(AC = \sqrt{l^2 + b^2}\).

Notice that ACEG is a rectangle with length AC and breadth = \(CE = h\), so the length of its diagonal \(AE = \sqrt{AC^2 + CE^2}\).

\[\Rightarrow AE = \sqrt{\left(\sqrt{l^2 + b^2}\right)^2 + h^2} = \sqrt{l^2 + b^2 + h^2}\]

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

It is the length of the longest rod that can be placed in the cuboid.

Teacher's Note

Understanding cuboid dimensions helps in practical applications like calculating storage space in boxes, designing rooms, and determining how much material is needed for packaging.

Cube

It is a particular case of a cuboid where length = breadth = height.

Let \(a\) be the length of an edge of a cube.

Volume

Volume of a cube = \(a^3\)

Surface Area

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

Lateral Surface Area

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

Length of Diagonal

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

Hence, the length of a diagonal of a cube with edge \(a = \sqrt{3}a\).

Teacher's Note

Cubes appear frequently in everyday life, from dice and storage boxes to building blocks, making these formulas essential for calculating their storage capacity and material requirements.

Example 1

The length, breadth and height of a cuboid are 30 cm, 25 cm and 20 cm respectively. Find its

(i) volume (ii) surface area (iii) lateral surface area (iv) length of a diagonal.

Solution

(i) Volume of the cuboid = \((30 \times 25 \times 20) cm^3 = 15000 cm^3\).

(ii) Surface area of the cuboid = \(2(lb + lh + bh)\)

= \(2(30 \times 25 + 30 \times 20 + 25 \times 20) cm^2\)

= \(3700 cm^2\).

(iii) Lateral surface area = \(2(l + b) \times h\)

= \(2(30 + 25) \times 20 cm^2 = 2200 cm^2\).

(iv) Length of a diagonal = \(\sqrt{l^2 + b^2 + h^2} = \sqrt{30^2 + 25^2 + 20^2} cm\)

= \(\sqrt{1925} cm = 5\sqrt{77} cm\).

Example 2

If the length of a diagonal of a cube is \(8\sqrt{3} cm\), find its surface area and volume.

Solution

Let the length of an edge of the cube be \(a\) cm, then the length of its diagonal = \(\sqrt{3}a cm\).

According to given information, \(\sqrt{3}a = 8\sqrt{3} \Rightarrow a = 8\)

The surface area of the cube = \(6a^2 cm^2\)

= \((6 \times 8 \times 8) cm^2 = 384 cm^2\).

The volume of the cube = \(a^3 cm^3\)

= \((8 \times 8 \times 8) cm^3 = 512 cm^3\).

Example 3

If the surface area of a cube is 2166 cm², find

(i) the length of a diagonal (ii) volume of the cube.

Solution

Let the length of an edge of the given cube be \(a\) cm.

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

According to given information, \(6a^2 = 2166 \Rightarrow a^2 = 361\)

\[\Rightarrow a = \sqrt{361} = 19\]

(i) The length of diagonal of the cube = \(\sqrt{3}a cm = 19\sqrt{3} cm\)

= \((19 \times 1.73) cm = 32.9 cm\).

(ii) The volume of cube = \(a^3 cm^3\)

= \((19 \times 19 \times 19) cm^3 = 6859 cm^3\).

Example 4

The area of three faces of a box are 120 cm², 72 cm² and 60 cm². What is the volume of the box?

Solution

Let the dimensions of the box be \(l\) cm, \(b\) cm and \(h\) cm.

According to given information,

\(l \times b = 120\)

\(l \times h = 72\)

\(b \times h = 60\)

Multiplying these equations, we get

\(l^2 \times b^2 \times h^2 = 120 \times 72 \times 60\)

\[\Rightarrow l \times b \times h = \sqrt{120 \times 72 \times 60} = \sqrt{60^2 \times 12^2}\]

\(\Rightarrow l \times b \times h = 60 \times 12 = 720\)

Hence, the volume of the box = 720 cm³.

Example 5

The length, breadth and height of a cuboid are in the ratio 7:6:5. If the surface area of the cuboid is 1926 cm², find

(i) volume (ii) length of a diagonal of the cuboid.

Solution

As the length, breadth and height of a cuboid are in the ratio 7:6:5, let its length = 7x cm, breadth = 6x cm and height = 5x cm.

Surface area = \(2(7x \times 6x + 7x \times 5x + 6x \times 5x) cm^2\)

= \((2 \times 107) x^2 cm^2 = 214 x^2 cm^2\).

According to given information, \(214 x^2 = 1926\)

\[\Rightarrow x^2 = 9 \Rightarrow x = 3\]

Length = \((7 \times 3) cm = 21 cm\), breadth = 18 cm and height = 15 cm

(i) The volume of the cuboid = \((21 \times 18 \times 15) cm^3 = 5670 cm^3\).

(ii) The length of a diagonal = \(\sqrt{21^2 + 18^2 + 15^2} cm\)

= \(\sqrt{441 + 324 + 225} cm\)

= \(\sqrt{990} cm = 3\sqrt{110} cm\).

Example 6

The internal dimensions of a rectangular room are 6 m, 5 m and 3-5 m. It has two doors of size 1-2 m by 2 m and three windows of size 1 m by 1-9 m. The walls of the room are to be papered with a wall paper of width 70 cm. Find the cost of the paper at the rate of 6.50 per metre.

Solution

The surface area of the walls of the room = \(2(l + b) \times h\)

= \(2(6 + 5) \times 3.5 m^2\)

= \((22 \times 3.5) m^2 = 77 m^2\).

Surface area of 2 doors = \((2 \times 1.2 \times 2) m^2 = 4.8 m^2\).

Surface area of 3 windows = \((3 \times 1 \times 1.9) m^2 = 5.7 m^2\).

Surface area to be papered = surface area of walls - surface area of doors and windows

= \(77 m^2 - (4.8 m^2 + 5.7 m^2) = 77 m^2 - 10.5 m^2 = 66.5 m^2\).

Width of the wall paper = 70 cm = 0-7 m.

The length of the wall paper required = \(\frac{66.5}{0.7} m = 95 m\).

The cost of the wall paper = \(95 \times 6.50) = 617.50\).

Example 7

The external dimensions of an open rectangular wooden box are 98 cm by 84 cm by 77 cm. If the wood is 2 cm thick all around, find

(i) the internal dimensions of the box.

(ii) the capacity of the box.

(iii) the volume of the wood used in making the box.

(iv) the weight of the box in kilogram correct to 1 decimal place, given that 1 cm³ of wood weighs 0-8 gm.

Solution

(i) The internal dimensions of the box are:

length = 98 cm - 4 cm = 94 cm,

breadth = 84 cm - 4 cm = 80 cm,

height = 77 cm - 2 cm = 75 cm.

(Since the box is open at the top, so only the thickness of bottom is to be reduced from the height.)

(ii) The capacity of the box = internal volume of the box

= \((94 \times 80 \times 75) cm^3 = 564000 cm^3\).

(iii) External volume of the box = \((98 \times 84 \times 77) cm^3 = 633864 cm^3\)

The volume of the wood used in making the box

= external volume - internal volume

= \(633864 cm^3 - 564000 cm^3 = 69864 cm^3\).

(iv) The weight of the box = the weight of the wood

= \((69864 \times 0.8) gm = 55891.2 gm\)

= \(\frac{55891.2}{1000} kg = 55.8912 kg = 55.9 kg\).

Teacher's Note

These practical problems demonstrate how cuboid formulas apply to real-world scenarios like wallpapering rooms and designing wooden boxes, helping us calculate material costs and storage capacity efficiently.

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