ICSE Class 8 Maths Mensuration Chapter 33 Volume and Surface Area of Cuboids

Volume And Surface Area Of Cuboids

Cubes and Cuboids

Volume

Total Surface Area

Lateral Surface Area

Let us first recall the units of volume and their conversions.

1 cm³ = 1 cm × 1 cm × 1 cm = 10 mm × 10 mm × 10 mm = 1000 mm³

1 dm³ = 1 dm × 1 dm × 1 dm = 10 cm × 10 cm × 10 cm = 1000 cm³ = 1000000 mm³ = 1 l

1 m³ = 1 m × 1 m × 1 m = 100 cm × 100 cm × 100 cm = 1000000 cm³ = 1000 l = 1 kl

1 dcm³ = 10 m × 10 m × 10 m = 1000 m³

Consider diagonal DG of square DCGH. \[DG^2 = DH^2 + HG^2 = l^2 + l^2 = 2l^2\]

\[\Rightarrow DG = \sqrt{2l}\]

Now consider rectangle AFGD where AD = FG = l and GD = AF = \(\sqrt{2l}\)

FD is a diagonal of rectangle AFGD.

\[FD^2 = DG^2 + FG^2\]

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

\[\Rightarrow FD^2 = 2l^2 + l^2 = 3l^2\]

\[\Rightarrow FD = \sqrt{3l}\]

Length of the diagonal of a cube = \(\sqrt{3l}\)

Cubes

Formulae for a cube with length l

Area of one surface = l²

Total surface area of a cube = 6l²

Lateral surface area of cube = 4l²

Volume of a cube = l³

Diagonal of a Cube

A cube has four diagonals. The opposite vertices of a cube are joined by the diagonals of the cube. In the cube shown in Figure 33.1, the four diagonals are AG, BH, FD, and EC. Consider diagonal FD. It does not lie in the same plane as any of the six surfaces of the cube.

Cuboids

Formulae for a cuboid with length l, breadth b, and height h

Total surface area of a cuboid = 2(lh + bh + lh)

Lateral surface area of a cuboid = 2h(l + b)

Volume of a cuboid = l × b × h

Diagonal of a Cuboid

A cuboid has four diagonals. The opposite vertices of a cuboid are joined by the diagonals of a cuboid. In the cuboid shown in Figure 33.2, the four diagonals are AG, BH, CE, and DF.

Consider diagonal EG in rectangle EFGH.

\[EG^2 = EH^2 + HG^2 = b^2 + l^2\]

\[\Rightarrow EG = \sqrt{b^2 + l^2}\]

Now consider diagonal CE in rectangle ACGE.

\[CE^2 = CG^2 + EG^2\]

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

\[= h^2 + b^2 + l^2\]

\[\Rightarrow CE = \sqrt{l^2 + b^2 + h^2}\]

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

Teacher's Note

Understanding how to calculate volume and surface area helps in real-world scenarios like determining how much paint is needed for a room or how much water a tank can hold.

Example 1: If the volume of a cube is 1728 cm³, find its total surface area and lateral surface area.

Given volume of cube = l³ = 1728 cm³

\[\Rightarrow l = \sqrt[3]{1728} \text{ cm}\]

\[= \sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3} \text{ cm}\]

\[= \sqrt[3]{2 \times 2 \times 3 \times 2 \times 2 \times 3 \times 2 \times 2 \times 3} \text{ cm}\]

\[= 2 \times 2 \times 3 \text{ cm}\]

Thus, the length of the cube is 12 cm.

Total surface area of the cube = 6l² = 6 × 12² = 6 × 144 = 864 cm²

Lateral surface area of the cube = 4l² = 4 × 12² = 4 × 144 = 576 cm²

Example 2: The length, breadth, and height of a cuboid are in the ratio 7 : 4 : 3. If the total surface area of the cuboid is 5978 cm², find its volume.

Let the length, breadth, and height of the cuboid be 7x cm, 4x cm, and 3x cm, respectively.

Given total surface area = 2(lh + bh + lh) = 5978 cm²

\[\Rightarrow 2\{(7x \times 4x) + (4x \times 3x) + (7x \times 3x)\} = 5978 \text{ cm}^2\]

\[\Rightarrow 2(28x^2 + 12x^2 + 21x^2) = 5978 \text{ cm}^2\]

\[\Rightarrow 2 \times 61x^2 = 5978 \text{ cm}^2 \Rightarrow 122x^2 = 5978 \text{ cm}^2\]

\[\Rightarrow x^2 = \frac{5978}{122} = 49 \text{ cm}^2 \Rightarrow x = \sqrt{49} = 7 \text{ cm}\]

Thus, the length = 7 × 7 = 49 cm, breadth = 4 × 7 = 28 cm, and height = 3 × 7 = 21 cm

Volume of cuboid = 49 cm × 28 cm × 21 cm = 28812 cm³ = 28.812 dm³

Teacher's Note

When dealing with ratios and surface areas, students learn how to work backwards from given measurements to find unknown dimensions - a skill useful in construction and design.

Example 3: A conference hall is 35.5 m long, 19.4 m wide, and 6 m high. Its ceiling is covered with sound absorbing material. Find out how much it would cost to:

(i) cover its floor with a wall-to-wall carpet at Rs 78.50 per sq. m

(ii) paint its walls at Rs 47.50 per sq. m

Area of floor of hall = length × breadth = 35.5 m × 19.4 m = 688.7 m²

At Rs 78.50 per sq. m, covering the entire floor with carpet would cost 688.7 × 78.50 = Rs 54062.95

The lateral surface area of the hall or the area of its four walls = 2h(l + b) = 2 × 6(35.5 + 19.4) m² = 12 × 54.9 = 658.8 m²

Painting the walls of the hall at Rs 47.50 per m² would cost = 47.50 × 658.8 = Rs 31293.00

Example 4: A car mechanic, wishing to collect distilled water, set up a tray 1 metre in length and breadth on the roof top and connected a pipe to drain the tray in a cuboidal tin below that was

50 cm long, 30 cm wide, and 30 cm high. If 3 cm of rain fell during the day, what was the height of the water that collected in the mechanic's tin?

Rainfall is recorded in terms of height of water collected, irrespective of how wide or narrow a container is.

The area of the tray on the roof = 100 cm × 100 cm = 10000 cm²

Height of rainwater collected in tray = 3 cm

Thus, volume of rainwater collected = 3 × 10000 = 30000 cm³ or 30 l

When 30 l of water is drained into the empty tin, volume of water in tin = l × b × h = 30000 cm³

\[\Rightarrow 50 \text{ cm} \times 30 \text{ cm} \times h = 30000 \text{ cm}^3\]

\[\Rightarrow h \times 1500 \text{ cm}^2 = 30000 \text{ cm}^3\]

\[\Rightarrow h = \frac{30000}{1500} = 20 \text{ cm}\]

Thus, the height of the water collected in the mechanic's tin = 20 cm

Example 5: The exterior of an empty wooden box measures 90 cm in length, 72 cm in breadth, and 60 cm in height. If the wood is 3 cm thick all around, find the volume of wood used to make the box. If 1 cc of wood weighs 0.09 g find the weight of the empty wooden box.

Volume of the outer cuboid = 90 cm × 72 cm × 60 cm = 388800 cm³

If the open box was seen from top it would look like the figure shown in Figure 33.3 As the wood is 3 cm thick on all sides, the length of the inner cuboid = 90 cm - 3 cm - 3 cm = 84 cm

Breadth of the inner cuboid = 72 cm - 3 cm - 3 cm = 66 cm

Similarly as the top and bottom of the box are also 3 cm thick,

height of the inner cuboid = 60 cm - 3 cm - 3 cm = 54 cm

Thus, the volume of the inner cuboid = 84 cm × 66 cm × 54 cm = 299376, cm³

Volume of wood = Volume of outer cuboid - Volume of inner cuboid = 388800 - 299376 = 89424 cm³

Thus, 89424 cm³ of wood was used to make the box.

Given 1 cm³ of wood weighs 0.09 g, 89424 cm³ of wood weighs 89424 × 0.09 = 8048.16 g

Thus, the weight of the empty wooden box is 8 kg 48.16 g.

Teacher's Note

Calculating the volume of wood in a hollow box demonstrates how geometry applies to manufacturing and quality control in woodworking industries.

Example 6: The length, breadth, and height of a hall are in the ratio 2 : 2 : 1. If the lateral surface area of the hall is 1152 m², find the length of its diagonal.

Lateral surface area = 2h(l + b)

Let the length, breadth, and height of the hall be 2x, 2x, and x respectively.

\[\Rightarrow (2 \times x)(2x + 2x) = 1152 \text{ m}^2\]

\[\Rightarrow 2x \times 4x = 1152 \text{ m}^2\]

\[\Rightarrow 8x^2 = 1152 \text{ m}^2\]

\[\Rightarrow x = \sqrt{\frac{1152}{8}} = 12 \text{ m}\]

Thus, length = 2 × 12 = 24 m, breadth = 24 m, and height = 12 m.

Length of diagonal = \(\sqrt{l^2 + b^2 + h^2}\) = \(\sqrt{24^2 + 24^2 + 12^2}\) = \(\sqrt{576 + 576 + 144}\) = \(\sqrt{1296}\) = 36 m

Thus, the length of the hall's diagonal is 36 m.

Teacher's Note

Finding diagonal lengths in real architectural spaces helps engineers understand the structural distances needed for support beams and structural planning.

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