ICSE Class 8 Maths Mensuration Chapter 02 Volume and Surface Area of Cuboids

Volume and Surface Area of Cuboids

Definitions and Units

The volume of a solid figure is the space occupied by it. Volume is measured in cubic units. The common units of volume and the corresponding units of length are given in the following table.

Table 2.1 Units of length and volume

Note The litre (L) is a unit commonly used for measuring the capacity of vessels or the volume of a liquid.

1 L = 1 cubic decimetre (dm³) = 1000 cm³, 1 mL = 1 cm³ or 1 cc

The surface area of a solid is the sum of the areas of the plane or curved faces of the solid. It is measured in square units, such as the square centimetre (cm²) and square metre (m²).

Cuboid

A cuboid is a solid figure bounded by six rectangular faces. The adjacent faces are mutually perpendicular and the opposite faces have the same dimensions. A cuboid has eight vertices (A, B, C, D, E, F, G, H) and 12 edges (AB, BC, CD, DA, EF, FG, GH, HE, AH, DE, CF, BG).

The volume of a cuboid is the product of its length, breadth and height. Denoting the volume, length, breadth and height by V, l, b and h respectively, we have

\[V = l \times b \times h, \quad l = \frac{V}{bh}, \quad b = \frac{V}{lh} \quad \text{and} \quad h = \frac{V}{lb}\]

The surface area of a cuboid is the sum of the surface areas of its six rectangular faces, which works out to the following.

The surface area of a cuboid = 2(lb + bh + hl)

The lateral surface area or the area of the four walls of a cuboid works out to.

The area of the four walls = 2(l + b)h = perimeter of the floor - height

Diagonal of a Cuboid

A diagonal of a cuboid is a line segment joining two vertices which are not on the same face. A cuboid has four diagonals (also called principal diagonals), namely HC, AF, BE and DG. All these diagonals are equal in length. Let us find the length of HC.

In the rectangle HFCA, HC is the diagonal.

\[\therefore \quad HC^2 = HF^2 + CF^2.\]

Now, HF is the diagonal of the rectangle EFGH.

\[\therefore \quad HF^2 = EF^2 + EH^2 = l^2 + b^2.\]

So, \(HC^2 = l^2 + b^2 + CF^2 = l^2 + b^2 + h^2\).

\[\therefore \quad HC = \sqrt{l^2 + b^2 + h^2}.\]

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

Example

The dimensions of a cuboid are 10 cm by 9.5 cm by 8 cm. Find (i) its volume, (ii) its surface area, (iii) the surface area of the four walls, and (iv) the length of a diagonal.

Solution

Here, l = 10 cm, b = 9.5 cm, h = 8 cm.

(i) The volume of the cuboid = l - b - h = 10 - 9.5 - 8 cm³ = 760 cm³.

(ii) Its surface area = 2(lb + bh + hl) = 2(10 - 9.5 + 9.5 - 8 + 8 - 10) cm² = 2 - 251 cm² = 502 cm².

(iii) The surface area of the four walls = 2(l + b)h = 2(10 + 9.5) - 8 cm² = 312 cm².

(iv) The length of a diagonal = \(\sqrt{l^2 + b^2 + h^2} = \sqrt{10^2 + 9.5^2 + 8^2}\) cm = \(\sqrt{254.25}\) cm = 15.9 cm (approximately).

Cube

A cube is a solid bounded by six square faces. Its adjacent faces are perpendicular to each other and all its 12 edges are equal in length. So, a cube is a cuboid in which length = breadth = height.

The volume of a cube is the cube of the length of its side. Denoting the volume of a cube by V and its length by a,

\[V = a^3 \quad \text{and} \quad a = \sqrt[3]{V}\]

The surface area of a cube is the sum of the areas of its six square faces or 6 - (length of an edge)². Denoting the surface area by S,

\[S = 6a^2 \quad \text{and} \quad a = \sqrt{\frac{1}{6}S}\]

The area of the four walls (lateral surface area) = 4 - (length of an edge)² = 4a²

Since a cube is a special cuboid in which l = b = h = a, the length of a diagonal of a cube = \(\sqrt{l^2 + b^2 + h^2} = \sqrt{3a^2} = \sqrt{3}a\).

The length of a diagonal = \(\sqrt{3}a\)

Example

The side of a cube is 6 cm. Find (i) its volume, (ii) its surface area, (iii) its lateral surface area, and (iv) the length of its diagonal.

Solution

(i) The volume of the cube = (length of a side)³ = (6 cm)³ = 216 cm³.

(ii) Its surface area = 6 - (length of a side)² = 6 - (6 cm)² = 216 cm².

(iii) The lateral surface area of the cube = 4 - (length of a side)² = 4 - (6 cm)² = 144 cm².

(iv) The length of a diagonal = \(\sqrt{3}\) - length of a side = 6\(\sqrt{3}\) cm.

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