Volume and Surface Area of Cuboid

What is Cuboid?

A cuboid is a solid with six rectangular plane faces, for example, a brick or a matchbox. Each of these is made up of six plane faces which are rectangular. Remember that since a square is a special case of a rectangle, a cuboid may have square faces too.

The figure below shows two cuboids.

$Cuboid Solid Figures$

Consider the cuboid on the left. It has

1. Six rectangular faces, namely ABCD, EFGH, ABGF, CDEH, ADEF and BGHC. Its opposite faces are congruent.

2. Twelve edges, namely AB,BC, CD, DA, FG, HE, EF, AF, BG, CH and DE. The edges AB, CD, FG, EH are equal; the edges BC, AD, GH, EF are equal; the edges AF, BG, CH, DE are equal.

3. Eight Corners (or vertices), namely A, B, C, D, E, F, G and H.

4. Three dimensions: Length (l) = FE, breadth (b) = FG and height (h) = AF.

5. Four diagonals, namely AH, FC, BE and GD which are all equal. These are line segments joining opposite corners (not on the same face).

Note: The dimensions of a cuboid are a cm × b cm × c cm means the length = a cm, breadth = b cm and height = c cm.

Volume of a Cuboid (V) = l × b × h

Total surface Are of a Cuboid (S) = 2(lb + bh +hl)

Diagonal a Cuboid (d) = $\sqrt{\mathrm{l^{2} + b^{2} + h^{2}}}$

Where l = Length, b = breadth and h = height.

$Volume and Surface Area of Cuboid$

Area of the Four Walls of a Room (Lateral Surface Area of a Cuboid)

Rooms area examples of cuboids.

Are of the four walls of a room = sum of the four vertical (or lateral) faces

= 2(l + b)h

Where l = Length, b = breadth and h = height.

$Lateral Surface Area of a Cuboid$

Problems on Volume and Surface Area of Cuboid:

1. A cuboid has three mutually perpendicular edges measuring 5 cm, 4 cm and 3 cm. Find (i) its volume, (ii) its surface area, and (iii) the length of the diagonal.

Solution:

Three mutually perpendicular edges are the length, breadth and height.

Length = l = 5 cm, breadth = b = 4 cm, height = h = 3 cm.

$Problems on Volume and Surface Area of Cuboid$

Therefore, (i) Volume = l × b × h = 5 × 4 × 3 cm³ = 60 cm³;

(ii) Surface area = 2(lb + bh + hl) = 2(5 × 4 + 4 × 3 + 3 × 5) cm²

= 2(20 + 12 + 15) cm²

= 94 cm²;

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

= $\sqrt{\mathrm{5^{2} + 4^{2} + 3^{2}}}$ cm

= $\sqrt{50}$ cm

= 5√2 cm.

2. The length, breadth and volume of a cuboid are 8 cm, 6 cm and 192 cm³ respectively. Find its (i) height, (ii) surface area, and (iii) lateral surface area.

Solution:

Let the height = h.

Then, volume = l × b × h

⟹ 192 cm³ = 8 cm × 6 cm × h

⟹ h = $\frac{192 cm^{3}}{8 × 6 cm^{2}}$

⟹ h = $\frac{192 cm^{3}}{48 cm^{2}}$

⟹ h = 4 cm.

Therefore, (i) height = 4 cm.

(ii) Surface area = 2(lb + bh + hl)

= 2(8 × 6 + 6 × 4 + 4 × 8) cm²

= 2(48 + 24 + 32) cm²

= 208 cm²

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

= 2(8 + 6) × 4 cm²

= 2(14) × 4 cm²

= 28 × 4 cm²

= 112 cm²

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