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.

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.

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.

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.

Therefore, (i) Volume = l × b × h = 5 × 4 × 3 cm^{3} = 60 cm^{3};

(ii) Surface area = 2(lb + bh + hl) = 2(5 × 4 + 4 × 3 + 3 ×
5) cm^{2}

=
2(20 + 12 + 15) cm^{2}

=
94 cm^{2};

(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 cm^{3} respectively. Find its (i)
height, (ii) surface area, and (iii) lateral surface area.

**Solution:**

Let the height = h.

Then, volume = l × b × h

⟹ 192 cm^{3} = 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) cm^{2}

=
2(48 + 24 + 32) cm^{2}

=
208 cm^{2}

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

= 2(8 + 6) × 4 cm^{2}

= 2(14) × 4 cm^{2}

= 28 × 4 cm^{2}

= 112 cm^{2}

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