**Cuboids :**

It is a three dimensional solid having six rectangular faces.

Example: Bricks, Books etc.,

Let l, b and h be the length, breadth and height of a cuboid respectively. To find the

total surface area, we split the faces into three pairs.

(i) The total area of the front and back faces is

lh + lh = 2lh square units.

(ii) The total area of the side faces is

bh + bh = 2bh square units.

(iii) The total area of the top and bottom faces is

lb + lb = 2lb square units.

The Lateral Surface Area (L.S.A)

= 2( l + b)h square units.

The Total Surface Area (T.S.A)

= 2( lb + bh + lh ) square units.

If the length, breadth and height of a cuboid are l, b and h respectively, then the volume V of the cuboid is given by the formula

V = l ⋅ b ⋅ h cubic units

**Example 1 :**

Find the total surface area of a cuboid whose length, breadth and height are 20 cm, 12 cm and 9 cm respectively.

**Solution :**

Given that l = 20 cm, b = 12 cm, h = 9 cm

T.S.A = 2 (lb + bh + lh)

= 2[(20 ⋅ 12) + (12 ⋅ 9) + (20 ⋅ 9)]

= 2(240 + 108 + 180)

= 2 (528)

= 1056 cm^{2}

**Example 2 :**

Find the L.S.A of a cuboid whose dimensions are given by 3m ⋅ 5m ⋅ 4m

**Solution :**

Given that l = 3 m, b = 5 m, h = 4 m

L.S.A = 2h (l + b)

= 2(4) (3 + 5)

= 8 (8)

= 64 sq. m

Hence the required lateral surface area is 64 sq. m.

**Example 3 :**

Find the volume of a cuboid whose dimensions are given by 11 m, 10 m and 7 m.

**Solution :**

Given that l = 11 m, b = 10 m, h = 7 m

Volume of cuboid = lbh

= 11 ⋅ 10 ⋅ 7

= 770 cu.m.

**Example 4 :**

Two cubes each of volume 216 cm^{3} are joined to form a cuboid as shown in the figure.

Find the T.S.A of the resulting cuboid.

**Solution :**

Let the side of each cube be a. Then a^{3} = 216

a = ∛216 = 6 cm

Now the two cubes of side 6 cm are joined to form a cuboid.

So,

l = 6 + 6 = 12 cm, b = 6 cm, h = 6 cm

Total surface area = 2 (lb + bh + lh)

= 2 [(12 ⋅ 6) + (6 ⋅ 6) + (12 ⋅ 6)]

= 2 [72 + 36 + 72]

= 2 ⋅ 180

= 360 cm^{2}

Hence the required total surface area is 360 cm^{2}

After having gone through the stuff given above, we hope that the students would have understood "Cuboid".

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