Formulas given below can be used to find volume, lateral surface area and total surface of a cuboid.

**Volume = l x w x h** cubic units

**Lateral Surface Area = 2h(l + w)** sq. units

**Surface Area = 2(lw + wh + hl)** sq.units

**Problem 1 :**

Find the total surface area and the lateral surface area of a cuboid whose length is 20 cm, width is 15 cm and height is 8 cm.

**Solution :**

Total surface area is

= 2(lw + wh + hl)

Substitute.

= 2[20(15) + 15(8) + 8(20)]

= 2 [300 + 120 + 160]

= 2(580)

= 1160 cm^{2}

Lateral surface area is

= 2h(l + w)

Substitute.

= 2(8)(20 + 15)

= 16(35)

= 560 cm^{2}

**Problem 2 :**

The dimensions of a cuboidal box are 6 m × 400 cm × 1.5 m. Find the cost of painting its entire outer surface at the rate of $22 per square meter.

**Solution :**

From the given information,

length (l) = 6 m

width (w) = 400 cm = 400/100 m = 4 m

height = 1.5 m

Outer surface area has six sides.

Then, the required area is

= 2(lw + wh + hl)

Substitute.

= 2[6(4) + 4(1.5) + 1.5(6)]

= 2[24 + 6 + 9]

= 2[39]

= 78 m^{2}

Cost of painting the surface is $22 per m^{2}.

So, the required cost is

= 78(22)

= $1716

**Problem 3 :**

The dimensions of a hall is 10 m × 9 m × 8 m. Find the cost of white washing the walls and ceiling at the rate of $8.50 per square meter.

**Solution :**

From the given information,

length (l) = 10 m

width (w) = 9 m

height = 8 m

Area of white washing is

= 2h(l + w) + lw

Substitute.

= 2(8)(10 + 9) + 10(9)

= 16(19) + 90

= 304 + 90

= 394 m^{2}

Cost of white washing is $8.50 per m^{2}.

So, the required cost is

= 394(8.50)

= $3349

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