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A closed wooden box is in the form of a cuboid. Its length, breadth and height are $$5 \ m$$, $$3 \ m$$ and $$150 \ cm$$ respectively. Find the total surface area and the cost of painting its entire outer surface at the rate of $$₹25$$ per $$m^2$$. Solution:

Length of the box $$=$$ $$5 \ m$$

Breadth of the box $$=$$ $$3 \ m$$

Height of the box $$=$$ $$150 \ cm$$ $$=$$ $$\frac{150}{100}$$ $$=$$$$1.5 \ m$$

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

$$=$$ $$2 ((5 \times 3) + (3 \times 1.5) + (5 \times 1.5))$$

$$=$$ $$2 (15 + 4.5 + 7.5)$$

$$=$$ $$2 (27)$$

$$=$$ $$54 \ m^2$$

Total surface area of the box is $$54 \ m^2$$.

Cost of painting per $$m^2$$ $$=$$ $$₹25$$

Cost of painting for $$54 \ m^2$$:

$$=$$ $$54 \times 25$$

$$=$$ $$1350$$

Cost of painting the entire outer surface area of the box is $$₹1350$$.

Important!
The units of length, breadth and height should be the same while calculating surface area of the cuboid.