Think of a stack of sheets, in the shape of a cuboid. The dimensions of each sheet in the stack are \(l \times b\) while the total height of the stack is \(h\). How much volume does this stack occupy? The volume will depend on two factors:
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the dimensions of the sheets. The larger the area A of the sheets, the larger will be the volume.
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the height \(h\) of the stack: the larger the height, the larger the volume
Thus, we can say that the volume of the cuboid is
\[V = A \times h = l \times b \times h\]
For example, for a cuboid of dimensions 5 cm × 4 cm × 3 cm, its volume is
\[V = {\rm{ }}60{\rm{ }}c{m^3}\]
For a cube of side \(l\) units, its volume will be
\[V = {l^3}\,{\rm{unit}}{{\rm{s}}^3}\]
Example 1: The dimensions of the base of a cuboid are 5 cm × 3 cm, while its height is 2 cm. Determine the total SA (surface area), lateral SA and volume of the cuboid.
Solution: We have:
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Total SA = \(2\left( {lb + bh + lh} \right) = 62\,{\rm{c}}{{\rm{m}}^2}\)
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Lateral SA = \(2h\left( {l + b} \right) = 32\,{\rm{c}}{{\rm{m}}^2}\)
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Volume = \(lbh = 30\,{\rm{c}}{{\rm{m}}^3}\)
Example 2: The volume of a cuboid is 0.16 liters, or 0.16 L. The base dimensions of the cuboid are 8 cm × 5 cm. What is the total SA of the cuboid?
Solution: Let the height of the cuboid be \(h\) cm. We have:
\[\begin{align}&V = lbh\\ &\Rightarrow \,\,\,0.16\,{\rm{L = 160 c}}{{\rm{m}}^3} = 8\,{\rm{cm}}\, \times \,5\,{\rm{cm}}\, \times \,h\,{\rm{cm}}\\& \Rightarrow \,\,\,h = 4\end{align}\]
The total surface area is
\[\begin{align}&S = 2\left( {lb + bh + lh} \right)\,{\rm{c}}{{\rm{m}}^2}\\&\,\,\,\, = 2\left( {40 + 20 + 32} \right)\,{\rm{c}}{{\rm{m}}^2}\\&\,\,\,\, = 184\,{\rm{c}}{{\rm{m}}^2}\end{align}\]
Example 3: A cube whose volume is 1 L has a total SA of _____ cm2.
Solution: If the length of each side of the cube is \(x\) cm, we have:
\[\begin{align}&{x^3}\,{\rm{c}}{{\rm{m}}^3} = 1\,{\rm{L}} = 1000\,{\rm{c}}{{\rm{m}}^3}\\ &\Rightarrow \,\,\,x = 10\end{align}\]
The total SA of the cube is \(6{l^2}\,{\rm{c}}{{\rm{m}}^2}\) or 600 cm2.
Example 4: The bases of two cuboidal containers A and B have the dimensions 10 cm x 8 cm and 15 cm x 10 cm respectively. Water is filled in A up to a height of 15 cm. Water from A is then poured into B completely. What will be the height of the water in B?
Solution: Let the height of the water in B be \({h_B}\) cm. As water from A is fully transferred to B, we have:
\[\begin{align}&{\rm{Water\,\, volume\, in\, A = Water\,\, volume\, in \,B}}\\& \Rightarrow \,\,\,10 \times 8 \times 15 = 15 \times 10 \times {h_B}\\ &\Rightarrow \,\,\,{h_B} = 8\end{align}\]
The height of the water in B is 8 cm.
Example 5: A cuboidal box has base dimensions 80 cm x 40 cm and has a volume of 160 L. It needs to be painted with a special kind of paint on all the sides except its bottom. The cost of this special paint is INR 6000/m2 of area. Find the cost of this paint job.
Solution: Let the height of the box be \(h\) cm. We have:
\[\begin{align}&{\rm{80}}\,{\rm{cm}}\, \times \,40\,{\rm{cm}}\, \times \,h\,{\rm{cm}} = 160\,{\rm{L}} = 160000\,{\rm{c}}{{\rm{m}}^3}\\ &\Rightarrow \,\,\,h = 50\end{align}\]
Now, we will find the combined area of all the faces of the cuboid excluding the bottom. If we denote this by S, we have:
\[\begin{align}&S = lb + 2h\left( {l + b} \right)\\&\,\,\,\, = 15200\,{\rm{c}}{{\rm{m}}^2}\\&\,\,\,\, = 1.52\,{{\rm{m}}^2}\end{align}\]
Thus, the total cost C of the paint job will be
\[\begin{align}&C = {\rm{INR}}\,6000/{{\rm{m}}^2}\,\, \times \,\,1.52{{\rm{m}}^2}\\&\,\,\,\,\, = {\rm{INR}}\,9120\,\end{align}\]