Question

The area of base of the rectangle is \[6500\text{ c}{{\text{m}}^{2}}\] and the volume of water contained in it is 2.6 cubic metre. Find the depth of water?

Answer 1

Hint: The volume of any 3D figure is equal to the product of area of base of 3D figure and height of 3D figure. So, the ratio of volume of 3D figure and area of base of 3D figure is equal to height of 3D figure. In the same way, the ratio of volume of 3D figure and height of 3D figure is equal to area of base of 3D figure.

Complete step-by-step solution -
Before solving the question, we should mention the given data in the question. In the question, it is given the area of base of the rectangle is \[6500\text{ c}{{\text{m}}^{2}}\] and the volume of water contained in it is 2.6 cubic metre. Let us assume a tank as shown in below figure.
 

Area of base of the rectangle = \[6500\text{ c}{{\text{m}}^{2}}\]
We know that 1 m =100 cm.
So, area of base of the rectangle = \[6500\text{ }{{\left( \dfrac{1}{100} \right)}^{2}}{{m}^{2}}=6500\left( \dfrac{1}{10000} \right){{m}^{2}}=\left( \dfrac{65}{100} \right){{m}^{2}}\]
\[\text{Volume of Tank = Area of base of Tank x Depth of Tank}\]
Let us assume the depth of Tank is equal to “d”.
\[\text{Volume of Tank = Area of base of Tank x d}\]
\[\Rightarrow 2.6\text{ }{{\text{m}}^{3}}=\left( \dfrac{65}{100} \right){{m}^{2}}\text{ x d}\]
By cross multiplication,
d = \[\dfrac{2.6\text{ }{{\text{m}}^{3}}}{\left( \dfrac{65}{100} \right)\text{ }{{\text{m}}^{2}}}\]
d = \[\dfrac{2.6\text{ }}{\left( \dfrac{65}{100} \right)\text{ }}m\]
d = \[\text{(2}\text{.6)}\left( \dfrac{100}{65} \right)m\]
d = \[\left( \dfrac{26}{10} \right)\left( \dfrac{100}{65} \right)m\]
d = \[\left( \dfrac{2600}{650} \right)m\]
d = \[\dfrac{260}{65}m\]
d= \[\text{4 m}\]
Hence, the depth of water is equal to 4 m.

Note: In this problem, the base of the rectangle is given in \[c{{m}^{2}}\] units and volume in \[{{m}^{3}}\] units. Before solving the problem, we have to convert both the base of the rectangle and volume of the tank into the same units. These small mistakes give wrong answers. So, before solving this type of problem we should ensure that all the given data must be converted into identical units.