# Ex.13.5 Q4 Surface Areas and Volumes Solution - NCERT Maths Class 10

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## Question

In one fortnight of a given month, there was a rainfall of $$10 \,\rm{cm}$$ in a river valley. If the area of the valley is $$7280\, \rm{km^2},$$ show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each $$1072\,\rm{km}$$ long, $$75\,\rm{m}$$ wide and $$3\, \rm{m}$$ deep.

## Text Solution

What is known?

In one fortnight there was a rainfall of $$10\,\rm{ cm}$$ in a river valley of area $$7280 \,\rm{km^2}$$

What is unknown?

The total rainfall was approximately equivalent to the addition to the normal water of three rivers each $$1072\,\rm{ km}$$ long, $$75 \rm{m}$$ wide and $$3 \rm{m}$$ deep.

Reasoning:

Since there was a rainfall of $$10\,\rm{cm}$$ in a river valley of area $$7280\,\rm{ km^2}$$ the volume of the rainfall will be calculated by

Volume of the rainfall $$=$$ Area of the river valley $$\times$$ height of rainfall in the river valley

Since the dimensions of three rivers are known, we can calculate the volume by

Volume of the river \begin{align} = \rm{length \times breadth \times height}\end{align}

Steps:

Area of the Valley

\begin{align}A &= 7280k{m^2} \\&= 7280 \times 1000000{m^2} \\&= 7.28 \times {10^9}{m^2}\end{align}

Height of rainfall in a fortnight

\begin{align}h& = 10cm\\&= \frac{{10}}{{100}}m \\&= 0.1m\end{align}

Volume of the rainfall $$=$$ Area of the river valley $$\times$$ height of rainfall in the river valley

\begin{align}&= 7.28 \times {10^9}{m^2} \times 0.1m\\& = 7.28 \times {10^8}{m^3}\end{align}

Length of river, \begin{align}l = 1072km = 1072 \times 1000m = 1.072 \times {10^6}m\end{align}

Width of river,$$b = 75m$$

Depth of river,$$h = 3m$$

Volume of $$3$$ rivers $$= 3$$ $$\times$$ volume of $$1$$ river

\begin{align}&= 3lbh\\&= 3 \times 1.072 \times 106m \times 75m \times 3m\\&= 723.6 \times {10^6}{m^3}\\&= 7.236 \times {10^8}{m^3}\end{align}

Since$$7.236 \times {10^8}{m^3}$$ is approximately equivalent to $$7.28 \times {10^8}{m^3}$$

Therefore, we can say that total rainfall in the valley was approximately equivalent to the addition of normal water of three rivers each $$1072 \,\rm{km}$$ long, $$75 \,\rm{m}$$ wide and $$3\,\rm{m}$$ deep.

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