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

#### Solution

Area of the valley = 7280 km^{2}

If there was a rainfall of 10 cm in the valley then amount of rainfall in the valley = Area of the valley × 10 cm

Amount of rainfall in the valley = 7280 km^{2} × 10 cm

`=7280×(1000m)^2×10/100m`

`=7280×10^5m^3`

`=7.28×10^8m3`

Length of each river, *l* = 1072 km = 1072 × 1000 m = 1072000 m

Breadth of each river, *b* = 75 m

Depth of each river, *h* = 3 m

Volume of each river = *l* × *b* × *h*

= 1072000 × 75 × 3 m^{3}

= 2.412 × 10^{8 }m^{3}

Volume of three such rivers = 3 × Volume of each river

= 3 × 2.412 × 10^{8} m^{3}^{ }

= 7.236 × 10^{8 }m^{3}

Thus, the total rainfall is approximately same as the volume of the three river