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 97280 \(km^2\), 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.

From the question, it is clear that

Total volume of 3 rivers
= 3 × [(Surface area of a river) × Depth]

Given,

Surface area of a river
= [\( 1072 × \frac{75}{1000} \)] km

And,

Depth = \( \frac{3}{1000} \) km

Now, volume of 3 rivers
= 3 × [1072 × \( \frac{75}{1000} \)] × \( \frac{3}{1000} \)

= 0.72 km³

Now, volume of rainfall in one fortnight (14 days)
= total surface area × total height of rain

\( = \ 9780 \ × \ \frac{10}{100 \ × \ 1000} \)

= 9.7 km³

Volume of rainfall in 1 day
= \( \frac{9.7}{14} = 0.694 \ km^3 \)

For the total rainfall to be approximately equivalent to the addition to the normal water of three rivers, the volume of rainfall has to be equal to volume of 3 rivers.

Since 0.694 km³ is nearly equivalent to 0.72 km³,

We can say that the total rainfall was approximately equivalent to the addition to the normal water of the three rivers.