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.

(N/A) Area of the valley $= 7280 \, km^2 = 7280 \times (1000 \, m)^2 = 7.28 \times 10^9 \, m^2$.
Rainfall depth $= 10 \, cm = 0.1 \, m$.
Total volume of rainfall $= \text{Area} \times \text{Depth} = 7.28 \times 10^9 \, m^2 \times 0.1 \, m = 7.28 \times 10^8 \, m^3$.
Volume of one river $= \text{length} \times \text{width} \times \text{depth} = 1072 \, km \times 75 \, m \times 3 \, m = 1072000 \, m \times 75 \, m \times 3 \, m = 2.412 \times 10^8 \, m^3$.
Volume of three such rivers $= 3 \times 2.412 \times 10^8 \, m^3 = 7.236 \times 10^8 \, m^3$.
Since $7.28 \times 10^8 \, m^3 \approx 7.236 \times 10^8 \, m^3$,the total rainfall is approximately equivalent to the volume of three such rivers.