There is a square field of area N square metres. A cylindrical ditch of radius 7 metres and depth 4 metres is dug, and the earth is taken out and spread over the remaining part of the square field, the height of square field goes up by 1.54 metres. What is the value of N?

Given:

There is a square field of area N square metres. A cylindrical ditch of radius 7 metres and depth 4 metres is dug, and the earth is taken out and spread over the remaining part of the square field, the height of the square field goes up by 1.54 metres.
To do:

We have to find the value of $N$.
Solution:
 Volume of the cylindrical ditch $= \pi r^2h$

$= \frac{22}{7}\times7^2\times4$

$= 616\ m^2$

Area of the circular hole on the square field $=\pi r^2$

Area of the remaining field $= (N −\pi r^2)$

$= (N − \frac{22}{7}\times7^2)$

$=N-154$

Volume of the earth dug out $=$ Area of the field formed after digging out $\times$ Height

$616=(N-154)\times1.54$

$(N-154)=\frac{616}{1.54}$

$N= 400+154$ 

$N=554\ m^2$ 

The value of $N$ is $554\ m^2$.