A field is in the form of a rectangle of length $18\ m$ and width $15\ m$. A pit, $7.5\ m$ long, $6\ m$ broad and $0.8\ m$ deep, is dug in a comer of the field and the earth taken out is spread over the remaining area of the field. Find out the extent to which the level of the field has been raised.

Given:

A field is in the form of a rectangle of length $18\ m$ and width $15\ m$. A pit, $7.5\ m$ long, $6\ m$ broad and $0.8\ m$ deep, is dug in a comer of the field and the earth taken out is spread over the remaining area of the field.

To do:

We have to find the height of the field raised.

Solution:

Length of the field $(L) = 18\ m$

Width of the field $(B) = 15\ m$

Length of the pit $(l) = 7.5\ m$

Breadth of the pit $(b) = 6\ m$

Depth of the pit $(h) = 0.8\ m$

Therefore,

Volume of the earth dugout $= lbh$

$= 7.5 \times 6 \times 0.8$

$= 45 \times 0.8$

$= 36\ m^3$

Total area of the field $= L \times B$

$= 18 \times 15$

$= 270\ m^2$

Area of the pit $= lb$

$= 7.5 \times 6$

$= 45\ m^2$

Remaining area of the field excluding pit $= 270 - 45$

$= 225\ m^2$

Let the height of the field raised be $h$.

This implies,

$225 \times h = 36$

$h = \frac{36}{225}$

$h= 0.16\ m$

$h= 16\ cm$

Hence the level of the field raised is $16\ cm$.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

24 Views

Kickstart Your Career

Get certified by completing the course

Get Started