Question

A rectangular tank measuring 5m X 4.5m X 2.1m is dug in the centre of the field measuring 13.5m by 2.5m. The earth dug out is evenly spread over the remaining portion of the field. How m

Answer 1

$\blue{Question}$

A rectangular tank measuring 5m X 4.5m X 2.1m is dug in the centre of the field measuring 13.5m by 2.5m. The earth dug out is evenly spread over the remaining portion of the field. How much land is raised?

$\huge\purple{\underline{\underline{\pink{Ans}\red{wer:-}}}}$

$\sf{Field \ is \ raises \ by \ 4.2 \ m}$

$\sf\orange{Given:}$

$\sf{For \ rectangular \ tank,}$

$\sf{\implies{Length (l)=5 \ m}}$

$\sf{\implies{Breadth (b)=4.5 \ m}}$

$\sf{\implies{Height (h)=2.1 \ m}}$

$\sf{For \ rectangular \ field,}$

$\sf{\implies{Length (l)=13.5 \ m}}$

$\sf{\implies{Breadth (b)=2.5 \ m}}$

$\sf\pink{To \ find:}$

$\sf{Land \ raised.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Volume \ of \ cuboid=lbh}$

$\sf{...formula}$

$\sf{\therefore{Volume \ of \ tank=5\times4.5\times2.1}}$

$\sf{\therefore{Volume \ of \ tank=47.25 \ m^{3}}}$

$\sf{\therefore{Earth \ dug \ out=47.25 \ m^{3}}}$

______________________________________

$\sf{Area \ of \ rectangle=lb}$

$\sf{... formula}$

$\sf{\therefore{Area \ of \ field=13.5\times2.5}}$

$\sf{\therefore{Area \ of \ field=33.75 \ m^{2}}}$

$\sf{Area \ of \ tank=5\times4.5}$

$\sf{Area \ of \ tank=22.5 \ m^{2}}$

$\sf{Area \ of \ field \ without \ tank}$

$\sf{=33.75-22.5}$

$\sf{=11.25 \ m^{2}}$

____________________________________

$\sf{Volume \ of \ field \ after \ spread \ of \ earth}$

$\sf{becomes \ 47.25 \ m^{2}}$

$\sf{Volume \ of \ cuboid=lbh}$

$\sf{47.25=11.25\times \ h}$

$\sf{\therefore{h=\frac{47.25}{11.25}}}$

$\sf{\therefore{h=4.2 \ m}}$

$\sf\purple{\tt{\therefore{Field \ is \ raises \ by \ 4.2 \ m}}}$