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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

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\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}}}



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