Water tank

A 288 hectoliter of water was poured into the tank with dimensions 12 m and 6 m bottom and 2 m depth. What part of the volume of the tank water occupied? Calculate the surface of tank wetted with water.

Result

p =  20 %
A =  86.4 m2

Solution:

a=12 m b=6 m c=2 m S=a⋅ b=12⋅ 6=72 m2 V=S⋅ c=72⋅ 2=144 m3 V1=288/10=1445=28.8 m3 p=100⋅ V1/V=100⋅ 28.8/144=20=20% c1=V1/S=28.8/72=25=0.4 ma=12 \ \text{m} \ \\ b=6 \ \text{m} \ \\ c=2 \ \text{m} \ \\ S=a \cdot \ b=12 \cdot \ 6=72 \ \text{m}^2 \ \\ V=S \cdot \ c=72 \cdot \ 2=144 \ \text{m}^3 \ \\ V_{1}=288 / 10=\dfrac{ 144 }{ 5 }=28.8 \ \text{m}^3 \ \\ p=100 \cdot \ V_{1}/V=100 \cdot \ 28.8/144=20=20 \% \ \\ c_{1}=V_{1}/S=28.8/72=\dfrac{ 2 }{ 5 }=0.4 \ \text{m}
A=S+c1⋅ (2⋅ a+2⋅ b)=72+0.4⋅ (2⋅ 12+2⋅ 6)=4325=86.4 m2A=S + c_{1} \cdot \ (2 \cdot \ a+2 \cdot \ b)=72 + 0.4 \cdot \ (2 \cdot \ 12+2 \cdot \ 6)=\dfrac{ 432 }{ 5 }=86.4 \ \text{m}^2

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Jimbob
This question wording make no sense in brainhole.

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