The external dimensions of an open concrete tank are 100 cm long, 75cm wide and 50 cm deep. If the concrete is 0.02 m thick, find the volume of concrete used

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You need to determine the internal dimensions of the tank, converting to cm the given thickness of 0.02 m such that:

height = `100 - 2 = 98 cm`

wide = `75 - 2*2 = 71 cm`

deep = `50 - 2*2 = 46 cm`

Hence, you need to evaluate the volume of the tank using external dimensions and then, you need to find the volume of the tank using internal dimensions such that:

`V_1= 100*75*50 = 375000 cm^3`

`V_2 = 98*71*46 = 320068 cm^3`

You need to subtract `V_2` from `V_1` to find the volume of concrete used such that:

`V = V_1 - V_2 => V = 375000 - 320068 = 54932 cm^3`

**Hence, evaluating the volume of concrete used yields `V = 54932 cm^3.` **

External Length of the tank (L1)=100cm, Width(B1)=75cm and Height (H1)=50cm width of the concrete (w)=0.02m=2cm

External volume of the tank=L1*W1*H1=100*75*50=375000 cubic cm

Internal Length (excluding concrete) of the tank = 100-2*2 =100-4=96cm[Since concrete occupy 2cm from both side length wise therefore 2cm is subtracted from both side]

Internal width (excluding concrete) of the tank = 75-2*2 =75-4=71cm [since contrecte occupy 2cm from both side width wise therefore 2cm is subtracted from both side in the width]

Internal Height (excluding concrete) of the tank = 50-2 =50-2=48cm [since tank is open thus width of the concrete will be at the bottom only]

Internal volume of the tank = 96*71*48 cubic cm =327168 cubic cm

Volume of theconcrete = external volume - internal volume

= 375000 - 327168 = 47832 cubic cm

Hence Volume of the concrete used = 47832 cubic cm <--answer

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