SOLUTION: A cylindrical gasoline tank, lying horizontally, 0.90 m in diameter and 3 m long is filled to a depth of 0.60 m. How many gallons of gasoline does it contain?

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Question 1161721: A cylindrical gasoline tank, lying horizontally, 0.90 m in diameter and 3 m long is filled to a depth of 0.60 m. How many gallons of gasoline does it contain?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The radius of the circle that is the base of that tank cylinder part is 0.9m%2F2=0.45m .
The cross section of that tank (with the fuel inside) looks like this:
The area of the tank cross section is pi%2A%280.45m%29%5E2=approx0.6362m%5E2
We want to know the area of the cross section of the fuel in the tank.
We could calculate it as the area of the isosceles triangle between the blue and green lines plus
the area of the circle sector below the green lines and above the red tank bottom.
We know that the fuel level is 0.6m-0.45m=0.15m
From that we can find the measure of the acute angle A , because
sin%28A%29=0.15m%2F0.45m=1%2F3 ,
and the calculator tells me that A=19.47%5Eo (approximately)
That allow me to calculate the other angle of the triangle as
180%5Eo-19.47%5Eo-19.47%5Eo=141.06%5Eo
and the angle of the circle sector below the green line as
360%5Eo-141.06%5Eo=218.94%5Eo .
The area of the triangle can be calculated as
%281%2F2%29%2A%280.45m%29%2A%280.45m%29%2Asin%28141.06%5Eo%29=approximately0.5%2A0.2025%2A0.6285=approximately0.0636m%5E2
The area of the circle sector below the green lines is
%28218.94%5Eo%2F360%5Eo%29%2A0.6362m%5E2=0.3869m%5E2 .
The total area of the fuel cross section is
0.3869m%5E2%2B0.0636m%5E2=0.4505m%5E2 .
Multiplying that area times the 3m length of the tank, we get the volume in cubuc meters:
V=%283m%29%2A%280.4505m%5E2%29=1.3515m%5E3
The computer tells me that 1m%5E3 is 264.172gallons , so
V=%281.3515m%5E3%29%2A%28264.172gal%2Fm%5E3%29=highlight%28357gal%29 . (reasonably rounded).