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Simple Geometry problem is stumping me... 
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#1
Apr412, 09:00 PM

P: 57

1. The problem statement, all variables and given/known data
A cylindrical tank of gas with a radius is viewed from the side (so that it looks like a circle, not a rectangle) and has a radius of 1m. Through a hole in the top, a vertical rod is lowered to touch the bottom of the tank. When the rod is removed, the gasoline level in the tank can be read from the mark on the rod. Where on the rod would the mark be if the tank was 3/4 full? 2. Relevant equations A = ∏r^{2} 3. The attempt at a solution I attached a picture (in paint) of my attempts, which mainly consist of some trig, but I can't seem to put it all together 


#2
Apr412, 09:21 PM

P: 61

The rod is lowered to touch the bottom. It is vertical, not horizontal. And are you sure no other quantities are given? 


#3
Apr512, 01:31 PM

P: 41

Have you covered the area of a sector of a circle?



#4
Apr512, 02:00 PM

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Simple Geometry problem is stumping me...
If the cylinder has radius R and the stick reads that height of liquid is h, Then drawing radii from the center of the tank to the points on the tank where the top of the liquid is gives two right triangles with hypotenuse r and one leg of length r h. The other leg has length, from the Pythagorean theorem, [itex]\sqrt{r^2 h^2}[/itex]. you can calculate the central angle (between the two radii) as [itex]2 arccos((rh)/r)[/itex]. The crosssection area, as seen on your picture, from the end, of those two right triangles is [itex]2(1/2)(rh)\sqrt{(r^2 h^2}= (rh)\sqrt{r^2 h^2}[/itex]. The crosssection area of the region above the two radii is
[tex]\frac{arccos((rh)/r)}{\pi} r[/tex] The total crosssection area, above the liquid, is the sum of those and the cross section area of the liquid is the area of the circle, [itex]\pi r^2[/itex], minus that. Finally, the volume of the liquid is that cross section area time the length of the cylinder. 


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