# Simple Geometry problem is stumping me...

by PotentialE
Tags: gasoline tank, geometry, trigonometry
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 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, $\sqrt{r^2- h^2}$. you can calculate the central angle (between the two radii) as $2 arccos((r-h)/r)$. The cross-section area, as seen on your picture, from the end, of those two right triangles is $2(1/2)(r-h)\sqrt{(r^2- h^2}= (r-h)\sqrt{r^2- h^2}$. The cross-section area of the region above the two radii is $$\frac{arccos((r-h)/r)}{\pi} r$$ The total cross-section area, above the liquid, is the sum of those and the cross section area of the liquid is the area of the circle, $\pi r^2$, minus that. Finally, the volume of the liquid is that cross section area time the length of the cylinder.