Question 1044466
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Consider a vertical cross-section of your cone.  The 20 cm radius of the base, the 100 cm height, and a generator of the cone form a right triangle.  You do not know the radius of the circular surface of the water, but you do know that the distance from the top of the water to the vertex of the cone must be 75 cm (100 minus 25).  So that 75 cm portion of the height of the cone, the unknown radius, and a portion of the generator of the cone form a right triangle that is similar to the 100 cm by 20 cm triangle described earlier.  Hence, you can solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{75}{100}\ =\ \frac{x}{20}]


for *[tex \Large x] to find the radius of the surface of the water from which the area of the surface of the water follows directly using:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  A\ =\ \pi r^2]


The volume of the portion of the cone that contains water is equal to the total volume of the cone minus the volume of that part of the cone that does not contain water.  Use:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  V\ =\ \frac{1}{3}\pi r^2h]


with the appropriate selections for *[tex \Large r] and *[tex \Large h] to calculate the necessary conical volumes.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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