Question 310726
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Since the slant height is 12 and the radius of the base is 8, the height is 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{144\ -\ 64}\ =\ \sqrt{80}\ =\ 4\sqrt{5}]


Because the slant height (a generator of the cone), the radius of the base, and the height form a right triangle.


The plane that intersects the cone creates two solids, one is a cone that is half the height of the original cone, and therefore, by similar triangles is half the slant height and half the radius of the original cone.  The other solid is a frustum of a cone.


The volume of the frustum, which is the volume the problem wants us to compute, is the difference between the volume of the original cone and the smaller, half-height cone that was cut off.


The volume of a cone is one-third the volume of a cylinder with the same base radius and same height, so:


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


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_{large}\ =\ \frac{\pi (8)^2(4\sqrt{5})}{3}]


And


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_{small}\ =\ \frac{\pi (4)^2(2\sqrt{5})}{3}]


Finally:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_{large}\ -\ V_{small}\ =\ \frac{256\pi\sqrt{5}}{3}\ -\ \frac{32\pi\sqrt{5}}{3}\ =\ \frac{224\pi\sqrt{5}}{3}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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