(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A solid cone is obtained by connecting every point of a plane region S with a vertex not in the plane of S. Let A denote the area of S, and let h denote the altitude of the cone. Prove that:

(a) The cross-sectional area cut by a plane parallel to the base and at a distance t from the vertex is (t/h)^{2}A, where 0 ≤ t ≤ h, and

(b) the volume of the cone is Ah/3.

3. The attempt at a solutionBefore I attempt this, I'd just like to know whether or not it is reasonable to assume that S is circular. This would make the A equal to pi*(radius of S)^{2}. I think

the limit as t goes to h of (t/h) is 1. Then, the change in radius of the circle is given by (radius of S)(t/h). This means that the area of the circle given the change in t is

pi*(radius of S)^{2}(t/h)^{2}, which is what I wanted to show. (I find this to be the most unconvincing part of my post)

In order to find the volume, I integrate. I have

∫(0≤t≤h) (pi*(radius of S)^{2}(t/h)^{2})dt

(pi*(radius of S)^{2}/h^{2}* (t)^{3}/3, evaluated from 0 to h. So I have pi*(radius of S)^{2}/h^{2}* h^{3}= 1/3 * pi * (radius of S)^{2}* h,

which is also what I wanted.

Look okay?

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# Homework Help: Proving Properties of a Cone

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