Just because I am a smarty pants I'll answer the question from above:

A right circular cone has a base of radius 1 and a height of 3. A cube is inscribed in the cone so that the one face of the cube is contained in the base of the cone. What is the length of an edge of the cube?

To make this a little more informative I'll try to give an insight into the thought processes involved in answering the question.

First we draw a picture (always the first step!)

This is hard on E2 so just imagine a cone upside down on x-y-z axes cutting at 1 on the x and y axes and 3 on the z.

Now remember something you learnt in multivariate calculus or just think about it. The equation of this cone is z = 3(1 - sqrt(x^2+y^2)).

Now the sticking point (where the cube hits the cone and so can't get bigger) is going to be on the top face of the cube at the corners. Think about this for a while and make sure you agree (draw some pictures).

So now if we let the side length be 2s (2 times s just to make the maths easier) by rearranging our cone equation, the equation of the level surfaces on z = 2s

2s = 3(1 - sqrt(x^2+y^2))
x^2+y^2 = {(3-2s)/3}^2

so a circle with radius (3-2s)/3

but we want to fit a square inside this circle. By Pythagoras' Theorem the diagonal is sqrt(s^2+s^2) = sqrt(2)*s

so these must be equal (3-2s)/3 = sqrt(2)*s
3 = 2s + 3*sqrt(2)*s
s = 3/(2+3*sqrt(2))

now this is almost our answer but we need to express in the same terms as the answer above: so rationalise the denominator

s = 3(2-3*sqrt(2))/(2+3*sqrt(2))(2-3*sqrt(2))
s = (6-9*sqrt(2))/(4-9*2)
s = (9*sqrt(2)-6)/14 (multiply top and bottom by -1)

and we said side length = 2s = (9*sqrt(2)-6)/7 as required.

So there you are: these questions definitely aren't impossible but I'm sure most of them require more knowledge than this one.