Geometry help, please? ?

Question

A circular cone with a base of radius 1 and height 3 has a cube inscribed in it; thus, one face of the cube is contained in the base of the cone. Determine the side and length of the cube.

Answer 1

Diagonal across face of inscribed cube is equal to diameter of cone's base (2). therefore side of cube is equal to the square root of 2.

Answer 2

Draw this from the side. It will look like an iscosceles triangle standing on its base, with a square standing on the base and touching the other two sides.

The height of the triangle is 3.

The width of the base of the triangle is 2.

Let x be the side of the square.

The small triangle above the square has a height of (3-x).

The small triangle is the same shape as the big triangle, so its base divided by its height is the same as the base of the big triangle divided by the big triangle's height.

Understand so far?

So:

x / (3-x) = 2 / 3

The rest is easy.

Answer 3

If the cube, with side length s, is placed so it is flat against the base of the cone, then at a height of s, its upper face's diagonals (of length s√2) will just touch the cone (Note that the upper face's edges do not touch the cone (except their endpoints)).  This establishes similar right triangles:

The first is due to the apex of the cone and a radius (from its base) => legs 1 and 3.

The second is from the apex of the cone and half of the upper face's diagonal => the corresponding legs are s/√2 and (3-s)

Thus,

(3-s) / (s/√2) = 3 / 1 so

3 - s = 3s/√2

3√2 = s(3 + √2)

3√2 (3 - √2) = s (3 + √2)(3 - √2) = s (9 - 2) = 7s

So s = (9√2 - 6) / 7