<p>There is a cube with a volume of 8 inscribed in a sphere, and each vertex is tangent to the sphere. What is the diameter of the sphere?</p>
<p>I got 3 srt 2 which is right... but I used pythagorus therum or watever his name is thre times (so 3(2^2) = sqrt 12 then simplified). Do you need to do that with 3d objects?</p>
<p>Sounds like you answered your own question. Of course you need to do that. That's how you solved the problem (and is the best (only?) way to solve the problem. Uh...</p>
<p>Isn't square root of 12 2 srt 3?</p>
<p>That's what I got. Basically just draw the cube and then use the super pythagorean theorem (for 3d shapes) and then use the diagonal of one side (2 srt 2)squared + the base side (2 squared) = x squared, where x is the diameter of the sphere.</p>
<p>12 = 2 sqrt 3</p>
<p>Yes you need the pythagorean theorem. The diameter of the cube is just the diagonal of the cube, which is the hypotenuse of the right triangle formed with the diagonal of the base and the height of the cube. Diagonal of base = 2 sqrt 2 because of 45-45-90 triangle properties.</p>
<p>Hence
(2 sqrt 2)^2 + (2)^2 = d^2
d^2 = 4*2 + 4
d^2 = 12
d = sqrt 12
d = 2 sqrt 3</p>
<p>Hey Jamesford, would you mind explaining how you knew to setup (2 sqrt 2)^2 + (2)^2 = d^2? I understand that 2 root 2 is a diagonal of a face but after that I am completely lost as to how you setup the equation. Thanks.</p>
<p>EDIT: I just googled and found the formula d = square root of 3s^2 which is the same thing as s root 3, the correct answer.</p>
<p>^That's just the pythagorean Theorem, in which 2 sqrt 2 is one of the sides of the triangle made (It is also the diagonal of a base of the cube), 2 is the other side of the triangle, and d is basically the hypotenuse of the triangle. He set up the equation like this:
a^2 + b^2= c^2, where a=2sqrt2, b=2, and c= diagonal of cube (diameter of circle)</p>
<p>BTW, it should help making a diagram/drawing if you haven't already</p>
<p>A helpful and easy to derive and memorize formula:
in a LxWxH rectangular prizm with the main diagonal D
D^2 = L^2 + W^2 + H^2.</p>
<p>In a cube AxAxA then
D^2 = 3A^2, and
D = A sqrt(3).</p>