How can I prove the space diagonal of a cuboid using the sum of squares?

[MathematicalPhysicist]

im trying to proove that d_abc=(a^2+b^2+c^2)^0.5
that the space diagonal of a cuboid equals the square root of the sum of the squares of the side lengths.
here is the picture of cuboid and the diagonals:
http://mathworld.wolfram.com/PerfectCuboid.html

now here's what i did:
{y}={the part of the space diagonal from the face diagonal d_bc till it touches c}
{z}={the rest of the sapce diagonal}
y^2=a^2+x^2
z^2=(c-x)^2+b^2
z+y=d_abc=(a^2+x^2)^0.5+[(c-x)^2+b^2]^0.5
from here I am stuck, can someone help me understand what should i do.

p.s- it's not hw.

 

[jcsd]

From pythagorus's theorum, we know that:

{d_{ab}}^{ 2} = a^2 + b^2

Also from pythagorus's theroum we known that:

{d_{abc}}^2 = {d_{ab}}^2 + c^2

simply substitue the first equation into the second and find the root and you get:

d_{abc} = \sqrt{a^2 + b^2 + c^2

 

[jcsd]

PS. the above holds true for any cuboid, a perfect cuboid isonly a special cae where all the edges and diagonals are intergers.