Question

Find the volume of a regular hexahedron if one of the diagonals of its faces is 8 root 2 inches.

Answer 1

The answer is 114sqrt{6} in³

A regular hexahedron is actually a cube.

Diagonal of a cube D is a hypotenuse of a right triangle which other two legs are face diagonal (f) and length of a side (a):
D² = f² + a²

Face diagonal is a hypotenuse of a right triangle which sides are a and a:
f² = a² + a² =2a²

D² = f² + a²
f² = 2a²

D² = 2a² + a² = 3a²
D = √3a² = √3 * √a² = √3 * a = a√3

Volume of a cube with side a is: V = a³
D = a√3
⇒ a = D/√3

V = a³ = (D/√3)³

We have:
D = 8√2 in


`V= ((D)/( √(3) ) )^(3) =( (8 √(2) )/( √(3) ) )^(3)=( (8 √(2) * √(3) )/( √(3)* √(3)) )^(3) =( (8 √(2*3) )/( √(3*3)) )^(3) =( \frac{8 √(6) }{ \sqrt{3^(2) }} )^(3) =( (8 √(6) )/( 3) )^(3) = \\ \\ = ( 8^(3) *( √(6) )^(3))/(3^(3)) = \frac{512* \sqrt{6 ^(2) }* √(6) }{27} = (512*6* √(6) )/(27) = 114sqrt{6}`