Volume of cube C is L³ = [tex](4\sqrt{2}) ^{3} = 128\sqrt{2}[/tex]
Step-by-step explanation:
We have , Cube A is inscribed in sphere B, which is inscribed in cube C. Side of cube A have length 4 . Since , cube A is inscribed in sphere B , half of Diagonal length of cube A will be the radius of sphere B . We can find it by Pythagoras Theorem, as
[tex]Diagonal^{2} = length^{2} + length^{2}\\Diagonal^{2} = 4^{2} + 4^{2} \\ Diagonal^{2} = 32 \\[/tex]
Taking square root both sides, [tex]diagonal = 4\sqrt[]{2}[/tex],
Radius of sphere B is half of diagonal i.e. r = [tex]2\sqrt[]{2}[/tex].
Now , length of cube C is twice the radius of sphere B i.e. L = 2r = [tex]4\sqrt[]{2}[/tex].
∴Volume of cube C is L³ = [tex](4\sqrt{2}) ^{3} = 128\sqrt{2}[/tex]