[tex]Diagonala\;paralelopipedului\;=\sqrt{L^2+l^2+h^2}\;\;la\;\;cub\;\;L=l=h\;\;\\deci\;\;\\
-\;D_{cub}=\sqrt{3\cdot\;1^2}=1m\cdot\sqrt3\;;\\
-\;A_t=6\cdot{l}^2=6\cdir1=6m^2\;\\
-\;R_{sfera}=1m:2=0,5m \\
\Rightarrow\;\;V_{sfera}=\frac{4}{3}\cdot\pi\cdot0,5^3=
=\frac{4\cdot0,125\cdot\pi}{3}=\frac{\pi}{6}m^3[/tex]
