Fie d- lungimea diagonalei paralelipipedului și a, b, c dimensiunile lui.
[tex]\it d^2=a^2+b^2+c^2 =(10\sqrt3)^2\Rightarrow a^2+b^2+c^2=300|_{\cdot2}\Rightarrow \\ \\ \Rightarrow 2a^2+2b^2+2c^2=600\ \ \ \ \ (1)\\ \\ \mathcal{A}_{t}=2ab+2bc+2ca = 600\ \ \ \ (2)[/tex]
[tex]\it (1),(2)\Rightarrow 2a^2+2b^2+2c^2=2ab+2bc+2ca \Rightarrow\\ \\\Rightarrow a^2+a^2+b^2+b^2+c^2+c^2-2ab-2bc-2ca=0 \Rightarrow\\ \\\Rightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0 \Rightarrow\\ \\ \Rightarrow (a-b)^2+(b-c)^2+(c-a)^2=0 \Rightarrow\begin{cases}\it (a-b)^2=0\Rightarrow a-b=0 \Rightarrow a=b\\ \\ \it (b-c)^2=0\Rightarrow b=c=0 \Rightarrow b=c\\ \\ \it (c-a)^2=0\Rightarrow c-a=0 \Rightarrow c=a\end{cases}[/tex]
[tex]\it Deci,\ a = b = c \Rightarrow paralelipipedul\ este\ cub.\ \ Fie\ l-\ muchia\ cubului.\\ \\ Diagonala\ cubului\ este\ d=\ell\sqrt3 = 10\sqrt3\Rightarrow\ell=10cm\\ \\ \mathcal{V}=\ell^3=10^3=1000\ cm^3[/tex]
