[tex]Notez~dimensiunile~paralelipipedului~cu~a,b~respectiv~c. \\ \\ (AB=CD=A'B'=C'D'=a~;~AD=BC=A'D'=B'C'=b; \\ \\ AA'=BB'=CC'=DD'=c).~Evident:~a,b,c\ \textgreater \ 0. \\ \\ Din~teorema~lui~Pitagora~aplicata~in~triunghiurile~dreptunghice \\ \\ \Delta~ABC,~\Delta~ABB'~si~\Delta ADD',~avem: \\ \\ AB^2+BC^2=AC^2 \Leftrightarrow a^2+b^2=40^2. \\ \\ AB^2+BB'^2=AB'^2 \Leftrightarrow a^2+c^2=(2 \sqrt{377})^2. \\ \\ AD^2+DD'^2=AD'^2 \Leftrightarrow b^2+c^2=30^2.[/tex]
[tex]Deci: \\ \\ a^2+b^2=1600 \\ \\ a^2+c^2=1508 \\ \\ b^2+c^2=900 \\ \\ Prin~insumarea~acestor~trei~relatii,~obtinem: \\ \\ 2(a^2+b^2+c^2)=4008 \Rightarrow a^2+b^2+c^2=2004. \\ \\ a^2+b^2+c^2=2004 \Rightarrow a^2=2004-(b^2+c^2) ~;~b^2=2004-(a^2+c^2), \\ \\ si,~respectiv~c^2=2004-(a^2+b^2).[/tex]
[tex]Asadar:~ \\ \\ a^2=2004-900 \Leftrightarrow a^2=1104 \Rightarrow \boxed{a=4 \sqrt{69}~(cm)}. \\ \\ b^2=2004-1508 \Leftrightarrow b^2=496 \Rightarrow \boxed{b=4 \sqrt{31}~(cm)}. \\ \\ c^2=2004-1600 \Leftrightarrow c^2=404 \Rightarrow \boxed{c=2 \sqrt{101}~(cm)}. \\ \\ Dimensiunile~paralelipipedului~dreptunghic~sunt~4 \sqrt{69}~cm; \\ \\ 4 \sqrt{31}~cm~si~2 \sqrt{101}~cm.[/tex]
