[tex]\displaystyle\\
\text{Din desen se vede ca }~BC = AD ~\Longrightarrow~\text{Trapezul este isoscel.}\\\\
NB = \frac{AB-CD}{2}= \frac{102-48}{2}=\frac{54}{2}= \boxed{27~cm}\\\\
AN = 102-27 = \boxed{75~cm} \\\\
\text{In triunghiurile: }\Delta CND \text{ si } \Delta DAN \text{ avem:}\\\\
\sphericalangle DCN = \sphericalangle NDA = 90°\\\\
\sphericalangle CDN = \sphericalangle DNA ~~\text{(alterne interne)}\\\\
\Longrightarrow~~\text{Cazul UU de asemanare}\\\\
\text{Scriem rapoartele de asemanare:}[/tex]
[tex]\displaystyle\\
\frac{DC}{ND} = \frac{ND}{AN}\\\\
ND^2 = AN\times DC = 75 \times 48 = 3600\\\\
ND = \sqrt{3600} = \boxed{60~cm} \\\\
\text{Pe NC il calculam cu T.P. din }\Delta CDN.\\\\
NC= \sqrt{ND^2-CD^2}= \sqrt{60^2-48^2}= \\\\
=\sqrt{3600-2304}=\sqrt{3600-2304}=\sqrt{1296}=\boxed{36~cm}\\\\
\text{Pe BC il calculam cu T.P. din }\Delta NBC.\\\\
BC = \sqrt{NB^2 + NC^2} = \sqrt{27^2 + 36^2} =\\\\
=\sqrt{729 + 1296} =\sqrt{2025} = \boxed{45~cm}[/tex]
