[tex]\displaystyle\\
\texttt{Problema 1), ~Fig. 1}\\\\
AC = BC \sin 30^o = BC \times \frac{1}{2}= \boxed{\frac{BC}{2}}\\\\
AB = BC\cos 30^o = BC\times \frac{\sqrt{3}}{2} =\boxed{\frac{BC\sqrt{3}}{2}}\\\\
P = AB+BC+AC= \frac{BC\sqrt{3}}{2}+BC+\frac{BC}{2}=\\\\
=BC\left(\frac{\sqrt{3}}{2}+1+\frac{1}{2}\right)=BC\left(\frac{\sqrt{3}}{2}+ \frac{2}{2} +\frac{1}{2}\right)=\\\\
=\boxed{BC\times \frac{3+\sqrt{3}}{2}}
[/tex]
[tex]\displaystyle\\
P=120~cm\\\\
BC\times \frac{3+\sqrt{3}}{2}=120~cm\\\\
BC= \frac{~~~~120~~~~}{\dfrac{3+\sqrt{3}}{2}} =120\times \frac{2}{3+\sqrt{3}} = \\\\
=120\times \frac{2(3-\sqrt{3})}{9-3}= \frac{120\times2(3-\sqrt{3})}{6}=\\\\=20\times2(3-\sqrt{3})=\boxed{40(3-\sqrt{3})~cm}\\\\
AB=\frac{BC\sqrt{3}}{2}=\frac{40(3-\sqrt{3})\times\sqrt{3}}{2}=\\\\
=20(3\sqrt{3}-3)=\boxed{60(\sqrt{3}-1)~cm}\\\\
AC = \frac{40(3-\sqrt{3})}{2}=\boxed{20(3-\sqrt{3})}[/tex]
[tex]\displaystyle\\
\texttt{Problema 2), ~Fig. 2}\\\\
\frac{AC}{BD} = \frac{3}{4} \\\\
\frac{AO}{BO} = \frac{ \frac{AC}{2}}{ \frac{BD}{2}} =\frac{AC}{2}\times \frac{2}{BD} = \frac{AC}{BD} = \frac{3}{4} \\\\
\Longrightarrow \text{In }~\Delta AOB \text{reportul catetelor = 3/4.}\\\\
AO= 3k~~~si~~~BO = 4k\\\\
AB= \sqrt{AO^2+BO^2}= \sqrt{(3k)^2+(4k)^2}= \sqrt{9k^2+16k^2}=\\\\
=\sqrt{9k^2+16k^2}=\sqrt{25k^2}=5k
[/tex]
[tex]\displaystyle\\
P=a\\\\
\Longrightarrow ~AB = \frac{P}{4} = \frac{a}{4} \\\\
5k \cdots \cdots \cdots \frac{a}{4}~~~(AB)\\\\
3k\cdots \cdots \cdots AO\\\\
4k\cdots \cdots \cdots BO\\\\
AO = \frac{3k}{5k}\times \frac{a}{4} =\boxed{\frac{3a}{20}} \\\\
BO = \frac{4k}{5k}\times \frac{a}{4} =\boxed{\frac{4a}{20}} \\\\
\text{Calculam diagonalele:}\\\\
AC = 2AO = 2\times\frac{3a}{20}=\boxed{\frac{3a}{10}}\\\\
BD=2BO=2\times\frac{4a}{20}=\boxed{\frac{2a}{5}}[/tex]
