[tex]\displaystyle \\
a) \\
P_{\Delta BCD} = 3\times CD=3\times4=\boxed{12 ~cm}\\\\
b) \\
\ \textless \ CAD=90^o~si~CD=4~cm\\
\Longrightarrow ~\Delta ACD~\texttt{este triunghi dreptunghic isoscel cu baza} = 4~cm. \\ \\
\Longrightarrow~AC=AD=\frac{CD}{\sqrt{2}}=\frac{4}{ \sqrt{2}} =\frac{4\sqrt{2}}{2} =2\sqrt{2}~cm\\\\
Aria~\Delta ABC=\frac{AC \times AD}{2}=\frac{2\sqrt{2}\times2\sqrt{2}}{2}=\frac{8}{2}=\boxed{4~cm^2}\\
\texttt{Aria laterala a piramidei } = 3 \times Aria~ \Delta ABC = 4 \times 3=\boxed{12~cm^2}[/tex]
Vezi imaginea pe care am atasat-o apoi citeste punctul c).
[tex]\displaystyle c)\\
\texttt{In triunghiul BCD ducem perpendiculara }BM \perp CD,~M\in CD.\\
\texttt{BM este inaltime si mediana in }\Delta BCD.\\\\
BM=\frac{CD \sqrt{3}}{2}=\frac{4 \sqrt{3} }{2}=\boxed{2\sqrt{3}~cm}\\\\
\texttt{NM este egal cu o treime din BM}\\\\
NM=\frac{1}{3}\times 2\sqrt{3}=\boxed{\frac{2\sqrt{3}}{3} ~cm}\\\\
\texttt{Calculam AM din triunghiul dreptung ACM in care:}\\\\
CM=\frac{CD}{2}=\frac{4}{2}=\boxed{2~cm} = cateta\\
AC=\boxed{2 \sqrt{2}~cm} = ipotenuza [/tex]
[tex]\displaystyle\\
AM=\sqrt{AC^2-CM^2}=\\\\
=\sqrt{(2\sqrt{2})^2-2^2}=\sqrt{(8-4}=\sqrt{(4}=\boxed{2~cm}\\\\
\texttt{Pe AN o aflam di triunghiul dreptunghic AMN in care:}\\\\
MN=\boxed{\frac{2\sqrt{3}}{3}~cm }=cateta\\\\
AM=\boxed{2~cm }=ipotenuza\\\\
AN=\sqrt{AM^2 - MN^2}=\sqrt{2^2-\Big(\frac{2\sqrt{3}}{3}\Big)^2}=\\\\
=\sqrt{4-\frac{12}{9}}=\sqrt{4-\frac{4}{3}}=\sqrt{\frac{12-4}{3}}=\sqrt{\frac{8}{3}}=\frac{2\sqrt2 }{\sqrt3}=\boxed{\frac{2\sqrt6 }{3}~cm}[/tex]
[tex]\displaystyle \\
V_{ABCD}=\frac{\texttt{Aria bazei ori Inaltimea}}{3}\\\\
=\frac{\frac{CD\times BM}{2}\times AN}{3}=\frac{\frac{4\times\frac{4 \sqrt{3}}{2}}{2}\times \frac{2\sqrt{6}}{3}}{3}=\frac{\frac{16\sqrt{3}}{4}\times \frac{2\sqrt{6}}{3}}{3}=\\\\
=\frac{4\sqrt{3}\times \frac{2\sqrt{6}}{3}}{3}=\frac{8\sqrt{18}}{9}= \frac{8\times 3\sqrt{2}}{9}=\boxed{\frac{8\sqrt{2}}{3}} \approx \frac{8 \times 1,41}{3} \approx \boxed{3,76~cm^3} \ \\ \\
1~cm^3 = 1~ml \\
\boxed{3,76~cm^3 = 3,76~ml \ \textless \ 4~ml} [/tex]