Question

hei :) imi explicati si mie cum se face problema?

Answer 1

   
[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]