Răspuns:
[tex]P=16+4\sqrt{5}~\text{cm}[/tex]
[tex]A=32~\text{cm}^2[/tex]
Explicație pas cu pas:
Ipoteza:
[tex]AB\parallel CD\\BD\perp BC\\BD=4\sqrt{5}~\text{cm}\\\tan{\sphericalangle BCD}=2[/tex]
Aflăm [tex]BC[/tex]:
[tex]\triangle BCD\text{ este dreptunghic }(BD\perp BC) \Rightarrow \tan{\sphericalangle BCD}=\frac{\text{cat. op}}{\text{cat. al.}}\Rightarrow \\\Rightarrow 2=\frac{BD}{BC}\Rightarrow 2=\frac{4\sqrt{5}}{BC}\Rightarrow 2BC=4\sqrt{5}\Rightarrow BC=\frac{4\sqrt{5}}{2}\Rightarrow \\\Rightarrow BC=2\sqrt{5}~\text{cm}[/tex]
Aflăm [tex]CD[/tex] (ipotenuza [tex]\triangle BCD[/tex]):
[tex]CD^2=BD^2+BC^2 \Rightarrow CD=\sqrt{BD^2+BC^2}\\\Rightarrow CD=\sqrt{(4\sqrt{5})^2+(2\sqrt{5})^2}=\sqrt{16\cdot5+4\cdot5}=\sqrt{100}\Rightarrow\\\Rightarrow CD=10\text{ cm}[/tex]
Aflăm [tex]\sin{\sphericalangle BCD}[/tex]:
[tex]\sin{\sphericalangle BCD}=\frac{\text{cat. op}}{\text{ip.}}\\\Rightarrow \sin{\sphericalangle BCD}=\frac{BD}{CD}=\frac{4\sqrt{5}}{10}=\frac{2\sqrt{5}}{5}[/tex]
Creăm segmentul [tex]BF[/tex] care este înălțimea trapezului (și a [tex]\triangle BCD[/tex]):
[tex]\text{Fie $\{F\}$ un punct din dreapta $CD$ astfel \^{i}nc\^{a}t $BF \perp CD$.}\\[/tex]
Aflăm [tex]BF[/tex] în [tex]\triangle BCF[/tex]:
[tex]\sin{\sphericalangle BCF}=\sin{\sphericalangle BCD}~(\sphericalangle BCF=\sphericalangle BCD)\\\\\sin{\sphericalangle BCF}=\frac{\text{cat. op}}{\text{ip.}}\\\Rightarrow \frac{2\sqrt{5}}{5}=\frac{BF}{BC} \Rightarrow \frac{2\sqrt{5}}{5}=\frac{BF}{2\sqrt{5}} \Rightarrow (2\sqrt{5})\cdot (2\sqrt{5})=5BF \Rightarrow\\\Rightarrow BF=\frac{20}{5}=4~\text{cm}[/tex]
Aflăm [tex]CF[/tex]:
[tex]\tan{\sphericalangle BCF}=\frac{BF}{CF}\Rightarrow \\\Rightarrow 2=\frac{4}{CF}\Rightarrow 2CF=4\Rightarrow\\\Rightarrow CF=2~\text{cm}[/tex]
Aflăm [tex]AB[/tex]:
[tex]AB\parallel CD \Rightarrow CD=AB+(CD-AB)\Rightarrow\\\Rightarrow 10=AB+2CF\Rightarrow 10=AB+2\cdot2\Rightarrow\\\Rightarrow AB=6~\text{cm}[/tex]
Aflăm perimetrul și aria trapezului:
[tex]P_\text{trapez is.}=AB+CD+2BC\\\Rightarrow P_{ABCD}=6+10+2\cdot2\sqrt{5}=\\=16+4\sqrt{5}~\text{cm}[/tex]
[tex]A_\text{trapez}=\frac{\text{baza}_\text{mic\u{a}}+\text{baza}_\text{mare}}{2}\cdot\text{\^{i}n\u{a}l\cb{t}imea}\\\Rightarrow A_{ABCD}=\frac{AB+CD}{2}\cdot BF\Rightarrow\\\Rightarrow A_{ABCD}=\frac{6+10}{2}\cdot4=8\cdot4=32~\text{cm}^2[/tex]
