Rezolvarea 1:
Consider ca cerinta: "calulati AB ori AC"
inseamna "calulati AB sau AC"
[tex]Se ~da: \\
\Delta ABC ~ cu ~\ \textless \ A = 90^o \\
BC= 20 ~dm \\
AD = 0,5~m = 5 ~dm \\
Se ~cere: \\
AB ~~sau ~~AC \\ \\
Rezolvare: \\
\text{Vom calcula BD si CD (ambele sunt necunoscute)} \\
\text{Consideram ca: } BD \ \textless \ CD \\
\text{Din teorema inaltimii, obtinem prima ecuatie: } \\
AD^2 = BD \times CD ~~~ dar~~~AD = 5 ~dm\\
\boxed{BD \times CD = 25 } ~~~\texttt{Asta e prima ecuatie.} \\
BD + CD = BC ~~~ dar~~~BC = 20 ~dm\\
\boxed{BD + CD = 20}~~~\texttt{Asta e a doua ecuatie.}
[/tex]
[tex]\texttt{Deoarece avem produsul si suma necunoscutelor, } \\
\texttt{vom transforma sistemul de ecuatii intr-o } \\
\texttt{ecuatiei de gradul doi, dupa schema: } \\
x^2 - Sx + P=0 ~~~\text{unde S = Suma iar P = Produsul} \\
\texttt{Ecuatia este: } \\
x^2 - 20x +25 =0 \\ \\
\displaystyle \\
x_{12}= \frac{20 \pm \sqrt{400 - 100} }{2}= \frac{20 \pm \sqrt{300} }{2}= \frac{20 \pm 10\sqrt{3} }{2} = 10+5\sqrt{3} [/tex]
[tex]CD = x_1 = \boxed{10 + 5\sqrt{3} ~cm }\\
BD = x_2 = \boxed{10 - 5\sqrt{3} ~cm} \\ \\
\texttt{Acum putem calcula una din catete, cu teorema catetei. } \\ \\
AB^2 = BC \times BD \\
AB^2 = 20 \times (10 - 5\sqrt{3}) = (200 - 100\sqrt{3}) ~cm \\
AB = \sqrt{200 - 100\sqrt{3}} =\sqrt{100(2 - \sqrt{3})}=\boxed{10\sqrt{2 - \sqrt{3}} ~cm}[/tex]
Rezolvarea 2:
Consider ca cerinta: "calulati AB ori AC"
inseamna "calulati AB inmultit cu AC"
.
[tex]\displaystyle\bf\\Se~da~\Delta ABC~cu~\sphericalangle A=90^o\\BC=20~dm\\AD=0,\!5~m=5~dm\\Se~cere:\\AB~inmultit~cu~AC\\\\Rzolvare:\\\\Folosim~formula:\\\\\textbf{Produsul catetelor = produsul dintre ipotenuza si inaltimea ei.}\\\\AB\times AC=BC\times AD=20\times 5=\boxed{\bf100}[/tex]
