Răspuns :
Explicație pas cu pas:
a) A (ABC) = (c1xc2) /2
A (ABC) = (AB x AC) /2= (3 x AC)/2
6 = (3 x AC)/2 => 12 = 3 x AC
AC = 4 cm
b) se folosește T. Pitagora =>BC^2=AC^2 + AB^2
BC^2 = 16 + 9 = 25
BC = 5 cm
P(ABC) = 3 + 4 +5 = 12 cm
c) Fie M aparține CB. AM perpendicular pe CB.
Exprimăm aria ABC în două moduri:
(C1 x C2) /2 = (h x b) /2
(AC x AB) /2 = (AM x CB) /2
Simplificam cu 2
4 x 3 = AM x 5
AM = 12/5 cm
[tex]\displaystyle\bf\\Se~da:\\\\\Delta ABC~dreptunghic~cu~m(\sphericalangle A)=90^o\\\\AB=3~cm\\\\Aria~\Delta ABC=6~cm^2\\\\Se~cere:\\a)~AC=?\\\\b)~Perimetrul~\Delta ABC=?\\\\c)~Inaltimea~AD=?\\\\Rezolvare: (Vezi~desenul~atasat.)\\\\a)\\A_{\Delta ABC}=\frac{AB\times AC}{2}=6~cm^2\\\\\frac{AB\times AC}{2}=6\\\\\frac{3\times AC}{2}=6\\\\3\times AC=6\times2\\3\times AC=12\\\\AC=\frac{12}{3}=4\\\\\boxed{\bf AC=4~cm}[/tex]
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[tex]\displaystyle\bf\\b)\\BC=\sqrt{AB^2+AC^2}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\\\\\boxed{\bf BC=5~cm}\\\\P_{\Deelta ABC}=AB+AC+BC=3+4+5=12~cm\\\\\boxed{\bf P_{\Deelta ABC}=12~cm}\\\\c)\\Inaltimea~ipotenuzei~AD=\frac{AB\times AC}{BC}=\frac{3\times4}{5}=\frac{12}{5}=2,\!4~cm\\\\\boxed{\bf AD=2,\!4~cm}[/tex]