Pai,avand in vedere ca avem raportul [tex] \frac{AB}{AC} = \frac{2}{3} [/tex] => [tex] \frac{AB}{AC} = \frac{2}{3} [/tex]<=>[tex] \frac{AB}{2} = \frac{AC}{3} [/tex]=K => AB=2K si AC=3K.
Apoi aplicam pitagora in triunghiul ABC:
[tex]( 2K)^{2}+ (3K)^{2}=12 \sqrt{13} cm
[/tex]
[tex] 4K^{2}+ 9K^{2}=144 X 13 [/tex]
[tex] 13K^{2}=1872 \\ K^{2}=144=\ \textgreater \ K= \sqrt{144} =\ \textgreater \ K=12
[/tex]
AB=2 X 12=24 cm
AC=3 X 12=36 cm
Aria lui ABC=[tex] \frac{AB X AC}{2} [/tex]=[tex] \frac{24 X 36}{2} =432 cm^{2} [/tex]
Asta e tot :) este corect
