AB/3=AC/4=BC/5=k --> AB=3k; AC=4k; BC=5k;
scriem teorema catetei pentru cele doua catete AB si AC:
AB^2=BD*BC --> (3k)^2=BD*5k --> BD=9k^2/(5k)=9k/5;
AC^2=CD*CB --> (4k)^2=CD*5k --> CD=16k^2/(5k)=16k/5;
scriem teorema inaltimii in ΔABC:
AD^2=DB*DC --> 24^2=(9k/5)*(16k/5) --> 576=(9k*16k)/25 --> 576*25=(9*16)*k^2 -->
--> k^2=(576*25)/(9*16) --> k=(24*5)/(3*4)=2*5=10;
AB=30; AC=40; BC=50 --> Perimetrul= 30+40+50=120 cm;
Aria= (BC*AD)/2=(50*24)/2=25*24=150 cm^2
:D