Question

Fie triunghiu oarecare ABC cu AB=24, AC=32 si BC=48 ,iar M apartine AB astfel incat MB supra AM=3supra5. Daca MN paralel cu BC, N apartine AC atunci calculati perimentrul trapezului MBCN

Answer 1

MB/AM=3/5=>MB=3AM/5

AB=AM+MB=AM+3AM/5=8AM/5

AB=24=>8AM/5=24=>8AM=120=>AM=15

MB=3AM/5=3*15/5=9

MN||BC=>(Teorema fundamentala a asemanarii):AM/AB=AN/AC=>AN=AM*AC/AB=15*32/24=20

AC=AN+NC=>NC=AC-AN=32-20=12

MN||BC=>(Teorema fundamentala a asemanarii):AM/AB=MN/BC=>MN=AM*BC/AB=15*48/24=30

P_MBCN=MB+BC+CN+NM=9+48+12+30=99