Daca ABCD-trapez isoscel⇒AD=BC=24 cm
cos(∠(ABC))=BC/AB
1/2=24/AB
AB=2·24
AB=48 cm
Ducem DE si CF⊥AB
cos(∠(DAE))= AE/AD
1/2=AE/24
AE=24/2
AE=12 cm
Analog si pentru FB
EF=AB-AE-FB
EF=48-12-12
EF=36-12
EF=24 cm⇒DC=24 cm
P=AB+BC+CD+AD
P=48+24+24+24
P=72+24+24
P=96+24
P=120 cm
ΔAED-dr
m(∠(AED))=90°
⇒ED²=AD²-AE²
ED²=576-144
ED=√432
ED=12√3 cm
A=(b+B)·h/2
A=(24+48)·12√3/2
A=52·6√3
A=312√3 cm²
b) ΔACB-dr
m(∠(ACB))=90°
⇒AC²=AB²-BC²
AC²=2304-576
AC=√1728
AC=24√3 cm
Analog si pentru BD⇒BD=24√3 cm
Sper ca te-am ajutat. Succes la scoala!