Răspuns:
Explicație pas cu pas:
Aflam laturile ΔMAC. Se stia ca AB=4√2cm.
In ΔAA'M, dreptunghic in A', AA'=4√2cm, A'M=(1/2)·A'D'=(1/2)·4√2=2√2cm. Atunci AM²=AA'²+A'M²=(4√2)²+(2√2)²=32+8=40=4·10, deci AM=2√10cm.
AC=AB√2=4√2·√2=4·2=8cm.
Oblica MF⊥AC, F∈AC, atunci si EF⊥AC, unde EF=pr(ABC)MF.
E=pr(AD)M. BD⊥AC, deci EF║BD si EF=(1/2)·OD=(1/2)·(1/2)·BD=2cm
ME=AA'=4√2cm. Din ΔMEF, dreptunghic in E, MF²=ME²+EF²=(4√2)²+2²=32+4=36, deci MF=√36=6cm.
Atunci Aria(ΔMAC)=(1/2)·AC·MF=(1/2)·8·6=24cm²
