AO=6⇒AC=12⇒(Pitagora) BC= √(12²-(6√3)²)=6
A dreptunghiului +AB*BC=6√3 86=36√3, cerinta
b) BM=OB(simetrice)=DO (doiagonalele dreptunghiului se injumatatersc)⇒DM=3BM⇔BM/DM=1/3
dar BM/DM=BS/DC ( Thales)
BS/6√3=1/3⇒BS=2√3, cerinta
c) OC=BC=OB=6,ΔBOC echilateral⇒mas∡(CBM)=180°-60°=120°
dar OB=BM (constructie) ⇒ΔCMB isoscel⇒mas ∡BCM=(180°-120°):2=30°
⇒mas ∡(ACM)=69°+30°=90° ⇔ΔACM dreptunghic in C
CS= √BC²+BS²= √(6²+(2√3)²)=4√3
MS/CS=BM/BD=1/2 (Thales)⇒MS=2√3
deci CM=CS+SM=4√3+2√3=6√3
AC=12
Cum ACM dreptunghic in C
AM²=AC²+CM²
AM²=12²+ (6√3)²=144+108=252
AM=√252=6√7
Deci Perimetrul ACM=12+6√3+6√7, cerinta
