daca faci desenul obtii 2 triunghiuri isoscele in care CD_|_AB⇒punctul M
AM=AO-7=18 cm
BM=25+7=32 cm
daca AB diametru⇒ΔACB dreptunghic cu C=90*
th inaltimii⇒CM²=AM.MB⇒CM²=18.32=576
CM=24=MD
A(CDB)=48.32/2=768 cm²
A(ADC)=48.18/2=432 cm²
pt perimetru cu Pitagora⇒AC²=18²+24²⇒AC=30
si CB²=CM²+MB²⇒CB=40
P=2.30+2.40=140 cm