trapezul ortodiagonal -diagonale perpendiculare
notam trapezul ABCD
ducem CE_ I _AB
AE=40
EB=90-40=50
teorema inaltimii
CE=√40×50=√200=10√2
in triunghiul CEB aplicam Pitagora
CB=√(10√2)²+50²=√(200+250)=√450=15√2 cm
A=(B+b)×h/2=(40+90)×10√2/2=650√2 cm²
P=90+40+10√2+15√2=130+25√2=5(26+5√2) cm
