Question

In trapezul dreptunghic ABCD, cu AB//CD, m(a)=90°, ACΠBD={O} a)Perimetrul si Aria lui ABCD b) AC=? si BD=? c) OA=? si OB=?

Answer 1

ABCD-trapez dreptunghic
Ducem  CE⊥AB
CE=AB
In triunghiul CEB
M(CEB)=90⇒EB²=CB²-CE²⇒EB²=72-36=√36=6
AB(este format)=AE+EB
EB=6
AE║DC⇒AE=DC=6⇒AB=12
a)Pabcd=AB+BC+CD+DA=12+6√2+6+6=24+6√2=6(√2+40
Aabcd=(B+b)*h/2=(12+6)*6/2=108/2=54
b)In triunghiul ACE⇒AC²=CE²+AE²=6√2
In triunghiul BDA⇒BD²=DA²+AB²=144+36=√180=6√5
c)OA-h in triunghiul dreptunghic DAB⇒AO=DA*AB/DB=6*12/6√5=12√5/5
Cu aria in doua moduri aflam OB
Aacb=CE*AB/2=OB*CA/2(2 cu 2 se simplifica)⇒BO=CE*AB/CA⇒BO=6*12/6√2⇒BO=12√2/2⇒BO=6√2