a) aria trapez At
At=(AB+DC) x AD/2 = (30+10) x 10√3 / 2
At = 200√3
b) perimetrul P
CB=√(CE^2 + (AB-AE)^2) = √(300+400)
CB=10√7
P=10+10√7+ 30+10√3
P=40+10(√7 + √3)
c)
tr ADC dreptunghic
AC=√(AD^2+DC^2)=√(100+300)
AC=20
AC=2DC ⇒∡DAC=30grade (intr-un tr. dreptunghic cateta care se opune la 30 grade e 1/2 din ipotenuza)
tr ADB dreptunghic
DB=√(AD^2+AB^2)
DB=20√3
DB=2AD ⇒ ∡DBA = 30 grade
in concluzie:
∡DBA = 30, ⇒∡BDC = 30 grade (alterne interne)
∡DAC=30 ⇒ ∡CAB=60 ⇒∡ACD=60 (alterne interne)
in triunghiul DOC ∡ DOC=180-30-60 = 90 grade, O este AC∩DB
