trapezul ABCD are latura BC neparalela
Fie CF perpendiculara din C pe AB
trg FCB aplic T Pitagora
BC^2=CF^2+FB^2
CF=9 cm, BF=20 cm - 8 cm= 12 cm
BC^2=144+81=225
⇒ BC= 15 cm
perimetrul = 15+20+9+8=52 cm
b) Distanta de la A la CD este AD=9 cm
c) trg EAB asemenea cu trg ECD
CD/AB=EC/EB=ED/AE
scriu proportii derivate cu termeni schimbati
EC/EB-EC=ED/AE-ED=CD/AB-CD
EC/15=ED/9=8/12=2/3
EC=15*2/3=10 cm
ED=9*2/3=6 cm
aria trg EAB=AE8AB/2=15*20/2=150 cm^2