ABCD trapez
...............................
Ad=8cm (inaltime)
se construieste CE ⊥ AB, unde CE=AD=h=8cm.
avem ΔBEC dreptunghic in E.
EB²= 10²-8²
EB²= 36⇒ EB=√36= 6cm.
AB=AE+EB, dar AE=DC
18= 6+AE⇒ AE= 18-6
AE= 12cm=DC
P= AB+BC+CD+AD
P= 18+10+12+8
P= 48cm.
d₁= AC.
avem ΔADC dreptunghic in D.
AC²= 8²+12²
AC²= 64+144
AC= √208= 4√13 cm.
d₂= DB
avem DAB dreptunghic in A
DB²= 8²+18²
DB²= 64+324 ⇒ DB= √388
DB= 2√97 cm.
