Notam trapezul ABCD cu baza mica AB=8 cm,inaltimea AD=9cm si latura neparalele BC=15cm.
Ridicam inaltimea BP perpendicular pe DC formanduse un dreptunghi ABDP⇒PB=AD=9cm
DP=AB=8cm
Conform teoremei lui Pitagora in ΔBPC⇒PC²=BC²-PB²=15²-9²=(15-9)(15+9)=6·24=144⇒PC=√144=12 cm
DC(baza mare)=DP+PC=12+8=20 cm
A=[tex] \frac{(B+b).h}{2} [/tex]=[tex] \frac{(20+8).9}{2} [/tex]=[tex] \frac{28.9}{2} [/tex]=126 cm²
