MN=3AB
VEZI FIGURA
MN linie mij in trapez⇒MN=(AB+DC)/2=(AB+25)/2
3AB=(AB+25)/2⇒6AB-AB=25⇒AB=5⇒
MN=3·5=15
<DBC=<DAB=90
<DBA=<<BDC (alterne interne)⇒ΔADB≈ΔDBC⇒
AB/DB=AD/BC=DB/DC⇒5/DB=DB/25⇒DB²=5·25⇒DB=5√5
in triung dreptunghic DAB ⇒cf pitagora:
AD²=DB²-AB²=125-25=100
AD=10 (INALTIMEA TRAPEZULUI)