In triunghiul drept ADB, m ADB=90 grade, mA=60⇒mB=30, AB=30⇒th 30-60-90 AD=15
AM bisectoare⇒mDAM=30
In triunghiul ADM, mD=90, mA=30, AD=15, notez AM=X si ⇒ DM=x/2 conf th 30-60-90
Aplici pitagora in ΔADM⇒AM²=AD²+DM²
X²=225+X²/4
3X²=225*4
X²=75*4
X=10√3=AM
Aria= AB*DN
DN-inaltimea trapezului
Aplici th inaltimii pt aflarea lui DN
DN=(AD*DB)/AB
Intr-un paralelogram diagonalele se injumatatesc, DB=2*DM=2*5√3=10√3
AD=15*10√3:30=5√3
Aria= 30*5√3=150√3
