Fie ABCD trapezul cu baza mica AB si baza mare DC. Construim inaltimile Am si BN perpendiculare pe DC.
ΔAMD drpetunghic in M are m(<ADM)=60=>m(<DAM)=30.Aplicand teorema unghiului de 30 grade in ΔAMD=>DM=AD/2=6 si AM=AD/2√3=3√3.
AM=BN=3√3
MN=AB=10
In triunghiul ΔBNC dreptunghic in N aplicam teorema lui Pitagora.
[tex]NC=\sqrt{BC^2-BN^2}=\sqrt{(6\sqrt{3})^2-(3\sqrt{3})^2}}=9\\
A_{ABCD}=\frac{(B+b)\cdot h}{2}= \frac{(22+10)\cdot3\sqrt{3}}{2} =48\sqrt{3}[/tex]
