m(∡ADC)=90° ; m(∡ABC)=30° ; m(∡DAB)=90° =>m(∡BCD)=150°
m(∡ACB)=90°=>ΔACB-> dreptunghic
In ΔACB , m(∡ACB)=90°=> sin B =AC/AB<=> sin 30°=6/AB<=>1/2=6/AB=>AB=12
m(∡ACD)=m(∡DCB)-m(∡ACB)=150°-90°=60°
In ΔADC , m(∡ADC)=90°=> cos C=DC/AC=>cos 60°=DC/6<=>1/2=DC/6=>6/2=3
In ΔADC , m(∡ADC)=90°=>sin C=AD/AC<=>sin 60°=AD/6<=>√3/2=AD/6=>AD=3√3
A=(AB+DC)*AD/2=(12+3)*3√3/2=15*3√3/2=45√3/2