[math]A_{ABCD}=\frac{\left(AD+BC\right)\cdot h}{2}[/math][br]If [math]CE\perp AD\Rightarrow CE=h\Rightarrow DE=\left(5-3\right):2=1[/math], then [math]tan60^{\circ}=\frac{h}{DE}\Rightarrow h=\sqrt{3}[/math][br][math]A_{ABCD}=\left(5+3\right)\cdot\sqrt{3}:2=4\sqrt{3}[/math]

AD: [math]y-y_A=m\left(x-x_A\right)\Rightarrow y=\sqrt{3}x[/math][br]BC: [math]y-y_B=m\left(x-x_B\right)\Rightarrow y=\sqrt{3}\left(x-2\right)\Rightarrow y=\sqrt{3}x-2\sqrt{3}[/math]