smc

Denote by $x$ the length of the shorter base. Thus, the height of the trapezoid is $$\begin{array}{r}\sqrt{1^2-{\left(\frac{x}{2}\right)}^2}.\end{array}$$ Thus, the area of the trapezoid is $$\begin{array}{rl}\frac{1}{2}\left(x+2x\right)\sqrt{1^2-{\left(\frac{x}{2}\right)}^2}& =\frac{3}{4}\sqrt{x^2\left(4-x^2\right)}\\ & \le \frac{3}{4}\frac{x^2+\left(4-x^2\right)}{2}\\ & =\boxed{\frac{3}{2}},\end{array}$$ where the inequality follows from the AM-GM inequality and it is binding if and only if $x^2=4-x^2$.