Given the above coordinate system, we have for the diagonals, with a bit of analytic geometry: $$ \overline{AD} \; : \; \begin{cases} x = -a/2 + (b/2+a/2)t \\ y = mt \quad ; \quad 0 \le t \le 1 \end{cases} \quad \Longrightarrow \quad x-(a/2+b/2)y/m=-a/2 \\ \overline{BC} \; : \; \begin{cases} x = +a/2 - (b/2+a/2)t \\ y = mt \quad ; \quad 0 \le t \le 1 \end{cases} \quad \Longrightarrow \quad x+(a/2+b/2)y/m=+a/2 $$ The diagonals are perpendicular to each other if the dot product of the normals of the line segments $\overline{AD}$ and $\overline{BC}$ is zero: $$ \left(1,-\frac{a/2+b/2}{m}\right)\cdot\left(1,+\frac{a/2+b/2}{m}\right) = 1-\frac{(a/2+b/2)^2}{m^2} = 0 \\ \Longrightarrow \quad m = \frac{a+b}{2} $$ $$ \mbox{Area} = m^2 = \left(\frac{a+b}{2}\right)^2 $$