Find the area of an isosceles trapezoid $ABCD$ $(AB\parallel CD)$ with height $m$ and perpendicular diagonals $(AC\perp BD)$ Find the area of an isosceles trapezoid $ABCD$ $(AB\parallel CD)$ with height $m$ and perpendicular diagonals $(AC\perp BD)$.
The area of $ABCD$ is given by $$S_{ABCD}=\dfrac{a+b}{2}.h=\dfrac{a+b}{2}.m$$
where $a$ and $b$ are the two bases $(a>b)$. So we have to find $a+b$ in terms of $m$ or maybe even $\dfrac{a+b}{2}$. Any help would be appreciated. Thank you in advance!
 A: The trapezoid is isoceles, hence $\,\overline{AC} = \overline{BD}\,$ in:

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:
$$
\left(1,-\frac{a/2+b/2}{m}\right)\cdot\left(1,+\frac{a/2+b/2}{m}\right) = 1-\frac{(a/2+b/2)^2}{m^2} = 0
\\ \Longrightarrow \quad m = \frac{a+b}{2}
$$
$$
\mbox{Area} = m^2 = \left(\frac{a+b}{2}\right)^2
$$
A: COMMENT.-The problem seems indeterminate because in the area between two parallels whose distance is equal to $m$, if only the fact that they are perpendicular diagonals is given as additional data and the problem were determined then the answer would be equal to $m^2$ because a square side $m$ satisfies the conditions of the problem. But it does not appear to be the case for any other trapezoid as shown in the accompanying figure.

