A line segment with a length of 24 makes a 90-degree angle with one of the legs of an isosceles trapezoid. What is the area of this Trapezoid? 

Given that $ABCD$ is an isosceles trapezoid and that $|EB|=24$, $|EC|=26$, and m(EBC)=$90^o$. Find $A(ABCD)= ?$

From the pythagorean theorem, I can find that $|BC|=|AD|=10$. Then, I can find the area of the triangle $EBC$. But after this point, I can't progress any further. How do I find the area of this trapezoid?
 A: Option 1. Consider the isosceles trapezoid with $BE=BD$ as diagonal:
$\hspace{3cm}$
We find:
$$BF=\frac{2S_{\Delta BCD}}{CD}=\frac{240}{26}=\frac{120}{13};\\
CF=\sqrt{BC^2-BF^2}=\sqrt{100-\frac{120^2}{13^2}}=\frac{50}{13};\\
S_{ABCD}=\frac{AB+CD}{2}\cdot BF=\frac{(26-2\cdot CF)+26}{2}\cdot \frac{120}{13}=\\
=\frac{34560}{169}\approx \color{red}{204.5}.\\
$$
Option 2. Consider the point $E$ as a midpoint:
$\hspace{3cm}$
We find:
$$EH=\sqrt{BE^2+BH^2}=\sqrt{24^2+5^2}=\sqrt{601};\\
BI=\frac{2S_{\Delta BEH}}{EH}=\frac{2\cdot 60}{\sqrt{601}};\\
S_{ABCD}=EH\cdot BF=\sqrt{601}\cdot \frac{4\cdot 60}{\sqrt{601}}=\color{red}{240}.$$
Conclusion: The trapezoid is not unique.
A: The triangle $BCE$ is uniqely determined, but other points $D$ and $A$ are not so this trapezoid does not have fixed area, it depend on $A$ (and then is $D$ determined also).
A: As a concrete example of how the figure is not determined:


*

*One option is that $A$ and $E$ coincide, and $ABCD$ is a rectangle of area $24\cdot 10=240$.

*Another option is that $E$ and $D$ coincide, in which case the area of the trapezoid is $\frac{10\cdot 24}{26}(26-\frac{10\cdot 10}{26}) \approx 204$.
