euclidean-geometrygeometryquadrilateral
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?
Let $\measuredangle FEH=\measuredangle EAF=\alpha.$
Thus, by your work and by law of sines we obtain: $$\frac{x}{\sin{\alpha}}=\frac{4}{\sin108^{\circ}}$$ and $$\frac{x}{\sin108^{\circ}}=\frac{1}{\sin\alpha},$$ which gives $$x^2=4$$ and $$x=2.$$
Construction: Drop perpendiculars from points $A,B,E$, to base $CD$.
Let $$CH=GC=3x \;\; \text{and} \;\; AH=BG=3h$$
Since $\triangle BGC\sim \triangle EFC$, $$EF=2h\;,\;FD=2x \;,\; GF=x.$$
Using the Pythagorean Theorem, in $\triangle EFC$ and $\triangle EFD$,
$$\begin{align*}
x^2+h^2&=1\\
(4x+2.6)^2+4h^2&=25
\end{align*}$$
Solving the above equations, we have $(x,h)=\left(\dfrac{2\sqrt{13\sqrt{109}-95}}{15},\dfrac{2\sqrt{109}-13}{15}\right)$.
The base length is $6x+2.6$ and height is $3h$.
$$\;\;\;\text{Base}\approx 5.752244$$
$$\text{Height}\approx 2.552613 $$
Best Answer
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.