What is the area of triangle AFE? If ED = 23 , and the value of the side of the  square ABCD is a multiple of 11, what is the area of the red triangle AFE?! Find the very shortest way to solve this puzzle and use only basic geometry, trigonometry is not allowed.

 A: Let $AB = 11x$. Triangles EDF and EAB are similar, so:
$\dfrac{ED}{EA} = \dfrac{DF}{AB}$
$\dfrac{23}{23 + 11x} = \dfrac{DF}{11x}$
$DF = \dfrac{253x}{23 + 11x}$
The area of $\triangle AFE$ is thus
$$\dfrac{1}{2} \cdot EA \cdot DF = \dfrac{1}{2} \cdot (23 + 11x) \cdot \dfrac{253x}{23 + 11x} = \dfrac{253}{2}x$$
A: let us consider one simple situation,suppose AB=33; you can check that 33 is multiple of 11,33/11=3,and also we know that DF/FC=1/2.if we denote DF by x,then FC=2*x so
x+2*x=33,   3*x=33  x=11;(sorry in first coment instead of DC should be FC)  so  DF=11;  length of AE=AD+DE or 33+23=56,so are of AEF=1/2*DF*AE`=1/2*56*11=28*11=308
A: Assume a side of the square is $3\times11=33$, then $$AE=AD+DE=23+33=56\quad\text{and}\quad \text{AREA}=\frac{AE\times DF}{2}=\frac{56\times11}{2}=308$$
Assume a side of the square is $4\times11=44$, then $$AE=AD+DE=23+44=67\quad\text{and}\quad \text{AREA}=\frac{AE\times DF}{2}=\frac{67\times11}{2}=368.5$$
Assume a side of the square is $5\times11=55$, then $$AE=AD+DE=23+55=78\quad\text{and}\quad \text{AREA}=\frac{AE\times DF}{2}=\frac{78\times11}{2}=429$$
A side of the square appears to be $3$-$4$ times $11$ but we cannot know except that the steps between area examples are $60.5$.
We know that a side is at least twice a multiple of $11$ or the diagonal $FB$ would be vertical but assume a side of the square is $1\times11=11$, then $$AE=AD+DE=23+11=34\quad\text{and}\quad \text{AREA}=\frac{AE\times DF}{2}=\frac{34\times11}{2}=187$$
This means that, given a side $AD=11x, \text{AREA}=187+60.5(x-1)$. For for example, if $x=3$, then $\text{AREA}=187+60.5(2)=187+121=308$ as shown above.
