i need help

Label the diagram as shown. Let $\angle FCE = \theta$ and $EC = x$.

Let H be a point on AC such that FH is perpendicular to AC. Then $FH = 8$.

Note that $CE = FD = HA = x$. Then $CH = 10 - x$. 

$CF = \sqrt{8^2 + (10 - x)^2} = \sqrt{x^2 - 20x + 164}$ by using Pythagoras theorem in $\triangle CFH$.

In $\triangle ECG$, $\sin \theta = \dfrac{EG}{EC} = \dfrac1x$.

In $\triangle FHC$, $\sin \theta = \dfrac{FH}{CF} = \dfrac8{\sqrt{x^2 -20x+164}}$.

So we have $\dfrac1x = \dfrac8{\sqrt{x^2 - 20x + 164}}$. Squaring gives $\dfrac1{x^2} = \dfrac{64}{x^2 - 20x + 164}$.

Rearranging, 

$64x^2 = x^2 - 20x + 164\\ 63x^2 + 20x - 164 = 0\\ x = \dfrac{-20 \pm \sqrt{20^2 - 4(63)(-164)}}{2(63)}\\ x = \dfrac{8 \sqrt{163} - 10}{63}\text{ (reject negative root)}$

Hence, the area is $\dfrac{8x}2 = \dfrac{32 \sqrt{163} - 40}{63}$.