Find the area of ​the shaded region $ABCE$

Solution 1:

Using power of point $C$, $CE = 5$.
Also note that $\triangle AOE \sim \triangle DCE$ and hence $DE = 2 AE$. So applying Pythagoras in $\triangle ADE$, $AE^2 + (2AE)^2 = 5^2 \implies AE^2 = 5 = AD$
Also $\frac{AH}{AE} = \frac{AE}{AD} \implies AH = 1$
$S_{\triangle AEB} = \frac 12 \cdot AH \cdot AB = \frac 52$

Shaded area is,
$ \displaystyle \frac 12 S_{ABCD} - S_{\triangle AEB} = \frac {25}{2} - \frac 52 = 10$

Solution 2:

Let $x=AF=FE$ (tangents to semicircle). Then in right $\triangle FBC$ by Pythagoras theorem, $$(5+x)^2=(5-x)^2+5^2\Rightarrow x=5/4$$

Drop $EP \perp BF$, which you have taken $h$. $\triangle EPF \sim \triangle CBF$, so $$\frac{EP}{CB}=\frac{EF}{CF}\Rightarrow h=1$$

Thus area of shaded region is $[ACB]-[AEB]=12.5-2.5=10$

Remark : $\triangle CBF$ turns out to be a $3:4:5$ triangle, always present in your posts. :)