Find the area of ​the shaded region $ABCE$

Question

For reference:

The side of square $ABCD$ measures $5$m. Calculate the area of ​​the shaded region.(Answer:$10\ \mathrm{m^2}$)

My progress:

$S_{ABCE} = S_{ABCD} - S_{ABE}-S_{ACD}=25 -S_{ABE}-12,5$

$S_{ABCE} = 12.5 - S_{ABE}$

$S_{BCE}\ (\text{isosceles})$

$S_{ABE} = 2.5h$

$S_{BEC}=2.5h_1$

$S_{AEC}=2.5\sqrt2\cdot h_2$

$CE^2 = CG\cdot CA \implies 5^2 = CG\cdot 5\sqrt2$

$\therefore CG = 2 5\sqrt2=AG$

I'm not seeing a way out for this...???

Answer 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$

Answer 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. :)