Find the area of ​the shaded region $CEOD$.

Question

For reference:

In figure $O$ and $O_1$ are centers, $\overset{\LARGE{\frown}}{AO_1}=\overset{\LARGE{\frown}}{O_1B}$. If $AD = 4\sqrt2$. Calculate the area of ​​the shaded region. (Answer: $2(4-\sqrt2)$)

My progress:

$\angle AOD = 90^\circ$

$[ECOD] = [AOD] - [AEC]$

$\triangle AEC \sim \triangle AOD \implies \dfrac{AE}{AO} = \dfrac{AC}{4\sqrt2} = \dfrac{EC}{OD}$

$\dfrac{[AEC]}{[AOD]}=\dfrac{AC\cdot AE}{4\sqrt2\cdot AO}$

$[ECOD] = \dfrac{OD+EC}{2}\cdot OE$

$[AOD] = \dfrac{AO\cdot OD}{2}$

$[AEC] =\dfrac{AE\cdot EC}{2} $

I couldn't see more...???

Answer 1

Let $OO_1=R$. Then $AO_1=R\sqrt 2$ (right angle isosceles triangle). Since $O_1A=O_1D$ (radii of the big circle), you have $OD=R+R\sqrt 2$. In $\triangle ADO$ you can apply Pythagoras:$$AD^2=AO^2+DO^2=R^2+R^2(1+\sqrt 2)^2$$so $$32=R^2(1+1+2+2\sqrt 2)$$simplify it to get $$R=\sqrt{16-8\sqrt 2}$$and $$OD=\sqrt{16+8\sqrt 2}$$ Now draw the perpendicular from $O$ to $AC$ in $P$. $\triangle APO\sim\triangle AOD$. $$AP=\frac{AC}2$$ and $$\frac{AP}{AO}=\frac{AO}{AD}$$ Therefore $$AC=\frac{2R^2}{4\sqrt 2}=4\sqrt 2-4$$ You can now continue how you started with similar triangles to get $AE$ and $EC$. Can you take it from here?

Answer 2

Let $r$ be the radius of the smaller circle, so the radius of the larger circle is $r\sqrt{2}$.

Firstly, $$\angle AO_1O=45^o\implies \angle ADO=22\frac12^o$$ So $$\angle DAO=67\frac12^o=\angle ACO$$ $$\implies \angle AOC=45^o\implies CE=EO=\frac{r}{\sqrt{2}}$$

The shaded area is then $$T=\frac12EO(CE+DO)=\frac12\cdot\frac{r}{\sqrt{2}}(\frac{r}{\sqrt{2}}+r+r\sqrt{2})$$ $$\implies T=\frac14r^2(3+\sqrt{2})$$

However, due to Pythagoras, $$r^2+(r\sqrt{2}+r)^2=(4\sqrt{2})^2$$ $$\implies r^2=\frac{32}{4+2\sqrt{2}}=8(2-\sqrt{2})$$

Hence $$T=\frac14(3+\sqrt{2})\cdot8(2-\sqrt{2})=2(4-\sqrt{2})$$