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…???
Best Answer
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 $$T=\frac14(3+\sqrt{2})\cdot8(2-\sqrt{2})=2(4-\sqrt{2})$$