Q:求EFGH面積

:::spoiler 答案(防暴雷按鈕)
$$ \frac{r^2(3+\pi-3\sqrt{3})}{3} $$

<!--:::spoiler 詳細題解-->
:::info
> $\overline{A H}=\overline{B H}=\overline{A B}=r$
> 代表$\triangle{ABH}$為正三角形
> $\angle{BAH}=\angle{HBA}=60^{\operatorname{\omicron}}$
> $\angle{BAD}=90^{\operatorname{\omicron}}$
> $\angle{BAD}-\angle{BAH}=\angle{HAD}=30^{\operatorname{\omicron}}$
> 同理可證$\angle{BAG}=30^{\operatorname{\omicron}}$
> $\angle{BAD}-\angle{BAG}-\angle{HAD}=\angle{GAH}=30^{\operatorname{\omicron}}$

過$H$作一直線$\overline{H I}$垂直於$\overline{A G}$，交於點$I$
$\angle{IAH}=30^{\operatorname{\omicron}}$
$\angle{AHI}=60^{\operatorname{\omicron}}$
$\angle{HIA}=90^{\operatorname{\omicron}}$
$\overline{A H}=r$
由上述四點可得$\overline{H I}=\frac{r}{2}$
$\triangle{AGH}面積=\frac{\overline{A G}}{2}\times\overline{A G}\times\frac{1}{2}=\frac{\overline{A G}^2}{4}$

$\angle{GAH}=30^{\operatorname{\omicron}}$
$\overline{A G}=r$
$⌔GAH面積=r^2\pi\times\frac{30^{\operatorname{\omicron}}}{360^{\operatorname{\omicron}}}=r^2\pi\times\frac{1}{12}$
$弓形GH面積=⌔GAH面積-\triangle{AGH}面積=\frac{r^2\pi}{12}-\frac{\overline{A G}^2}{4}=\frac{r^2\pi}{12}-\frac{\overline{A G}^2\times3}{12}=\frac{r^2\pi-r^2\times3}{12}=\frac{r^2(\pi-3)}{12}$

作線段$\overline{A C}$，並作交點$I$於$\overline{G H}$上
$\because\angle{IAH}=\frac{\angle{GAH}}{2}=15^{\operatorname{\omicron}}$
$\quad\overline{A H}=r$
$\quad\angle{IAH}=15^{\operatorname{\omicron}}$
$\therefore\overline{H I}=r\times\frac{1}{\sqrt6+\sqrt2}$
$\quad\overline{G H}=\frac{2r}{\sqrt6+\sqrt2}=\frac{2r(\sqrt6-\sqrt2)}{6-2}=\frac{(\sqrt6-\sqrt2)r}{2}$
$\quad\Box{EFGH}面積=\overline{G H}^2=\frac{r^2(8-4\sqrt{3})}{4}$

$EFGH面積=弓形GH面積\times4+\Box{EFGH}=\frac{r^2(\pi-3)}{12}\times4+\frac{r^2(8-4\sqrt{3})}{4}$
$=\frac{r^2(\pi-3)\times4}{12}+\frac{r^2(8-4\sqrt{3})\times3}{12}=\frac{r^2(\pi-3)\times4+r^2(8-4\sqrt{3})\times3}{12}=\frac{r^2(12+4\pi-12\sqrt{3})}{12}=\frac{r^2(3+\pi-3\sqrt{3})}{3}$
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