An Area Problem

Since $\angle AOB = 150^\circ$ , $\angle BAO = 15^\circ$ and $\angle OAD = 60^\circ - 15^\circ = 45^\circ$ . Therefore, $\angle AOD = 90^\circ$ and $AD = \sqrt 2 r$ and the side length of equilateral $\triangle ABC$ , $AC = \sqrt 2 r + \sqrt r$ . Area of $\triangle ABC$ , $[ABC] = \dfrac {\sqrt 3(\sqrt 2r+\sqrt r)^2}4$ .

Since $\triangle BCD$ has the same height as $\triangle ABC$ , so its area $[BCD] = \dfrac {\sqrt 3(\sqrt 2r+\sqrt r)^2}4 \times \dfrac {\sqrt r}{\sqrt 2 r + \sqrt r} = \dfrac {r\sqrt {6r} + \sqrt 3 r}4$ .

And area of $\triangle BDF$ , $[BDF] = [BCD]-[CDF] = \dfrac {r\sqrt {6r} + \sqrt 3 r}4 - \dfrac {\sqrt 3 r}4 = \dfrac {r\sqrt {6r}}4$ .

From $AB = AC$ we have:

$\begin{aligned} 2 r \sin 75^\circ & = \sqrt 2 r + \sqrt r \\ 2r \times \frac {\sqrt 3+1}{2\sqrt 2} & = \sqrt 2 r + \sqrt r \\ \left(\frac {\sqrt 3+1}{\sqrt 2} - \sqrt 2\right)r & = \sqrt r \\ \implies \sqrt r & = \frac {\sqrt 2}{\sqrt 3 - 1} = \frac {\sqrt 3+1}{\sqrt 2} \end{aligned}$

Therefore $[BDF] = \dfrac {r\sqrt{6r}}4 = \left(\dfrac {\sqrt 3+1}{\sqrt 2}\right)^3 \times \dfrac {\sqrt 6}4 = \dfrac {9+5\sqrt 3}4 \approx \boxed{4.415}$

$\angle OAB=\dfrac{180^\circ-150^\circ}{2}=15^\circ$ Hence $\angle OAD=45^\circ=\angle ODA\Rightarrow \angle AOD=180^\circ-(45^\circ+45^\circ)=90^\circ$ Similarly $\angle BOF=90^\circ$ , $\angle DOF=360^\circ-150^\circ-90^\circ-90^\circ=30^\circ$

Since $|AD|=|BF|=r \sqrt{2}$ ( $45^\circ-45^\circ-90^\circ$ triangle) , $DFC$ is equilateral . Applying Law of cosines on $ODF$ :

$r=r^{2}(2-2cos30^\circ)$

$\dfrac{1}{r}=2-\sqrt{3}$

$r=2+\sqrt{3}$

Area of $BDF$ is $\dfrac{1}{2}|BF|h$ . and $h=\dfrac{\sqrt{3r}}{2}$ (heigth of equilateral $DFC$ ) , $|BF|=r\sqrt{2}$

Hence the area is $\dfrac{r\sqrt{6r}}{4}$ putting $r=2+\sqrt{3}$ we get $\dfrac{(2+\sqrt{3})\sqrt{12+2\sqrt{27}}}{4}$

(using $\sqrt{a+b+2\sqrt{ab}}= \sqrt{a}+\sqrt{b}$ ) , $\dfrac{(2+\sqrt{3})\sqrt{12+2\sqrt{27}}}{4}=\dfrac{(2+\sqrt{3})(3+\sqrt{3})}{4}$

$=\boxed{\dfrac{9+5\sqrt{3}}{4}\approx 4.415}$

another solution: Since $\angle DBF=15^\circ$ and $|BD|=r\sqrt{3}$ ( $120^\circ-30^\circ-30^\circ$ triangle) area is $\dfrac{1}{2}\sin{15^\circ}\cdot r\sqrt{3}\cdot r\sqrt{2}$ put $r=2+\sqrt{3}$ and $\sin{15^\circ}=\dfrac{\sqrt{6}-\sqrt{2}}{4}$ and you will get : $\boxed{\dfrac{9+5\sqrt{3}}{4}}$

$\angle~BAC=60....Equ.~\Delta.\\ ~~~~ \\ \angle~AOB=150. ~~~~OA=OB=r.\\ So~in~isosceles~\Delta~OAB,~~ \angle~BAO=\dfrac{180-150} 2=15.\\ ~~~\\ So~\Delta~sides~AB=AC=say~S=2rCos15..........(1)\\ \therefore~\angle~OAD=60-15=45.\\ But~in~\Delta~OAD,~OA=OD=r.\\ So~\angle~DOA=90,~~~and~~~AD=\sqrt 2 r........(2).\\ ~~~\\ But~AD=AC- \sqrt r= 2rCos15-\sqrt r....By~(1)\\ So~ \sqrt 2 r= 2rCos15-\sqrt r. \\ Solving~r=3.7321.\\ ~~\\ Area~\Delta~BDF=1/2*\sqrt r*FB*Sin120\\ \because~of~symmetry~FB=AD=\sqrt 2 r\\ So~area~=\Large~\color{#D61F06}{~4.41515}.$