$\begin{aligned}OC & =\frac{1}{2}AC\\
& =\frac{1}{2}\sqrt{2}
\end{aligned}
$

$\begin{aligned}TO & =AC\\
& =1
\end{aligned}
$

$\begin{aligned}TC & =\sqrt{TO^{2}+OC^{2}}\\
& =\sqrt{1^{2}+\left(\frac{1}{2}\sqrt{2}\right)^{2}}\\
& =\frac{1}{2}\sqrt{6}
\end{aligned}
$

$\begin{aligned}OB & =\frac{1}{2}BD\\
& =\frac{1}{2}\sqrt{2}
\end{aligned}
$

$PO=\frac{1}{2}OT=DH=\frac{1}{2}$

$\begin{aligned}PB & =\sqrt{PO^{2}+OB^{2}}\\
& =\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\sqrt{2}\right)^{2}}\\
& =\frac{\sqrt{3}}{2}\\
& =\frac{1}{2}\sqrt{3}
\end{aligned}
$

$\begin{aligned}AE & =CG\\
& =4
\end{aligned}
$

$\begin{aligned}AT & =\frac{1}{2}GE\\
& =\frac{1}{2}\sqrt{58}
\end{aligned}
$

$\begin{aligned}TE & =\sqrt{AT^{2}+AE^{2}}\\
& =\sqrt{\left(\frac{1}{2}\sqrt{58}\right)^{2}+4^{2}}\\
& =\frac{1}{2}\sqrt{122}
\end{aligned}
$

$AT=\frac{1}{2}AC=\frac{1}{2}\sqrt{58}$

$PT=\frac{1}{2}CG=2$

$AP=\sqrt{AT^{2}+TP^{2}}$

$=\sqrt{\left(\frac{1}{2}\sqrt{58}\right)^{2}+2^{2}}$

$=\frac{1}{2}\sqrt{74}$

Garis potong $=UV=BC=3.$

$\triangle XPE\sim\triangle XAP$

$\Rightarrow\frac{AX}{PX}=\frac{AC}{PE}=\frac{\sqrt{2}}{\frac{1}{4}\sqrt{2}}=4$

$\Rightarrow AX:PX=4:1$$\Rightarrow AX=\frac{4}{5}AP$

$\mathbf{\Rightarrow}AX=\frac{4}{5}\sqrt{PE^{2}+AC^{2}}$$=\frac{4}{5}\times\frac{3}{4}\sqrt{2}=\frac{3}{5}\sqrt{2}$

$\triangle YEG\sim\triangle YCG$

$\Rightarrow YE:YC=EG:QC$

$=\sqrt{2}:\frac{1}{2}\sqrt{2}$

$=4:1$

$EC\rightarrow$$\sqrt{AC^{2}+AE^{2}}$$=\sqrt{(\sqrt{2})+1}$$=\sqrt{3}$

$\Rightarrow YC=\frac{1}{5}EC=\frac{1}{5}\sqrt{3}$

$\Rightarrow\frac{AX}{PX}=\frac{AC}{PE}=\frac{\sqrt{2}}{\frac{1}{4}\sqrt{2}}=4$

$\Rightarrow AX:PX=4:1\Rightarrow AX=\frac{4}{5}AP$

$\triangle YEG\sim\triangle YCG\Rightarrow\frac{UX}{AX}=\frac{AE}{AP}$$\Rightarrow UX=\frac{AX}{AP}\times PE=\frac{4}{5}$

$\frac{AU}{AX}=\frac{PE}{AP}$$\Rightarrow AU=\frac{AX}{AP}\times PE=\frac{1}{5}\sqrt{2}$

stewart $\sqrt{2}\cdot BU^{2}=\frac{4}{5}\sqrt{2}(1)^{2}$$+\frac{1}{5}\sqrt{2}(1)^{2}-\frac{4}{5}\sqrt{2}\cdot\frac{1}{5}\sqrt{2}\cdot\sqrt{2}$

$\sqrt{2}BU^{2}=\sqrt{2}-\frac{8}{25}\sqrt{2}=\frac{17}{25}\sqrt{2}$$\Rightarrow BU^{2}=\frac{17}{25}$

$BX^{2}=\sqrt{BU^{2}+UX^{2}}$

$=\sqrt{\frac{17}{25}+\frac{16}{25}}$

$=\frac{1}{5}\sqrt{33}$

$\triangle YEG\sim\triangle YCG$

$\Rightarrow YE:YC=EG:QC$

$=\sqrt{2}:\frac{1}{2}\sqrt{2}$

$=4:1$

$EC=\sqrt{AC^{2}+AE^{2}}$

$=\sqrt{(\sqrt{2})+1}$

$=\sqrt{3}$

$\Rightarrow YC=\frac{1}{5}EC=\frac{1}{5}\sqrt{3}$

Karena simetris, $CY=EX$$\Rightarrow XY=CE-CY-EX$$=CE-2CY$$=\sqrt{3}-\frac{2}{5}\sqrt{3}$$=\frac{3}{5}\sqrt{3}$

$\triangle YEG\sim\triangle YCG$

$\Rightarrow YE:YC=EG:QC$

$=\sqrt{2}:\frac{1}{2}\sqrt{2}$

$=4:1$

$EC=\sqrt{AC^{2}+AE^{2}}$

$=\sqrt{(\sqrt{2})+1}$

$=\sqrt{3}$

$\Rightarrow YC=\frac{1}{5}EC=\frac{1}{5}\sqrt{3}$

$\triangle CYV\sim\triangle CEA\Rightarrow\frac{YV}{YC}=\frac{AE}{CE}$

$YV=\frac{YC}{CE}\times AE=\frac{1}{5}$

$\begin{aligned}HF & =BD\\
& =\sqrt{AB^{2}+AD^{2}}\\
& =\sqrt{1^{2}+1^{2}}\\
& =\sqrt{2}
\end{aligned}
$

$\begin{aligned}DO & =\frac{1}{2}\times BD\\
& =\frac{1}{2}\sqrt{2}
\end{aligned}
$

$\triangle RHF\sim\triangle ROD$$\Rightarrow\frac{HR}{RO}=\frac{HF}{DO}=2$$\Rightarrow HR=\frac{2}{3}HO$

$HO=\sqrt{HD^{2}+DO^{3}}=\frac{\sqrt{6}}{2}$$\Rightarrow HR=\frac{\sqrt{6}}{3}=\frac{1}{3}\sqrt{6}$

$\begin{aligned}HF & =BD\\
& =\sqrt{AB^{2}+AD^{2}}\\
& =\sqrt{1^{2}+1^{2}}\\
& =\sqrt{2}
\end{aligned}
$

$\begin{aligned}DO & =\frac{1}{2}\times BD\\
& =\frac{1}{2}\sqrt{2}
\end{aligned}
$

$VR:WR=DO:HF=1:2$$\Rightarrow VR=\frac{1}{3}VW=\frac{1}{3}$$\Rightarrow WR=\frac{2}{3}$

$\triangle HWR\sim\triangle ADH$$\Rightarrow HW:WR$$\Rightarrow OD:DH=\frac{1}{2}\sqrt{2}$

$\Rightarrow HW=\frac{1}{2}\sqrt{2}WR=\frac{1}{3}\sqrt{2}$

$\Rightarrow WF=\frac{2}{3}\sqrt{2}$

Stewart: $\sqrt{2}GW^{2}=\frac{1}{3}\sqrt{2}(1)^{2}$$+\frac{2}{3}\sqrt{2}(1)^{2}-\frac{1}{3}\sqrt{2}\cdot\frac{2}{3}\sqrt{2}\cdot\sqrt{2}$

$=\sqrt{2}-\frac{4}{9}\sqrt{2}=\frac{5}{9}\sqrt{2}$

$\Rightarrow GW^{2}=\frac{5}{9}$

$GR^{2}=\sqrt{GW^{2}+WR^{2}}$

$=\sqrt{\frac{5}{9}+(\frac{2}{3})^{2}}$

$=1$

Titik tembus AG pada BCHG O

Garis potong BCHE dan AXG $\rightarrow$OX

OX memotong EH pada M (titik tembusnya)

Maka, M pertengahan EH

Karena HB $\perp$ ACH dan $\frac{EM}{EH}=\frac{1}{2}=\frac{EU}{EB}$$\Rightarrow MU\,||\, HB$ sehingga MU$\perp$ACH. (U pusat persegi ABFE).

$MU=\frac{1}{2}HB=\frac{1}{2}\sqrt{3}$

Karena $BD||l_{1}\Rightarrow BD\,||$ bidang melalui $l_{1}\&F$

$\Rightarrow$Jarak $T^{1}$ ke bidang tersebut = jarak garis BD ke bidang tersebut = Jarak O ke bidang tersebut = X

$AO=OC=\frac{1}{2}\sqrt{2}$

$OT=\sqrt{1^{2}-(\frac{1}{2}\sqrt{2})^{2}}=\frac{1}{2}\sqrt{2}$

$FO=\frac{1}{2}TO=\frac{1}{4}\sqrt{2}$

$AF=\sqrt{(\frac{1}{2}\sqrt{2})^{2}+(\frac{1}{4}\sqrt{2})^{2}}=\frac{\sqrt{10}}{4}$

$2\times[AOF]=AF\cdot X$$=OF\cdot AO$

$\begin{aligned}X & =\frac{OF\cdot AO}{AF}\\
& =\frac{\frac{1}{4}\sqrt{2}\cdot\frac{1}{2}\sqrt{2}}{\frac{1}{4}\sqrt{10}}\\
& =\frac{1}{\sqrt{10}}\\
& =\frac{1}{10}\sqrt{10}
\end{aligned}
$

Bidang memuat DF $\rightarrow$ TDHFB

Perpotongan TDHFB dan TGE (TAEGC)$\rightarrow$TV (V pusat EFGH)

P $\rightarrow$perpotongan TV dan DF $\Rightarrow$P pusat ACGE

Bidang memuat CE $\rightarrow$TAEGC

Perpotongan TAEGC dan AIJ$\rightarrow$X

Perpotongan AX dan CE $\rightarrow$Q

AE=TU=UV=1, UP=PV=$\frac{1}{2}$

AU=UC=EV=VG=$\frac{1}{2}\sqrt{2}$

EX=XV=$\frac{1}{4}\sqrt{2}$

$Y\rightarrow AX\cap CG$

YG:AE=AG$\cdot$EX=$3:1$$\Rightarrow$YG=$2$

$\Rightarrow$CY=$4$

$\Rightarrow$EQ:QE=AE:CY=1:4

$\Rightarrow$EQ=$\frac{1}{5}$EC=$\frac{1}{5}\sqrt{3}$

EP=$\frac{1}{2}$EC=$\frac{1}{2}\sqrt{3}$

$\Rightarrow$PQ=EP-EQ=$\frac{3}{10}\sqrt{3}$

Karena $PC=\frac{1}{2}\sqrt{3},$ $TC=\frac{1}{2}\sqrt{6}$$\Rightarrow$TP=$\frac{3}{2}$

Maka $\mbox{TP}{}^{2}=\mbox{TP}^{2}+\mbox{PC}^{2}$ sehingga $\angle TCP=90^{\circ}.$

$\Rightarrow$Tinggi $\triangle$TPQ = TC.

$\Rightarrow$Luas $\triangle$TPQ=$\frac{1}{2}\times PQ\times TP$

$=\frac{1}{2}\times\frac{3}{10}\sqrt{3}\times\frac{1}{2}\sqrt{6}$

$=\frac{9}{40}\sqrt{2}$.