Perhatikan gambar berikut :

$AC=CF=AF=$ Diagonal Bidang $=10\sqrt{2}$

Perhatikan bahwa segitiga ACF adalah segitiga sama sisi.

Jika FO adalah jarak titik F ke garis AC, maka :

$\begin{aligned}FO & =\sqrt{AF^{2}-AO^{2}}\\
& =\sqrt{\left(10\sqrt{2}\right)^{2}-\left(5\sqrt{2}\right)^{2}}\\
& =\sqrt{200-50}\\
& =\sqrt{150}\\
& =5\sqrt{6}\, cm
\end{aligned}
$

Perhatikan gambar berikut :

$\begin{aligned}BQ & =\sqrt{AQ^{2}+AB^{2}}\\
& =\sqrt{2^{2}+6^{2}}\\
& =\sqrt{4+36}\\
& =\sqrt{40}\\
& =2\sqrt{10}
\end{aligned}
$

Perhatikan segitiga BAQ :

Luas segitiga BAQ adalah :

$\frac{1}{2}\cdot AB\cdot AQ=\frac{1}{2}\cdot BQ\cdot AA’$

$\begin{aligned}AA’ & =\frac{AB.AQ}{BQ}\\
& =\frac{6.2}{2\sqrt{10}}\\
& =\frac{6}{\sqrt{10}}.\frac{\sqrt{10}}{\sqrt{10}}\\
& =\frac{6}{10}\sqrt{10}\\
& =\frac{3}{5}\sqrt{10}
\end{aligned}
$

Perhatikan gambar berikut :

$\begin{aligned}AP & =PG\\
& =\sqrt{PH^{2}+HG^{2}}\\
& =\sqrt{2^{2}+4^{2}}\\
& =\sqrt{20}
\end{aligned}
$

$BG=$Diagonal bidang $=4\sqrt{2}$

$\begin{aligned}BP & =\sqrt{AP^{2}+AB^{2}}\\
& =\sqrt{\left(\sqrt{20}\right)^{2}+4^{2}}\\
& =\sqrt{20+16}\\
& =6
\end{aligned}
$

Perhatikan Segitiga BPG. Jika PO adalah jarak P ke garis BG maka :

$PO^{2}=PO^{2}$

$BP^{2}-BO^{2}=PG^{2}-GO^{2}$

$6^{2}-x^{2}=\left(\sqrt{20}\right)^{2}-\left(4\sqrt{2}-x\right)^{2}$

$36-x^{2}=20-\left(32-8\sqrt{2}x+x^{2}\right)$

$48=8\sqrt{2}x$

$x=\frac{48}{8\sqrt{2}}=\frac{6}{\sqrt{2}}$

$\Rightarrow x=\frac{6}{\sqrt{2}}.\frac{\sqrt{2}}{\sqrt{2}}=3\sqrt{2}$

$\begin{aligned}PO^{2} & =36-x^{2}\\
& =36-\left(3\sqrt{2}\right)^{2}\\
& =36-18\\
& =18
\end{aligned}
$

$\begin{aligned}PO & =\sqrt{18}\\
& =3\sqrt{2}
\end{aligned}
$

Jadi jarak titik P ke garis BG adalah $3\sqrt{2}.$

Perhatikan gambar berikut :

Misalkan panjang rusuk kubus adalah a cm

HF (Diagonal Bidang) = a$\sqrt{2}$cm

DF (Diagonal Ruang) = a$\sqrt{3}$ cm

Misalkan HH’ adalah jarak dari titik H ke DF

Perhatikan segitiga DHF yang siku-siku di H.

Luas segitiga DHF.

$\frac{1}{2}\cdot DH\cdot HF=\frac{1}{2}\cdot DF\cdot HH’$

$\begin{aligned}HH’ & =\frac{DH\cdot HF}{DF}\\
& =\frac{a\cdot a\sqrt{2}}{a\sqrt{3}}\\
& =\frac{1}{3}a\sqrt{6}
\end{aligned}
$

Perhatikan gambar berikut :

Perhatikan segitiga TAC.

$\begin{aligned}AB & =BC\\
& =CD\\
& =AD\\
& =8\, cm
\end{aligned}
$

$\begin{aligned}TA & =TB\\
& =TC\\
& =TD\\
& =8\sqrt{2}
\end{aligned}
$

AC adalah diagonal bidang. AC = 8$\sqrt{2}$ cm

$\begin{aligned}AO & =OC\\
& =4\sqrt{2}\, cm
\end{aligned}
$

$\begin{aligned}TO & =\sqrt{AT^{2}-AO^{2}}\\
& =\sqrt{\left(8\sqrt{2}\right)^{2}-\left(4\sqrt{2}\right)^{2}}\\
& =\sqrt{128-32}\\
& =\sqrt{96}\\
& =4\sqrt{6}
\end{aligned}
$

Misalkan AA’ adalah jarak dari A ke garis TC.

Perhatikan segitiga TAC :

Luas segitiga TAC :

$\frac{1}{2}\cdot AC\cdot TO=\frac{1}{2}\cdot TC\cdot AA’$

$\begin{aligned}AA’ & =\frac{AC.TO}{TC}\\
& =\frac{8\sqrt{2}\cdot4\sqrt{6}}{8\sqrt{2}}\\
& =4\sqrt{6}
\end{aligned}
$

Perhatikan gambar berikut :

$\begin{aligned}PG & =\sqrt{PH^{2}+HG^{2}}\\
& =\sqrt{2^{2}+4^{2}}\\
& =\sqrt{20}
\end{aligned}
$

$BG=$Diagonal bidang $=4\sqrt{2}$

$\begin{aligned}BP & =\sqrt{AP^{2}+AB^{2}}\\
& =\sqrt{\left(\sqrt{20}\right)^{2}+4^{2}}\\
& =\sqrt{20+16}\\
& =6
\end{aligned}
$

Perhatikan Segitiga BPG. Jika PO adalah jarak P ke garis BG maka :

$PO^{2}=PO^{2}$

$BP^{2}-BO^{2}=PG^{2}-GO^{2}$

$6^{2}-x^{2}=\left(\sqrt{20}\right)^{2}-\left(4\sqrt{2}-x\right)^{2}$

$36-x^{2}=20-\left(32-8\sqrt{2}x+x^{2}\right)$

$\begin{aligned}48 & =8\sqrt{2}x\\
\Rightarrow x & =\frac{6}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\\
& =3\sqrt{2}
\end{aligned}
$

$\begin{aligned}PO^{2} & =36-x^{2}\\
& =36-\left(3\sqrt{2}\right)^{2}\\
& =36-18\\
& =18
\end{aligned}
$

$\begin{aligned}PO & =\sqrt{18}\\
& =3\sqrt{2}
\end{aligned}
$

Jadi jarak titik P ke garis BG adalah $3\sqrt{2}.$

Perhatikan gambar berikut :

Misalkan panjang rusuk kubus adalah a cm

HF (Diagonal Bidang) $=a\sqrt{2}$ cm

DF (Diagonal Ruang) $=a\sqrt{3}$ cm

Misalkan HH’ adalah jarak dari titik H ke DF

Perhatikan segitiga DHF yang siku-siku di H.

Luas segitiga DHF.

$\frac{1}{2}\cdot DH\cdot HF=\frac{1}{2}\cdot DF\cdot HH’$

$\begin{aligned}HH’ & =\frac{DH\cdot HF}{DF}\\
& =\frac{a\cdot a\sqrt{2}}{a\sqrt{3}}\\
& =\frac{1}{3}a\sqrt{6}
\end{aligned}
$

Perhatikan gambar berikut :

Perhatikan segitiga TAC.

$\begin{aligned}AB & =BC\\
& =CD\\
& =AD\\
& =8\, cm
\end{aligned}
$

$\begin{aligned}TA & =TB\\
& =TC\\
& =TD\\
& =8\sqrt{2}
\end{aligned}
$

AC adalah diagonal bidang. AC $=8\sqrt{2}$ cm

$\begin{aligned}AO & =OC\\
& =4\sqrt{2}\, cm
\end{aligned}
$

$\begin{aligned}TO & =\sqrt{AT^{2}-AO^{2}}\\
& =\sqrt{\left(8\sqrt{2}\right)^{2}-\left(4\sqrt{2}\right)^{2}}\\
& =\sqrt{128-32}\\
& =\sqrt{96}\\
& =4\sqrt{6}
\end{aligned}
$

Misalkan AA’ adalah jarak dari A ke garis TC.

Perhatikan segitiga TAC :

Luas segitiga TAC :

$\frac{1}{2}\cdot AC\cdot TO=\frac{1}{2}\cdot TC\cdot AA’$

$\begin{aligned}AA’ & =\frac{AC\cdot TO}{TC}\\
& =\frac{8\sqrt{2}\cdot4\sqrt{6}}{8\sqrt{2}}\\
& =4\sqrt{6}
\end{aligned}
$

Perhatikan gambar berikut :

PB tegak lurus bidang BCGF, BQ tegak lurus PB

$\begin{aligned}BQ & =\sqrt{FB^{2}+FQ^{2}}\\
& =\sqrt{a^{2}+\left(2a\right)^{2}}\\
& =\sqrt{5a^{2}}\\
& =a\sqrt{5}
\end{aligned}
$

$\begin{aligned}PQ & =\sqrt{BQ^{2}+PB^{2}}\\
& =\sqrt{\left(a\sqrt{5}\right)^{2}+\left(2a\right)^{2}}\\
& =\sqrt{5a^{2}+4a^{2}}\\
& =\sqrt{9a^{2}}\\
& =3a
\end{aligned}
$

$V=s^{3}=1l$

$1l=1000\, cm^{3}$

$\Rightarrow s=10\, cm$

Perhatikan gambar berikut :

Perhatikan bahwa AC = AF = CF = Diagonal ruang $=10\sqrt{2}$ cm

Dengan demikian segitiga ACF sama sisi. Oleh karena itu garis FO tegak lurus terhadap garis AC

$\begin{aligned}BO & =\frac{1}{2}BD\\
& =5\sqrt{2}\, cm
\end{aligned}
$

FO adalah jarak F ke garis AC. Perhatikan segitiga FBO

$\begin{aligned}FO & =\sqrt{FB^{^{2}}+OB^{2}}\\
& =\sqrt{\left(10\right)^{2}+\left(5\sqrt{2}\right)^{2}}\\
& =\sqrt{100+50}\\
& =\sqrt{150}\\
& =5\sqrt{6}\, cm
\end{aligned}
$

Perhatikan gambar berikut :

$\begin{aligned}QE & =ER\\
& =a
\end{aligned}
$

$\begin{aligned}QR & =\sqrt{QE^{2}+ER^{2}}\\
& =\sqrt{a^{2}+a^{2}}\\
& =a\sqrt{2}
\end{aligned}
$

Perhatikan segitiga PQR , siku siku di Q :

$\begin{aligned}PR & =\sqrt{PQ^{2}+QR^{2}}\\
& =\sqrt{a^{2}+\left(a\sqrt{2}\right)^{2}}\\
& =\sqrt{3a^{2}}\\
& =a\sqrt{3}
\end{aligned}
$

Perhatikan gambar berikut :

Perhatikan segitiga BKG tegak lurus di K.

Misalkan Jarak titik K ke garis BG adalah KI

Misalkan panjang rusuk adalah 4 cm

$\begin{aligned}EK & =KG\\
& =\frac{1}{2}EG\\
& =\frac{4}{2}\sqrt{2}\\
& =2\sqrt{2}\, cm
\end{aligned}
$

$\begin{aligned}BK & =\sqrt{BF^{2}+FK^{2}}\\
& =\sqrt{4^{2}+\left(2\sqrt{2}\right)^{2}}\\
& =\sqrt{16+8}\\
& =2\sqrt{6}
\end{aligned}
$

Perhatikan segitiga BKG luasnya adalah :

$\frac{1}{2}\cdot BK\cdot KG=\frac{1}{2}\cdot BG\cdot KI$

$\begin{aligned}KI & =\frac{BK.KG}{BG}\\
& =\frac{\left(2\sqrt{6}\right)\left(2\sqrt{2}\right)}{4\sqrt{2}}\\
& =\sqrt{6}
\end{aligned}
$

Jadi jarak dari titik K ke garis BG adalah $\sqrt{6}$ cm

Perhatikan gambar berikut :

Perhatikan segitiga PBQ adalah segitiga sama kaki dimana PB = QB

Misalkan jarak dari titik B ke garis PQ adalah BR

$\begin{aligned}PR & =RQ\\
& =\frac{1}{2}PQ
\end{aligned}
$

$\begin{aligned}PB & =\sqrt{AB^{2}+AP^{2}}\\
& =\sqrt{4^{2}+\left(2\sqrt{2}\right)^{2}}\\
& =\sqrt{16+8}\\
& =\sqrt{24}\\
& =2\sqrt{6}\, cm
\end{aligned}
$

$\begin{aligned}PQ & =\sqrt{PH^{2}+HQ^{2}}\\
& =\sqrt{\left(2\sqrt{2}\right)^{2}+\left(2\sqrt{2}\right)^{2}}\\
& =\sqrt{8+8}\\
& =\sqrt{16}\\
& =4\, cm
\end{aligned}
$

$\begin{aligned}PR & =RQ\\
& =\frac{1}{2}PQ\\
& =2
\end{aligned}
$

$\begin{aligned}BR & =\sqrt{PB^{2}+PR^{2}}\\
& =\sqrt{\left(2\sqrt{6}\right)^{2}-2^{2}}\\
& =\sqrt{24-4}\\
& =\sqrt{20}\\
& =2\sqrt{5}\, cm
\end{aligned}
$

Jadi jarak titik B ke garis PQ adalah $2\sqrt{5}$ cm

Perhatikan gambar berikut :

$CH$ (Diagonal Bidang) $=12\sqrt{2}$ cm

$\begin{aligned}CK & =\sqrt{BK^{2}+BC^{2}}\\
& =\sqrt{6^{2}+12^{2}}\\
& =\sqrt{36+144}\\
& =\sqrt{180}\\
& =6\sqrt{5}\, cm
\end{aligned}
$

$KD=CK$

$\begin{aligned}KH & =\sqrt{DH^{2}+KD^{2}}\\
& =\sqrt{12^{2}+\left(6\sqrt{5}\right)^{2}}\\
& =\sqrt{144+180}\\
& =\sqrt{324}\\
& =18\, cm
\end{aligned}
$

Perhatikan segitiga KCH

$\begin{aligned}\cos\angle KCH & =\frac{CK^{2}+CH^{2}-KH^{2}}{2\cdot CH\cdot KH}\\
& =\frac{\left(6\sqrt{5}\right)^{2}+\left(12\sqrt{2}\right)^{2}-18^{2}}{2\cdot6\sqrt{5}\cdot12\sqrt{2}}\\
& =\frac{180+288-324}{144\sqrt{10}}\\
& =\frac{144}{144\sqrt{10}}\\
& =\frac{1}{\sqrt{10}}
\end{aligned}
$

KO adalah jarak dari K ke garis CH

$\sin\angle KCH=\frac{3}{\sqrt{10}}$

$\frac{KO}{CK}=\frac{3}{\sqrt{10}}$

$\begin{aligned}KO & =CK\cdot\frac{3}{\sqrt{10}}\\
& =6\sqrt{5}\cdot\frac{3}{\sqrt{10}}\\
& =\frac{18}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\\
& =9\sqrt{2}
\end{aligned}
$

Perhatikan gambar berikut :

Misalkan jarak dari E ke garis BT adalah EE’

$BE$(Diagonal Bidang) $=6\sqrt{2}$ cm

$\begin{aligned}BT & =\sqrt{BC^{2}+CT^{2}}\\
& =\sqrt{6^{2}+3^{2}}\\
& =\sqrt{36+9}\\
& =\sqrt{45}\\
& =3\sqrt{5}\, cm
\end{aligned}
$

$\begin{aligned}BT & =FT\\
& =3\sqrt{5}\, cm
\end{aligned}
$

$\begin{aligned}ET & =\sqrt{EF^{2}+FT^{2}}\\
& =\sqrt{6^{2}+\left(3\sqrt{5}\right)^{2}}\\
& =\sqrt{36+45}\\
& =\sqrt{81}\\
& =9
\end{aligned}
$

Perhatikan segitiga BET

$\begin{aligned}\cos\alpha & =\frac{BT^{2}+ET^{2}-BE^{2}}{2\cdot BT\cdot ET}\\
& =\frac{\left(3\sqrt{5}\right)^{2}+9^{2}-\left(6\sqrt{2}\right)^{2}}{2\cdot3\sqrt{5}\cdot9}\\
& =\frac{45+81-72}{54\sqrt{5}}\\
& =\frac{54}{54\sqrt{5}}\\
& =\frac{1}{\sqrt{5}}
\end{aligned}
$

sin$\alpha=\frac{2}{\sqrt{5}}$…(1)

Perhatikan segitiga BET

sin$\alpha=\frac{EE’}{ET}=\frac{EE’}{9}$…(2)

substitusikan pers (1) ke pers (2) sehingga :

$\frac{2}{\sqrt{5}}=\frac{EE’}{9}$

$\begin{aligned}EE’ & =\frac{18}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}}\\
& =\frac{18}{5}\sqrt{5}
\end{aligned}
$

Jadi jarak dari E ke garis BT adalah $\frac{18}{5}\sqrt{5}$.