sebuah kubus memiliki panjang rusuk (seperti gambar diatas) luas permukaan kubus tersebut adalah, dengan caranya thanks

[tex]Rusuk\ Kubus=Sisi\ Persegi \\ \\ Rusuk= \frac{1}{2- \sqrt{2} } \\ \\ Luas\ Persegi \\ \\ =(sisi)^{2} \\ \\ =(\frac{1}{2- \sqrt{2} } )^{2} =\frac{1^{2}}{(2- \sqrt{2})^{2}} \\ \\ =\frac{1}{2^{2}- (2.(2).(\sqrt{2}))+(\sqrt{2})^{2}} \\ \\ =\frac{1}{4- 4\sqrt{2}+2}\\ \\ =\frac{1}{6- 4\sqrt{2}}[/tex]
[tex]=\frac{1}{6- 4\sqrt{2}}. \frac{(6+ 4\sqrt{2})}{(6+ 4\sqrt{2})} \\ \\ = \frac{6+ 4\sqrt{2}}{6^{2}+((6).(4\sqrt{2}))-((6).(4\sqrt{2}))+((-4\sqrt{2}).(4\sqrt{2}))}} \\ \\= \frac{6+ 4\sqrt{2}}{36+0+((-16).(2))}\\ \\= \frac{6+ 4\sqrt{2}}{36+0-32}\\ \\= \frac{6+ 4\sqrt{2}}{4}[/tex]

Luas permukaan kubus 

[tex]L = (6).(\frac{6+ 4\sqrt{2}}{4}) \\ \\ L=\frac{(6.6)+ (4.6)\sqrt{2}}{4}\\ \\ L=\frac{36+ 24\sqrt{2}}{4} \\ \\ L=\frac{36}{4}+ \frac{24}{4}\sqrt{2}} \\ \\ L=9+6\sqrt{2}[/tex]