AC = 8
h = 8/2=4
4^2 + 4^2 = CB^2
16+16=CB^2
CB=√32 = 4√2
[tex]AB = \frac{h\sqrt{3} }{2}+h = 2\sqrt{3}+4[/tex]
[tex]Ob = AC + CB + AB = 8+4\sqrt{2} + 2\sqrt{3}+4 = 12+4\sqrt{2} + 2\sqrt{3}[/tex]
Szczegółowe wyjaśnienie:
z wlasnosci trojkatow o katach 90, 60, 30 i 90, 45, 45
