Obliczmy bok AB
[tex] {3}^{2} + {10}^{2} = {x}^{2} \\ 9 + 100 = {x}^{2} \\ x = \sqrt{109} [/tex]
Obliczmy bok BC
[tex] {6}^{2} + {6}^{2} = {x}^{2} \\ 36 + 36 = {x}^{2} \\ x = \sqrt{72} = \sqrt{36 \times 2} = 6 \sqrt{2} [/tex]
Obliczmy bok CA
[tex] {3}^{2} + {4}^{2} = {x}^{2} \\ 9 + 16 = {x}^{2} \\ x = \sqrt{25} = 5[/tex]
Zatem obwód
[tex] \sqrt{109} + 6 \sqrt{2} + 25[/tex]
Myślę że pomogłem ;)
