Sprawdźmy po kolei
A
[tex] {4}^{2} + {8}^{2} = {x}^{2} \\ 16 + 64 = {x}^{2} \\ 80 = {x}^{2} \\ x = \sqrt{80} [/tex]
B
[tex] {8}^{2} + {16}^{2} = {x}^{2} \\ 64 +256 = {x}^{2} \\ 320 = {x}^{2} \\ x = \sqrt{320} [/tex]
C
[tex] {5}^{2} + { \sqrt{15} }^{2} = {x}^{2} \\ {x}^{2} = 25 + 15 \\ {x}^{2} = 40 \\ x = \sqrt{40} [/tex]
D
[tex] { \sqrt{32} }^{2} + { \sqrt{10} }^{2} = {x}^{2} \\ 32 + 10 = {x}^{2} \\ x = \sqrt{42} [/tex]
Odpowiedź: B
