Question

Utilizando o teorema de pitagoras, determine o valor de x

Answer 1

a)

${5}^{2} = {x}^{2} + {4}^{2} \\ 25 = {x}^{2} + 16 \\ 25 - 16 = {x }^{2} \\ 9 = {x}^{2} \\ {x}^{2} = 9 \\ x = \sqrt{9 } \\ x = 3$

b)

${x}^{2} = {1}^{2} + {(2 \sqrt{2)} }^{2} \\ {x}^{2} = 1 + 8 \\ {x}^{2} = 9 \\ x = \sqrt{9 } \\ x = 3$