Question

Utilizando o Teorema de Pitágoras, calcule o valor de x:​

Answer 1

a)

${10}^{2} = {8}^{2} + {x}^{2} \\ 100 = 64 + {x}^{2} \\ 100 - 64 = {x}^{2} \\ 36 = {x}^{2} \\ {x}^{2} = 36 \\ x = \sqrt{36 } \\ x = 6$

b)

${x}^{2} = {3}^{2} + {3}^{2} \\ {x}^{2} = 9 + 9 \\ {x}^{2} = 18 \\ x = \sqrt{18 } \\ x = 3 \sqrt{2}$

c)

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

d)

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

Answer 2

Resposta: a) x = ±6 ; b) x = $\sqrt{18}$ ; c) x = ±5 ; d) x = ±3

Explicação passo a passo:

a) 10² = 8² + x²

   100 = 64 + x²

   64 + x² = 100

           x² = 100 - 64

           x² = 36

           x = $\sqrt{36}$

           x = ±6

b) x² = 3² + 3²

   x² = 9 + 9

   x² = 18

   x = $\sqrt{18}$

c) x² = 4² + 3²

   x² = 16 + 9

   x² = 25

   x = $\sqrt{25}$

   x = ±5

d) 5² = x² + 4²

   25 = x² + 16

   x² + 16 = 25

   x² = 25 - 16

   x² = 9

   x = $\sqrt{9}$

   x = ±3