Question

resolva problemas com triângulos (avançado)

Determine o valor de x:

Answer 1

Triângulo ABC

${x}^{2} = {3}^{2} + {3}^{2} \\ {x}^{2} = 1 \times {3}^{2} + 1 \times {3}^{2} \\ {x}^{2} = (1 + 1) \times {3}^{2} \\ {x}^{2} = 2 \times {3}^{2} \\ {x}^{2} = 2 \times 9 \\ {x}^{2} = 18 \\ x = \sqrt{18} \\ x = \sqrt{9 \times 2} \\ x = 3 \sqrt{2}$

AC = $3 \sqrt{2}$

Vamos descobrir o valor de x

Lembrando que CE = x

$(3 \sqrt{2} + x {)}^{2} = (3 + x {)}^{2} + {13}^{2} \\ (3 \sqrt{2} {)}^{2} + 2 \times 3 \sqrt{2} x + {x}^{2} = (3 + x {)}^{2} + {13}^{2} \\ 9 \times 2 + 2 \times 3 \sqrt{2} x + {x}^{2} = (3 + x {)}^{2} + {13}^{2} \\ 18 + 6 \sqrt{2} x + {x}^{2} = (3 + x {)}^{2} + {13}^{2} \\ 18 + 6 \sqrt{2} x + {x}^{2} = {3}^{2} + 2 \times 3x + {x}^{2} + {13}^{2} \\ 18 + 6 \sqrt{2} x + x {}^{2} = 9 + 6x + {x}^{2} + {13}^{2} \\ 18 + 6 \sqrt{2} x = 9 + 6x + 169 \\ 18 + 6 \sqrt{2} x = 178 + 6x \\ 6 \sqrt{2} x - 6x = 178 - 18 \\ 6 \sqrt{2} x - 6x = 160 \\ \frac{6 \sqrt{2} x - 6x}{2} = \frac{160}{2} \\ 3 \sqrt{2} x - 3x = 80 \\ (3 \sqrt{2} - 3)x = 80 \\ \boxed{x = \frac{80}{3 \sqrt{2} - 3} }$

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