Question

Utilizando o teorema de Pitágoras determine o valor de x no triângulo retângulo

Answer 1

$\large a) ~ os~ lados ~do ~tri\hat{a}ngulo ~ \Rightarrow ~20, ~~16, ~ ~12 \\\\\\\large b) ~ os~ lados ~do ~tri\hat{a}ngulo ~ \Rightarrow ~3\sqrt{5}, ~~6, ~~3$

Utilizando o teorema de Pitágoras determine o valor de x no triângulo retângulo

a)

$h^2 = c^2 + c^2\\\\20^2 = (4x)^2 + (3x)^2\\\\400 = 16x^2 + 9x^2\\\\400 = 25x^2\\\\25x^2 = 400\\\\\\x^2 = \dfrac{400}{25} \\\\\\x^2 = 16\\\\\\x = \sqrt{16} \\\\x = 4$

Substituir o valor de x no triângulo:

$C_b = 4x\\C_b = 4 ~. ~4\\C_ b = 16$

$C_c = 3x\\C_c = 3 ~. ~4\\C_ c = 12$

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b)

$(3\sqrt{5})^2 = 6^3 + x^2\\\\(3\sqrt{5})^2 = 6^3 + x^2\\\\9 ~.~5 = 36 + x^2\\\\45 = x^2 + 36\\\\x^2 + 36 = 45\\\\x^9 = 45 - 36\\\\x = \sqrt{9} \\\\x = 3$

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