Question

Calcule o valor de x nos triângulos retângulos.

Answer 1

$a)\\\\Teorema \: de \: Pitagoras: a^{2} = b^2 + c^2\\\\2x^{2} = x^{2} + (3\sqrt{3})^2\\\\2x^2 - x^2 = (3\sqrt{3})^2\\\\x^2 = (9\sqrt{9})\\\\x^2 = 9.3\\\\x^2 = 27\\\\x = \sqrt{27}\\\\x = 3\sqrt{3}$

$b)\\\\Teorema \: de \: Pitagoras: a^2 = b^2 + c^2\\\\(x + 5)^2 = 6^2 + (x + 3)^2\\\\x^2 + 10x + 25 = 36 + x^2 + 6x + 9\\\\x^2 - x^2 + 10x - 6x = 36 + 9 - 25\\\\10x - 6x = 20\\\\4x = 20\\\\x = \frac{20}{4}\\\\x = 5$

Espero ter ajudado!