Question

Calcule o valor de x dos triângulos abaixo:

(foto a cima)​

Answer 1

Resposta:

A) 13

B) $\sqrt{13}$ ou, resolvendo a raiz, 3,6055512754639892931192212674705

C) $\sqrt{18}$ ou, resolvendo a raiz, 4,2426406871192851464050661726291

D) $\sqrt{18}$ ou, resolvendo a raiz, 4,2426406871192851464050661726291

E) -3 ou 6

F) 3

Explicação passo-a-passo:

A) x² = 5² + 12²

x² = 25 + 144

x = raiz(169)

x = 13

B) x² = 2² + 3²

x² = 4 + 9

x = raiz(13)

x = 3,6055512754639892931192212674705

C)

$x^2 = (2\sqrt{2})^2 + \sqrt{10}^2\\x^2 = 8 + 10\\x = \sqrt{18}\\x = 4,2426406871192851464050661726291$

D) 6²= x² + x²

36 = 2x²

36/2 = x²

18 = x²

x = raiz(18)

x = 4,2426406871192851464050661726291

E)

(x + 9)² = (2x)² + (x+3)²

x² + 18x + 81 = 4x² + x² + 6x + 9

Subtraindo os valores

12x + 72 = 4x²

-4x² + 12x + 72 = 0

Resolvendo por bhaskara

$delta = 12^2 - 4 * -4 * 72\\delta = 144 + 16 * 72\\delta = 144 + 1152\\delta = 1296\\x = \frac{-12 +- \sqrt{1296} }{(2 * -4)}\\\\x = \frac{-12 +- 36}{-8}\\\\x1 = \frac{-12 + 36}{-8}\\\\x1 = \frac{24}{-8} = -3\\\\x2 = \frac{-12 - 36}{-8}\\\\x2 = \frac{-48}{-8} = 6$

Portanto X pode ser -3 ou 6

F)

$\sqrt{26}^2 = \sqrt{17} ^2 + x^2$

Cancela expoente com raiz e tem

26 = 17 + x²

x² = 26 - 17

x² = 9

x = 3