Question

galera e urgente..

 Utilizando
o Teorema de Pitágoras, determine o valor de x nos triângulos retângulos:

Answer 1

A) $20 ^{2}$ + (4x)$^{2}$ + (3x)$^{2}$
400 = 16$x^{2}$ + 9$x^{2}$
25$x^{2}$ = 400
$x^{2}$ = 400/25
$x^{2} = 16$
$x = \sqrt{16}$
x = 4

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

C) (x + 1)$^{2}$ = ($\sqrt{7}$)$^{2}$ + $x^{2}$
$x^{2} + 2x+2 = 7+ x^{2}$
$x^{2} - x^{2} +2x = 7-2$
2x = 5
x = 5/2
x = 2,5

D) ($3\sqrt{2}$)$^{2}$ = $x^{2} + x^{2}$
18 = $2 x^{2}$
$x^{2}$ = 18/2
$x^{2}$ = 9
x = $\sqrt{9}$
x = 3