Question

utilizando o teorema de Pitágoras, determine o valor re x nos triângulos abaixo​

Answer 1

Explicação passo-a-passo:

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

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

$c) {(x + 1)}^{2} = {x}^{2} + {( \sqrt{7} )}^{2} \\ {x}^{2} + 2x + 1 = {x}^{2} + 7 \\ {x}^{2} + 2x - {x}^{2} = 7 - 1 \\ 2x = 6 \\ x = \frac{6}{2} \\ \\ x = 3$

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