Question

utilize o teorema de pitagoras ,determinando o valor de x nos triangulos

Answer 1

Explicação:

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

logo x'=+4 e x"=-4

${a}^{2} = {b}^{2} + {c}^{2} \\ (3 \sqrt{5}) {}^{2} = (6) {}^{2} + (x) {}^{2} \\ 9\sqrt{25} = 36 + {x}^{2} \\ 9.5 = 36 + {x}^{2} \\ 45 = 36 + {x}^{2} \\ 45 - 36 = {x}^{2} \\ 9 = {x}^{2} \\ x = \frac{ + }{ - } \sqrt{9} \\ x = \frac{ + }{ - } 3$

logo x'=+3 e x"=-3