Question

calcule o valor de X nos triângulos retângulos​

Answer 1

Aplicando o teorema de Pitágoras:

$a){x}^{2} = {6}^{2} + {8}^{2} \\ {x}^{2} = 36 + 64 \\ {x}^{2} = 100 \\ x = \sqrt{100} \\ x = 10$

$b) {x}^{2} = {9}^{2} + {12}^{2} \\ {x}^{2} = 81 + 144 \\ {x}^{2} = 225 \\ x = \sqrt{225} \\ x = 15$

$c) {5}^{2} = {x}^{2} + {4}^{2} \\ 25 = {x}^{2} + 16 \\ {x}^{2} = 9 \\ x = \sqrt{9} \\ x = 3$

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