Question

Aplicando o teorema de Pitágoras determine "X" nos triangulos e retângulos

Answer 1

a)

${x}^{2} + {5}^{2} = {( \sqrt{29} )}^{2} \\ {x}^{2} + 25 = 29 \\ {x}^{2} = 29 - 25 \\ {x}^{2} = 4 \\ x = \sqrt{4}$

$x = 2$

b)

${x}^{2} = {24}^{2} + {32}^{2} \\ {x}^{2} = 576 + 1024 \\ {x}^{2} = 1600$

$x = \sqrt{1600} \\ x = 40$

c)

${x}^{2} = {( \sqrt{10} )}^{2} + {( \sqrt{10}) }^{2} \\ {x}^{2} = 10 + 10 \\ {x}^{2} = 20 \\ x = \sqrt{20} \\ x = 2 \sqrt{5}$

d)

${x}^{2} + {24}^{2} = {25}^{2} \\ {x}^{2} + 576 =625 \\ {x}^{2} = 625 - 576 \\ {x}^{2} = 49 \\ x = \sqrt{49} \\ x = 7$