Question

teorema de pitagoras

Answer 1

Resposta:

a)

${x}^{2} = {5}^{2} + {12}^{2} \\ {x}^{2} = 25 + 144 \\ {x}^{2} = 169 \\ x = \sqrt{169 } \\ x = 13$

b)

${x}^{2} = {2}^{2} + {3}^{2} \\ {x}^{2} = 4 + 9 \\ {x}^{2} = 13 \\ x = \sqrt{13}$

c)

${x}^{2} = { (\sqrt{10} )}^{2} + ( {2 \sqrt{2} })^{2} \\ {x}^ {2} = 10 + 4 \times 2 \\ {x}^{2} = 10 + 8 \\ x = \sqrt{18} \\ x = \sqrt{9 \times 2} \\ x = 3 \sqrt{2}$

d)

${6}^{2} = {x}^{2} + {x}^{2} \\ 36 = 2 {x}^{2} \\ {x}^{2} = \frac{36}{2} \\ {x}^{2} = 18 \\ x = 3 \sqrt{2}$

e)

${(x + 9)}^{2} = {(x + 3)}^{2} + {(2x)}^{2} \\ {x}^{2} + 18x + 81 = {x}^{2} + 6x + 9 + 4 {x}^{2} \\ {x }^{2} + 18x - {x}^{2} - 6x - 4 {x}^{2} = 9 - 81 \\ - 4 {x}^{2} + 12x = - 72 \\ - {x}^{2} + 3x = - 18 \\ - {x}^{2} + 3x + 18 = 0 \\ \\ d = {3}^{2} - 4 \times ( - 1) \times 18 \\ d = 9 + 72 \\ d = 81 \\ \\ x = \frac{ - 3 + \sqrt{81} }{2 \times ( - 1)} \: \: ou \: \: x = \frac{ - 3 - \sqrt{81} }{2 \times ( - 1)} \\ \\ x = - \frac{ - 3 + 9}{2} \: \: ou \: \: x = - \frac{ - 3 - 9}{2} \\ \\ x = - \frac{6}{2} \: \: ou \: \: x = \frac{12}{2} \\ \\ x = - 3 \: \: ou \: \: x = 6$

Nesse caso, por se tratar de medidas, x>0, portanto x=6

f)

$( { \sqrt{26} })^{2} = ( { \sqrt{17} })^{2} + {x}^{2} \\ 26 = 17 + {x}^{2} \\ {x}^{2} = 26 - 17 \\ {x}^{2} = 9 \\ x = \sqrt{9} \\ x = 3$