Question

URGENTE!!!

Calcule o valor de x aplicando o Teorema de Pitágoras a2 = b2 + c2 :​

Answer 1

Resposta:

a)

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

b)

$6^{2}=x^{2} +x^{2} \\ 36=2.x^{2} \\36/2=x^{2} \\18=x^{2} \\\sqrt{18}=x\\ x=4,2$

c)

$x^{2} =6^{2} + \sqrt{11}^{2} \\x^{2} = 36 + 11\\x= \sqrt{47}\\\\x=6,85$

d)

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

e)

$10^{2} =x^{2} +4^{2} \\100=x^{2} + 16\\100-16=x^{2} \\84=x^{2} \\\sqrt{84}=x\\ 9,1=x$

f)

$(x+4)^{2}=(x+2)^{2} +x^{2} \\x^{2} + 16 = x^{2} + 4 + x^{2} \\x^{2} +16=2x^{2} +4\\x^{2} -2x^{2} = 4 -16\\-x^{2} = -12 (-1)\\x^{2} =12\\x=\sqrt{12}\\ x=3,46$