Question

Aplicando Pitágoras, encontre o x nos triângulos abaixo

Answer 1

Resposta:

a) $x=13$

b) $x=\sqrt{13}$

c) $x=4$

d) $x=3\sqrt{2}$

e) $x=7,47 \\x=3+2\sqrt{5}$

f) $x=3$

Explicação passo-a-passo:

a) $x^{2} =12^{2} +5^{2}$

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

b) $x^{2} =2^{2} +3^{2}$

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

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

$x=\sqrt{10+(4*2)}$

$x=\sqrt{16}\\ x=4$

d)

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

decompondo o numero 18

$x=\sqrt{3^{2} *2}$

$x=3\sqrt{2}$

e) $(x+9)^{2} =2x^{2} +(x+3)^{2}$

desenvolvendo os dois produtos notaveis da soma: $(a + b)^{2} = (a + b)*(a + b) = \\\\= a^{2} + 2*a*b + b^{2}$

$(x+9)^{2}=x^{2} +2*x*9+9^{2}= x^{2} +18x+81$

$(x+3)^{2} =x^{2} +2*x*3+3^{2} = x^{2} +6x+9$

resolvendo:

$x^{2} +18x+81=2x^{2} +x^{2} +6x+9$

$-2x^{2} +12x+72=0$

resolvendo a equação do 2 grau

$x'=\frac{-12+8\sqrt{5} }{2*(-2)}=-1,47$

$x"=\frac{-12-8\sqrt{5} }{2*(-2)}=+7,47$

f) $(\sqrt{26}) ^{2} = (\sqrt{17}) ^{2}+x^{2}$

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