Question

1) Aplicando o teorema de Pitágoras, determine a medida x indicada em cada um dos triângulos retângulos.​

Answer 1

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$\tt a)~\sf x^2=6^2+8^2\\\sf x^2=36+64\\\sf x^2=100\\\sf x^2=\sqrt{100}\\\sf x=10\\\tt b)~\sf x^2=5^2+12^2\\\sf x^2=25+144\\\sf x^2=169\\\sf x=\sqrt{169}\\\sf x=13\\\tt c)~\sf x^2=9^2+12^2\\\sf x^2=81+144\\\sf x^2=225\\\sf x=\sqrt{225}\\\sf x=15\\\tt d)~\sf 3^2+x^2=5^2\\\sf x^2+9=25\\\sf x^2=25-9\\\sf x^2=16\\\sf x=\sqrt{16}\\\sf x=4\\\tt e)~\sf x^2+9^2=15^2\\\sf x^2+81=225\\\sf x^2=225-81\\\sf x^2=144\\\sf x=\sqrt{144}\\\sf x=12$

$\tt f)~\sf x^2+21^2=29^2\\\sf x^2+441=841\\\sf x^2=841-441\\\sf x^2=400\\\sf x=\sqrt{400}\\\sf x=20$