Question

Portanto, as medidas dos lados do triângulo retângulo são 12 cm, 16 cm e 20 cm
através do Teorema de Pitágoras encontre o valor do x.

Answer 1

$\large\boxed{\begin{array}{l}\rm a)~\sf x^2=5^2+12^2\\\sf x^2=25+144\\\sf x^2=169\\\sf x=\sqrt{169}\\\sf x=13\\\rm b)~\sf x^2=2^2+3^2\\\sf x^2=4+9\\\sf x^2=13\\\sf x=\sqrt{13}\\\rm c)~\sf x^2=(2\sqrt{2})^2+(\sqrt{10})^2\\\sf x^2=4\cdot2+10\\\sf x^2=8+10\\\sf x^2=18\\\sf x=\sqrt{18}\\\sf x=\sqrt{3^2\cdot2}\\\sf x=3\sqrt{2}\\\rm d)~\sf x^2+x^2=6^2\\\sf 2x^2=36\\\sf x^2=\dfrac{36}{2}\\\\\sf x^2=18\\\sf x=\sqrt{18}=\sqrt{3^2\cdot2}=3\sqrt{2}\end{array}}$

$\boxed{\begin{array}{l}\rm e)~\sf(x+9)^2=(2x)^2+(x+3)^2\\\sf x^2+18x+81=4x^2+x^2+6x+9\\\sf x^2-4x^2-x^2+18x-6x+81-9=0\\\sf -4x^2+12x+72=0\div(-4)\\\sf x^2-3x-18=0\\\sf \Delta=b^2-4ac\\\sf\Delta=(-3)^2-4\cdot1\cdot(-18)\\\sf\Delta=9+72\\\sf\Delta=81\\\sf x =\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(-3)\pm\sqrt{81}}{2\cdot1}\\\\\sf x=\dfrac{3\pm9}{2}\begin{cases}\sf x_1=\dfrac{3+9}{2}=\dfrac{12}{2}=6\\\sf x_2=\dfrac{3-9}{2}=-\dfrac{6}{2}=-3\end{cases}\\\\\sf como\,x>0,ent\tilde ao~x=6\end{array}}$

$\large\boxed{\begin{array}{l}\rm f)~\sf x^2+(\sqrt{17})^2=(\sqrt{26})^2\\\sf x^2+17=26\\\sf x^2=26-17\\\sf x^2=9\\\sf x=\sqrt{9}\\\sf x=3\end{array}}$