Question

Sabendo que AB // CD, determina x e y.​

Answer 1

Utilizando semelhança de triângulos, temos:

$\frac{CE}{CD}~=~\frac{AE}{AB}\\\\\\\frac{8-x}{4}~=~\frac{x+(8-x)}{6}\\\\\\\frac{8-x}{4}~=~\frac{8}{6}\\\\\\Multiplicando~Cruzado\\\\\\6~.~(8-x)~=~4~.~8\\\\\\48-6x~=~32\\\\\\6x~=~48-32\\\\\\x~=~\frac{16}{6}\\\\\\\boxed{x~=~\frac{8}{3}}$

Vamos agora utilizar o teorema de Pitágoras para determinar BE:

$BE^2~=~AB^2+AE^2\\\\\\BE^2~=~6^2+(x+8-x)^2\\\\\\BE^2~=~36+8^2\\\\\\BE~=~\sqrt{36+64}\\\\\\BE~=~\sqrt{100}\\\\\\\boxed{BE~=~10}$

Por fim, para determinar "y", utilizamos novamente a semelhança de triângulos:

$\frac{BE}{AB}~=~\frac{DE}{CD}\\\\\\\frac{10}{6}~=~\frac{y}{4}\\\\\\Multiplicando~Cruzado\\\\\\6~.~y~=~10~.~4\\\\\\y~=~\frac{40}{6}\\\\\\\boxed{y~=~\frac{20}{3}}$

Answer 2

Utilizando semelhança de triângulos, temos:

\frac{CE}{CD}~=~\frac{AE}{AB}\\\\\\\frac{8-x}{4}~=~\frac{x+(8-x)}{6}\\\\\\\frac{8-x}{4}~=~\frac{8}{6}\\\\\\Multiplicando~Cruzado\\\\\\6~.~(8-x)~=~4~.~8\\\\\\48-6x~=~32\\\\\\6x~=~48-32\\\\\\x~=~\frac{16}{6}\\\\\\\boxed{x~=~\frac{8}{3}}

Vamos agora utilizar o teorema de Pitágoras para determinar BE:

BE^2~=~AB^2+AE^2\\\\\\BE^2~=~6^2+(x+8-x)^2\\\\\\BE^2~=~36+8^2\\\\\\BE~=~\sqrt{36+64}\\\\\\BE~=~\sqrt{100}\\\\\\\boxed{BE~=~10}

Por fim, para determinar "y", utilizamos novamente a semelhança de triângulos:

\frac{BE}{AB}~=~\frac{DE}{CD}\\\\\\\frac{10}{6}~=~\frac{y}{4}\\\\\\Multiplicando~Cruzado\\\\\\6~.~y~=~10~.~4\\\\\\y~=~\frac{40}{6}\\\\\\\boxed{y~=~\frac{20}{3}}