Question

Me ajudem pfv, não consigo fazer essas questão de forma alguma

Answer 1

Segue a figura

Temos que CD vale

$CD=CE+DE$, então basta encontrarmos CE e DE

Note que os triangulos ACE e DEB são semelhantes pelo caso AA(angulo, angulo) aplicando então a semelhança obtemos

$\dfrac{x}{13-x}=\dfrac{3}{2}\\\\2x=3(13-x)\\\\2x=39-3x\\\\5x=39\\\\x=\dfrac{39}{5}$

logo o lado DE vale

$DE=13-\dfrac{39}{5}\\\\DE=\dfrac{13\cdot5-39}{5}\\\\DE=\dfrac{26}{5}$

Aplicando o teorema de Pitágoras no triangulo

ACE, obtemos

$3^2+CE^2=\left(\dfrac{39}{5}\right)^2\\\\9+CE^2=\dfrac{1521}{25}\\\\CE^2=\dfrac{1521}{25}-9\\\\CE^2=\dfrac{1521-25\cdot9}{25}\\\\CE=\sqrt{\dfrac{1296}{25}}\\\\\boxed{CE=\dfrac{36}{5}}$

Aplicando o teorema de pitagoras no triangulo

DBE, obtemos

$2^2+DE^2=\left(\dfrac{26}{5}\right)^2\\\\4+DE^2=\dfrac{676}{25}\\\\DE^2=\dfrac{676}{25}-4\\\\DE^2=\dfrac{676-4\cdot25}{5}\\\\DE=\sqrt{\dfrac{576}{25}}\\\\\boxed{DE=\dfrac{24}{5}}$

Por fim, fazendo $CD=CE+DE$, temos

$CD=\dfrac{36}{5}+\dfrac{24}{5}\\\\CD=\dfrac{36+24}{5}\\\\CD=\dfrac{60}{5}\\\\\boxed{\boxed{\boxed{CE=12cm}}}$