Given the figure below, find EF if CA = (6n) and EF =

Given:
[tex]\begin{gathered} CA=6n \\ EF=n+8 \end{gathered}[/tex]To find:
The length of EF.
Explanation:
According to the figure,
We can write,
[tex]\frac{BE}{BC}=\frac{EF}{CA}[/tex]Substituting the given values, we get
[tex]\begin{gathered} \frac{x}{2x}=\frac{n+8}{6n} \\ \frac{1}{2}=\frac{n+8}{6n} \\ 6n=2n+16 \\ 4n=16 \\ n=4 \end{gathered}[/tex]Therefore, the length of EF will be,
[tex]EF=n+8=4+8=12[/tex]Final answer:
The length of EF will be 12.