ABCD is a rectangle with AB=8 inches and BC=6 inches. CE is drawn parallel to both CD and BC at C. If EC=4 inches, find the length of AE.

Question

I know this is a fairly simple question, I got the answer to either be $2\sqrt{41}$ or $6\sqrt{5}$, however its a multiple choice and the answers are either $10.77, 9.17, 11.22,$ and $10$. I'm not sure if I'm missing something, if by some chance the selection of answers is wrong I don't exactly have the credibility to prove it though, so I really need an expert on this.

Answer 1

AC and CE are the legs of a right triangle where AE is the hypotenuse. $|AC|^2=|AB|^2+|BC|^2=64+36=100$.
So $|AE|^2=|AC|^2+|EC|^2=116$.
Therefore $|AE|=10.77$.

Answer 2

Since $EC\perp BC$, $EC\perp DC$ and $DC\cap BC=\{C\},$ we obtain that $EC\perp(ABCD),$ which gives $EC\perp AC$.

Id est, by the Pythagoras's theorem twice we obtain: $$AE=\sqrt{EC^2+AC^2}=\sqrt{EC^2+AB^2+BC^2}=\sqrt{4^2+8^2+6^2}=\sqrt{116}.$$