Let $AB$ be a segment of length $26$, and let points $C$ and $D$ be located on $AB$ such that $AC=1$ and $AD=8$. Let $E$ and $F$ be points on one of the semicircles with diameter $AB$ for which $EC$ and $FD$ are perpendicular to $AB$. Find $EF.$
$\text{(A) }5 \qquad \text{(B) }5 \sqrt{2}  \qquad \text{(C) }7 \qquad \text{(D) }7 \sqrt{2} \qquad \text{(E) }12$
Solution
We can solve this with some simple coordinate geometry. Let $A$ be the origin at let $AB$ be located on the positive $x-$axis. The equation of semi-circle $AB$ is $(x-13)^2+y^2=13^2, y \geq 0.$ Since $E$ and $F$ are both perpendicular to $C$ and $D$ respectively, they must have the same $x -$ coordinate. Plugging in $1$ and $8$ into our semi-circle equation gives us $y=5$ and $y=12$ respectively. The distance formula on $(1, 5)$ and $(8, 12)$ gives us our answer of $\sqrt{(1-8)^2 + (5-12)^2}=\sqrt{2(7^2)}=\boxed{\textbf{(D) } 7\sqrt{2}}.$