Feisty Geometry

The circumference of the circle is 36pi.

Arc BD is 100 degrees as is arc AC. Arc CD and AB are therefore both 80 degree arcs.

80/360 x 36 pi = $8\pi$.

This assumes that the latex symbol you used (\widehat, something I've never seen used outside of noting angles) means what I assume it to mean: arc.

I drew one line down from b to d to make a 90 degree angle.

90+50=140

180-140=40^o

and that was CPhill's way, but it was wrong...

Here's my attempt: 

Because $AB \parallel CD$, $\angle DAB = \angle ADC = 50^\circ$.

Now, draw line segment $CB$, and label the intersection point $E$.

Note that $\triangle ECD$ is isosceles, so $\angle ECD = 50^ \circ$. This means that $\angle ECD = \angle AEB = 80^ \circ$

So, the length of $\overset{\large\frown}{AB} = {80 \over 360} \times 2 \times 18 \times \pi = \color{brown}\boxed{8 \pi }$

here

By symmetry, $\widehat{BD}=\widehat{CA}=100^\circ$. Furthermore,$\widehat{AB}=\widehat{CD}$ , so$[360^\circ=\widehat{AB}+\widehat{BD}+\widehat{DC}+\widehat{CA}=2\widehat{AB}+200^\circ$Therefore the arc $\widehat{AB}$ measures $80^\circ$. Since the diameter of the circle is $36''$, the length of the arc is $\frac{80}{360}(\pi\cdot36)=\boxed{8\pi}\text{~inches}.$

Hmm  the thing is, it may be talking about the major arc instead of the minor arc. 

Edit: Nevermind, I thought nerdiest was saying that 8pi was wrong. Sorry! :(

Thank you all for your answers, I was kinda stumped when I reached CPhill's answer, and it was wrong.

TYSM

Do you understand now nerdiest? 

You are a good student here and you never should be worried about saying clearly when you do, or do not fully understand.

I put an emphasis on the work 'clearly'

Thanks to the three of you who have helped here :)