In the circle below, \(\overline{AB} \| \overline{CD}\). \(\overline{AD}\) is a diameter of the circle, and \(AD = 36^{\prime \prime}\). What is the number of inches in the length of \(\widehat{AB}\)? Express your answer in terms of \(\pi\).
CPhill had a try at this, but then again, he's human, and the answer he posted was incorrect.
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=40o
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 \(\frac{80}{360}(\pi\cdot36)=\boxed{8\pi}\text{~inches}.\)
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 :)