Is there a question?
LOL Dan told me to wait, and it should reappear. Nice to see some eager helpers here! Okay here we go... gonna try one more time
You got this circle:
https://i.imgsafe.org/e92d938fb2.jpg
The lines from A to B and C to D are parallel.
How can you show that the area of the trapezoid ABCD can be expressed as:
Area = 1/2 R^2 sin theta - 1/2r^2 sin 2theta
see attached diagram ...
Area triangle adb = $\dfrac{1}{2}r^2 \sin(2\theta-180) = \dfrac{1}{2}r^2\bigg[\sin(2\theta)\cos(180)-\cos(2\theta)\sin(180)\bigg] = \dfrac{1}{2}r^2\bigg[-\sin(2\theta)\bigg] = -\dfrac{1}{2}r^2 \sin(2\theta)$
Area triangle bdc = $\dfrac{1}{2}r^2 \sin(180-\theta) = \dfrac{1}{2}r^2\bigg[\sin(180)\cos{\theta}-\cos(180)\sin{\theta}\bigg] = \dfrac{1}{2}r^2\sin{\theta}$
sum of the areas = $\dfrac{1}{2}r^2\sin{\theta} - \dfrac{1}{2}r^2 \sin(2\theta)$
I replied to your post but it too disappeared.
I gave two valuable webpages: circular segments & circular sectors.
Use Skeeter's diagram and the pages are try to work it out for your self.
Thanks everyone for your timely help! And to imagine I did a math minor in university. My profs are rolling in their graves I bet! I used to be pretty good at math back in high school and university. I even took intermediate calculus back in the day, but well that was like 3 decades ago so I've forgotten a few things... sigh... I think I even posted in the wrong section... maybe this is more Geometry? Anyways, have a nice weekend!