Thread: Trapezoid within a Circle Area Question

1. Re: Trapezoid within a Circle Area Question

Is there a question?

2. Re: Trapezoid within a Circle Area Question

We do not do homework.

3. Re: Trapezoid within a Circle Area Question

I've learned that as a newb on the forum not to edit a post more than once, or it vanishes!

4. Re: Trapezoid within a Circle Area Question

What post.
We haven't seen a problem from you.
Please post the problem you want help with...you hockey puck

5. Re: Trapezoid within a Circle Area 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

6. Re: Trapezoid within a Circle Area Question

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 = $\dfrac{1}{2}r^2\sin{\theta} - \dfrac{1}{2}r^2 \sin(2\theta)$