Largest Quadrateleral in a Dome Really stuck on a geomtery problem. I will do by best to explain.
Imagine a large sphere.
Imagine I cut/slice that sphere at any point other than middle. In other words the cut produces a flat circular base which is less than the actual diameter of the sphere. Imagine slicing an orange from the top portions of the orange than straight down the middle.
So here is my question: from the base we made the slice, which is smaller than the actual diameter of the sphere, what is the dimensions of the largest quadrilateral (by area) that I can fit into the dome which has 4 corners touching the dome of the sliced sphere but maintains its base flat on the cut which was made. 
I believe if the cut was made straight down the middle the quad would have the shape of a square, but since this is not a cut down the middle it must be a rectangle. But what would the dimensions in relations to the shape of the arc OR length of the base?
I would really appreciate your help! Thank you!
 A: If I have understood correctly, this would be the cross-section through the "sphere with a base."

Let $\alpha$ be the angle that defines the position of the cut. This is a given constant.
Let $\theta$ be the angle that defines the other vertices of the quadrilateral.
The base of the trapezium formed is $2r \sin \alpha$. The side parallel to the base is $2r \sin \theta$. The distance between the two sides is $r\left(\cos \alpha + \cos \theta\right)$.
Area of trapezium is $A=\frac{\left (2r \sin \alpha+ 2r \sin \theta \right)\left(r\left(\cos \alpha + \cos \theta\right)\right)}2$
$A=r^2\left (\sin \alpha+ \sin \theta \right)\left(\cos \alpha + \cos \theta\right)$
$A=r^2\left (\sin \alpha \cos \alpha + \sin \alpha \cos \theta+ \sin \theta \cos \alpha  + \sin \theta \cos \theta\right)$
$A=r^2\left (\frac 1 2 \sin 2\alpha +\sin \left(\alpha +  \theta \right) +  \frac 1 2 \sin 2\theta \right)$
$\frac {dA}{d\theta}=r^2\left (\cos \left(\alpha +  \theta \right) +  \cos 2\theta \right)$
$A$ is maximum when $\frac {dA}{d\theta}=0$
$\cos \left(\alpha +  \theta \right) +  \cos 2\theta =0$
$\cos P + \cos Q=0$ implies one of the following:
1) $Q=\pi-P$ or
2) $Q=\pi+P$ or
3) $Q=2\pi-P$ or
4) $Q=2\pi+P$
1) $\alpha + \theta + 2\theta = \pi$
$3\theta=\pi-\alpha$
$\theta=\frac{\pi-\alpha}3$
etc.
