Five points on a plane determine a conic section. Which kind of conic section depends on the relative location of the points. But to try this obvious solution first requires having coordinates of the points.
I started with:
A(0,0)
B(cos(150),sin(150))
D(5+4*cos(36),4*sin(36))
E(5,0)
Then a rotation transform to make BD parallel to the x-axis yielded
A(0,0)
B(-0.42039517082,0.90734111576)
D(2.79572035881,0.90734111576)
E(4.08871727719,-2.87791435369)
Then C more readily is solvable to yield (0.41032733443,2.72665420571)
The Mathworld webpage http://mathworld.wolfram.com/ConicSection.html has an order 6 matrix to find the coefficients.
The math gets messy but ultimately yields 1.43596209878*x^2 + 1.69972929265*xy + y^2 - 4.95311121512*x - 2.76738775333y = 0.
To check if this is an ellipse, calculate the discriminant b^2-4*a*c. An ellipse will have a negative discriminant. In this case the discriminant is -2.85476872685. The circumconic is an ellipse!
This ellipse has the center (1.82235862151, -0.16506428867). Its area is 15.93390872003. For comparison, the area of ABCDE is 10.26246999038, approximately 64.4% of the area of the ellipse.
This half of the problem alone is at least D4, especially considering that the much earlier
Ellipses was rated D5 and wanted the same thing for just a triangle.
Edited on December 4, 2016, 11:57 pm
Circumellipse (Part A)
