A pentagon has vertices A, B, C, D and E, where ABCE is a square of side length 2 units, and CDE is an equilateral triangle.
Points A, B and D lie on the circumference of a circle G.
What is the exact area of the part of circle G that lies outside pentagon ABCDE?
Divide the diagram into two halves, divided by the line through A and the midpoint of BD. Call that midpoint M.
AM = sqrt(3)+2
BM = 1
Angle BAM = arctan(1/(sqrt(3)+2))
The apex angle of the isosceles triangle BAD:
Angle BAD = 2*arctan(1/(sqrt(3)+2)); call this angle A.
The circumradius equals a/(2*sin(A)).
Side a is length 2 (because it's BD).
The circumradius is 2/(2*sin(2*arctan(1/(sqrt(3)+2)))) (see Wikipedia for the formula based on a side and its opposite angle)
The sought area is pi*(2/(2*sin(2*arctan(1/(sqrt(3)+2)))))^2 - (4 + sqrt(3)).
That's approximately 6.8343198067903.
BTW, the circumradius is 2, making the diameter 4.
Edited on May 5, 2024, 2:08 pm
|
|
Posted by Charlie
on 2024-05-05 14:06:25 |
solution
