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?
ABCE are at (±1, ±1)
D is at (0, 1 + √3)
Center of circle is at (0,h) where h is the solution to:
1 + √3  h = √(1 + (h+1)^2)
1 + 3 + h^2 + 2(√3  h  √3h) = 1 + (h+1)^2
1 + 3 + 2√3  2(1+√3)h= 2h + 2
2 + 2√3 = 2h + 2(1+√3)h
1+√3 = (2+√3)h
h = (1+√3) / (2+√3) = (1+√3)*(2√3) = √31
r = 1 + √3  (√31) = 2
Area circle = 4pi approx 12.5663706143592
Area pentagon = 4 + √3 approx 5.73205080756888
Circle outside pentagon = 4pi  4  √3 approx 6.8343198067903

Posted by Larry
on 20240505 23:36:29 