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) = √3-1
r = 1 + √3 - (√3-1) = 2
Area circle = 4pi approx 12.5663706143592
Area pentagon = 4 + √3 approx 5.73205080756888
Circle outside pentagon = 4pi - 4 - √3 approx 6.8343198067903
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Posted by Larry
on 2024-05-05 23:36:29 |