A triangle has corners at (4 ,4 )(4,4), (8 ,2 )(8,2), and (3 ,1 )(3,1). What is the area of the triangle's circumscribed circle?

1 Answer
Jun 5, 2018

{325 pi}/49 325π49

Explanation:

The radius of the circumcircle equals the product of the sides of the triangle divided by four times the area of the triangle.

This is more useful squared:

r^2 = {a^2 b^2 c^2}/{16 \ text{area}^2}

Archimedes' Theorem says

16 \ text{area}^2 = 4a^2b^2-(c^2-a^2-b^2)^2

That all we need to do our problem.

a^2 = (4-8)^2+(4-2)^2=20

b^2 = (8-3)^2+(2-1)^2=26

c^2=(4-3)^2+(4-1)^2=10

r^2 = {20(26)(10)}/{ 4(10)(20)-(26-20-10)^2} = 325/49

I permuted a^2, b^2 and c^2, necessarily permissible in an area formula.

Anyway the area of the circumcircle is of course pi r^2 =

{325 pi}/49