A triangle has corners at $(9 ,4 )$ , $(3 ,2 )$ , and $(5 ,8 )$ . What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-06-29T07:56:15

$color(orange)("Area of Circum circle " A_R = pi R^2 = 39.27 " sq units"$

![http://mathibayon.blogspot.com/2015/01/ derivation-of-formula-for-radius-of-circumcircle.html

$"Area of Triangle " = A_T = (a b c) / (4 R)$

$A (9,4), B (3,2), C(5,8)$

$a = sqrt((5-3)^2 + (8-2)^2) = sqrt 40 = 6.3246$

$b = sqrt((5-9)^2 + (8-4)^2) = sqrt 32 = 5.6569$

$c = sqrt((9-3)^2 + (2-2)^2) = sqrt 40 = 6.3246$

$"Semi perimeter of the triangle " s = (a + b + c ) / 2 = 9.153$

$A_T = sqrt s (s-a) (s-b) (s - c))$

$A_T = sqrt(9.153 (9.153 - 6.3246) (9.153 - 5.6569) (9.153 - (6.3246)) = 16$

$R = (a b c ) / (4 * A_T) = (sqrt40 * sqrt32* sqrt40) / (4 * 16) = 3.5355$

$color(orange)("Area of Circum circle " A_R = pi R^2 = pi * 3.5355^2 = 39.27 " sq units"$

A triangle has corners at (9 ,4 )(9 ,4 ), (3 ,2 )(3 ,2 ), and (5 ,8 )(5 ,8 ). What is the area of the triangle's circumscribed circle?