A triangle has corners at (9 ,3 ), (4 ,6 ), and (2 ,4 ). What is the area of the triangle's circumscribed circle?

1 Answer
Jan 25, 2018

Solving Eqns (1), (2), we get the circum centerr Ocolor(blue)(43/8, 21/8)

Area of the circum-circle R_c = color(brown)(41.7232)

Explanation:

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A (9,3), D (4,6), C (2,4) are the three points given and the corresponding sides are a, d, c.

Slope of line segment a_m = (6-4) / (4-2) = 1

Slope of perpendicular bisector passing through F

F_m = - (1/ a_m) = -1

Mid point of DC = F has coordinates

#F ( (4+2)/2, (6+4)/2) = F(3,5)

Equation of FO where O is the circumcenter

y - 5 = -1 (x - 3)

color(red)(y + x = 8) Eqn (1)

Slope of line segment c_m = (6-3) / (4-9) = -(3/5)

Slope of perpendicular bisector passing through B

B_m = - (1/ c_m) = 5/3

Mid point of AD = B has coordinates

#B ( (9+4)/2, (3+6)/2) = F(13/2,9/2)

Equation of BO where O is the circumcenter

y - 9/2 = (5/3) (x - 13/2)

2y - 9 = (5/3) (2x - 13)

color(red)(6y - 10x = -38) Eqn (2)

Solving Eqns (1), (2), we get the coordinates of circum centerr O

color(blue)(43/8, 21/8)

we can get the radius of the circum-circle by finding the distance of O from any one of the three vertices.

Radius R_c = OA = sqrt((9-(43/8))^2 + (3-(21/8)^2)) = color(green)(3.6443)

Area of circum-circle A_c = pi R_c^2 = pi * (3.6443)^2 = color(brown)(41.7232)