Question

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

Answer 1

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

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

Explanation:

`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)`