Question

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

Answer 1

Area of the circum-circle `=color(green)48.05`

Explanation:

ABC is the triangl and D,E,F are the midpoints of AB,BC & CA respectively.

Point D has coordinates as `((3+5)/2,(8+9)/2)=(4,17/2)`
Similarly, E coordinates `((5+8)/2,(9+2)/2)=(13/2,11/2)`
F coordinates `((8+3)/2,(2+8)/2)=(11/2,5)`

Slope of AB `=(9-8)/(5-3)=1/2`
Slope of `C’D=-2` where C’ is the circum center and C’D is perpendicular bisector of AB.
Circum-center C’ is the meeting point of perpendicular bisectors of the Sides.
Equation of C’D is `y-(17/2)=-2*(x-4)`
`2y-17=-4x+16`
`color(red)(2y+4x=33)`. Eqn (1)

Slope of BC `=9-2/5-8=-(7/3)`
Slope ofC’E perpendicular bisector of BC is `=3/7`
Equation of C’E is `y-(11/2)=(3/7)(x-(13/2))`
`14y-77=6x-39`
`color(red)(14y-6x=38)`. Eqn (2)

Solving equations (1) & (2) will give point C’.
`6y+12x=99`
`28y-12x=76`
Adding both equations,
`34y=175`
`y=175/34`
`4x=33-2(175/34)==33-(175/17)=(561-175)/68=386/68=193/34`
Coordinates of C’ is `color (red)(193/34,175/34)`

Radius of circum-circle is C’A=C’B=C’C
C’A `=sqrt((3-(193/34))^2+(8-(175/34)^2)`
`=sqrt(7.16+8.14)=color(blue)3.91`

Verification:
C’B `=sqrt(5-(193/34)^2+(9-(175/34))^2)`
`=sqrt(0.46+14.85)=color(blue)(3.91)`

Area of circle `=pir^2=(22/7)*(3.91)^2= color(green)(48.05)`