Question

A triangle has corners at `(7 ,7 )`, `(1 ,3 )`, and `(6 ,5 )`. What is the area of the triangle's circumscribed circle?

Answer 1

Area of the circum circle is `158.503`

Explanation:

Midpoint of corners `(7,7) and (1,3)` is given by
`((7+3)/2,(7+1)/2)`
`= (5,4)`
Slope of the line between the corners `(7,7) and (1,3)` is given by
`(3-7)/(1-7)`
`=(-4)/(-6)`
`=2/3`
Slope of the perpendicular bisector for the line between the corners `(7,7) and (1,3)` is given by
`=-1/(2/3)`
`=-3/2`
We have the equation of the perpendicular bisector for the line between the corners `(7,7) and (1,3)` given by
Point `(5,4)`
Slope `-3/2`
point slope form is
`(y-4)/(x-5)=-3/2`

Midpoint of corners `(1,3) and (6,5)` is given by
`((1+6)/2,(3+5)/2)`
`= (3.5,4)`
Slope of the line between the corners `(1,3) and (6,5)` is given by
`(5-3)/(6-1)`
`=(2)/(5)`
`=2/5`
Slope of the perpendicular bisector for the line between the corners `(1,3) and (6,5)` is given by
`=-1/(2/5)`
`=-5/2`
We have the equation of the perpendicular bisector for the line between the corners `(1,3) and (6,5)` given by
Point `(3.5,4)`
Slope `-5/2`
point slope form is
`(y-4)/(x-3.5)=-5/2`

Solving for the center of the circumcircle
`(y-4)/(x-5)=-3/2`
`2(y-4)=-3(x-5)`
`2y-8=-3x+15`
`2y+3x-15-8=0`
`3x+2y-23=0`

`(y-4)/(x-3.5)=-5/2`
`2(y-4)=-5(x-3.5)`
`2(y-4)+5(x-3.5)=0`
`2y-8+5x-17.5=0`
`5x+2y-24.5=0`

We have,
`3x+2y-23=0`
`5x+2y-24.5=0`
Eliminating y
`3x-5x-23+24.5=0`
`-2x+1.5=0`
`2x=1.5`
`x=0.75`
Substituting
`3(0.75)+2y-23=0`
`2.25+2y-23=0`
Simplifying
`2y=23-2.25`
`2y=20.75`
`y=10.375`
Center for the circum circle has the coordinates
`(0.75,10.375)`
Radius r=distance from center to vertex
`(0.75,10.375)----(7,7)`
Area of the circum circle is
`A = pir^2`
`pi=3.141`
`r^2=(7-0.75)^2+(7-10.375)^2`
`r^2=(6.25)^2+(-3.375)^2`
`r^2=50.453`
Hence, Area is
`A=3.141*50.453`
Area of the circum circle is `158.503`