Midpoint of corners (7,7) and (1,3)(7,7)and(1,3) is given by
((7+3)/2,(7+1)/2)(7+32,7+12)
= (5,4)=(5,4)
Slope of the line between the corners (7,7) and (1,3)(7,7)and(1,3) is given by
(3-7)/(1-7)3−71−7
=(-4)/(-6)=−4−6
=2/3=23
Slope of the perpendicular bisector for the line between the corners (7,7) and (1,3)(7,7)and(1,3) is given by
=-1/(2/3)=−123
=-3/2=−32
We have the equation of the perpendicular bisector for the line between the corners (7,7) and (1,3)(7,7)and(1,3) given by
Point (5,4)(5,4)
Slope -3/2−32
point slope form is
(y-4)/(x-5)=-3/2y−4x−5=−32
Midpoint of corners (1,3) and (6,5)(1,3)and(6,5) is given by
((1+6)/2,(3+5)/2)(1+62,3+52)
= (3.5,4)=(3.5,4)
Slope of the line between the corners (1,3) and (6,5)(1,3)and(6,5) is given by
(5-3)/(6-1)5−36−1
=(2)/(5)=25
=2/5=25
Slope of the perpendicular bisector for the line between the corners (1,3) and (6,5)(1,3)and(6,5) is given by
=-1/(2/5)=−125
=-5/2=−52
We have the equation of the perpendicular bisector for the line between the corners (1,3) and (6,5)(1,3)and(6,5) given by
Point (3.5,4)(3.5,4)
Slope -5/2−52
point slope form is
(y-4)/(x-3.5)=-5/2y−4x−3.5=−52
Solving for the center of the circumcircle
(y-4)/(x-5)=-3/2y−4x−5=−32
2(y-4)=-3(x-5)2(y−4)=−3(x−5)
2y-8=-3x+152y−8=−3x+15
2y+3x-15-8=02y+3x−15−8=0
3x+2y-23=03x+2y−23=0
(y-4)/(x-3.5)=-5/2y−4x−3.5=−52
2(y-4)=-5(x-3.5)2(y−4)=−5(x−3.5)
2(y-4)+5(x-3.5)=02(y−4)+5(x−3.5)=0
2y-8+5x-17.5=02y−8+5x−17.5=0
5x+2y-24.5=05x+2y−24.5=0
We have,
3x+2y-23=03x+2y−23=0
5x+2y-24.5=05x+2y−24.5=0
Eliminating y
3x-5x-23+24.5=03x−5x−23+24.5=0
-2x+1.5=0−2x+1.5=0
2x=1.52x=1.5
x=0.75x=0.75
Substituting
3(0.75)+2y-23=03(0.75)+2y−23=0
2.25+2y-23=02.25+2y−23=0
Simplifying
2y=23-2.252y=23−2.25
2y=20.752y=20.75
y=10.375y=10.375
Center for the circum circle has the coordinates
(0.75,10.375)(0.75,10.375)
Radius r=distance from center to vertex
(0.75,10.375)----(7,7)(0.75,10.375)−−−−(7,7)
Area of the circum circle is
A = pir^2A=πr2
pi=3.141π=3.141
r^2=(7-0.75)^2+(7-10.375)^2r2=(7−0.75)2+(7−10.375)2
r^2=(6.25)^2+(-3.375)^2r2=(6.25)2+(−3.375)2
r^2=50.453r2=50.453
Hence, Area is
A=3.141*50.453A=3.141⋅50.453
Area of the circum circle is 158.503158.503
