A triangle has corners at #(4 ,4 )#, #(8 ,9 )#, and #(3 ,1 )#. What is the area of the triangle's circumscribed circle?

Question

A triangle has corners at #(4  ,4   )#, #(8   ,9   )#, and #(3   ,1  )#.  What is the area of the triangle's circumscribed circle?

Noah Atkinson · Accepted Answer

Area of circumscribed circle is #584.99#

If the sides of a triangle are #a#, #b# and #c#, then the area of the triangle #Delta# is given by the formula

#Delta=sqrt(s(s-a)(s-b)(s-c))#, where #s=1/2(a+b+c)#

and radius of circumscribed circle is #(abc)/(4Delta)#

Hence let us find the sides of triangle formed by #(4,4)#, #(8,9)# and #(3,1)#. This will be surely distance between pair of points, which is

#a=sqrt((8-4)^2+(9-4)^2)=sqrt(16+25)=sqrt41=6.4031#

#b=sqrt((3-8)^2+(1-9)^2)=sqrt(25+64)=sqrt89=9.4340# and

#c=sqrt((3-4)^2+(1-4)^2)=sqrt(1+9)=sqrt10=3.1623#

Hence #s=1/2(6.4031+9.4340+3.1623)=1/2xx18.9994=9.4997#

and #Delta=sqrt(9.4997xx(9.4997-6.4031)xx(9.4997-9.4340)xx(9.4997-3.1623)#

= #sqrt(9.4997xx3.0966xx0.0657xx6.3374)=sqrt12.2482=3.4997#

And radius of circumscribed circle is

#(6.4031xx9.4340xx3.1623)/(4xx3.4997)=13.6458#

And area of circumscribed circle is #3.1416xx(13.6458)^2=584.99#

Violet Aldrich · Answer

undefined To find the area of the circumscribed circle of a triangle, we need to find the radius of the circle, which is the distance from the center of the circle to any of the triangle's vertices. Then, we can use the formula for the area of a circle, which is π times the square of the radius.

Given the coordinates of the vertices of the triangle, we can use the formula for the circumcenter of a triangle to find its center, which is the center of the circumscribed circle.

Let's denote the vertices of the triangle as A (4, 4), B (8, 9), and C (3, 1). The circumcenter (O) of the triangle is the intersection point of the perpendicular bisectors of its sides.

First, let's find the equations of the perpendicular bisectors of two sides. Then, we'll find their intersection point, which is the center of the circumscribed circle.

1. Perpendicular bisector of side AB:
   Midpoint of AB = ((x1 + x2) / 2, (y1 + y2) / 2)
                   = ((4 + 8) / 2, (4 + 9) / 2)
                   = (6, 6.5)
   Slope of AB = (y2 - y1) / (x2 - x1)
               = (9 - 4) / (8 - 4)
               = 5 / 4
   Slope of perpendicular bisector = -1 / (slope of AB) = -4 / 5
   Using the point-slope form of a line: y - y1 = m(x - x1), where (x1, y1) is the midpoint of AB,
   Equation of the perpendicular bisector of AB is:
   y - 6.5 = (-4 / 5)(x - 6)

2. Perpendicular bisector of side BC:
   Midpoint of BC = ((x2 + x3) / 2, (y2 + y3) / 2)
                   = ((8 + 3) / 2, (9 + 1) / 2)
                   = (5.5, 5)
   Slope of BC = (y3 - y2) / (x3 - x2)
               = (1 - 9) / (3 - 8)
               = -8 / -5
               = 8 / 5
   Slope of perpendicular bisector = -1 / (slope of BC) = -5 / 8
   Using the point-slope form of a line: y - y1 = m(x - x1), where (x1, y1) is the midpoint of BC,
   Equation of the perpendicular bisector of BC is:
   y - 5 = (-5 / 8)(x - 5.5)

Now, we solve the system of equations formed by the two perpendicular bisectors to find the coordinates of the circumcenter (O).

Solving these equations, we get the coordinates of the circumcenter (O) as (5.625, 7.25).

Now, the radius of the circumscribed circle is the distance from the circumcenter to any vertex of the triangle. Let's take vertex A (4, 4).

Radius (r) = √((x2 - x1)^2 + (y2 - y1)^2)
           = √((5.625 - 4)^2 + (7.25 - 4)^2)
           ≈ √(1.390625 + 11.5625)
           ≈ √12.953125
           ≈ 3.598

Finally, the area of the circumscribed circle is π times the square of the radius.

Area = π * r^2
     ≈ π * (3.598)^2
     ≈ π * 12.94
     ≈ 40.59 square units.