A triangle has corners at #(9 ,7 )#, #(3 ,1 )#, and #(5 ,2 )#. What is the area of the triangle's circumscribed circle?

2 Answers
Apr 28, 2017

9225/8

Explanation:

In order to find the area of the circumscribed circle, we need to find its radius. There is a formula to do this, although it is a little troublesome.

The formula for the circumradius is: R= abc/4A
where R is the circumradius, a,b,c are the sides of the triangle, and A is the area of the triangle.

For a proof of the formula, go to this site: https://artofproblemsolving.com/wiki/index.php?title=Circumradius

So we need to find the area of the triangle and its side lengths.

To find the area of the triangle, we use the shoelace formula. If you don't know what that is, go to: https://brilliant.org/wiki/triangles-calculating-area/#area-of-triangles-shoelace-formula (you may also learn some other methods to calculate the area of a triangle)

So A = #1/2|9*1-3*7+3*2-1*5+5*7-2*9|# = #1/2|-12+1+17|# = #1/2*6# = 3

Next, find each of the side lengths using the distance formula:
a = #sqrt((5-3)^2+(2-1)^2) = sqrt(4+1) = sqrt5#
b = #sqrt((9-5)^2+(7-2)^2) = sqrt(16+25) = sqrt41#
c = #sqrt((9-3)^2+(7-1)^2) = sqrt(36+36) = sqrt72#

Not nice numbers but still. abc = #sqrt(5*41*72) = sqrt(5*2952) = sqrt14760#

R = #sqrt(14760)/12#

A (circle) = #pi*R^2# = #pi* 14760/144 = 9225/8#

Hope that helps!

Apr 28, 2017

The area of the circumscribed circle is:

#A = (205pi)/2#

Explanation:

Shift the given points so that one of them is the origin:

#(3-3,1-1)to (0,0)#
#(9-3,7-1)to (6,6)#
#(5-3,2-1)to (2,1)#

Use the equation of a circle #(x-h)^2+(y-k)=r^2# and the 3 new points to write 3 equations:

#h^2+k^2=r^2" [1]"#
#(6-h)^2+(6-k)^2=r^2" [2]"#
#(2-h)^2+(1-k)^2=r^2" [3]"#

Substitute the left side of equation [1] into equations [2] and [3]:

#(6-h)^2+(6-k)^2=h^2+k^2" [4]"#
#(2-h)^2+(1-k)^2=h^2+k^2" [5]"#

Expand the squares:

#36-12h+h^2+36-12k+k^2=h^2+k^2" [6]"#
#4-4h+h^2+1-2k+k^2=h^2+k^2" [7]"#

Subtract #h^2+k^2# from both sides:

#36-12h+36-12k=0" [8]"#
#4-4h+1-2k=0" [9]"#

Divide the first equation by -12 and multiply the second by -1:

#h+k=6" [10]"#
#4h+2k=5" [11]"#

Multiply equation [10] by -2 and add to equation [11]:

#2h = -7#

#h = -7/2#

Use equation [10] to solve for k:

#-7/2 + k = 6#

#k = 19/2#

Use equation [1] to find the value of r^2:

#r^2 = (-7/2)^2+(19/2)^2#

#r^2 = (-7/2)^2+(19/2)^2#

#r^2 = 205/2#

The area of a circle is:

#A = pir^2#

Therefore, the area of the circumscribed circle is:

#A = (205pi)/2#