A triangle has corners at #(3 ,3 )#, #(2 ,9 )#, and #(8 ,4 )#. What is the area of the triangle's circumscribed circle?

Question

A triangle has corners at #(3  ,3   )#, #(2   ,9   )#, and #(8   ,4  )#.  What is the area of the triangle's circumscribed circle?

Evelyn Arboleda · Accepted Answer

Area of the circumcircle = 49.5193

Circumcenter of a triangle is a point where all the three perpendicular bisectors of the sides concur. Distance between the vertices and the Circumcenter is the radius of the circumcircle.

Steps to find the area of the circumcircle
1. Find and calculate the mid point of the sides AB, BC, CA.
2. Calculate the slope of the particular line.
3. By using the midpoint and the slope, find the equation of the perpendicular bisector.
4. Find the equation of other perpendicular bisectors.
5. Solve 2 bisector equations to find out the intersection point (circumcenter.
6. Find the distance between circumcenter and one vertex of the triangle to find the radius of the circumcircle.
7. Having known the radius, area of the circumcircle is given by formula #pir^2#.

Claire Akin · Answer

undefined To find the area of the circumscribed circle of a triangle, you can use the formula $ A = \frac{{abc}}{{4R}} $, where $ A $ is the area of the triangle, $ a $, $ b $, and $ c $ are the lengths of the sides of the triangle, and $ R $ is the radius of the circumscribed circle.

First, calculate the lengths of the sides of the triangle using the distance formula:

1. Side $ a $ between (3, 3) and (2, 9):
   $ a = \sqrt{(2 - 3)^2 + (9 - 3)^2} = \sqrt{(-1)^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37} $

2. Side $ b $ between (3, 3) and (8, 4):
   $ b = \sqrt{(8 - 3)^2 + (4 - 3)^2} = \sqrt{5^2 + 1^2} = \sqrt{25 + 1} = \sqrt{26} $

3. Side $ c $ between (2, 9) and (8, 4):
   $ c = \sqrt{(8 - 2)^2 + (4 - 9)^2} = \sqrt{6^2 + (-5)^2} = \sqrt{36 + 25} = \sqrt{61} $

Now, calculate the area of the triangle using Heron's formula:

$$ s = \frac{{a + b + c}}{2} = \frac{{\sqrt{37} + \sqrt{26} + \sqrt{61}}}{2} $$

$$ A = \sqrt{s(s - a)(s - b)(s - c)} $$

Where $ s = \frac{{\sqrt{37} + \sqrt{26} + \sqrt{61}}}{2} $.

Once you have the area of the triangle, you can find the radius of the circumscribed circle using the formula $ A = \frac{{abc}}{{4R}} $ and then the area of the circumscribed circle using the formula for the area of a circle $ \pi R^2 $.