A triangle has corners at (3 ,7 ), (2 ,5 ), and (8 ,4 ). What is the area of the triangle's circumscribed circle?

2 Answers
Nov 7, 2016

Area of the triangle's circumscribed circle: color(green)(2.65 " sq.units") (approx.)

Explanation:

Perhaps there is a simpler way, but here is how I would do it:

Part 1: Find the Lengths of the Three Sides
Let a be the side (3,7):(2,5)
color(white)("XXX")abs(a)=sqrt((2-3)^2+(5-7)^2)=sqrt(1+4)=sqrt(5)

Let b be the side (2,5):(8,4)
color(white)("XXX")abs(b)=sqrt((8-2)^2+(4-5)^2)=sqrt(36+1)=sqrt(37)

Let c be the side (8,4):(3,7)
color(white)("XXX")abs(c)=sqrt((3-8)^2+(7-4)^2)=sqrt(25+9)=sqrt(34)

Part 2: Find the Perimeter, Semi-Perimeter, and Area of the Triangle
(I will need the help of a calculator here)
Perimeter: p=sqrt(5)+sqrt(37)+sqrt(34)~~14.1497824026

Semi-perimeter: s=p/2~~7.0748912013

Area of triangle (using Heron's formula):
color(white)("XXX")"Area_triangle=sqrt(s(s-a)(s-b)(s-c))~~6.5

Part 3: Find the Radius and Area of the Circumscribed Circle
(more calculator usage)
Radius of Circumscribed Circle: r=("Area"_triangle)/s=0.9187420435
(ask as a separate question if you are unfamiliar with this relation).

Area of Circle: "Area"_circ = pir^2 =2.6517773377

Nov 7, 2016

A = (6290pi)/676

Explanation:

The standard form for the equation of a circle is:

(x - h)^2 + (y - k)^2 = r^2

where (h, k) is the center and r is the radius

Because the triangles vertices are points on the circumscribed circle, we can use the 3 points and the standard form to write 3 equations:

(3 - h)^2 + (7 - k)^2 = r^2" [1]"
(2 - h)^2 + (5 - k)^2 = r^2" [2]"
(8 - h)^2 + (4 - k)^2 = r^2" [3]"

We have 3 equations and we can use them to find the values of, h, k, and r^2.

Temporarily eliminate r^2 by setting the left side of equation [1] equal to the left side of equation [2] and the same for left side of equation [3]:

(3 - h)^2 + (7 - k)^2 = (2 - h)^2 + (5 - k)^2" [1 = 2]"
(3 - h)^2 + (7 - k)^2 = (8 - h)^2 + (4 - k)^2" [1 = 3]"

Expand the squares, using the pattern, (a - b)^2 = a^2 - 2ab + b^2:

9 -6h + h^2 + 49 -14k + k^2 = 4 -4h + h^2 + 25 -10k + k^2
9 -6h + h^2 + 49 -14k + k^2 = 64 -16h + h^2 + 16 -8k + k^2

Please notice that for every h^2 and k^2 on the left there is a corresponding one on the right so they cancel:

9 -6h + 49 -14k = 4 -4h + 25 -10k
9 -6h + 49 -14k = 64 -16h + 16 -8k

Combine all of the constant terms into a single term on the right:

- 6h - 14k = - 4h- 10k - 29
- 6h- 14k =- 16h- 8k + 22

Combine all of the h terms into a single term on the right:

- 14k = 2h- 10k - 29
- 14k =- 10h- 8k + 22

Combine all of the k terms into a single term on the left:

-4k = 2h - 29" [4]"
-6k = -10h + 22" [5]"

Multiply equation [4] by -3 and equation [5] by 2 and add them:

12k - 12k = -6h - 20h + 87 + 44

0 = -26h + 131

h = 131/26

Substitute 131/26 for h in equation [4]:

-4k = 2(131/26) - 29

-4k = -492/26

k = 123/26

Substitute the values for h and k into equation [3]

(8 - 131/26)^2 + (4 - 123/26)^2 = r^2

(77/26)^2 + (-19/26)^2 = r^2

r^2 = 6290/676

The area of the circle is A = pir^2

A = (6290pi)/676