Question

A triangle has corners at `(4 ,6 )`, `(8 ,9 )`, and `(3 ,5 )`. What is the area of the triangle's circumscribed circle?

Answer 1

Area of circumscribed circle is `1610.07` sq.unit .

Explanation:

The three corners are `A (4,6) B (8,9) and C (3,5)`

Distance between two points `(x_1,y_1) and (x_2,y_2)` is

`D= sqrt ((x_1-x_2)^2+(y_1-y_2)^2`

Side `AB= sqrt ((4-8)^2+(6-9)^2)=sqrt(25) = 5.0` unit

Side `BC= sqrt ((8-3)^2+(9-5)^2)=sqrt(41) ~~6.40`unit

Side `CA= sqrt ((3-4)^2+(5-6)^2)=sqrt(2) ~~ 1.41`unit

Area of Triangle is

`A_t = |1/2(x_1(y_2−y_3)+x_2(y_3−y_1)+x_3(y_1−y_2))|`

`A_t = |1/2(4(9−5)+8(5−6)+3(6−9))|` or

`A_t = |1/2(16-8-9)| = 1/2*1 =0.5` sq.unit.

Radius of circumscribed circle is `R=(AB*BC*CA)/(4*A_t)` or

`R=(5*sqrt(41)*sqrt(2))/(4*0.5) ~~ 22.64` unit

Area of circumscribed circle is `A_c=pi*R^2=pi*22.64^2`

`~~1610.07`sq.unit [Ans]

Answer 2

`{1025 pi }/2`

Explanation:

Allow me to illustrate the proper method; no square roots needed. Please contrast this with the other answer, which is a typical good answer.

Archimedes' Theorem says for a triangle with sides `a,b,c:`

`16 \ text{area}^2 = 4a^2b^2-(c^2-a^2-b^2)^2`

The radius of the circumcircle equals the product of the sides of the triangle divided by four times the area of the triangle. This is more useful squared:

`r^2 = {a^2 b^2 c^2}/{16 \ text{area}^2} = {a^2 b^2 c^2}/{ 4a^2b^2-(c^2-a^2-b^2)^2}`

That all we need to do our problem. `(4,6),(8,9),(3,5)`

`a^2 = (4-8)^2+(6-9)^2=25`

`b^2 = (8-3)^2+(9-5)^2=41`

`c^2=(4-3)^2+(6-5)^2=2`

# text{circumcircle area} = pi r^2 = pi {25(41)(2)}/{4(25)(2)-(41-25-2)^2 } = {1025 pi }/2 #