Question

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

Answer 1

Area of the triangle's circumscribed circle`= (53pi)/9` units.

Explanation:

First of all, we have to find out circumcenter (G) of the triangle.

`G=(3+5+8)/3,(8+9+5)/3`

`G=(16)/3,(22)/3`

For area of the triangle's circumscribed circle, we have to calculate radius of the circumscribed circle. And, it is equal to the distance between G and any of the vertices of the triangle.

`R= sqrt((16/3 -3)^2+(22/3 -8)^2`

`R= sqrt(((16-9)/3)^2+((22-24)/3)^2`

`R= sqrt(((7)/3)^2+((-2)/3)^2`

`R= sqrt(49/9+4/9`

`R= sqrt(53/9`

`R= sqrt(53)/3` units

Area of the triangle's circumscribed circle`= piR^2`

Area of the triangle's circumscribed circle`= pixx53/9`

Area of the triangle's circumscribed circle`= (53pi)/9` units.

Answer 2

#{ 2125 pi }/242 #

Explanation:

There are no square roots needed. Please contrast this with the other answer, 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}`

`a^2,b^2 and c^2` are available directly from `(3,8),(5,9),(8,5)`

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

`b^2 =(5-8)^2+(9-5)^2=25`

`c^2=(8-3)^2+(5-8)^2=34`

# text{circumcircle area} = pi r^2 = pi {(5)(25)(34) }/{ 4(5)(25) - (34-5-25)^2 } ={2125 pi }/242 #

It appears me or the other answer is wrong.

Check: Alpha `quad` I'm right `quad sqrt`