Question

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

Answer 1

`12.5pi~~39.3`

Explanation:

Given `A(5,1), B(2,7) and C(7,2)`

By the distance formula, we get :
`AB^2=(2-5)^2+(7-1)^2=45`
`AC^2=(7-5)^2+(2-1)^2=5`
`BC^2=(7-2)^2+(2-7)^2=50`
`=> BC^2=AB^2+AC^2`
This means that `DeltaABC` is a right triangle, with `BC` as the hypotenuse.
In a right triangle, the hypotenuse is the circum-diameter,
`=>` the circumradius `r` of `DeltaABC =(BC)/2`

Therefore, the area of the circumscribed circle of `DeltaABC`
`=pir^2=pi(BC)^2/4=pixx50/4=12.5pi`