Question

A triangle has corners at `(3 ,5 )`, `(4 ,9 )`, and `(4 ,2 )`. What is the area of the triangle's circumscribed circle?

Answer 1

Area circumscribed circle `=(85pi)/2=133.5176878" "`square units

Explanation:

For any plane triangle, the `color(red)("Sine Law")` states that

`a/sin A=b/sin B=c/sin C`

The radius `R` of the triangle's circumscribed circle is given

`R=a/(2*sin A)=b/(2*sin B)=c/(2*sin C)`

If we let the vertices as `A(3, 5)` and `B(4, 9)` and `C(4, 2)` then we have

`R=(BC)/(2*sin A)=(AC)/(2*sin B)=(AB)/(2*sin C)`

We only need to use

`R=(BC)/(2*sin A)`

Angle A can be determined through the slopes
`m_(AB)=(9-5)/(4-3)=4` and `m_(AC)=(5-2)/(3-4)=-3`

Angle `A=tan^-1 ((m_(AB)-m_(AC))/(1+m_(AB)*m_(AC)))`

Angle `A=tan^-1 ((4-(-3))/(1+4*(-3)))`

Angle `A=tan^-1 (7/(-11))=147.5288077^@`

We can now compute for the radius `R`

`R=(BC)/(2*sin A)=7/(2*sin 147.5288077^@)`

`R=6.519202405`

Compute the area of the triangle's circumscribed circle

Area `=piR^2`

Area `=pi(6.519202405)^2`

Area `=42.5 pi`
Area `=133.5176878" "`square units

God bless....I hope the explanation is useful.