Question

A triangle has corners at `(2 ,8 )`, `(3 ,6 )`, and `(4 ,7 )`. What is the area of the triangle's circumscribed circle?

Answer 1

`"The area of the triangle's circumscribed circle is :"`
`Delta=piR^2=pi*(sqrt50/6)^2=pi(50/36)~~4.3633 ,sq.units`

Explanation:

Let `triangle ABC` be the triangle with corners at

`A(2,8)B(3,6) and C(4,7)`

Using Distance formula ,we get

`a=BC=sqrt((7-6)^2+(4-3)^2)=sqrt(1+1)=sqrt2`

`b=CA=sqrt((8-7)^2+(2-4)^2)=sqrt(1+4)=sqrt5`

`c=AB=sqrt((8-6)^2+(2-3)^2)=sqrt(4+1)=sqrt5`

Using cosine Formula ,we get

`cosB=(c^2+a^2-b^2)/(2ca)=(5+2-5)/(2sqrt5sqrt2)=2/(2sqrt10)=1/sqrt10`

We know that,

`sin^2B=1-cos^2B`

`=>sin^2B=1-1/10=9/10`

`=>sinB=3/(sqrt10)to[because Bin(0 ^circ,180^circ)]`

Using sine formula:we get

`b/sinB=2R=>R=b/(2sinB)`

`=>R=sqrt5/(2(3/sqrt10))=(sqrt50)/(6)~~1.1785`

So , the area of the triangle's circumscribed circle is:

`Delta=piR^2=pi*(sqrt50/6)^2=pi(50/36)~~4.3633 ,sq.units`