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#