How to find circle area circumscribed to the triangle with the sides:#a=4sqrt3,b=3sqrt2,c=sqrt66#?

1 Answer
Jun 24, 2017

Here in #DeltaABC#

#a=4sqrt3,b=3sqrt2,c=sqrt66#

We see #a^2+b^2=(4sqrt3)^2+(3sqrt2)^2=48+18=66=(sqrt66)=c^2#

So #c# is the hypotenuse of #DeltaABC# and #/_C=90^@#

Hence side #c# will be diameter of the circle circumscribing #DeltaABC#. So it is obvious the radius of the circle circumscribing #DeltaABC# will be #r=c/2#

Hence the area of the circle circumscribing #DeltaABC# is

#Area =pir^2=(pic^2)/4=22/7xx66/4=363/7=51 6/7# squnit