A triangle has corners at $(6 ,3 )$ , $(5 ,4 )$ , and $(3 ,2 )$ . What is the area of the triangle's circumscribed circle?

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Answer 1 · 2017-02-20T17:03:37

$2.5pi.$

Name the points $A(6,3), B(5,4) and C(3,2).$

By the Distance Formula,

$AB^2=(6-5)^2+(3-4)^2=1+1=2,$

$BC^2=4+4=8, &, AC^2=10 rArr AB^2+BC^2=AC^2.$

This means that, $Delta ABC" is a right "Delta,$ having

Hypotenuse $AC.$

Knowing that, in the right $Delta$ , the hypotenuse is the

circum-diameter, we find that, the cicrumradius $R$ of

$DeltaABC$ is, $R=(AC)/2.$

Therefore, the Area of the Circumscribed Circle of $DeltaABC$

$=piR^2=pi(AC)^2/4=pi(10/4)=2.5pi.$

Enjoy Maths.!

A triangle has corners at (6 ,3 )(6 ,3 ), (5 ,4 )(5 ,4 ), and (3 ,2 )(3 ,2 ). What is the area of the triangle's circumscribed circle?