A triangle has corners at $(3 ,6 )$ , $(2 ,9 )$ , and $(8 ,4 )$ . What is the area of the triangle's circumscribed circle?

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Answer 1 · 2016-07-12T17:46:35

Area of the Circumcircle $=(61*29*5*pi)/(26*13)~=82.17$ sq.unit

Let $A(3,6), B(2,9), C(8,4)$ be the vertices of $DeltaABC.$

In the usual notation, we know by Trigo., that, $abc=4RDelta....(i).$

Here, $a^2=BC^2=(2-8)^2+(9-4)^2=36+25=61$ .

$b^2=AC^2=(3-8)^2+(6-4)^2=25+4=29$ .

$c^2=AB^2=(3-2)^2+(6-9)^2=1+9=10$ .

Also, $Delta=1/2|D|,$ where, $D=|(3,6,1),(2,9,1),(8,4,1)|$ ,

$=3(9-4)-6(2-8)+1(8-72)=3(5)-6(-6)-64=15+36-64=-13$ ,

$:. Delta=(1/2)|-13|=13/2.$

Thus, $a^2=61, b^2=29, c^2=10, Delta=13/2$ . Sub.ing, in $(i)$ ,

$sqrt(61*29*10)=4R*13/2=26R rArr R=sqrt(61*29*10)/26.$

Therefore, Area of the Circumcircle of $DeltaABC=piR^2=(61*29*10*pi)/(26*26)=(61*29*5*pi)/(26*13)~=82.17$ sq.unit

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A triangle has corners at (3 ,6 )(3 ,6 ), (2 ,9 )(2 ,9 ), and (8 ,4 )(8 ,4 ). What is the area of the triangle's circumscribed circle?