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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