A triangle has corners at #(9 ,3 )#, #(4 ,1 )#, and #(2 ,4 )#. What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-07-11T10:48:03

The area of the triangle's circumscribed circle is#=Delta=41.0096#sq.units

Let , #triangleABC" be the triangle with corners at "#

#A(9,3) , B(4,1) and C(2,4).#

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#a=BC=sqrt((4-2)^2+(1-4)^2)=sqrt(4+9)=sqrt13#

#b=CA=sqrt((9-2)^2+(3-4)^2)=sqrt(49+1)=sqrt50#

#c=AB=sqrt((9-4)^2+(3-1)^2)=sqrt(25+4)=sqrt29#

#cosA=(b^2+c^2-a^2)/(2bc)=(50+29-13)/(2sqrt50sqrt29)=33/(sqrt1450#

We know that,

#sin^2A=1-cos^2A#

#=>sin^2A=1-1089/1450=361/1450#

#=>sinA=19/sqrt1450to[because Ain(0 ^circ,180^circ)]#

Using sine formula:we get

#a/sinA=2R=>R=a/(2sinA)#

#=>R=sqrt13/(2 (19/sqrt1450))=(sqrt13xxsqrt1450)/(2*19)~~3.6130#

So , the area of the triangle's circumscribed circle is:

#Delta=piR^2=pi*(3.6130)^2~~41.0096 ,sq.units#
....................................................................................................

Note:

If we take, #R=3.61# then #A=40.94#