A triangle has corners at #(5 ,8 )#, #(2 ,6 )#, and #(7 ,3 )#. What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-08-11T16:03:37

The area of the triangle's circumscribed circle is:
#Delta~~23.2499 ,sq.units#

Let , #triangle ABC# be the triangle with corners at

#A(5,8), B(2,6) and C(7,3) .#

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#a=BC=sqrt((7-2)^2+(3-6)^2)=sqrt(25+9)=sqrt34#

#b=CA=sqrt((5-7)^2+(8-3)^2)=sqrt(4+25)=sqrt29#

#c=AB=sqrt((5-2)^2+(8-6)^2)=sqrt(9+4)=sqrt13#

#cosB=(c^2+a^2-b^2)/(2ca)=(13+34-29)/(2sqrt13sqrt34)=18/(2sqrt442)=9/sqrt442#

We know that,

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

#=>sin^2B=1-9/442=433/442#

#=>sinB=(sqrt(433/442))to[because Bin(0 ^circ,180^circ)]#

Using sine formula:we get

#b/sinB=2R=>R=b/(2sinB)#

#=>R=sqrt29/(2(sqrt(433/442)))~~2.75#

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

#Delta=piR^2=pi*(sqrt29/(2(sqrt(433/442))))^2=pi((29xx442)/(4 xx433))#

#Delta~~23.2499 ,sq.units#