A triangle has corners at #(9 ,5 )#, #(2 ,3 )#, and #(7 ,4 )#. What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-07-23T07:26:48

The area of the triangle's circumscribed circle is:
#Delta=piR^2=pi*(5.9238)^2~~110.24 ,sq.units#

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

#A(9,5) , B(2,3) and C(7,4)#

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#a=BC=sqrt((2-7)^2+(3-5)^2)=sqrt(25+4)=sqrt29#

#b=CA=sqrt((9-7)^2+(5-4)^2)=sqrt(4+1)=sqrt5#

#c=AB=sqrt((9-2)^2+(5-3)^2)=sqrt(49+4)=sqrt53#

#cosB=(c^2+a^2-b^2)/(2ca)=(53+29-5)/(2sqrt53sqrt29)=77/(2sqrt1537)#

We know that,

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

#=>sin^2B=1-5929/(4xx1537)=219/6748#

#=>sinB=sqrt219/(2sqrt1537)to[because Bin(0 ^circ,180^circ)]#

Using sine formula:we get

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

#=>R=sqrt5/(2xx(sqrt219/(2sqrt1537)))=(sqrt5xxsqrt1537)/(sqrt219)~~5.9238#

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

#Delta=piR^2=pi*(5.9238)^2~~110.24 ,sq.units#