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 $A=40.94$

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