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:

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

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