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:

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

$Delta~~23.2499 ,sq.units$

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