A triangle has corners at $(7 ,4 )$ , $(3 ,6 )$ , and $(4 ,8 )$ . What is the area of the triangle's circumscribed circle?

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Answer 1 · 2016-07-03T13:57:41

Area of the crcm.crcl. of $Delta=pi*R^2=25pi/4~=19.625.$

Call the vertices $A(7,4), B(3,6)$ and $C(4,8).$

Let $c$ denote the length of side $AB$ , then, by Dist. Formula,

$c^2=(7-3)^2+(4-6)^2=16+4=20.$

Similarly, $a^2=1+4=5$ and, $b^2=9+16=25$

$:. b^2=c^2+a^2$

Therefore, $DeltaABC$ is a right $/_^(ed) Delta$ with $b=AC$ as its hypo.

Accordingly, its circumcentre is the mid-pt. $M$ of hypo. $AC$ , and as such, its cicm.radi. $R=b/2=5/2.$

Finally, the reqd. Area of the crcm.crcl. of $Delta ABC=pi*R^2=25pi/4~=19.625.$

A triangle has corners at (7 ,4 )(7 ,4 ), (3 ,6 )(3 ,6 ), and (4 ,8 )(4 ,8 ). What is the area of the triangle's circumscribed circle?