A triangle has corners at $(4 ,6 )$ , $(2 ,9 )$ , and $(7 ,5 )$ . What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-07-11T10:21:54

The area of the triangle's circumscribed circle $=Delta=85.4329$ sq. units.

Let $triangleABC " be the triangle with corners at"$

$A(4,6) ,B(2,9) and C(7,5)$

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$a=BC=sqrt((2-7)^2+(9-5)^2)=sqrt(25+16)=sqrt41$

$b=CA=sqrt((4-7)^2+(6-5)^2)=sqrt(9+1)=sqrt10$

$c=AB=sqrt((4-2)^2+(6-9)^2)=sqrt(4+9)=sqrt13$

$cosB=(c^2+a^2-b^2)/(2ca)=(13+41-10)/(2sqrt13sqrt41)=22/(sqrt533$

We know that,

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

$=>sin^2B=1-484/533=49/533$

$=>sinB=7/sqrt533to[because Bin(0 ^circ,180^circ)]$

Using sine formula:we get

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

$=>R=sqrt10/(2 (7/sqrt533))=(sqrt10xxsqrt533)/(2*7)~~5.2148$

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

$Delta=piR^2=pi*(5.2148)^2~~85.4329 ,sq.units$

A triangle has corners at (4 ,6 )(4 ,6 ), (2 ,9 )(2 ,9 ), and (7 ,5 )(7 ,5 ). What is the area of the triangle's circumscribed circle?