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

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Answer 1 · 2018-02-02T12:29:49

Area of triangle's circumscribed circle is #19.63# sq.unit.

Vertices of triangle are #A(6,4), B(7,6) , (3,8)#

Side #AB=a=sqrt((6-7)^2+(4-6)^2)=sqrt5 ~~2.24#

Side #BC=b=sqrt((7-3)^2+(6-8)^2)= sqrt20 ~~ 4.47#

Side #CA=c=sqrt((3-6)^2+(8-4)^2)=sqrt 25 = 5.0#

Semi perimeter of triangle #S=(2.24+4.47+5.0)/2=5.855#

Area of the triangle #A_t=sqrt(s(s-a)(s-b)(s-c))#

#=sqrt(5.555(5.855-2.24)(5.855-4.47)(5.855-5.0))#

# = sqrt25.06 ~~5.0# sq.unit.

Circumscribed circle radius is #R=(a*b*c)/(4.A_t)#

#R=(2.24*4.47*5.0)/(4*5.0) ~~ 2.5# unit

Area of circumscribed circle is #A_c =pi*R^2=pi*2.5^2~~19.63#.

Area of circumscribed circle is #19.63# sq.unit. [Ans]