A triangle has corners at #(6 ,4 )#, #(8 ,2 )#, and #(3 ,1 )#. What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-01-25T12:08:52

Area of circumscribed circle is #20.42# sq.unit.

The three corners are #A (6,4) B (8,2) and C (3,1)#

Distance between two points #(x_1,y_1) and (x_2,y_2)# is

#D= sqrt ((x_1-x_2)^2+(y_1-y_2)^2#

Side #AB= sqrt ((6-8)^2+(4-2)^2)=sqrt(8) ~~ 2.83#unit

Side #BC= sqrt ((8-3)^2+(2-1)^2)=sqrt(26) ~~5.10#unit

Side #CA= sqrt ((3-6)^2+(1-4)^2)=sqrt(18) ~~ 4.24#unit

Area of Triangle is #A_t = |1/2(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))|#

#A_t = |1/2(6(2−1)+8(1−4)+3(4−2))|# or

#A_t = |1/2(6-24+6)| = |-6| =6.0# sq.unit.

Radius of circumscribed circle is #R=(AB*BC*CA)/(4*A_t)# or

#R=(sqrt(8)*sqrt(26)*sqrt(18))/(4*6) ~~ 2.55# unit

Area of circumscribed circle is #A_c=pi*R^2=pi*2.55^2~~20.42#

sq.unit [Ans]