A triangle has corners at #(5 ,1 )#, #(7 ,9 )#, and #(4 ,3 )#. What is the area of the triangle's circumscribed circle?

1 Answer
Harish Chandra Rajpoot
Jun 19, 2018

#=\frac{3825\pi}{16}\ unit^2#

Explanation:

Lengths #a, b \ & \ c# of sides of given triangle with vertices at #(5, 1)#, #(7, 9)# & #(4, 3)# are given as
#a=\sqrt{(5-7)^2+(1-9)^2}=2\sqrt{17}#
#b=\sqrt{(5-4)^2+(1-3)^2}=\sqrt{5}#
#c=\sqrt{(7-4)^2+(9-3)^2}=9\sqrt{5}#
The area (#\Delta#) of given triangle with vertices at #(5, 1)#, #(7, 9)# & #(4, 3)# are given as
#\Delta=1/2 |5(9-3)+7(3-1)+4(1-9)|#
#=1/2 |12|#
#=6#
Now, the radius (#R#) of circum-circle is given as follows
#R=\frac{abc}{4\Delta}#
#=\frac{2\sqrt{17}\cdot \sqrt5\cdot 9\sqrt5}{4cdot 6}#
#=\frac{15\sqrt{17}}{4}#
Hence the area of circumscribed circle
#=\pi R^2#
#=\pi (\frac{15\sqrt{17}}{4})^2#
#=\frac{3825\pi}{16}\ unit^2#
#=751.0369936\ unit^2#

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