A triangle has corners at (5 ,1 )(5,1), (7 ,9 )(7,9), and (4 ,3 )(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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