Question

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

Answer 1

`=\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`