# A triangle has corners at (3 ,8 ), (5 ,9 ), and (8 ,5 ). What is the area of the triangle's circumscribed circle?

Jun 1, 2018

Area of the triangle's circumscribed circle$= \frac{53 \pi}{9}$ units.

#### Explanation:

First of all, we have to find out circumcenter (G) of the triangle.

$G = \frac{3 + 5 + 8}{3} , \frac{8 + 9 + 5}{3}$

$G = \frac{16}{3} , \frac{22}{3}$

For area of the triangle's circumscribed circle, we have to calculate radius of the circumscribed circle. And, it is equal to the distance between G and any of the vertices of the triangle.

R= sqrt((16/3 -3)^2+(22/3 -8)^2

R= sqrt(((16-9)/3)^2+((22-24)/3)^2

R= sqrt(((7)/3)^2+((-2)/3)^2

R= sqrt(49/9+4/9

R= sqrt(53/9

$R = \frac{\sqrt{53}}{3}$ units

Area of the triangle's circumscribed circle$= \pi {R}^{2}$

Area of the triangle's circumscribed circle$= \pi \times \frac{53}{9}$

Area of the triangle's circumscribed circle$= \frac{53 \pi}{9}$ units.

Jun 5, 2018

{ 2125 pi }/242

#### Explanation:

Archimedes' Theorem says for a triangle with sides $a , b , c :$

$16 \setminus {\textrm{a r e a}}^{2} = 4 {a}^{2} {b}^{2} - {\left({c}^{2} - {a}^{2} - {b}^{2}\right)}^{2}$

The radius of the circumcircle equals the product of the sides of the triangle divided by four times the area of the triangle. This is more useful squared:

${r}^{2} = \frac{{a}^{2} {b}^{2} {c}^{2}}{16 \setminus {\textrm{a r e a}}^{2}} = \frac{{a}^{2} {b}^{2} {c}^{2}}{4 {a}^{2} {b}^{2} - {\left({c}^{2} - {a}^{2} - {b}^{2}\right)}^{2}}$

${a}^{2} , {b}^{2} \mathmr{and} {c}^{2}$ are available directly from $\left(3 , 8\right) , \left(5 , 9\right) , \left(8 , 5\right)$

${a}^{2} = {\left(3 - 5\right)}^{2} + {\left(8 - 9\right)}^{2} = 5$

${b}^{2} = {\left(5 - 8\right)}^{2} + {\left(9 - 5\right)}^{2} = 25$

${c}^{2} = {\left(8 - 3\right)}^{2} + {\left(5 - 8\right)}^{2} = 34$

 text{circumcircle area} = pi r^2 = pi {(5)(25)(34) }/{ 4(5)(25) - (34-5-25)^2 } ={2125 pi }/242

It appears me or the other answer is wrong.

Check: Alpha $\quad$ I'm right quad sqrt