Kepler Triangle

What is the area of the largest Kepler Triangle that can be inscribed in the circle whose equation is x^2 + y^2  = 9   ????

{A Kepler Triangle is a right triangle whose sides are in the ratio of 1 : √Phi : Phi....where Phi = [1 + √5] / 2 }

$$\boxed{a:b:c = 1:\sqrt{\varphi}:\varphi}$$

$$r_{circle} = \sqrt{9} = 3\qquad c = 2\cdot r_{circle} \\\\ \small{\text{$ \begin{array}{rcl} \sqrt{ \varphi } & \cdot & \dfrac{ \varphi }{ \sqrt{ \varphi } } = \varphi \\\\ b & \cdot & \dfrac { \varphi } { \sqrt{ \varphi } } = c \\\\ \textcolor[rgb]{1,0,0}{b} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{c \cdot \dfrac{ \sqrt{ \varphi } } { \varphi }}\\\\ \hline &\\ \end{array} $}}\\\\\\ \small{\text{$ \begin{array}{rcl} 1 & \cdot & \dfrac{ \varphi }{ 1 } = \varphi \\\\ a & \cdot & \dfrac{ \varphi }{ 1 } = c \\\\ \textcolor[rgb]{1,0,0}{a} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{\dfrac{ c } { \varphi } }\\\\ \hline &\\ \end{array} $}}\\\\$$

$$\small{\text{$ \begin{array}{rcl} A &=& \dfrac{a\cdot b}{2} \\\\ A &=& \dfrac{\dfrac{ c } { \varphi }\cdot c \cdot \dfrac{ \sqrt{ \varphi } } { \varphi } }{2} \\\\ A&=& \dfrac{c^2}{\varphi^2} \cdot \dfrac{ \sqrt{ \varphi } } { 2}\\\\ A &=& 2\cdot r^2 \cdot \dfrac{ \sqrt{ \varphi } } { \varphi^2 } \\\\ A &=& 18 \cdot \dfrac{ \sqrt{ \varphi } } { \varphi^2 } \\\\ A &=& 8.74562889162 \\\\ \end{array} $}}$$

(I realize that golden rectangle and kepler triangle are different,Thank you CPhill!)

Yep....correct....nice work, fiora and heureka.....!!!!!

Was it a test Chris ?

LOL!!!!!......nah.......it's just something I thought up that might make a nice problem.......

Did you already know the answer and how to get there ?

Yep......I used fiora's  approach  [more or less ].......but.....I like the way heureka did it, too.....