Where Do I Put This $\triangle$ ?

Let x be half of the length of the base. Let y be the length of the height.

By Pythagoras's Theorem, $x^2+y^2=36$ $x^2+(9-y)^2=81$ $\Rightarrow x^2+y^2-18y=0$ $\Rightarrow 36=18y$ $\Rightarrow y=2, x=4\sqrt{2}$

The area then is just $xy=2*4\sqrt{2}=8\sqrt{2} \approx 11.313708$

Let $\angle ABC=\angle ACB=\theta$ and $\angle BAC=\gamma$ . By Law of Sines we know that $2R=\dfrac{AB}{\sin \theta}$ , so $\sin \theta=\dfrac{AB}{2R}=\dfrac{6}{2(9)}=\dfrac{1}{3}$ . Now, we have that $2\theta+\gamma=\pi$ , then $\gamma=\pi-2\theta$ and $\sin \gamma=2\sin \theta \cos \theta$ . Find that $\cos\theta=\sqrt{1-\left(\dfrac{1}{3}\right)^2}=\dfrac{2\sqrt{2}}{3}$ . So $\sin \gamma=2\left(\dfrac{1}{3}\right)\left(\dfrac{2\sqrt{2}}{3}\right)=\dfrac{4\sqrt{2}}{9}$ .

Finally, $[ABC]=\dfrac{1}{2}AB \cdot AC \cdot \sin \gamma=\dfrac{1}{2}\cdot 6 \cdot 6 \cdot \dfrac{4\sqrt{2}}{9}=8\sqrt{2} \approx 11.314$ .

Assume a circle centered at origin with radius = 9

Equation is x^2 + y^2 = 81 (Eqn 1)

Place vertex A at (0,9)

Since AB and AC = 6 they will not reach the ends of a diameter but touch the circle above the diameter.

Assume another circle centered at (0,9) with radius = 6

Equation is x^2 + (y-9)^2 = 36 (Eqn 2)

The two circle intersect at B and C, the vertices of the triangle.

Subtracting the two equations :

 y^2 - (y-9)^2 = 45  which yields y = 7

Substituting back into first equation results in x = +/- 5.66

Therefore the base of the triangle is 2 * 5.66 = 11.32

The height is 9 - 7 = 2

So the area is 0.5 x b x h = 0.5 x2x11.32 = 11.32 cm

R= a b c/4(area) ..... so 9=36 c/4(area).......area=c so 0.5 * c * h=area.......so h=2......and so c/2=(6^2 - 2^2)^(0.5)=(32)^(0.5).... so c=area=2 (32)^(0.5)=8*(2)^(0.5)