Maximum area of a isosceles triangle in a circle with a radius r As said in the title, I'm looking for the maximum area of a isosceles triangle in a circle with a radius $r$.

I've split the isosceles triangle in two, and I solve for the area $A=\frac{bh}{2}$*. I have made my base $x$, and solve for the height by using the Pythagorean theorem of the smaller triangle (seen in picture). 

$h=r+\sqrt{r^2-x^2}$



So my the formula, I think, for both triangles should be $A=x(r+\sqrt{r^2-x^2})$

But after I solved for the derivative, when put "$= 0$", and checked on my calculator, I got the maximum to be about $4.3301r$, which differs a lot from my book's answer of $\frac{3\sqrt{3}}{4}r^2$*. Is my formula for the area right? Am I going about this the wrong way, or is it just my derivative that is wrong? Thanks in advance 

*
 Edited from original post

$A=rx+x\sqrt{r^2-x^2}$

$A'=r+x(\frac{1}{2})(r^2-x^2)^{-\frac{1}{2}}(-2x)+\sqrt{r^2-x^2}$

$r+\sqrt{r^2-x^2}=\frac{x^2}{\sqrt{r^2-x^2}}$

$r\sqrt{r^2-x^2}+(r^2-x^2)=x^2$

$r^2+r\sqrt{r^2-x^2}=2x^2$

$r^2(r^2-x^2)=(2x^2-r^2)^2$

$r^4-r^2x^2=4x^2-4x^2r^2+r^4$

$4x^4=3x^2r^2$

$x=\frac{\sqrt{3}}{2}r$
Let $\theta$ be one-half of the vertex angle (less than a right angle) of the isosceles triangle. 


  Exercise: Show that the area of the inscribed triangle is 
  $A(\theta) = \dfrac{h b}{2} = \dfrac{(r + r \cos\theta)}{2} (2 r \sin\theta) = r^2 \,(1+\cos\theta) \, \sin\theta $


Differentiating $A$ and setting it to $0$, you will be left with the problem of solving
$\cos\theta = -\cos2 \theta.$ 


  Exercise: Show that the only $\theta$ in $\left(0, \dfrac{\pi}{2}\right)$ that works is $\dfrac{\pi}{3}$.


Plugging these into $A$ we get the answer: $\dfrac{3\sqrt 3 r^2}{4}.$
Could it be possible that the 2 rs in the drawing are not equal and the other r (diagonal) could be longer than the vertical r in the drawing (or vice versa). In such case, the pythagorean equation h=r + sqrt(r^2 -x^2) is not applicable.