Answer
Height = $\frac{3r}{2}$
Base = $\sqrt 3{r}$
Work Step by Step
The area of the triangle is
$A(x)$ = $\frac{1}{2}(2t)(r+x)$ = $t(r+x)$ = $\sqrt {r^{2}-x^{2}}(r+x)$
$A'(x)$= $r\left(\frac{-2x}{2\sqrt {r^{2}-x^{2}}}\right)+\sqrt {r^{2}-x^{2}}+x\left(\frac{-2x}{2\sqrt {r^{2}-x^{2}}}\right)$
= $\frac{-rx+(r^2-x^2)-x^2}{\sqrt {r^{2}-x^{2}}}$
=$-\frac{2x^2+rx-r^2}{\sqrt {r^{2}-x^{2}}}$
$A'(x)$ = $0$
$2x^2+rx-r^2$ = $0$
$(2x-r)(x+r)$ = $0$
$x$ = $\frac{r}{2}, -r$
$A(r)$ = $0$ = $A(-r)$
The maximum occurs where $x$ = $\frac{r}{2}$
so the triangle has the height $r+\frac{r}{2}$ = $\frac{3r}{2}$ and base $2\sqrt {{r^{2}-(\frac{r}{2})^{2}}}$ = $\sqrt 3{r}$.
