# $ABC$ is an isosceles triangle inscribed in a circle of radius r. If $AB=AC$ and $h$ is the altitude from $A$ to $BC$, then the triangle $ABC$ has perimeter $P=2(\sqrt{2hr-h^2})+\sqrt{2hr})$ and area $A$=_______also $\lim\limits_{h\to 0}\large\frac{A}{p^3}=$_______

$\begin{array}{1 1}(a)\;A=h\sqrt{2rh-h^2},\lim\limits_{h\to 0}\large\frac{A}{p^3}=\frac{1}{128r}\\(b)\;A=h\sqrt{2rh-h^3},\lim\limits_{h\to 0}\large\frac{A}{p^3}=\frac{1}{28r}\\(c)\;A=h\sqrt{2rh^2-h},\lim\limits_{h\to 0}\large\frac{A}{p^3}=\frac{2}{128r}\\(d)\;A=3h\sqrt{2rh-h^2},\lim\limits_{h\to 0}\large\frac{A}{p^3}=\frac{3}{128r}\end{array}$

In $\Delta ABC,AB=AC$
$AD \perp BC$
$OA=OB=OC=r$
Now BD=$\sqrt{BO^2-OD^2}$
$\Rightarrow \sqrt{r^2-(h-r)^2}$
$\Rightarrow \sqrt{2rh-h^2}$
$BC=2\sqrt{2rh-h^2}$
Area of $\Delta ABC=\large\frac{1}{2}$$\times BC\times AD \Rightarrow h\sqrt{2rh-h^2} Also \lim\limits_{h\to 0}\large\frac{A}{p^3}=$$\lim\limits_{h\to 0}\large\frac{h(\sqrt{2rh-h^2}}{8(\sqrt{2rh-h^2}+\sqrt{2rh})^3}$
$\Rightarrow \lim\limits_{h\to 0}\large\frac{h^{3/2}(\sqrt{2r-h}}{8h^{3/2}(\sqrt{2r-h}+\sqrt{2r})^3}$
$\Rightarrow \lim\limits_{h\to 0}\large\frac{(\sqrt{2r-h}}{8(\sqrt{2r-h}+\sqrt{2r})^3}$
$\Rightarrow \lim\limits_{h\to 0}\large\frac{(\sqrt{2r}}{8(\sqrt{2r}+\sqrt{2r})^3}$
$\Rightarrow \lim\limits_{h\to 0}\large\frac{(\sqrt{2r}}{8.8.2r.\sqrt{2r}}$
$\Rightarrow \large\frac{1}{128 r}$
Hence (a) is the correct answer.