I am in trouble with the following question:
QUESTION
ABC is an isosceles triangle inscribed in a circle of radius $r$. If $AB=AC$ and $h$ is the altitude from $A$ to $BC$ and $p$ be the perimeter of $ABC$ then find out: $$\lim_{h\to0} \frac{\Delta}{p^3}$$ (where $\Delta$ is the area of triangle )
MY ATTEMPT
First, let us try to understand what's happening during the limiting process $h\to0$: The base $BC$ is translated vertically upwards to the apex $A$, all the while staying horizontal.
If I describe the various quantities playing a role here in terms of the half angle $\alpha$ at $A$ (which tends to ${\pi\over2}$ in the process). Then I obtained an expression $${\Delta \over p^3}=\Psi(\alpha)\ ,$$ and now I have to determine $\lim_{\alpha\nearrow{\pi\over2}}\Psi(\alpha)$, whereby $r$ is considered constant and $\Psi(\alpha)$ is a function of half-angle.
HENCE applying trigonometric identities will yield:
$$\lim_{\alpha\to\frac{\pi}{2}} \frac{h^2\tan\frac{\alpha}{2}}{[2h(\sec\frac{\alpha}{2}+\tan\frac{\alpha}{2})]^3}=\frac{h^2}{8h^3} \lim_{\alpha\to\frac{\pi}{2}}\frac{\tan\frac{\alpha}{2}}{[\sec\frac{\alpha}{2}+\tan\frac{\alpha}{2}]^3}$$
$$=\frac{1}{8h}\frac{1}{(\sqrt{2}+1)^3}=\frac{1}{h(56+40\sqrt{2})}$$
at $\lim_{h\to0}$ i can replace h by r (from figure h=AO=radius of circle)
HELP
Please let me know whether it is the correct answer or not. I am kind of worried as the answer given in my book is $\frac{1}{128r}$.
PLEASE EXPLAIN HOW TO GET THIS ANSWER
*I am sorry for any kind of mistake, I am in high school and this is my first week on MSE.
You can use the Pythagorean theorem in the triangle formed by $B$, $O$ and the centre of the circle to find $\overline{OB} = \sqrt{2rh -h^2}$ . Then use it again to obtain $\overline{AB} = \sqrt{2rh}$ . Now $\Delta = h \sqrt{2rh -h^2}$ and $p = 2 [\sqrt{2rh-h^2} + \sqrt{2rh}]$, so $$ \lim_{h \to 0} \frac{\Delta}{p^3} = \lim_{h \to 0} \frac{\sqrt{1-\frac{h}{2r}}}{16 r \left(1+\sqrt{1-\frac{h}{2r}}\right)^3} = \frac{1}{128 r} \, . $$
As for your own approach, there are two problems: As already mentioned in Caelan Ritter's comment, you should take the limit $\alpha \to \pi$ instead of $\alpha \to \frac{\pi}{2}$ . And since $h$ depends on $\alpha$, you cannot treat it as a constant in the limiting process.