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Factorization Method Form to Remove Indeterminate Form

ABC is an iso...

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.The triangle ABC has perimeter $P = 2 [\sqrt{(2 h r - h^{2})} + \sqrt{2 h r}]$ and A be the area of the triangle .Find $lim h \to 0 \frac{A}{P^{3}}$

\frac{1}{r}

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\frac{1}{64 r}

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\frac{1}{128} r

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\frac{1}{2 r}

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Solution

The correct option is C $\frac{1}{128} r$

In $△ A B C, A B = A C$
$A D ⊥ B C (D i s m i d p o i n t o f B C)$
Let r = radius of circumcircle
$\Rightarrow$ OA = OB = OC = r
Now BD = $\sqrt{B O^{2} - O D^{2}}$
$= \sqrt{r^{2} - (h - r)^{2}} = \sqrt{2 r h - h^{2}}$
$\Rightarrow B C = 2 \sqrt{2 r h - h^{2}}$
$∴$ Area of $△ A B C =$ $\frac{1}{2} \times B C \times A D$
$= h \sqrt{2 r h - h^{2}}$
Also $lim h \to 0 \frac{A}{p^{3}} = \frac{h \sqrt{2 r h - h^{2}}}{8 (\sqrt{2 r h - h^{2}} + \sqrt{2 h r})^{3}}$
$= lim h \to 0 \frac{h^{3 / 2} \sqrt{2 r - h}}{8 h^{3 / 2} (\sqrt{2 r - h} + \sqrt{2 r})^{3}}$
$= lim h \to 0 \frac{\sqrt{2 r - h}}{8 [\sqrt{2 r - h} + \sqrt{2 r}]^{3}}$
$= \frac{\sqrt{2 r}}{8 (\sqrt{2 r} + \sqrt{2 r})^{3}} = \frac{\sqrt{2 r}}{8.8.2 r \sqrt{2 r}} = \frac{1}{128 r}$

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