ABC is an isosceles triangle inscribed in a circle of radius r. If AB = AC and h

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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}}

Options:

A . 1r

B . 164r

C . 1128r

D . 12r

Answer: Option C
:
C

ABC Is An Isosceles Triangle Inscribed In A Circle Of Radius...

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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