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Area of Sector

ABC is an iso...

Question

$A B C$ is an isosceles triangle inscribed in a circle of radius r. If $A B = A C & h$ is the altitude from A to BC and P be the perimeter of ABC, then ${lim}_{h \to 0} \frac{Δ}{P^{3}}$ equals (where $Δ$ is the area of traingle)

\frac{1}{32 r^{3}}

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\frac{1}{64 r^{3}}

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\frac{1}{128 r^{3}}

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None of these

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Solution

The correct option is C $\frac{1}{128 r^{3}}$
$D C = \sqrt{r^{2} - {(h - r)}^{2}} = \sqrt{r^{2} - (h^{2} + r^{2} - 2 h r)} = \sqrt{2 h r - h^{2}} A r e a = \frac{1}{2} \times 2 D C \times h \Rightarrow h \sqrt{2 h r - h^{2}} A C = \sqrt{{D C}^{2} + h^{2}} = \sqrt{2 h r - h^{2} + h^{2}} = \sqrt{2 h r} P = 2 A C + B C = 2 \sqrt{2 h r} + 2 \sqrt{2 h r - h^{2}} \Rightarrow 2 \sqrt{h} (\sqrt{2 r} + \sqrt{2 r - h}) P^{3} = 8 h \sqrt{h} {(\sqrt{2 r} + \sqrt{2 r - h})}^{3} \frac{△}{P^{3}} = \frac{h \sqrt{2 h - h^{2}}}{8 h \sqrt{h} {(\sqrt{2 r} + \sqrt{2 r - h})}^{3}} {lim}_{h \to 0} \Rightarrow \frac{h \sqrt{h} \sqrt{2 - h}}{8 h \sqrt{h} {(\sqrt{2 r} + \sqrt{2 r - h})}^{3}} \Rightarrow \frac{\sqrt{2 - h}}{8 {(\sqrt{2 r} + \sqrt{2 r - h})}^{3}} \Rightarrow \frac{1}{8} \times \frac{\sqrt{2}}{{(2 \sqrt{2} r)}^{3}} \Rightarrow \frac{1}{8} \times \frac{\sqrt{2}}{16 \sqrt{2} r^{3}} \Rightarrow \frac{1}{128 r^{3}}$

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ABC is an isosceles triangle inscribed in a circle of radius r. If AB=AC& h is the altitude from A to BC and P be the perimeter of ABC, then limh→0ΔP3 equals (where Δ is the area of traingle)

$A B C$ is an isosceles triangle inscribed in a circle of radius r. If $A B = A C & h$ is the altitude from A to BC and P be the perimeter of ABC, then ${lim}_{h \to 0} \frac{Δ}{P^{3}}$ equals (where $Δ$ is the area of traingle)