ABC is an isosceles triangle inscribed in a c | vedclass

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(C) Let $M$ be the midpoint of $BC$. In the right triangle $ABM$,$AM = h$. Let $BM = x$. By the property of chords,$x^2 = h(2r - h)$,so $x = \sqrt{2rh

Vedclass · Accepted Answer

$\frac{1}{{128r}}$

Vedclass · Answer

(C) Let $M$ be the midpoint of $BC$. In the right triangle $ABM$,$AM = h$. Let $BM = x$. By the property of chords,$x^2 = h(2r - h)$,so $x = \sqrt{2rh - h^2}$.$AB = AC = \sqrt{h^2 + x^2} = \sqrt{h^2 + 2rh - h^2} = \sqrt{2rh}$.Perimeter $P = AB + AC + BC = 2\sqrt{2rh} + 2\sqrt{2rh - h^2}$.Area $\Delta = \frac{1}{2} 	imes BC 	imes AM = \frac{1}{2} 	imes (2\sqrt{2rh - h^2}) 	imes h = h\sqrt{2rh - h^2}$.We want to evaluate $\mathop {\lim }\limits_{h 	o 0} \frac{\Delta }{{{P^3}}} = \mathop {\lim }\limits_{h 	o 0} \frac{h\sqrt{2rh - h^2}}{[2\sqrt{2rh} + 2\sqrt{2rh - h^2}]^3}$.$= \mathop {\lim }\limits_{h 	o 0} \frac{h\sqrt{h}\sqrt{2r - h}}{8[\sqrt{2rh} + \sqrt{2rh - h^2}]^3} = \mathop {\lim }\limits_{h 	o 0} \frac{h^{3/2}\sqrt{2r - h}}{8(h^{1/2})^3[\sqrt{2r} + \sqrt{2r - h}]^3}$.$= \frac{\sqrt{2r}}{8[\sqrt{2r} + \sqrt{2r}]^3} = \frac{\sqrt{2r}}{8[2\sqrt{2r}]^3} = \frac{\sqrt{2r}}{8 	imes 8 	imes 2r 	imes \sqrt{2r}} = \frac{1}{128r}$.

$ABC$ is an isosceles triangle inscribed in a circle of radius $r$. If $AB = AC$ and $h$ is the altitude from $A$ to $BC$,and $P$ is the perimeter of $ABC$,then $\mathop {\lim }\limits_{h \to 0} \frac{\Delta }{{{P^3}}}$ equals (where $\Delta$ is the area of the triangle).

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