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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. If the ABC has perimeter p and area a then lim(h tending to 0) 512ra/p^3 equals to?


6 years ago

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### Answers : (1)

										Ans: 4Area of triangle:$a = h\sqrt{r^{2}- (h-r)^{2}}$Perimeter of triangle:$p = 2\sqrt{r^{2}- (h-r)^{2}}+2\sqrt{2hr}$$L = \lim_{h\rightarrow 0}\frac{512.r.a}{p^{3}}$$L = \lim_{h\rightarrow 0}\frac{512.r.h\sqrt{r^{2}-(h-r)^{2}}}{(2\sqrt{r^{2}- (h-r)^{2}}+2\sqrt{2hr})^{3}}$$L = \lim_{h\rightarrow 0}\frac{512.r.h\sqrt{2hr-h^{2}}}{(2\sqrt{2hr-h^{2}}+2\sqrt{2hr})^{3}}$$L = \lim_{h\rightarrow 0}\frac{512.r\sqrt{2r-h}}{(2\sqrt{2r-h}+2\sqrt{2r})^{3}}$$L = \frac{512r}{8.8(\sqrt{2r})^{2}} = 4$Thanks & RegardsJitender SinghIIT DelhiaskIITians Faculty

2 years ago

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