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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?

manav sharma , 14 Years ago
Grade 12
anser 1 Answers
Jitender Singh

Last Activity: 10 Years ago

Ans: 4
Area 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 & Regards
Jitender Singh
IIT Delhi
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