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Grade 12Differential Calculus

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

16 Years agoGrade 12

1 Answer

Jitender Singh

12 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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