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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 triangle ABC has perimeter $$P$$ and area $$\Delta $$ then $$\displaystyle \lim_{h\rightarrow 0}512\:r\:\frac{\Delta }{P^{3}}$$ is equal to


Solution

We have $$BC=2BD, AD=h$$ and $$OD=h-r$$.

$$\therefore \displaystyle BC=2\sqrt{r^{2}-(h-r)^{2}}=2\sqrt{2hr-h^{2}}$$

$$\Rightarrow \displaystyle AB=\sqrt{2hr-h^{2}+h^{2}}=\sqrt{2hr}$$

so that $$P=2AB+BC$$

$$=\displaystyle 2[\sqrt{2hr-h^{2}}+\sqrt{2hr}]$$

Also the area of $$\Delta ABC$$ is
$$\displaystyle \Delta =BD\cdot AD=h\sqrt{2hr-h^{2}}$$

$$\therefore \displaystyle \frac{\Delta }{P^{3}}=\frac{h\sqrt{2hr-h^{2}}}{8(\sqrt{2hr-h^{2}}+\sqrt{2hr})^{3}}$$

$$=\displaystyle \frac{h\sqrt{2hr-h}}{8(\sqrt{2hr-h}+\sqrt{2r})^{3}}$$

$$\Rightarrow \displaystyle \lim_{h\rightarrow 0}512r\frac{\Delta }{P^{3}}=512\:r\:\frac{\sqrt{2r}}{8(2\sqrt{2r})^{3}}=4$$

205109_198927_ans_2855fb38dcf0424db853b960897cbb07.png

Mathematics

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