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  • Suppose that circle of equal diameter are packed tighty in n rows inside an equilateral triangle. (The figure illustrates the case <img alt="\mathrm{n=4}" src="https://entrancecorner.oncodecog

Suppose that circle of equal diameter are packed tighty in n rows inside an equilateral triangle. (The figure illustrates the case \mathrm{n=4} ) If \mathrm{A}  is the area of the triangle and \mathrm{A_n}  is the total area occupied by the circles in n rows then   \mathrm{\operatorname{Lim}_{n \rightarrow \infty} \frac{A_n}{A} \text { e'quals }}

Option: 1

\frac{\pi}{\sqrt{3}}


Option: 2

\frac{\pi \sqrt{3}}{6}


Option: 3

\frac{\pi1}{2 \sqrt{3}}


Option: 4

\frac{\pi}{6}


Answers (1)

best_answer

Le'l rudius of each circle \mathrm{=r}  and side of triangle \mathrm{=a}

\mathrm{a=(n-2) 2 r+2\left(r+30^{\circ}\right)=r(2 n-4+2+2 \sqrt{3})=r(2 n-2+2 \sqrt{3}) }

\mathrm{ A=\frac{\sqrt{3}}{4} a^2=\frac{\sqrt{3}}{4} r^2 4(n-1+\sqrt{3})^2 \\ }

\mathrm{ A=\sqrt{3} r^2(n+\sqrt{3}-1)^2 }\mathrm{A_n-\pi^2(1+2+\ldots .+n)-\frac{n(n+1)}{2} \pi r^2 }

\mathrm{\operatorname{Lim}_{n \rightarrow \infty} \frac{A_e}{A}=\lim _{n \rightarrow \infty} \mathrm{\frac{2}{\sqrt{3} r^2(i r+\sqrt{3}-1)^2}} 

\mathrm{=\operatorname{Lim}_{n \rightarrow \infty} \frac{\pi}{2 \sqrt{3}} \frac{\left(1+\frac{1}{n}\right)}{\left(1+\frac{\sqrt{3}-1}{n}\right)^2}} \mathrm{=\frac{\pi}{2 \sqrt{3}} \frac{(1+0)}{(1+0)^2}=\frac{\pi}{2 \sqrt{3}}}.

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