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Question Number 28597 by beh.i83417@gmail.com last updated on 27/Jan/18

Commented by beh.i83417@gmail.com last updated on 27/Jan/18

$$\mathit{a}\mathit{l}\mathit{s}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{g}\mathit{e}\mathit{s}\mathit{t}\mathit{s}\mathit{q}\mathit{u}\mathit{a}\mathit{r}\mathit{e},\mathit{i}\mathit{n}\mathit{s}\mathit{c}\mathit{r}\mathit{i}\mathit{b}\mathit{e}\mathit{d}\mathit{i}\mathit{n}$$$$\mathit{e}\mathit{q}\mathit{u}\mathit{i}\mathit{l}\mathit{a}\mathit{t}\mathit{e}\mathit{r}\mathit{a}\mathit{l}\mathit{t}\mathit{r}\mathit{i}\mathit{a}\mathit{n}\mathit{g}\mathit{l}\mathit{e}\mathit{o}\mathit{f}\mathit{s}\mathit{i}\mathit{d}\mathit{e}=1\mathit{u}\mathit{n}\mathit{i}\mathit{t}.$$

Answered by mrW2 last updated on 29/Jan/18

$$max.equilateraltriangle:$$$$a=\frac{1}{\cos 15°}=\frac{4}{\sqrt{2}+\sqrt{6}}=\sqrt{6}-\sqrt{2}$$$$A=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}(8-4\sqrt{3})=2\sqrt{3}-3$$$$$$$$max.squareinequilateraltriangle:$$$$\frac{b}{1}=\frac{\frac{\sqrt{3}}{2}-b}{\frac{\sqrt{3}}{2}}=\frac{\sqrt{3}-2b}{\sqrt{3}}$$$$\Rightarrow b=\frac{\sqrt{3}}{2+\sqrt{3}}=2\sqrt{3}-3$$$$A=b^2=21-12\sqrt{3}$$

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