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Question Number 97051 by john santu last updated on 06/Jun/20

$$\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{perimeter}\mathrm{of}\mathrm{a}\mathrm{regular}$$$$\mathrm{dodecagon}\left(12\mathrm{sides}\right)\mathrm{whose}$$$$\mathrm{area}\mathrm{is}24+12\sqrt{3}?$$

Commented by bemath last updated on 06/Jun/20

$$\mathrm{let}\mathrm{the}\mathrm{sides}\mathrm{of}\mathrm{dodecagon}\mathrm{is}a$$$$area=12\times \frac{r^2}{2}\times \sin \left(\frac{{360}^o}{12}\right)$$$$24+12\sqrt{3}=6r^2\times \sin {30}^o$$$$24+12\sqrt{3}=3r^2\Rightarrow r=\sqrt{8+4\sqrt{3}}$$$$usingC\mathrm{osine}\mathrm{of}\mathrm{law}$$$$a=\sqrt{2r^2-2r^2\times \cos {30}^o}$$$$a=\sqrt{2r^2(1-\frac{1}{2}\sqrt{3})}$$$$a=\sqrt{2(8+4\sqrt{3})\left(\frac{2-\sqrt{3}}{2}\right)}$$$$a=\sqrt{(8+4\sqrt{3})(2-\sqrt{3})}$$$$a=\sqrt{16-8\sqrt{3}+8\sqrt{3}-12}=2$$$$sotheperimeterofdodecagon$$$$equalto24.$$

Commented by bobhans last updated on 06/Jun/20

$$\mathrm{good}$$

Answered by 1549442205 last updated on 06/Jun/20

$$\mathrm{The}\mathrm{angle}\mathrm{at}\mathrm{the}\mathrm{center}\mathrm{imtercept}\mathrm{sides}\mathrm{of}\mathrm{dodecagon}\mathrm{equal}$$$$\mathrm{to}\frac{360°}{12}=30°.\mathrm{Hence},\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{side}\mathrm{of}\mathrm{dodecagon}\mathrm{is}$$$$\mathrm{a}=2\mathrm{Rsin}15°=2\mathrm{R}\times \frac{1-\cos 30°}{2}=\mathrm{R}(1-\frac{\sqrt{3}}{2}).$$$$\mathrm{It}\mathrm{follows}\mathrm{that}\mathrm{the}\mathrm{perimeter}\mathrm{of}\mathrm{dodecagon}$$$$\mathrm{equal}\mathrm{to}12\times \mathrm{R}(1-\frac{\sqrt{3}}{2})\left(1\right).\mathrm{The}\mathrm{area}\mathrm{of}$$$$\mathrm{dodecagon}\mathrm{equal}\mathrm{to}{6\mathrm{R}}^2\sin 30°=24+12\sqrt{3}$$$$\Rightarrow \mathrm{R}=\sqrt{8+4\sqrt{3}}=\sqrt{2}\times \sqrt{{(\sqrt{3}+1)}^2}=\sqrt{6}+\sqrt{2}$$$$\mathrm{Replace}\mathrm{into}\left(1\right)\mathrm{we}\mathrm{get}\mathrm{the}\mathrm{perimeter}\mathrm{of}$$$$\mathrm{dodecagon}\mathrm{is}\mathrm{C}=6(\sqrt{6}+\sqrt{2})(2-\sqrt{3})=$$$$6(\sqrt{6}-\sqrt{2})$$$$\mathrm{Thus},\mathbf{C}=6(\sqrt{6}-\sqrt{2})$$

Commented by bobhans last updated on 06/Jun/20

$$\sin {}^2\left({15}^{\mathrm{o}}\right)=\frac{1}{2}-\frac{1}{2}\cos {30}^{\mathrm{o}}$$$$\mathrm{not}\sin {15}^{\mathrm{o}}=\frac{1-\cos {30}^{\mathrm{o}}}{2}$$

Commented by 1549442205 last updated on 07/Jun/20

$$\mathrm{Thank}\mathrm{you},\mathrm{i}\mathrm{am}\mathrm{sorry}\mathrm{done}\mathrm{a}\mathrm{mistake}.$$$$\mathrm{It}\mathrm{must}\mathrm{be}\sin 15°=\sqrt{\frac{1-\cos 30°}{2}}=\sqrt{\frac{2-\sqrt{3}}{4}}$$$$=\sqrt{\frac{{(\sqrt{3}-1)}^2}{8}}=\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{2}.\mathrm{Hence},$$$$\mathrm{a}=\mathrm{R}(\sqrt{6}-\sqrt{2})\Rightarrow \mathrm{C}=12\mathrm{R}(\sqrt{6}-\sqrt{2})$$$$=12(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})=48$$

Commented by john santu last updated on 07/Jun/20

$$\mathrm{but}\mathrm{the}\mathrm{answer}\mathrm{is}24\mathrm{sir}$$

Answered by john santu last updated on 07/Jun/20

Commented by bemath last updated on 07/Jun/20

$$\mathrm{i}\mathrm{agree}\mathrm{sir}.\mathrm{great}$$

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