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Question Number 33317 by mondodotto@gmail.com last updated on 14/Apr/18

$$\mathbf{find}\mathbf{the}\mathbf{radius}\mathbf{of}\mathbf{a}\mathbf{circle}$$$$\mathbf{wich}\mathbf{inscribes}\mathbf{an}\mathbf{equalateral}\mathbf{triangle}$$$$\mathbf{with}\mathbf{perimeter}\mathbf{of}24\mathbf{cm}$$

Answered by MJS last updated on 14/Apr/18

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{sure}\mathrm{if}\mathrm{you}\mathrm{mean}\mathrm{the}$$$$\mathrm{circumcircle}\mathrm{or}\mathrm{the}\mathrm{inscribed}$$$$\mathrm{circle}(\mathrm{my}\mathrm{english}\mathrm{might}\mathrm{not}\mathrm{be}$$$$\mathrm{good}\mathrm{enough})$$$$\mathrm{equilateral}\mathrm{triangle}a=b=c$$$$\mathrm{perimeter}=a+b+c=3a=24$$$$a=8$$$$\mathrm{in}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}\mathrm{the}$$$$\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circumcircle}$$$$R=\frac{2}{3}h\mathrm{and}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}$$$$\mathrm{inscribed}\mathrm{circle}\mathrm{is}r=\frac{1}{3}\Rightarrow R=2r$$$$a^2=h^2+{\left(\frac{a}{2}\right)}^2$$$$h=\sqrt{a^2-\frac{a^2}{4}}=\sqrt{\frac{3a^2}{4}}=a\frac{\sqrt{3}}{2}$$$$R=a\frac{\sqrt{3}}{3}=\frac{8\sqrt{3}}{3}\approx 4.62$$$$r=a\frac{\sqrt{3}}{6}=\frac{4\sqrt{3}}{3}\approx 2.31$$

Commented by mondodotto@gmail.com last updated on 15/Apr/18

$$\mathbf{sure}\mathbf{sir}\mathbf{thats}\mathbf{what}\mathbf{i}\mathbf{meant}\mathbf{thanks}\mathbf{for}\mathbf{your}\mathbf{help}$$

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