find-the-radius-of-a-circle-wich-inscribes-an-equalateral-triangle-with-perimeter-of-24cm-

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

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

Answered by MJS last updated on 14/Apr/18

I′m not sure if you mean the circumcircle or the inscribed circle (my english might not be good enough) equilateral triangle a=b=c perimeter=a+b+c=3a=24 a=8 in an equilateral triangle the radius of the circumcircle R=(2/3)h and the radius of the inscribed circle is r=(1/3) ⇒ R=2r a^2 =h^2 +((a/2))^2 h=(√(a^2 −(a^2 /4)))=(√((3a^2 )/4))=a((√3)/2) R=a((√3)/3)=((8(√3))/3)≈4.62 r=a((√3)/6)=((4(√3))/3)≈2.31

$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{the} \\ $$$$\mathrm{circumcircle}\:\mathrm{or}\:\mathrm{the}\:\mathrm{inscribed} \\ $$$$\mathrm{circle}\:\left(\mathrm{my}\:\mathrm{english}\:\mathrm{might}\:\mathrm{not}\:\mathrm{be}\right. \\ $$$$\left.\mathrm{good}\:\mathrm{enough}\right) \\ $$$$\mathrm{equilateral}\:\mathrm{triangle}\:{a}={b}={c} \\ $$$$\mathrm{perimeter}={a}+{b}+{c}=\mathrm{3}{a}=\mathrm{24} \\ $$$${a}=\mathrm{8} \\ $$$$\mathrm{in}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{the} \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circumcircle} \\ $$$${R}=\frac{\mathrm{2}}{\mathrm{3}}{h}\:\mathrm{and}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{inscribed}\:\mathrm{circle}\:\mathrm{is}\:{r}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:{R}=\mathrm{2}{r} \\ $$$${a}^{\mathrm{2}} ={h}^{\mathrm{2}} +\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${h}=\sqrt{{a}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}=\sqrt{\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{4}}}={a}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${R}={a}\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}=\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{3}}\approx\mathrm{4}.\mathrm{62} \\ $$$${r}={a}\frac{\sqrt{\mathrm{3}}}{\mathrm{6}}=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}\approx\mathrm{2}.\mathrm{31} \\ $$

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

sure sir thats what i meant thanks for your help

$$\boldsymbol{\mathrm{sure}}\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{thats}}\:\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{meant}}\:\boldsymbol{\mathrm{thanks}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{your}}\:\boldsymbol{\mathrm{help}} \\ $$

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