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Circle with equilateral triangle [on hold]


Geometry - Equilateral triangle covered with five circlesEquilateral triangle coveringArea of equilateral triangle from circumcircleCommon area between Circle and Equilateral triangleequilateral triangle and inscribed circle6 Circle in Equilateral TriangleHomework: Is the triangle an equilateral triangle?Area of enclosed overlapping circles within an equilateral trianglecircumcentre of equilateral triangleInscribe an equilateral triangle inside a triangle













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How to use property of equilateral triangle which is given in the question .






geometry triangle circle







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put on hold as off-topic by John Omielan, steven gregory, Eevee Trainer, user21820, José Carlos Santos 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Omielan, steven gregory, Eevee Trainer, user21820, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.





















    -1












    $begingroup$


    enter image description here



    How to use property of equilateral triangle which is given in the question .






    geometry triangle circle







    share|cite|improve this question











    $endgroup$



    put on hold as off-topic by John Omielan, steven gregory, Eevee Trainer, user21820, José Carlos Santos 2 days ago


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Omielan, steven gregory, Eevee Trainer, user21820, José Carlos Santos

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      -1












      -1








      -1


      1



      $begingroup$


      enter image description here



      How to use property of equilateral triangle which is given in the question .






      geometry triangle circle







      share|cite|improve this question











      $endgroup$




      enter image description here



      How to use property of equilateral triangle which is given in the question .






      geometry triangle circle




      geometry triangle circle






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      Dr. Mathva

      2,296526




      2,296526










      asked 2 days ago









      Abhinov SinghAbhinov Singh

      1354




      1354




      put on hold as off-topic by John Omielan, steven gregory, Eevee Trainer, user21820, José Carlos Santos 2 days ago


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Omielan, steven gregory, Eevee Trainer, user21820, José Carlos Santos

      If this question can be reworded to fit the rules in the help center, please edit the question.







      put on hold as off-topic by John Omielan, steven gregory, Eevee Trainer, user21820, José Carlos Santos 2 days ago


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Omielan, steven gregory, Eevee Trainer, user21820, José Carlos Santos

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Here's a picture, with a bunch of useful things drawn in and labeled:



          Figure 1



          The center of the circles are marked with dots; $O$ is the center of the yellow circle and $E$ is the center of the green circle.



          So now, what's the ratio of $ED$ to $AE$?






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            According to Jmerry's diagram,



            $$2R = AE + ED = r+ frac{r}{cos60°} = r+ frac{r}{1/2} = r+ 2r$$



            Thus,
            $$frac{R}{r}=frac{3}{2}$$ or
            $$frac{r}{R}=frac{2}{3}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Therefore $$frac{text{yellow}}{text{green}}=frac{R^2pi-r^2pi}{r^2pi}=frac{R^2pi}{r^2pi}-1=big(frac{R}{r}big)^2-1=big(frac{3}{2}big)^2-1=frac{5}{4}$$ Thus $$frac{text{green}}{text{yellow}}=frac{4}{5}$$
              $endgroup$
              – Dr. Mathva
              2 days ago




















            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Here's a picture, with a bunch of useful things drawn in and labeled:



            Figure 1



            The center of the circles are marked with dots; $O$ is the center of the yellow circle and $E$ is the center of the green circle.



            So now, what's the ratio of $ED$ to $AE$?






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Here's a picture, with a bunch of useful things drawn in and labeled:



              Figure 1



              The center of the circles are marked with dots; $O$ is the center of the yellow circle and $E$ is the center of the green circle.



              So now, what's the ratio of $ED$ to $AE$?






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Here's a picture, with a bunch of useful things drawn in and labeled:



                Figure 1



                The center of the circles are marked with dots; $O$ is the center of the yellow circle and $E$ is the center of the green circle.



                So now, what's the ratio of $ED$ to $AE$?






                share|cite|improve this answer









                $endgroup$



                Here's a picture, with a bunch of useful things drawn in and labeled:



                Figure 1



                The center of the circles are marked with dots; $O$ is the center of the yellow circle and $E$ is the center of the green circle.



                So now, what's the ratio of $ED$ to $AE$?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                jmerryjmerry

                13.4k1629




                13.4k1629























                    2












                    $begingroup$

                    According to Jmerry's diagram,



                    $$2R = AE + ED = r+ frac{r}{cos60°} = r+ frac{r}{1/2} = r+ 2r$$



                    Thus,
                    $$frac{R}{r}=frac{3}{2}$$ or
                    $$frac{r}{R}=frac{2}{3}$$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Therefore $$frac{text{yellow}}{text{green}}=frac{R^2pi-r^2pi}{r^2pi}=frac{R^2pi}{r^2pi}-1=big(frac{R}{r}big)^2-1=big(frac{3}{2}big)^2-1=frac{5}{4}$$ Thus $$frac{text{green}}{text{yellow}}=frac{4}{5}$$
                      $endgroup$
                      – Dr. Mathva
                      2 days ago


















                    2












                    $begingroup$

                    According to Jmerry's diagram,



                    $$2R = AE + ED = r+ frac{r}{cos60°} = r+ frac{r}{1/2} = r+ 2r$$



                    Thus,
                    $$frac{R}{r}=frac{3}{2}$$ or
                    $$frac{r}{R}=frac{2}{3}$$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Therefore $$frac{text{yellow}}{text{green}}=frac{R^2pi-r^2pi}{r^2pi}=frac{R^2pi}{r^2pi}-1=big(frac{R}{r}big)^2-1=big(frac{3}{2}big)^2-1=frac{5}{4}$$ Thus $$frac{text{green}}{text{yellow}}=frac{4}{5}$$
                      $endgroup$
                      – Dr. Mathva
                      2 days ago
















                    2












                    2








                    2





                    $begingroup$

                    According to Jmerry's diagram,



                    $$2R = AE + ED = r+ frac{r}{cos60°} = r+ frac{r}{1/2} = r+ 2r$$



                    Thus,
                    $$frac{R}{r}=frac{3}{2}$$ or
                    $$frac{r}{R}=frac{2}{3}$$






                    share|cite|improve this answer











                    $endgroup$



                    According to Jmerry's diagram,



                    $$2R = AE + ED = r+ frac{r}{cos60°} = r+ frac{r}{1/2} = r+ 2r$$



                    Thus,
                    $$frac{R}{r}=frac{3}{2}$$ or
                    $$frac{r}{R}=frac{2}{3}$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 2 days ago









                    Oscar Lanzi

                    13.1k12136




                    13.1k12136










                    answered 2 days ago









                    Mathew MahindaratneMathew Mahindaratne

                    562212




                    562212












                    • $begingroup$
                      Therefore $$frac{text{yellow}}{text{green}}=frac{R^2pi-r^2pi}{r^2pi}=frac{R^2pi}{r^2pi}-1=big(frac{R}{r}big)^2-1=big(frac{3}{2}big)^2-1=frac{5}{4}$$ Thus $$frac{text{green}}{text{yellow}}=frac{4}{5}$$
                      $endgroup$
                      – Dr. Mathva
                      2 days ago




















                    • $begingroup$
                      Therefore $$frac{text{yellow}}{text{green}}=frac{R^2pi-r^2pi}{r^2pi}=frac{R^2pi}{r^2pi}-1=big(frac{R}{r}big)^2-1=big(frac{3}{2}big)^2-1=frac{5}{4}$$ Thus $$frac{text{green}}{text{yellow}}=frac{4}{5}$$
                      $endgroup$
                      – Dr. Mathva
                      2 days ago


















                    $begingroup$
                    Therefore $$frac{text{yellow}}{text{green}}=frac{R^2pi-r^2pi}{r^2pi}=frac{R^2pi}{r^2pi}-1=big(frac{R}{r}big)^2-1=big(frac{3}{2}big)^2-1=frac{5}{4}$$ Thus $$frac{text{green}}{text{yellow}}=frac{4}{5}$$
                    $endgroup$
                    – Dr. Mathva
                    2 days ago






                    $begingroup$
                    Therefore $$frac{text{yellow}}{text{green}}=frac{R^2pi-r^2pi}{r^2pi}=frac{R^2pi}{r^2pi}-1=big(frac{R}{r}big)^2-1=big(frac{3}{2}big)^2-1=frac{5}{4}$$ Thus $$frac{text{green}}{text{yellow}}=frac{4}{5}$$
                    $endgroup$
                    – Dr. Mathva
                    2 days ago





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