Circle with equilateral triangle












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











$endgroup$

















    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











    $endgroup$















      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 hours ago









      Dr. Mathva

      2,104425




      2,104425










      asked 5 hours ago









      Abhinov SinghAbhinov Singh

      1394




      1394






















          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 hours ago













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3140839%2fcircle-with-equilateral-triangle%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            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 4 hours ago









                jmerryjmerry

                12.8k1628




                12.8k1628























                    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 hours 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 hours 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 1 hour ago









                    Oscar Lanzi

                    13k12136




                    13k12136










                    answered 3 hours 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 hours 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 hours 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 hours 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 hours ago




















                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3140839%2fcircle-with-equilateral-triangle%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    404 Error Contact Form 7 ajax form submitting

                    Image
                    0 The contact form does not work, and affects loading after one try. The chrome console shows the error message: POST http://example.com/wp-json/contact-form-7/v1/contact-forms/50/feedback 404 (Not Found) send @ jquery.js?ver=1.12.4:4 ajax @ jquery.js?ver=1.12.4:4 wpcf7.submit @ scripts.js?ver=5.0.3:346 (anonymous) @ scripts.js?ver=5.0.3:53 dispatch @ jquery.js?ver=1.12.4:3 r.handle @ jquery.js?ver=1.12.4:3 How to solve this issue? thanks in advance javascript wordpress contact-form-7 share | improve this question edited Jul 17 '18 at 5:39 Kiran Shahi 4,745 3 16 43 ...
                    Read more

                    How to resolve this name issue having white space while installing the android Studio.?

                    Image
                    1 SDK location should not contain whitespace, as this can cause problems with the NDK tools. It says that the path name should not contain any white spaces. But i cant rename it. There is no any rename option for that folder when right clicked. When i change the folder to which there is not any space, it says that "Target Folder is neither empty nor does it point to an existing SDK Installation." android android-studio sdk android-ndk android-sdk-tools share | improve this question edited Nov 25 '18 at 17:49 Elletlar 1,515 3 15 23 ...
                    Read more

                    C# WPF - Problem with Material Design Textbox

                    Image
                    1 I'm new to programming and started to make an application for school. For that I use C# WPF and for a good-looking Application i installed Material Design. First i started the Demo of Material Design, where you can see what things you can use etc. with the good to copy. So after a quick view I found a nice TextBox with a border and a Checkbox on top of it. The code to copy was this <StackPanel> <CheckBox x:Name="MaterialDesignOutlinedTextFieldTextBoxEnabledComboBox" IsChecked="True" Margin="32,0,0,8"> Enabled </CheckBox> <TextBox Style="{StaticResource MaterialDesignOutlinedTextFieldTextBox}" Margin="32,0,0,0" VerticalAlignment="Top" Height="100" AcceptsReturn=...
                    Read more