Given that $AB=12$, $CD=1$ and the semicircle is tangent to $BC$, find the radius of semicircle.












4












$begingroup$


I tried it by reflecting it horizontally, by making complete circle and using it as incircle. But, was not able to get an answer.
enter image description here






geometry euclidean-geometry triangle circle







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    4












    $begingroup$


    I tried it by reflecting it horizontally, by making complete circle and using it as incircle. But, was not able to get an answer.
    enter image description here






    geometry euclidean-geometry triangle circle







    share|cite|improve this question









    New contributor




    Sumit Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      4












      4








      4


      3



      $begingroup$


      I tried it by reflecting it horizontally, by making complete circle and using it as incircle. But, was not able to get an answer.
      enter image description here






      geometry euclidean-geometry triangle circle







      share|cite|improve this question









      New contributor




      Sumit Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      $endgroup$




      I tried it by reflecting it horizontally, by making complete circle and using it as incircle. But, was not able to get an answer.
      enter image description here






      geometry euclidean-geometry triangle circle




      geometry euclidean-geometry triangle circle






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      edited 9 hours ago









      user21820

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      asked 16 hours ago









      Sumit SinghSumit Singh

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          2 Answers
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          $begingroup$

          Let $K$ be a touching point to $CB$, the radius be equal to $x$ and $O$ be a center of the circle.



          Thus, $$CO=x+1,$$
          $$AC=2x+1,$$ $$CK=sqrt{(1+x)^2-x^2}=sqrt{2x+1}.$$
          Thus, since $Delta CKOsimDelta CAB,$ we obtain:
          $$frac{sqrt{2x+1}}{2x+1}=frac{x}{12}.$$
          Can you end it now?



          I got $x=4$.






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            We have $$CE^2+r^2=(1+r)^2$$ and $$AB=BE=12$$ and $$12^2+(2r+1)^2=(12+CE)^2$$
            Can you solve this system?






            share|cite|improve this answer









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              2 Answers
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              2 Answers
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              5












              $begingroup$

              Let $K$ be a touching point to $CB$, the radius be equal to $x$ and $O$ be a center of the circle.



              Thus, $$CO=x+1,$$
              $$AC=2x+1,$$ $$CK=sqrt{(1+x)^2-x^2}=sqrt{2x+1}.$$
              Thus, since $Delta CKOsimDelta CAB,$ we obtain:
              $$frac{sqrt{2x+1}}{2x+1}=frac{x}{12}.$$
              Can you end it now?



              I got $x=4$.






              share|cite|improve this answer









              $endgroup$


















                5












                $begingroup$

                Let $K$ be a touching point to $CB$, the radius be equal to $x$ and $O$ be a center of the circle.



                Thus, $$CO=x+1,$$
                $$AC=2x+1,$$ $$CK=sqrt{(1+x)^2-x^2}=sqrt{2x+1}.$$
                Thus, since $Delta CKOsimDelta CAB,$ we obtain:
                $$frac{sqrt{2x+1}}{2x+1}=frac{x}{12}.$$
                Can you end it now?



                I got $x=4$.






                share|cite|improve this answer









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  Let $K$ be a touching point to $CB$, the radius be equal to $x$ and $O$ be a center of the circle.



                  Thus, $$CO=x+1,$$
                  $$AC=2x+1,$$ $$CK=sqrt{(1+x)^2-x^2}=sqrt{2x+1}.$$
                  Thus, since $Delta CKOsimDelta CAB,$ we obtain:
                  $$frac{sqrt{2x+1}}{2x+1}=frac{x}{12}.$$
                  Can you end it now?



                  I got $x=4$.






                  share|cite|improve this answer









                  $endgroup$



                  Let $K$ be a touching point to $CB$, the radius be equal to $x$ and $O$ be a center of the circle.



                  Thus, $$CO=x+1,$$
                  $$AC=2x+1,$$ $$CK=sqrt{(1+x)^2-x^2}=sqrt{2x+1}.$$
                  Thus, since $Delta CKOsimDelta CAB,$ we obtain:
                  $$frac{sqrt{2x+1}}{2x+1}=frac{x}{12}.$$
                  Can you end it now?



                  I got $x=4$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 16 hours ago









                  Michael RozenbergMichael Rozenberg

                  103k1891195




                  103k1891195























                      3












                      $begingroup$

                      We have $$CE^2+r^2=(1+r)^2$$ and $$AB=BE=12$$ and $$12^2+(2r+1)^2=(12+CE)^2$$
                      Can you solve this system?






                      share|cite|improve this answer









                      $endgroup$


















                        3












                        $begingroup$

                        We have $$CE^2+r^2=(1+r)^2$$ and $$AB=BE=12$$ and $$12^2+(2r+1)^2=(12+CE)^2$$
                        Can you solve this system?






                        share|cite|improve this answer









                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          We have $$CE^2+r^2=(1+r)^2$$ and $$AB=BE=12$$ and $$12^2+(2r+1)^2=(12+CE)^2$$
                          Can you solve this system?






                          share|cite|improve this answer









                          $endgroup$



                          We have $$CE^2+r^2=(1+r)^2$$ and $$AB=BE=12$$ and $$12^2+(2r+1)^2=(12+CE)^2$$
                          Can you solve this system?







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 15 hours ago









                          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                          75.1k42865




                          75.1k42865






















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