Radius of a circumference having two chords

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In the image, the lenghts of the chords are $6$ and $8$, and the gap between the chords is $1$. Then the radius of the circumference is?



enter image description here



I drew the perpendicular diameters to the chords and tried to apply power of a point, but i didn't find anything. I need some hints.




geometry circle




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    up vote
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    down vote

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    In the image, the lenghts of the chords are $6$ and $8$, and the gap between the chords is $1$. Then the radius of the circumference is?



    enter image description here



    I drew the perpendicular diameters to the chords and tried to apply power of a point, but i didn't find anything. I need some hints.




    geometry circle




    share|cite|improve this question





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      In the image, the lenghts of the chords are $6$ and $8$, and the gap between the chords is $1$. Then the radius of the circumference is?



      enter image description here



      I drew the perpendicular diameters to the chords and tried to apply power of a point, but i didn't find anything. I need some hints.




      geometry circle




      share|cite|improve this question











      In the image, the lenghts of the chords are $6$ and $8$, and the gap between the chords is $1$. Then the radius of the circumference is?



      enter image description here



      I drew the perpendicular diameters to the chords and tried to apply power of a point, but i didn't find anything. I need some hints.





      geometry circle





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 14 at 19:05









      Rodrigo Pizarro

      696117




      696117




















          1 Answer
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          By Pythagoras,



          $$sqrtr^2-3^2-sqrtr^2-4^2=1$$



          then



          $$(r^2-3^2)-2sqrtr^2-3^2sqrtr^2-4^2+(r^2-4^2)=1,$$



          $$(2r^2-26)^2=4(r^2-9)(r^2-16),$$



          $$4r^2-100=0.$$






          share|cite|improve this answer





















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            1 Answer
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            active

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            1 Answer
            1






            active

            oldest

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            active

            oldest

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            active

            oldest

            votes








            up vote
            0
            down vote



            accepted










            By Pythagoras,



            $$sqrtr^2-3^2-sqrtr^2-4^2=1$$



            then



            $$(r^2-3^2)-2sqrtr^2-3^2sqrtr^2-4^2+(r^2-4^2)=1,$$



            $$(2r^2-26)^2=4(r^2-9)(r^2-16),$$



            $$4r^2-100=0.$$






            share|cite|improve this answer

























              up vote
              0
              down vote



              accepted










              By Pythagoras,



              $$sqrtr^2-3^2-sqrtr^2-4^2=1$$



              then



              $$(r^2-3^2)-2sqrtr^2-3^2sqrtr^2-4^2+(r^2-4^2)=1,$$



              $$(2r^2-26)^2=4(r^2-9)(r^2-16),$$



              $$4r^2-100=0.$$






              share|cite|improve this answer























                up vote
                0
                down vote



                accepted







                up vote
                0
                down vote



                accepted






                By Pythagoras,



                $$sqrtr^2-3^2-sqrtr^2-4^2=1$$



                then



                $$(r^2-3^2)-2sqrtr^2-3^2sqrtr^2-4^2+(r^2-4^2)=1,$$



                $$(2r^2-26)^2=4(r^2-9)(r^2-16),$$



                $$4r^2-100=0.$$






                share|cite|improve this answer













                By Pythagoras,



                $$sqrtr^2-3^2-sqrtr^2-4^2=1$$



                then



                $$(r^2-3^2)-2sqrtr^2-3^2sqrtr^2-4^2+(r^2-4^2)=1,$$



                $$(2r^2-26)^2=4(r^2-9)(r^2-16),$$



                $$4r^2-100=0.$$







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 14 at 19:13









                Yves Daoust

                111k665205




                111k665205






















                     

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