If a regular polygon has a fixed edge length, can I know how many edges it has by knowing the length from...












21












$begingroup$


So I wonder if there is a formula so that when there's a defined edge length, I can calculate a regular polygon's edges amount by knowing its length from corner to center, or vice versa.



So let's say our defined edge length is 1, then by knowing the length from corner to center is $frac{sqrt2}{2}$, I knows it's a square. And if the length is one, I know it's a hexagon.






polygons







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that if we join the center to two adjacent vertices of the polygon, we get an isosceles triangle with sides $l$, $r$ and $r$ where $l$ is side length and $r$ is the radius.
    $endgroup$
    – Faiq Irfan
    Jan 22 at 14:34
















21












$begingroup$


So I wonder if there is a formula so that when there's a defined edge length, I can calculate a regular polygon's edges amount by knowing its length from corner to center, or vice versa.



So let's say our defined edge length is 1, then by knowing the length from corner to center is $frac{sqrt2}{2}$, I knows it's a square. And if the length is one, I know it's a hexagon.






polygons







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that if we join the center to two adjacent vertices of the polygon, we get an isosceles triangle with sides $l$, $r$ and $r$ where $l$ is side length and $r$ is the radius.
    $endgroup$
    – Faiq Irfan
    Jan 22 at 14:34














21












21








21


2



$begingroup$


So I wonder if there is a formula so that when there's a defined edge length, I can calculate a regular polygon's edges amount by knowing its length from corner to center, or vice versa.



So let's say our defined edge length is 1, then by knowing the length from corner to center is $frac{sqrt2}{2}$, I knows it's a square. And if the length is one, I know it's a hexagon.






polygons







share|cite|improve this question











$endgroup$




So I wonder if there is a formula so that when there's a defined edge length, I can calculate a regular polygon's edges amount by knowing its length from corner to center, or vice versa.



So let's say our defined edge length is 1, then by knowing the length from corner to center is $frac{sqrt2}{2}$, I knows it's a square. And if the length is one, I know it's a hexagon.






polygons




polygons






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 21:26









costrom

4081518




4081518










asked Jan 22 at 14:29









Andrew-at-TWAndrew-at-TW

1147




1147








  • 1




    $begingroup$
    Note that if we join the center to two adjacent vertices of the polygon, we get an isosceles triangle with sides $l$, $r$ and $r$ where $l$ is side length and $r$ is the radius.
    $endgroup$
    – Faiq Irfan
    Jan 22 at 14:34














  • 1




    $begingroup$
    Note that if we join the center to two adjacent vertices of the polygon, we get an isosceles triangle with sides $l$, $r$ and $r$ where $l$ is side length and $r$ is the radius.
    $endgroup$
    – Faiq Irfan
    Jan 22 at 14:34








1




1




$begingroup$
Note that if we join the center to two adjacent vertices of the polygon, we get an isosceles triangle with sides $l$, $r$ and $r$ where $l$ is side length and $r$ is the radius.
$endgroup$
– Faiq Irfan
Jan 22 at 14:34




$begingroup$
Note that if we join the center to two adjacent vertices of the polygon, we get an isosceles triangle with sides $l$, $r$ and $r$ where $l$ is side length and $r$ is the radius.
$endgroup$
– Faiq Irfan
Jan 22 at 14:34










2 Answers
2






active

oldest

votes


















12












$begingroup$

Yes, and here is the logic of it.



To show how, I've drawn a square, pentagon and hexagon.



In each case, I've drawn the same construction lines - a circle that touches all their corners (we can do this with any regular polygon). I've labelled the length of the "defined edge" as $s$ and the length from corner to centre as $r$. I'm going to call the number of sides, $n$.



I've also marked a grey line that bisects the edge, from the centre. Because of the way I've positioned the line, it is a perpendicular bisector of the edge (crosses it at a right angle) so each half is a right angle triangle. The right angle triangle has one side $r$, and one side $sbig/2$, and I've labelled the angle these make at the centre, $A$.





Basic trigonometry says that for the right angle triangles, $$begin{align}
sin (A) &= frac{left[dfrac s2right]}r = frac s{2r}\
A &= sin^{-1} left(frac s{2r}right)
end{align}$$



But we also know that each edge, "takes up" $2times A$ degrees, and so $n$ sides will "take up" $2times Atimes n$ degrees. But all $n$ sides must take up 360 degrees, the number of degrees at the centre. So $2cdot Acdot n = 360$.



Now we can solve the problem



Since $$begin{align}
2An &= 360\
An &= 180\
sin^{-1}left(frac s{2r}right)cdot n &= 180\
n &= frac{180}{sin^{-1}left(dfrac s{2r}right)}
end{align}$$



Testing this with your square:



$s=1, r=frac{sqrt2}{2}$



$$implies n = frac{180}{sin^{-1}left(dfrac{1}{2cdotdfrac{sqrt2}{2}}right)} = frac{180}{45} = 4$$



So your example object was a square (4 sides).



Testing this with your hexagon:



$s=1, r=1$



$$implies n = frac{180}{sin^{-1}left(dfrac{1}{2cdot1}right)} = frac{180}{30} = 6$$



So your example object was a hexagon (6 sides).






share|cite|improve this answer











$endgroup$





















    36












    $begingroup$

    Yes. The quantity you are referring to is the radius of the polygon, and it has the formula
    $$r=frac{s}{2sinleft(frac{180}{n}right)}$$
    where $s$ is the side length of the polygon and $n$ is the number of sides. So given $r$ and $s$, you can simply solve the above equation for $n$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      This is correct as far as it goes. It's worth pointing out that when you solve for $n$ there's no guarantee that it will turn out to be an integer, and hence correspond to a regular polygon.
      $endgroup$
      – Ethan Bolker
      Jan 22 at 14:35






    • 6




      $begingroup$
      That's right. But if the given radius is indeed the radius of some regular polygon, then the solution $n$ will be an integer.
      $endgroup$
      – kccu
      Jan 22 at 15:48






    • 13




      $begingroup$
      Good answer. A minor nitpick: $sin(180^{color{blue}{circ{}}} / n)$
      $endgroup$
      – lastresort
      Jan 23 at 0:09













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    2 Answers
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    2 Answers
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    12












    $begingroup$

    Yes, and here is the logic of it.



    To show how, I've drawn a square, pentagon and hexagon.



    In each case, I've drawn the same construction lines - a circle that touches all their corners (we can do this with any regular polygon). I've labelled the length of the "defined edge" as $s$ and the length from corner to centre as $r$. I'm going to call the number of sides, $n$.



    I've also marked a grey line that bisects the edge, from the centre. Because of the way I've positioned the line, it is a perpendicular bisector of the edge (crosses it at a right angle) so each half is a right angle triangle. The right angle triangle has one side $r$, and one side $sbig/2$, and I've labelled the angle these make at the centre, $A$.





    Basic trigonometry says that for the right angle triangles, $$begin{align}
    sin (A) &= frac{left[dfrac s2right]}r = frac s{2r}\
    A &= sin^{-1} left(frac s{2r}right)
    end{align}$$



    But we also know that each edge, "takes up" $2times A$ degrees, and so $n$ sides will "take up" $2times Atimes n$ degrees. But all $n$ sides must take up 360 degrees, the number of degrees at the centre. So $2cdot Acdot n = 360$.



    Now we can solve the problem



    Since $$begin{align}
    2An &= 360\
    An &= 180\
    sin^{-1}left(frac s{2r}right)cdot n &= 180\
    n &= frac{180}{sin^{-1}left(dfrac s{2r}right)}
    end{align}$$



    Testing this with your square:



    $s=1, r=frac{sqrt2}{2}$



    $$implies n = frac{180}{sin^{-1}left(dfrac{1}{2cdotdfrac{sqrt2}{2}}right)} = frac{180}{45} = 4$$



    So your example object was a square (4 sides).



    Testing this with your hexagon:



    $s=1, r=1$



    $$implies n = frac{180}{sin^{-1}left(dfrac{1}{2cdot1}right)} = frac{180}{30} = 6$$



    So your example object was a hexagon (6 sides).






    share|cite|improve this answer











    $endgroup$


















      12












      $begingroup$

      Yes, and here is the logic of it.



      To show how, I've drawn a square, pentagon and hexagon.



      In each case, I've drawn the same construction lines - a circle that touches all their corners (we can do this with any regular polygon). I've labelled the length of the "defined edge" as $s$ and the length from corner to centre as $r$. I'm going to call the number of sides, $n$.



      I've also marked a grey line that bisects the edge, from the centre. Because of the way I've positioned the line, it is a perpendicular bisector of the edge (crosses it at a right angle) so each half is a right angle triangle. The right angle triangle has one side $r$, and one side $sbig/2$, and I've labelled the angle these make at the centre, $A$.





      Basic trigonometry says that for the right angle triangles, $$begin{align}
      sin (A) &= frac{left[dfrac s2right]}r = frac s{2r}\
      A &= sin^{-1} left(frac s{2r}right)
      end{align}$$



      But we also know that each edge, "takes up" $2times A$ degrees, and so $n$ sides will "take up" $2times Atimes n$ degrees. But all $n$ sides must take up 360 degrees, the number of degrees at the centre. So $2cdot Acdot n = 360$.



      Now we can solve the problem



      Since $$begin{align}
      2An &= 360\
      An &= 180\
      sin^{-1}left(frac s{2r}right)cdot n &= 180\
      n &= frac{180}{sin^{-1}left(dfrac s{2r}right)}
      end{align}$$



      Testing this with your square:



      $s=1, r=frac{sqrt2}{2}$



      $$implies n = frac{180}{sin^{-1}left(dfrac{1}{2cdotdfrac{sqrt2}{2}}right)} = frac{180}{45} = 4$$



      So your example object was a square (4 sides).



      Testing this with your hexagon:



      $s=1, r=1$



      $$implies n = frac{180}{sin^{-1}left(dfrac{1}{2cdot1}right)} = frac{180}{30} = 6$$



      So your example object was a hexagon (6 sides).






      share|cite|improve this answer











      $endgroup$
















        12












        12








        12





        $begingroup$

        Yes, and here is the logic of it.



        To show how, I've drawn a square, pentagon and hexagon.



        In each case, I've drawn the same construction lines - a circle that touches all their corners (we can do this with any regular polygon). I've labelled the length of the "defined edge" as $s$ and the length from corner to centre as $r$. I'm going to call the number of sides, $n$.



        I've also marked a grey line that bisects the edge, from the centre. Because of the way I've positioned the line, it is a perpendicular bisector of the edge (crosses it at a right angle) so each half is a right angle triangle. The right angle triangle has one side $r$, and one side $sbig/2$, and I've labelled the angle these make at the centre, $A$.





        Basic trigonometry says that for the right angle triangles, $$begin{align}
        sin (A) &= frac{left[dfrac s2right]}r = frac s{2r}\
        A &= sin^{-1} left(frac s{2r}right)
        end{align}$$



        But we also know that each edge, "takes up" $2times A$ degrees, and so $n$ sides will "take up" $2times Atimes n$ degrees. But all $n$ sides must take up 360 degrees, the number of degrees at the centre. So $2cdot Acdot n = 360$.



        Now we can solve the problem



        Since $$begin{align}
        2An &= 360\
        An &= 180\
        sin^{-1}left(frac s{2r}right)cdot n &= 180\
        n &= frac{180}{sin^{-1}left(dfrac s{2r}right)}
        end{align}$$



        Testing this with your square:



        $s=1, r=frac{sqrt2}{2}$



        $$implies n = frac{180}{sin^{-1}left(dfrac{1}{2cdotdfrac{sqrt2}{2}}right)} = frac{180}{45} = 4$$



        So your example object was a square (4 sides).



        Testing this with your hexagon:



        $s=1, r=1$



        $$implies n = frac{180}{sin^{-1}left(dfrac{1}{2cdot1}right)} = frac{180}{30} = 6$$



        So your example object was a hexagon (6 sides).






        share|cite|improve this answer











        $endgroup$



        Yes, and here is the logic of it.



        To show how, I've drawn a square, pentagon and hexagon.



        In each case, I've drawn the same construction lines - a circle that touches all their corners (we can do this with any regular polygon). I've labelled the length of the "defined edge" as $s$ and the length from corner to centre as $r$. I'm going to call the number of sides, $n$.



        I've also marked a grey line that bisects the edge, from the centre. Because of the way I've positioned the line, it is a perpendicular bisector of the edge (crosses it at a right angle) so each half is a right angle triangle. The right angle triangle has one side $r$, and one side $sbig/2$, and I've labelled the angle these make at the centre, $A$.





        Basic trigonometry says that for the right angle triangles, $$begin{align}
        sin (A) &= frac{left[dfrac s2right]}r = frac s{2r}\
        A &= sin^{-1} left(frac s{2r}right)
        end{align}$$



        But we also know that each edge, "takes up" $2times A$ degrees, and so $n$ sides will "take up" $2times Atimes n$ degrees. But all $n$ sides must take up 360 degrees, the number of degrees at the centre. So $2cdot Acdot n = 360$.



        Now we can solve the problem



        Since $$begin{align}
        2An &= 360\
        An &= 180\
        sin^{-1}left(frac s{2r}right)cdot n &= 180\
        n &= frac{180}{sin^{-1}left(dfrac s{2r}right)}
        end{align}$$



        Testing this with your square:



        $s=1, r=frac{sqrt2}{2}$



        $$implies n = frac{180}{sin^{-1}left(dfrac{1}{2cdotdfrac{sqrt2}{2}}right)} = frac{180}{45} = 4$$



        So your example object was a square (4 sides).



        Testing this with your hexagon:



        $s=1, r=1$



        $$implies n = frac{180}{sin^{-1}left(dfrac{1}{2cdot1}right)} = frac{180}{30} = 6$$



        So your example object was a hexagon (6 sides).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 23 at 19:15

























        answered Jan 23 at 10:39









        StilezStilez

        41129




        41129























            36












            $begingroup$

            Yes. The quantity you are referring to is the radius of the polygon, and it has the formula
            $$r=frac{s}{2sinleft(frac{180}{n}right)}$$
            where $s$ is the side length of the polygon and $n$ is the number of sides. So given $r$ and $s$, you can simply solve the above equation for $n$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              This is correct as far as it goes. It's worth pointing out that when you solve for $n$ there's no guarantee that it will turn out to be an integer, and hence correspond to a regular polygon.
              $endgroup$
              – Ethan Bolker
              Jan 22 at 14:35






            • 6




              $begingroup$
              That's right. But if the given radius is indeed the radius of some regular polygon, then the solution $n$ will be an integer.
              $endgroup$
              – kccu
              Jan 22 at 15:48






            • 13




              $begingroup$
              Good answer. A minor nitpick: $sin(180^{color{blue}{circ{}}} / n)$
              $endgroup$
              – lastresort
              Jan 23 at 0:09


















            36












            $begingroup$

            Yes. The quantity you are referring to is the radius of the polygon, and it has the formula
            $$r=frac{s}{2sinleft(frac{180}{n}right)}$$
            where $s$ is the side length of the polygon and $n$ is the number of sides. So given $r$ and $s$, you can simply solve the above equation for $n$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              This is correct as far as it goes. It's worth pointing out that when you solve for $n$ there's no guarantee that it will turn out to be an integer, and hence correspond to a regular polygon.
              $endgroup$
              – Ethan Bolker
              Jan 22 at 14:35






            • 6




              $begingroup$
              That's right. But if the given radius is indeed the radius of some regular polygon, then the solution $n$ will be an integer.
              $endgroup$
              – kccu
              Jan 22 at 15:48






            • 13




              $begingroup$
              Good answer. A minor nitpick: $sin(180^{color{blue}{circ{}}} / n)$
              $endgroup$
              – lastresort
              Jan 23 at 0:09
















            36












            36








            36





            $begingroup$

            Yes. The quantity you are referring to is the radius of the polygon, and it has the formula
            $$r=frac{s}{2sinleft(frac{180}{n}right)}$$
            where $s$ is the side length of the polygon and $n$ is the number of sides. So given $r$ and $s$, you can simply solve the above equation for $n$.






            share|cite|improve this answer









            $endgroup$



            Yes. The quantity you are referring to is the radius of the polygon, and it has the formula
            $$r=frac{s}{2sinleft(frac{180}{n}right)}$$
            where $s$ is the side length of the polygon and $n$ is the number of sides. So given $r$ and $s$, you can simply solve the above equation for $n$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 22 at 14:32









            kccukccu

            10.5k11228




            10.5k11228








            • 1




              $begingroup$
              This is correct as far as it goes. It's worth pointing out that when you solve for $n$ there's no guarantee that it will turn out to be an integer, and hence correspond to a regular polygon.
              $endgroup$
              – Ethan Bolker
              Jan 22 at 14:35






            • 6




              $begingroup$
              That's right. But if the given radius is indeed the radius of some regular polygon, then the solution $n$ will be an integer.
              $endgroup$
              – kccu
              Jan 22 at 15:48






            • 13




              $begingroup$
              Good answer. A minor nitpick: $sin(180^{color{blue}{circ{}}} / n)$
              $endgroup$
              – lastresort
              Jan 23 at 0:09
















            • 1




              $begingroup$
              This is correct as far as it goes. It's worth pointing out that when you solve for $n$ there's no guarantee that it will turn out to be an integer, and hence correspond to a regular polygon.
              $endgroup$
              – Ethan Bolker
              Jan 22 at 14:35






            • 6




              $begingroup$
              That's right. But if the given radius is indeed the radius of some regular polygon, then the solution $n$ will be an integer.
              $endgroup$
              – kccu
              Jan 22 at 15:48






            • 13




              $begingroup$
              Good answer. A minor nitpick: $sin(180^{color{blue}{circ{}}} / n)$
              $endgroup$
              – lastresort
              Jan 23 at 0:09










            1




            1




            $begingroup$
            This is correct as far as it goes. It's worth pointing out that when you solve for $n$ there's no guarantee that it will turn out to be an integer, and hence correspond to a regular polygon.
            $endgroup$
            – Ethan Bolker
            Jan 22 at 14:35




            $begingroup$
            This is correct as far as it goes. It's worth pointing out that when you solve for $n$ there's no guarantee that it will turn out to be an integer, and hence correspond to a regular polygon.
            $endgroup$
            – Ethan Bolker
            Jan 22 at 14:35




            6




            6




            $begingroup$
            That's right. But if the given radius is indeed the radius of some regular polygon, then the solution $n$ will be an integer.
            $endgroup$
            – kccu
            Jan 22 at 15:48




            $begingroup$
            That's right. But if the given radius is indeed the radius of some regular polygon, then the solution $n$ will be an integer.
            $endgroup$
            – kccu
            Jan 22 at 15:48




            13




            13




            $begingroup$
            Good answer. A minor nitpick: $sin(180^{color{blue}{circ{}}} / n)$
            $endgroup$
            – lastresort
            Jan 23 at 0:09






            $begingroup$
            Good answer. A minor nitpick: $sin(180^{color{blue}{circ{}}} / n)$
            $endgroup$
            – lastresort
            Jan 23 at 0:09




















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            What does “Dominus providebit” mean?

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            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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