Mixing
If a regular polygon has a fixed edge length, can I know how many edges it has by knowing the length from...
$begingroup$
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
edited Jan 22 at 21:26
costrom
4081518
asked Jan 22 at 14:29
Andrew-at-TWAndrew-at-TW
1167
$endgroup$
add a comment |
$begingroup$
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
edited Jan 22 at 21:26
costrom
4081518
asked Jan 22 at 14:29
Andrew-at-TWAndrew-at-TW
1167
$endgroup$
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
add a comment |
$begingroup$
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
edited Jan 22 at 21:26
costrom
4081518
asked Jan 22 at 14:29
Andrew-at-TWAndrew-at-TW
1167
$endgroup$
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
edited Jan 22 at 21:26
costrom
4081518
asked Jan 22 at 14:29
Andrew-at-TWAndrew-at-TW
1167
edited Jan 22 at 21:26
costrom
4081518
asked Jan 22 at 14:29
Andrew-at-TWAndrew-at-TW
1167
edited Jan 22 at 21:26
costrom
4081518
edited Jan 22 at 21:26
costrom
4081518
edited Jan 22 at 21:26
costrom
4081518
4081518
asked Jan 22 at 14:29
Andrew-at-TWAndrew-at-TW
1167
asked Jan 22 at 14:29
Andrew-at-TWAndrew-at-TW
1167
asked Jan 22 at 14:29
Andrew-at-TWAndrew-at-TW
1167
1167
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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).
answered Jan 23 at 10:39
StilezStilez
42129
$endgroup$
add a comment |
$begingroup$
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 $n$.
answered Jan 22 at 14:32
kccukccu
10.6k11229
$endgroup$
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
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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).
answered Jan 23 at 10:39
StilezStilez
42129
$endgroup$
add a comment |
$begingroup$
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).
answered Jan 23 at 10:39
StilezStilez
42129
$endgroup$
add a comment |
$begingroup$
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).
answered Jan 23 at 10:39
StilezStilez
42129
$endgroup$
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).
answered Jan 23 at 10:39
StilezStilez
42129
edited Jan 23 at 19:15
answered Jan 23 at 10:39
StilezStilez
42129
answered Jan 23 at 10:39
StilezStilez
42129
answered Jan 23 at 10:39
StilezStilez
42129
42129
add a comment |
add a comment |
$begingroup$
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 $n$.
answered Jan 22 at 14:32
kccukccu
10.6k11229
$endgroup$
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
add a comment |
$begingroup$
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 $n$.
answered Jan 22 at 14:32
kccukccu
10.6k11229
$endgroup$
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
add a comment |
$begingroup$
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 $n$.
answered Jan 22 at 14:32
kccukccu
10.6k11229
$endgroup$
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 $n$.
answered Jan 22 at 14:32
kccukccu
10.6k11229
answered Jan 22 at 14:32
kccukccu
10.6k11229
answered Jan 22 at 14:32
kccukccu
10.6k11229
answered Jan 22 at 14:32
kccukccu
10.6k11229
10.6k11229
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
add a comment |
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
add a comment |
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$begingroup$
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