If a regular polygon has a fixed edge length, can I know how many edges it has by knowing the length from...
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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 square 2 to 2, I knows it's a square. And if the length is one, I know it's a hexagon.
polygons
asked 26 mins ago
Andrew-at-TWAndrew-at-TW
284
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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 square 2 to 2, I knows it's a square. And if the length is one, I know it's a hexagon.
polygons
asked 26 mins ago
Andrew-at-TWAndrew-at-TW
284
New contributor
Andrew-at-TW is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
20 mins ago
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 square 2 to 2, I knows it's a square. And if the length is one, I know it's a hexagon.
polygons
asked 26 mins ago
Andrew-at-TWAndrew-at-TW
284
New contributor
Andrew-at-TW is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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 square 2 to 2, I knows it's a square. And if the length is one, I know it's a hexagon.
polygons
polygons
asked 26 mins ago
Andrew-at-TWAndrew-at-TW
284
New contributor
Andrew-at-TW is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 26 mins ago
Andrew-at-TWAndrew-at-TW
284
New contributor
Andrew-at-TW is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 26 mins ago
Andrew-at-TWAndrew-at-TW
284
New contributor
Andrew-at-TW is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 26 mins ago
Andrew-at-TWAndrew-at-TW
284
asked 26 mins ago
Andrew-at-TWAndrew-at-TW
284
284
New contributor
Andrew-at-TW is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Andrew-at-TW is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Andrew-at-TW is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
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
20 mins ago
add a comment |
$begingroup$
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
20 mins ago
$begingroup$
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
20 mins ago
$begingroup$
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
20 mins ago
add a comment |
1 Answer
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$begingroup$
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 23 mins ago
kccukccu
9,261924
$endgroup$
$begingroup$
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
19 mins ago
add a comment |
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1 Answer
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$begingroup$
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 23 mins ago
kccukccu
9,261924
$endgroup$
$begingroup$
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
19 mins ago
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 23 mins ago
kccukccu
9,261924
$endgroup$
$begingroup$
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
19 mins ago
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 23 mins ago
kccukccu
9,261924
$endgroup$
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 23 mins ago
kccukccu
9,261924
answered 23 mins ago
kccukccu
9,261924
answered 23 mins ago
kccukccu
9,261924
answered 23 mins ago
kccukccu
9,261924
9,261924
$begingroup$
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
19 mins ago
add a comment |
$begingroup$
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
19 mins ago
$begingroup$
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
19 mins ago
$begingroup$
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
19 mins ago
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
20 mins ago