Area of a concave quadrilateral
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I was asked a question by an engineering friend recently and was not certain of how to solve the problem. I know there are (somewhat complicated) formulae that compute the area of a quadrilateral (convex or concave) in terms of the sides and a diagonal or an angle, but is there a formula for the area of a concave quadrilateral in terms of just the side lengths?
My intuition is that the side lengths do not uniquely determine the quadrilateral, but I cannot prove this either. If the sides were specified by points on a coordinate system, I feel like the lengths would produce slopes, hence, angles, but I am not even certain this is true.
I have two questions: are the lengths of the sides enough to specify the quadrilateral? What is the area of the resulting figure? Any insight helps.
geometry
asked Nov 16 '16 at 7:48
MakenzieMakenzie
515
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add a comment |
$begingroup$
I was asked a question by an engineering friend recently and was not certain of how to solve the problem. I know there are (somewhat complicated) formulae that compute the area of a quadrilateral (convex or concave) in terms of the sides and a diagonal or an angle, but is there a formula for the area of a concave quadrilateral in terms of just the side lengths?
My intuition is that the side lengths do not uniquely determine the quadrilateral, but I cannot prove this either. If the sides were specified by points on a coordinate system, I feel like the lengths would produce slopes, hence, angles, but I am not even certain this is true.
I have two questions: are the lengths of the sides enough to specify the quadrilateral? What is the area of the resulting figure? Any insight helps.
geometry
asked Nov 16 '16 at 7:48
MakenzieMakenzie
515
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1
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are the lengths of the sides enough to specify the quadrilateralNo. Think of square vs. rhombus.
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– dxiv
Nov 16 '16 at 7:57
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But this quadrilateral is concave. Both a square and a rhombus are convex.
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– Makenzie
Nov 16 '16 at 8:05
1
$begingroup$
If the sides alone are not enough to specify a convex quadrilateral, why would that be any different for concave ones? For example, take something like this and play with it to make infinitely many quadrilaterals having the same sides but different shapes and areas, both convex and concave.
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– dxiv
Nov 16 '16 at 8:12
2
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Thank you! That answers my question.
$endgroup$
– Makenzie
Nov 16 '16 at 8:15
add a comment |
$begingroup$
I was asked a question by an engineering friend recently and was not certain of how to solve the problem. I know there are (somewhat complicated) formulae that compute the area of a quadrilateral (convex or concave) in terms of the sides and a diagonal or an angle, but is there a formula for the area of a concave quadrilateral in terms of just the side lengths?
My intuition is that the side lengths do not uniquely determine the quadrilateral, but I cannot prove this either. If the sides were specified by points on a coordinate system, I feel like the lengths would produce slopes, hence, angles, but I am not even certain this is true.
I have two questions: are the lengths of the sides enough to specify the quadrilateral? What is the area of the resulting figure? Any insight helps.
geometry
asked Nov 16 '16 at 7:48
MakenzieMakenzie
515
$endgroup$
I was asked a question by an engineering friend recently and was not certain of how to solve the problem. I know there are (somewhat complicated) formulae that compute the area of a quadrilateral (convex or concave) in terms of the sides and a diagonal or an angle, but is there a formula for the area of a concave quadrilateral in terms of just the side lengths?
My intuition is that the side lengths do not uniquely determine the quadrilateral, but I cannot prove this either. If the sides were specified by points on a coordinate system, I feel like the lengths would produce slopes, hence, angles, but I am not even certain this is true.
I have two questions: are the lengths of the sides enough to specify the quadrilateral? What is the area of the resulting figure? Any insight helps.
geometry
geometry
asked Nov 16 '16 at 7:48
MakenzieMakenzie
515
asked Nov 16 '16 at 7:48
MakenzieMakenzie
515
asked Nov 16 '16 at 7:48
MakenzieMakenzie
515
asked Nov 16 '16 at 7:48
MakenzieMakenzie
515
asked Nov 16 '16 at 7:48
MakenzieMakenzie
515
515
1
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are the lengths of the sides enough to specify the quadrilateralNo. Think of square vs. rhombus.
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– dxiv
Nov 16 '16 at 7:57
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But this quadrilateral is concave. Both a square and a rhombus are convex.
$endgroup$
– Makenzie
Nov 16 '16 at 8:05
1
$begingroup$
If the sides alone are not enough to specify a convex quadrilateral, why would that be any different for concave ones? For example, take something like this and play with it to make infinitely many quadrilaterals having the same sides but different shapes and areas, both convex and concave.
$endgroup$
– dxiv
Nov 16 '16 at 8:12
2
$begingroup$
Thank you! That answers my question.
$endgroup$
– Makenzie
Nov 16 '16 at 8:15
add a comment |
1
$begingroup$
are the lengths of the sides enough to specify the quadrilateralNo. Think of square vs. rhombus.
$endgroup$
– dxiv
Nov 16 '16 at 7:57
$begingroup$
But this quadrilateral is concave. Both a square and a rhombus are convex.
$endgroup$
– Makenzie
Nov 16 '16 at 8:05
1
$begingroup$
If the sides alone are not enough to specify a convex quadrilateral, why would that be any different for concave ones? For example, take something like this and play with it to make infinitely many quadrilaterals having the same sides but different shapes and areas, both convex and concave.
$endgroup$
– dxiv
Nov 16 '16 at 8:12
2
$begingroup$
Thank you! That answers my question.
$endgroup$
– Makenzie
Nov 16 '16 at 8:15
1
1
$begingroup$
are the lengths of the sides enough to specify the quadrilateral No. Think of square vs. rhombus.$endgroup$
– dxiv
Nov 16 '16 at 7:57
$begingroup$
are the lengths of the sides enough to specify the quadrilateral No. Think of square vs. rhombus.$endgroup$
– dxiv
Nov 16 '16 at 7:57
$begingroup$
But this quadrilateral is concave. Both a square and a rhombus are convex.
$endgroup$
– Makenzie
Nov 16 '16 at 8:05
$begingroup$
But this quadrilateral is concave. Both a square and a rhombus are convex.
$endgroup$
– Makenzie
Nov 16 '16 at 8:05
1
1
$begingroup$
If the sides alone are not enough to specify a convex quadrilateral, why would that be any different for concave ones? For example, take something like this and play with it to make infinitely many quadrilaterals having the same sides but different shapes and areas, both convex and concave.
$endgroup$
– dxiv
Nov 16 '16 at 8:12
$begingroup$
If the sides alone are not enough to specify a convex quadrilateral, why would that be any different for concave ones? For example, take something like this and play with it to make infinitely many quadrilaterals having the same sides but different shapes and areas, both convex and concave.
$endgroup$
– dxiv
Nov 16 '16 at 8:12
2
2
$begingroup$
Thank you! That answers my question.
$endgroup$
– Makenzie
Nov 16 '16 at 8:15
$begingroup$
Thank you! That answers my question.
$endgroup$
– Makenzie
Nov 16 '16 at 8:15
add a comment |
1 Answer
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Having the lengths of the sides is not enough to calculate the area of a polygon (except triangle). But if you have the coordinates of the vertices then you are able to calculate the area. Here is how the formula looks like:
answered Nov 16 '16 at 12:58
SeyedSeyed
6,84841424
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1
$begingroup$
“Having the lengths of the sides is not enough to calculate the area of a polygon.” Yes it is, when $n=3$.
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– symplectomorphic
Dec 28 '18 at 4:08
add a comment |
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$begingroup$
Having the lengths of the sides is not enough to calculate the area of a polygon (except triangle). But if you have the coordinates of the vertices then you are able to calculate the area. Here is how the formula looks like:
answered Nov 16 '16 at 12:58
SeyedSeyed
6,84841424
$endgroup$
1
$begingroup$
“Having the lengths of the sides is not enough to calculate the area of a polygon.” Yes it is, when $n=3$.
$endgroup$
– symplectomorphic
Dec 28 '18 at 4:08
add a comment |
$begingroup$
Having the lengths of the sides is not enough to calculate the area of a polygon (except triangle). But if you have the coordinates of the vertices then you are able to calculate the area. Here is how the formula looks like:
answered Nov 16 '16 at 12:58
SeyedSeyed
6,84841424
$endgroup$
1
$begingroup$
“Having the lengths of the sides is not enough to calculate the area of a polygon.” Yes it is, when $n=3$.
$endgroup$
– symplectomorphic
Dec 28 '18 at 4:08
add a comment |
$begingroup$
Having the lengths of the sides is not enough to calculate the area of a polygon (except triangle). But if you have the coordinates of the vertices then you are able to calculate the area. Here is how the formula looks like:
answered Nov 16 '16 at 12:58
SeyedSeyed
6,84841424
$endgroup$
Having the lengths of the sides is not enough to calculate the area of a polygon (except triangle). But if you have the coordinates of the vertices then you are able to calculate the area. Here is how the formula looks like:
answered Nov 16 '16 at 12:58
SeyedSeyed
6,84841424
edited Dec 28 '18 at 10:55
answered Nov 16 '16 at 12:58
SeyedSeyed
6,84841424
answered Nov 16 '16 at 12:58
SeyedSeyed
6,84841424
answered Nov 16 '16 at 12:58
SeyedSeyed
6,84841424
6,84841424
1
$begingroup$
“Having the lengths of the sides is not enough to calculate the area of a polygon.” Yes it is, when $n=3$.
$endgroup$
– symplectomorphic
Dec 28 '18 at 4:08
add a comment |
1
$begingroup$
“Having the lengths of the sides is not enough to calculate the area of a polygon.” Yes it is, when $n=3$.
$endgroup$
– symplectomorphic
Dec 28 '18 at 4:08
1
1
$begingroup$
“Having the lengths of the sides is not enough to calculate the area of a polygon.” Yes it is, when $n=3$.
$endgroup$
– symplectomorphic
Dec 28 '18 at 4:08
$begingroup$
“Having the lengths of the sides is not enough to calculate the area of a polygon.” Yes it is, when $n=3$.
$endgroup$
– symplectomorphic
Dec 28 '18 at 4:08
add a comment |
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are the lengths of the sides enough to specify the quadrilateralNo. Think of square vs. rhombus.$endgroup$
– dxiv
Nov 16 '16 at 7:57
$begingroup$
But this quadrilateral is concave. Both a square and a rhombus are convex.
$endgroup$
– Makenzie
Nov 16 '16 at 8:05
1
$begingroup$
If the sides alone are not enough to specify a convex quadrilateral, why would that be any different for concave ones? For example, take something like this and play with it to make infinitely many quadrilaterals having the same sides but different shapes and areas, both convex and concave.
$endgroup$
– dxiv
Nov 16 '16 at 8:12
2
$begingroup$
Thank you! That answers my question.
$endgroup$
– Makenzie
Nov 16 '16 at 8:15