Difference between revisions of "Shoelace Theorem"
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or as the special case of Green's Theorem | or as the special case of Green's Theorem | ||
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− | <math>A=\iint\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)dxdy=</math><font size="6">∳</font><math>(Ldx+Mdy)</math> | + | <math>\tilde{A}=\iint\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)dxdy=</math><font size="6">∳</font><math>(Ldx+Mdy)</math> |
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− | where <math>L=-y</math> and <math>M=0</math>. | + | where <math>L=-y</math> and <math>M=0</math> so <math>\tilde{A}=A</math>. |
==Proof 1== | ==Proof 1== |
Revision as of 06:40, 24 December 2020
The Shoelace Theorem is a nifty formula for finding the area of a polygon given the coordinates of its vertices.
Contents
Theorem
Suppose the polygon has vertices
,
, ... ,
, listed in clockwise order. Then the area (
) of
is
You can also go counterclockwise order, as long as you find the absolute value of the answer.
The Shoelace Theorem gets its name because if one lists the coordinates in a column,
and marks the pairs of coordinates to be multiplied,
the resulting image looks like laced-up shoes.
Other Forms
This can also be written in form of a summation
or in terms of determinants as
which is useful in the
variant of the Shoelace theorem,
or as the special case of Green's Theorem
∳
where and
so
.
Proof 1
Claim 1: The area of a triangle with coordinates ,
, and
is
.
Proof of claim 1:
Writing the coordinates in 3D and translating so that
we get the new coordinates
,
, and
. Now if we let
and
then by definition of the cross product
.
Proof:
We will proceed with induction.
By claim 1, the shoelace theorem holds for any triangle. We will show that if it is true for some polygon then it is also true for
.
We cut into two polygons,
and
. Let the coordinates of point
be
. Then, applying the shoelace theorem on
and
we get
Hence
As claimed.
~ShreyJ
Proof 2
Let be the set of points belonging to the polygon.
We have that
where
.
The volume form
is an exact form since
, where
Using this substitution, we have
Next, we use the theorem of Stokes to obtain
We can write
, where
is the line
segment from
to
. With this notation,
we may write
If we substitute for
, we obtain
If we parameterize, we get
Performing the integration, we get
More algebra yields the result
Proof 3
This is a very nice approach that directly helps in understanding the sum as terms which are areas of trapezoids.
See page 281 in this book (in the Polygon Area section.) https://cses.fi/book/book.pdf
(The only thing that needs to be modified in this proof is that one must shift the entire polygon up by k, until all the y coordinates are positive, but this term gets canceled in the resulting sum.)
Problems
Introductory
In right triangle , we have
,
, and
. Medians
and
are drawn to sides
and
, respectively.
and
intersect at point
. Find the area of
.
External Links
A good explanation and exploration into why the theorem works by James Tanton: [1] AOPS