Difference between revisions of "Shoelace Theorem"
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=== Introductory === | === Introductory === | ||
In right triangle <math>ABC</math>, we have <math>\angle ACB=90^{\circ}</math>, <math>AC=2</math>, and <math>BC=3</math>. [[Median]]s <math>AD</math> and <math>BE</math> are drawn to sides <math>BC</math> and <math>AC</math>, respectively. <math>AD</math> and <math>BE</math> intersect at point <math>F</math>. Find the area of <math>\triangle ABF</math>. | In right triangle <math>ABC</math>, we have <math>\angle ACB=90^{\circ}</math>, <math>AC=2</math>, and <math>BC=3</math>. [[Median]]s <math>AD</math> and <math>BE</math> are drawn to sides <math>BC</math> and <math>AC</math>, respectively. <math>AD</math> and <math>BE</math> intersect at point <math>F</math>. Find the area of <math>\triangle ABF</math>. | ||
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== External Links== | == External Links== |
Revision as of 00:03, 23 October 2018
The Shoelace Theorem is a nifty formula for finding the area of a polygon given the coordinates of its vertices.
Theorem
Suppose the polygon has vertices
,
, ... ,
, listed in clockwise order. Then the area of
is
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.
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
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