Shoelace Formula: Connecting the Area of a Polygon

Shoelace Formula: Connecting the Area of a Polygon with Vector Cross Product Author(s): Younhee Lee and Woong Lim Source: The Mathematics Teacher, Vol. 110, No. 8 (April 2017), pp. 631-636 Published by: National Council of Teachers of Mathematics Stable URL: http://www.jstor.org/stable/10.5951/mathteacher.110.8.0631 Accessed: 31-03-2017 17:43 UTC JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected].

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DELVING deeper Younhee Lee and Woong Lim

Shoelace Formula: Connecting the Area of a Polygon and the Vector Cross Product

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nderstanding how one representation connects to another and how the essential ideas in that relationship are generalized can result in a mathematical theorem or a formula. In this article, we demonstrate this process by connecting a vector cross product in algebraic form to a geometric representation and applying a key mathematical idea from the relationship to prove the Shoelace theorem. The Shoelace theorem gives a formula for finding the area of a polygon from the coordinates of its vertices. For example, the triangle with vertices A (x1, y1), B (x2, y2), and C (x3, y3) has its area determined by the following (Beyer 1978): area( ABC)

=

⎡ x 1 det ⎢ 1 2 ⎢ y1 ⎣

⎡ x x2 ⎤ ⎥ + det ⎢ 2 ⎢ y2 y2 ⎥ ⎦ ⎣

⎡ x x3 ⎤ ⎥ + det ⎢ 3 ⎢ y3 y3 ⎥ ⎦ ⎣

x1 ⎤ ⎥, y1 ⎥ ⎦

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1 2 Edited by Daniel Ness

1 Delving Deeper offers a forum for new insights into 2

mathematics, engaging secondary school teachers to extend their own content knowledge; it appears in every Mathematics Teacher. x1 issuexof x3 x1 Manuscripts 2 1for the department should be submitted via http:// = mt.mtsubmit.net. ↘ ↘ For more ↘ information on submitting 2 y2 http://www.nctm.org/mtcalls. y3 y1 y1 manuscripts, visit Department editors Daniel Ness, [email protected], St. John’s University, 1 3 −2 1 1Jamaica, New York; Nick Wasserman, wasserman@ = tc.columbia.edu, ↘ ↘ Teachers ↘ College, Columbia University, 2 4 Benjamin 2 Dickman, 0 benjamindickman@ New0 York; and gmail.com, The Hewitt School, New York

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x x x x x x 1 = det 1 2 + det 2 3 + det 3 1 , where det of a matrix 2 M denotes y1 y2 the determinant y2 y3 y3 y1 M. The area can also be written and represented as x x x x x x 1 follows:= det 1 2 + det 2 3 + det 3 1 , 2 y1 y2 y2 y3 y3 y1 1 x1 x2 x2 x3 x x 1 area( 2 ABC) = det + det + det 3 1 , 2 y1 y2 y2 y3 y3 y1 = 1 |x1y2 – x2y1 + x2y3 – x3y2 + x3y1 – x1y3| 12 2 = 1 |(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)| 2 1 2 x1 x2 x3 x1 1 =1 2 2 y1x y2x2 y3x3 y1x1 1 1 = 2 xthe x3 x1 inside absolute x1 y y y y 2 In this formula, “shoelace” 3 2 1 11 =1 is meant to represent the difference value bars = 2 between 2 theysums yof the ydiagonal y products. To 0 1 4 2 2 2 3 01 1 make sense1of the3 formula, we first explore the vec1 = product using triangles and quadrilaterals. tor cross 2 Then we the 01 43 argument 22 01 to generalize the rep1 extend = 11• for 4 + n-sided 3 • 2 + ( polygons. 2)• 0 3 • 0 + ( 2)• 4 + 1• 2 resentation = 2 2 0 4 2 0 CASE =1:1TRIANGLES 1• 4 + 3 • 2 + ( 2)• 0 3 • 0 + ( 2)• 4 + 1• 2 v w = 2 you find the area of a triangle given How would the coordinates of its three vertices? (See fig. 1.) A 1 1• 4 + 3 • 2 + ( 2)• 0 3 • 0 + ( 2)• 4 + 1• 2 v v=strategy typical would be to first draw the rectangle 2w = encompassing the given triangle. Visualizing the neighboring w v w =triangles (i.e., triangles ABF, ACD, and BCE), students can subtract the sum of their areas from the area of rectangle BEDF: vw vw

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area(rectangle) – area(surrounding) = area( ABC) 20 – (4 + 3 + 5) = 8 v vw w With the Shoelace formula, the calculation would w.v w be as follows: v w. v

Vol. 110, No. 8 • April 2017 | MATHEMATICS TEACHER 631 This content downloaded from 129.24.105.37 on Fri, © 312017 MarThe 2017 17:43:20 Copyright National CouncilUTC of Teachers of Mathematics, Inc. www.nctm.org. w.vw. Allmaterial use subject tobe http://about.jstor.org/terms All rights reserved. This may not copied or distributed electronically or in any other format without written permission from NCTM.

= 2 xdet yx y x+ det xy y + det y y , 2 1 y112 y22 31 y122 y133 y33 y11 1 1 x2 x32 x1 2 = x1 x2 x23 x1 x x1 2 1 y x2 x3 x1 y2 y3 1 =1 1 1y1 = x2 x3 x1 x2 x3 x1 = 1 12 1 y y2 y+3det = det + det , 2 1 2y1 y xy12 y31 2 yxx1 yx 2 1 y2 y yxy323 y1 xx 1 yx31 y1 3 area( 2ABC) y1 y2 2 1 12 23 1 1 = 1 3 2 1 = 2 11 y1 y2 2 yy3 yy1 =1 1 3 2 1 y3 y1 212 2 x23 x32 x11 1 x 21 0 4 2 0 11 1 1 3 2 1 =2 1= 1 2 = = 0 4 2 02 0 y 41 y23 02 2 10 x2 x3 2 yx x1 y 42 2 0 1 1 2 3 1 x2 x3 x1 1 1 3 1 11 x1 ==1 11• 4 + 3 • 2 + ( 2)• 0 3 •=0 + ( 2)• 4 +=1•1 2 2 = 22 12 y3 0 1y13 • 0 + (0 2)• 44+ 1•2 220 = 2 1•yy14 + 3 •yy22 + ( 2)• 40 2 0 y3 = y1• 0 +4(+ 2)• 4 + 1• 2 =8 2 1 2 12 1 43 + 3 • 22+ ( 2)• 1 0 = 13 • 1• 3 • 2 + ( 2)• 0 3 • 0 + ( 2)• 4 + 1• 2 1 2 Fig. 1 A rectangle encloses the triangle with known v w =x = x x x 1 2 3 1 1 the work 1 formula2 simplifies vertices.+ ( 2)• 4 + 1• 2 1 =1 this vAlthough w 4 + 3 =• 20 + (1•2)• 3 2 v 10 w ==4 1• 2 = 4 +03 • 2 +3(• 02)• 0 3 • 0 + ( 2)• 4 + 1• 2 2 1 3 2 1 1 involved in finding the area, it requires little mathv w = 2 2 v= 1 y y2 Toy3delveyinto = 2 reasoning. 1 1 ematical mathematics 4 2 10 v 2 0 beyond the illustrations, we connect the Shoelace 0 4 2 v0 = 1• 4 +v 3 •w2 = + ( 2)•v0 w =3v• 0 + ( 2)• 4 + 1• 2 w formula to the cross product defined on two vectors 2 3 2space.1 The cross product is a w 1 1 in three-dimensional 1operation 1• 4 + 3 • 2usually + ( 2)•w0 3v• 0 + ( 2)• 4defined + 1• 2 binary v== 2w 1• 4 + 3 • 2 + ( v2)•denoted 2)• 4v+ 1• 2w w0 = 3 • 0by+ (× and x 2following 0 4 2 0 in the form: v w x v w w w v w vv w = v x2, v3w1 – v1w3, v1w2 – v2w1) x v w = (v2w3 – v3w v 1 = 1• 4 + 3 • 2 +xx( 2)•v0 3v• 0w + ( 2)• 4 + 1• 2 v w v w.v =2(v1, v2, v3) and for = (w , w , w w 1 2 3). Its geometxx v x 2 2 2 x2gives + x3 the . cross prodx1 +2) ric w. interpretation (see xfig. x2 2 + xv32 . x12 +w. x perpendicular v w =as xa vector uctvw to both v and w. v w x 2wmagnitude 2x 2 of this vector equals the area of the The xv1 + x2 + x3 . x vx w v w = v w sin , v and w.2 parallelogram formed Fig. 2 The cross product of two 3D-vectors has a geometv by 2w 2 w. v v x + x + x . x w. v wthat the magnitude 1v 2 of a3vector x = (x , x , Note ric interpretation. 1 2 v w = v w sin , x12v+ xw2 2 +x x2 3+2 .x 2 + x 2 . 2 2 2 w 1 2 3 2 2 x x xv3)w. is denoted by =v w v w = v wv sinw ,= v w sinv , w 1w.+ x2 v+ x3 . Thus, 2 2 2 v w. noting v that w =sin v w≥sin 0. In, this way, the of , +w x12 + xvv2 w. vw v magnitude w = v w sin wx3 . x v v w , 2 2 2 v w = v w sin , the cross product (i.e., v1w3) + (v1w2 – v2w1) v w = v w sin , = v wv sin v 2 w 2 = (v w ) is equal to the area of v 2ww 3 – v3w2) 2 + (v32w1 – 2 x 2 +v x w 2 2 2 + x 2. 2 v w AB v w the parallelogram. v w w. v =w(v1 + v2 + v13 v)(w21 + w32w.+ w3 ) w. v v w v v w 2 v w = v w sin v , w 2 2 2 2 w. – (v1w1 + v2vw22 w The essential bridge + v3w3) x1 + x2 + x3 . v wfrom the calculation to the 2 2 2 2 2 v w AB v w 2 2 v w v w proof of the Shoelace formula is the concept of equiv = – (v 1v+ vw 2w2 + v3w3) . w. v 1ww v v AB AB AC v w. w. alence between the magnitude of the cross product AB v v w v 2 w 2v AB v w AB AC and the area of the parallelogram. To elaborate furThe expression inside the final parentheses, AB v v • w, v w. AB 2 v2 w. AC let us consider ther, three v1w1w. + v2w2 + v3w3,vrepresents the dot product of AC AB points (A, B, and C) in the v •ww, AC xy-plane and create two vectors ( AB and AC ) in two w. vectors v and w. The dot product, denoted by AC AB v2 w2 w. w. equal to vw.w cos , where is the angle three-dimensional AC v • w,v is also space with their z-component as AB AC v v w cos AB AC, y , 0), B(x , y , 0), and , zero. Then, given A(x between the two vectors. Hence, 1 1 2 2 AB v • w, w. AC AB C(x3, y3, 0), so that AB AC = (x2 – x1, y2 – y1, 0) and v vv • w w,cos v, • w, 2 2 2 2 AB w. v • w, v 2w v • 2w AC AB w. = 2v 2w = (x3 –AB x1, yAC we apply algebraic 3 – y= 1, 0), AC x )( ythe y1 ) ( y2 defiy1 )(x3 x1 ) 0, 0,(x 3 vv • w w ,v•w ACproduct to get2 the1 following w, =v vw cos 2 2 2 nition of cross result: AB AC 2 , w cos 2 v w 2 cos 2 v 2 =AC0,=0,(x y + x y + x y ) (x y + x y2 +3 x1x , = 2vv w2wcos , v w cosw. AB AC 0, 0,(x 1 2 2 2x13)( y33 1y1 ) (2y21 y13)(x AC v w v w cos v w =v w v • w v=• w, AB AC = 0, 0,(x x )( y y1 x ) )(( yy2 y y1 )(x 2 2 ) (3y2 x1y)1 )(x3 x1 ) AB AC 2 = 10, 0,(x 3 2 2 3 0, (x 1y y v w 2cos =2,22v 22w 1 cos y y+)(x x0,3 y30,(x ) x1(x +x y AB AC = 0, 0,(x2 2 1x=1)(0,y )2 + ( yxAC ) 2xy)( 2 2 1 AC 2 3 1 1 y 3 AB 3 v w = v w v • w 2 2 = v 2 w 22 = 0, 0, (x1AB y=2 +0,xAC y3 = + yx0, y0,(x ) y(x +y)x33 (x y22y+1y)=x1+ v2 w 22cos= 2v 2w2 2 12 2 cos 2 v • w, xy11)( (yxy3 2)y y+12x)(xy31) x13) 2 3 1 2 1 0,(x + x + x y 2 v w = vv w w = v •w 2 x1x wv vwvw •=w 2 2 y1y31AB (3y(yy+3xy 1yy)(x AB AC )((x y)2x)AC )(x AC == 0, 0,0,(x 0,(x =22 0,xx10,)( y2 y+ ) (x y) )+3 x 2y2 +y1 x3+ y3 )y + 2 ,w =2cos v w 1 110, 0,3(x sin . 1 3 2 1 1 2 32 2 31 11 3 3 2= 1 x 2vv 2w 2 2 2 w cos 22 =2 v2 w 2 v 2 = 0, 0, (x y + x y + x y ) (x2 y11 +2 x3 y22 +3 x 2 2 2 = v w sin . 2 2 =v w 1 cos 1 2 2 3 3 1 2 2 2 =y y0, 0,(x v =w vv= wv cos w v wv cos •w == 0, + +AB xAB (x(x +x 0,0,(x 0, (x11yy22++x2xy2 y xy y)AC )AC + xy y+ x+x xy)(y)y ) y1 ) ( y2 =v w 3 3 3 31 1 2 21 1 3 322 2 11 13 33 =2 v 2 w v2 w cosv w cos , AB AC AB 2 2 22 =2 v 2 2 22 w 1 cos AC 2 2 2 = v w sinv = . 22 =w v wv2 2w vcos 2 w AB ACAB = 0, 0, (x1AB y2 + xAC y + x3 y1 ) ( 2• w 2 3 = vWe w take cos sides of the 2 =1 v the wv square 1 cos AB AC =2root v 2 wof 2both 1 cos 22 2 2 2 2 w2 sin . w =v w ABof the parallelogram 2 =2 v equation, 2 2 v • w But AB AC is the area 2 2 2 v wvcos =w v2 1.w cos 2 2 AB =v w = sinv2= w .v sin 2 AB AC, formed by or twice AB the area of ABC. = v w sin . 2 2 2 ACAB AB and 2 2 2 2 2 =v w v w cos 2 Therefore, AB AC, = v =wv v sin ww= v.1 wcos sin , 2 2 AC, AB 2 2 AC, = v w 1 cos2 = v w sin 2 . AC, AB AC, area( ABC) AC, 2 2 v w 2 to represent the area of a parallelogram, is = which v w sin . AC, AC, 1 the product of the lengths of two adjacent sides and = |(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)|. 2 AC, the sine of the angle between them, AB

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AB AC 0 1 2 0 Make note (for later analysis) that the value inside 1 1 = the absolute value bars is exactly the z-component 2 2 5 0 2 5 of the cross product. To illustrate this connection between a vector 1 1 = 0 • 0 + ( 1)• 2 + 2 • 5 ( 1)• 5 + 2 • 0 + 0 • 2 2 cross product and the Shoelace formula onceAB more, 2 1 1. Think of the revisit the shaded triangle in figure = 6.5 2 three vertices of the triangle lying on the xy-plane in AB AC three-dimensional space: A (1, 0, 0), B (3, 4, 0), and C 0 2 5 0 (–2, 2, 0) (see fig. 3). This yields AB = (2, 4, 0) and 1 AC = (–3, 2, 0) for which the cross product AB AC = 2 is (0, 0, 16). This vector’s magnitude, 16, equals its 5 2 1 5 AC z-component, meaning that the area of ABC is 8. AB AC 0 1 12 0 So, is the Shoelace formula specific to triangles 2 • 5 + 5• 2 + 0 • 1 = 0 • 2 + 2 • 1+ 5• 5 1 = 2 only? Does it also work for quadrilaterals? AB AC What 2 0 = 3.5 2 5 about0 pentagons? 1 2 In what 0 follows, we extend 5 1 discussion to the case of quadrilaterals and our = 1 0 21 =1 0 •20 + ( 0 1)• 23+The 2 • 5 ( 1)• 5 + 2 • 0 + 0 • 2 attempt Fig. 1 form of2the 2 a 5more general AB ACmagnitude of this vector cross product equals the absolute value of its 1 5 to 0construct = 2 1 Shoelace theorem. z-component. 2 2 = 6.5 1 2 5 0 2 5 = 0 • 0 + ( 1)• 2 + 2 • 5 ( 1)• 5 + 2 • 0 + 0 • 2 AC (ABC) AD, + area (ACD) 2 1 area CASE 2: QUADRILATERALS 1 AB 1 AB ( 1)• = 0 • 0 + ( 1)• 2 + 2 • 5 2 = 6.5 0 2 5 05 + 2 • 0 + 0 • 2 Notice that the ABCD can be 2 quadrilateralAB 2 1 = decomposed into two triangles (see fig. 4, p. 634), 0 1 2 0 0 2 5 0 = 6.5 2 AC 1 andAB we can apply the Shoelace formula twice as5 AC 2 = 11 5 + 0 2 5AB 0 AC 1 follows: 2 2 = 5 0 2 5 5 2 1 5 1 0 2= 05• 2 + 20• 1+ 5• 5 2AB AC 2 • 5 + 5• 2 + 0 • 1 1 5 2 1 5 AB AC 2 AC ABC) area( 0 1 2 5 0 AB= 2 AC AC = 3.5 1 5 2 1 5 =1 = 0•2 + 2 • 1+ 5• 52 2 •05 + 5• 2 + 0 • 1 2 0 2AB 0AC 1 1 2 0 5 0 2 1 5 1 10 1 AB AC 1 2 0 = 2 • 5 + 5• 2 + 0 • 1 = 0 • 2 + 2 • 1+ 5• 5 = 3.5 AB AC =2 1 1 2 5 = 2 0 2 5 5 0 2 5 = 0 + ( 2)+ 2 + 25 ( 5)+ 0 + 10 + 0 2 = 3.5 0 1 2 0 2 5 0 2 5 2 0 AB= 1 1AC • 0 + ( 1)•120+ 2 • 5 1 ( 1)• = 10 5 + 2 • 0 +AC 0 • 2 AD, =2 0 0 • 0 +( = 1)• 2 + 2 • 5 ( 1)• 1 5+ 2• 0 + 0• 2 2 5 + ( 1)• 2 + 2 • 5 ( 1)• 5 + 2 • 0 + 0 • 2 0 22 5 = AB0 • 0AC 5 0 5 2 = 6.5 caveats AC= 6.5 0 1 Three 2 0 0seem 2relevant 5 when 0 attempting 1AD, = 6.5 1“shoelace” representations: (1) = 0 • 0 + ( 1)•12 + 2 • 5 ( 1)• 5 + 2 • 0 += 01• 2 to combine the + = 0 • 0 + ( 1)• 2AC + 2 • 5AD, ( 1)• 5 + 2 • 0 + 0 • 2 2 2 Coordinates of2the same point made up the first 25 0 2 0 5 2 1 5 5 0 2 5 0 2 5 0 = 6.5 01 1 =26.5 0 0 area( 1 ACD) 02 25 50 0 column in the formula for both triangles; (2) The 1= 1 = 1 + = 2 same2 orientation 0 10 = 2 0 5 2 0 of 5 vertices 0 (counterclockwise or 2 2 5 2 21 1 55 5 2 5 2 12 0 1 11 5 2 1 5 05 0 2 clockwise) applied to both triangles; (3) Although 5 0 5 = 5 0 2 = 1 + 1 0 2 5 2 the cross product is only defined for vectors in 1 0 • 2 + 2 • 1+ = 2 ••05 5+ + 5• 5•22 2+ +50 0 •• 1 1 0 5 2 0 5 22 5 1 2 5 1 5• 5 55 2 5 21 5• = 02 0 • 21+ 2 •=1+ 1 three dimensions, the “shoelace” notation was eas1 2 2 • 5 + 5• 2 + 0 • 1 = 0 • 2 + 2 • 1+ 5• 5 2 5 21 = = 3.55 1 5 2 1 5 2 ily extended to include additional columns. 0 2 ( 2)+ 5 2 + 25 0 2= 3.5 =1 0 + ( 5)+ 0 + 10 + 0 51 0 2 1 5 = 3.51 2 Generalizing this example, we’d like to write the = 2 + 0•1 = 0 • 2 + 2 • 1+1 5• 5 2 • 5 + 5• 2 2 • 5 + 5• 2 + 0 • 1 = 0 • 2 + 2 • 1+ 5• 5 = 10 2 Shoelace formula for quadrilaterals as follows: 1 5 0 2 1 5 2 = =AB 0 + (AC 2)+ 2 + 25 ( 5)+ 0 + 10 + 0 AB AC 3.5 2 AB 1 AC = 3.5 = 0 + ( 2)+ 2 + 25 area( ( 5)+ ABCD) 0 + 10 + 0 = 10 2 AC AD, That is, we can calculate the z-components of two AC AD, AB AC x2 x3 x4 x1 x1 = 10AD, and add cross products, AB AC and AC 1 half their absolute values to find the area of quadri= 0 1 0 2 x24 x 0AD, is 1 6.5 2 2+ 3.5 0 0 2 x1 5 5 x2 0 0 x3 AC lateral ABCD = 10.1 0 1 y1 1 y2 y3 y4 y1 1 AC AD, + 1 0 1 1 2 0 0 2 5 0 = = 2the calculations above,+some = In cancel 1 products 1 2 2= 2 22 + 1 0 5 1 out. For5 52 in the 2 y represen1 y 5 5 y 5example, 0 +22 2 • 5 and 5 –0 • 5 y y x2 y1 + x3 y2 + x4 y3 + x1 y4 0 1 2 0 • 20 5 2 •01 5 22 0 53 2= 54 x1 y22 1+ x21y3 + x35y4 + x4 y1 tation1 of ABC cancel out +0 22and –2 5 in0the 2 2 0 0 1 0 5 1 0 1 2 5 0 + = 1 1 0 1of ACD. 2 5As a result, 0 1the+sum can representation = 21 2 1 x2 y1 + x3 y2 + x4 y3 + x1 y4 = 5 5 0= 22 x11y221+ x22 y53 + x53 y4 +0x4 y1 = 2 5 this0 way: 22 1 be written 5 0 2 5 5 2 1 5 2 5 =5 0 2 1 5 0 2 1 5 0 1 2 5 02 5 0 2 1 5 AB AC 0 1 1 for A (x1, y1), B (x2, y2), C (x3, y3), and D (x4, y4). 1 0 + ( 2)+ 21+ 25 ( 5)+ 20 + 10 5+ 0 0 = = 2 0 + ( 2)+= 2 + 25 ( 5)+10 + 10 + 0 2 5 AC 2 + 25 ( 5)+ 0 + 10 + 0 + ( 2)+ 0 22 1 =5 0AB Vol. 110, No. 8 • April 2017 | MATHEMATICS TEACHER 633 AC AB 5 0 1 5 2 = = 10 10 This content downloaded from 129.24.105.37 on Fri, 31 Mar 2017 17:43:20 UTC 1 = 10 = 0 + ( 2)+ 21+ 25 ( 5)+ 0 + 10 + 0 All use subject to http://about.jstor.org/terms = 0 + ( 2)+ 2 + 25AC ( AB 5)+ 0 + 10 + 0 2 1 for counterclockwise A A A 2

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area(BCDA) = (–1)area( BCD) + area( BDA) area(CDAB) = area( CDA) + area( CAB) area(DABC) = area( DAB) + (–1)area( DBC)

Assumptions about appending triangles Assume, without loss of generality, that one of the triangles has its vertices labeled in a counterclockwise sequence. We can accomplish the

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AB AB AC AC and noting xthat x xn x1 1 n 1 AC AC AB AB z= p • area( A1An-1An) = p • (1/2)|z| = (1/2)z y1 yn 1 yn y1 AA AA 1 1for forcounterclockwise counterclockwise AA 1 1 n 1n 1 n n p p= the == xz1 component with yn 1 + xn 1 yn of + xthe y cross xn product y + x ycalcu+x y n 1 1 1 forclockwise clockwise AA AA AA n n 1 1 n 1 1 n 1n 1 n n lated as 1 1 for

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x2 x3 ... xn x1 x1 x11x1 xnx1n 1 xnxn x1x1 = z z= = 2 y y2 y3 ... yn y1 1 y1y1 yny1n 1 ynyn y1y1 1 = =x1=xy1n2y1n +x1 1+xynx2 1n+y1xny2+ny+x3 n+ xyn... y + xxn nyx11n y11y+1x+x2nyxy1nn+ y1nx+13 y +x21xy+1ny...n + x1 yn 1 1

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634 MATHEMATICS TEACHER | Vol. 110, No. 8 • April 2017

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Notice that choice of vertex A in figure 4 was somewhat fortuitous. The interiors of triangles ABC and ACD are disjoint, so area(ABCD) = area( ABC) + area( ACD). If we had chosen to begin with vertex B and tried to write area(BCDA) = area( BCD) + area( BDA), we would not have the same orientation for both triangles (and we would not have a true statement of area!). Requiring that both triangles have the same orientation (caveat 2 above), or that the quadrilateral ABCD be convex, would be sufficient but is not actually necessary. Notice that for the non-convex quadrilateral in figure 4, any of the following are valid:

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Fig. 5 The sign of a cross product’s z-coordinate is determined by the orientation of the triangle’s vertices in the xy-plane.

x1

x2 x3 x4 x1 1y3 y4 y1 = 2 y1 y2 y3 y4 1regardless of the first shoelace calculation, = x1 y2 + x2 y3 + x3 y4 + xvertex y x2 y1 + x3 y 4 1 2 subtraction (rather than chosen, if we allow for 1 = x of y the + x2 very3 + x3 y4 + x4 addition) of areas when the orientation 2 1 2 tices of the next triangle is clockwise (rather than counterclockwise). AB AC To explain the idea of signed areas, revisit figure 1 and confirm that while AB AC = (0, 0, 16) we find AC AB = (0, 0, –16). The magnitude of these vectors is just the absolute AC AB value of their z-components (because the three 1 for counterclockwise A1An 1An points were in the pxy-plane). Figure 5 shows = 1 as the forright clockwise A1An 1An what is commonly known hand 1 forrule counterclockw for cross products; in our contextpit= suggests that 1 for clockwise the area of a triangle with clockwise orientation xn 1 xnof thex1corx1 z-coordinate is the opposite of half the responding cross product. z= xn 1 xn x1 So, for a triangle with orieny1 counterclockwise yn 1 yn y1 = shoelace fortation, the absolute value bars in zthe mula do not change the calculation—essentially, = x1 yn 1 + xn 1 yn + xn yy11 xyn n1 y1 1 + xnyyn n 1 + |z| = z. For a triangle with clockwise orientation, = x1 yn 1 + xn 1 yn + xn y1 however, the z-component of the corresponding x2 = z. x3 In either ... xn x1 x1 so –|z| cross product is negative, 1 case, we see that admitting signed areas results = x2 x3 ... x1 in our being able to 2 drop 1y3 value y1 the yabsolute ... signs, yn y1 2 = and thus combine the shoelace representations. 2x x2 1 x3 x4 x1 y2 y3 ... 1 y1 1 = x1 y2 + x2 y3 + ... + xn y1 x2 y1 + x3 y2 + = CASE 3: n-SIDED POLYGONS 2 1 2 A rigorous yproof of y2the shoelace y3 yformula = y1 xfor y2the + x2geny3 + ... + xn y1 1 4 2 1 of eral case (see fig. 6) requires consideration the 1 signs of = thexquantity y + x2 yinside + x3 ythe + xabsolute y xvalues. y + x3When y2 + x4 y3 + x1 y 1 2 3 4 4 1 2 1 2 of the vertices a triangle are labeled in a counterclockwise orientation, the quantity inside the absolute value isxpositive of the two x2x2because x3x3 thex4cross x4 xproduct x x 1 1 1 1 vectors in the xy-plane results in a positive z-com11 AC = = AB ponent (see fig. 5). Clockwise orientation results in 22 y1y1 z-component. y2y2 y3y3 Wey4can y4 consider y1y1 a negative these two AC together AB situations by defining a function p, 11 = = x1xy12y+2 +x2xy23y+3 +x3xy34y+4 +x4xy41y1 x2xy21y+1 +x3xy32y+2 +x4xy43y+3 +x1xy14y4 22 1 for counterclockwise A1An 1An p= 1 for clockwise A1An 1An y1

(

Fig. 4 The quadrilateral can be decomposed into two triangles for which the Shoelace formula applies.

x4

) )( ( ) (

))

......

xnxn

x1x1

......

ynyn

y1y1

)

1 for counterclockwise A1An 1An

p=

1

for clockwise A1An 1An

xn 1 xn x1 x1 Now, we present a mathematical induction z = for the extension of the Shoelace formula to proof y1 polygon. yn 1 yn begin y1 by writing a general an n-sided We form: = x1 yn 1 + xn 1 yn + xn y1 xn 1 y1 + xn yn 1 + x1 yn area(n-sided polygon A1A2 . . . An)

(

) (

1 = 2

)

x1

x2

x3

...

xn

x1

y1

y2

y3 x1

... x2

yn ...

y1 xn

xn 1 2 1 1 = x y + x2 y3 + ... + xn y1 x2 y1 + x3 y2 + ... + x1 yn 2 2 1 2 y1 y2 ... yn 2 yn 1

(

We then append A1An-1An, adding or subtracting its area depending on whether the orientation of vertices is counterclockwise or clockwise. (See fig. 6.) In either case, we can add the appropriately signed area of A1An-1An as the z-component of the the x2 product ... associated xn 2 xnwith x1 triangle. x1 cross 1 That is, 1 2 y1 A .y.2 . A ... yn 2 yn 1 y1 area(A 1 2 n ) = area(A1A2 . . . An-1 ) + p • area( A1An-1An) x1 x1 x2 ... xn 2 xn 1 x1 1 x x ... x x x1 x1 1 2 n 2 n 1 1 21 y=1 y1 y2 ... yn 2 yn 1 y1 + p 2 2 y1 y2 ... yn 2 yn 1 y1 y1 x1 x2 ... xn 2 xn 1 x1 x1 1x xx x xx x...... xx xx1xn n1 1 xx1 1 xx1 1 1 11 1 22 n 1 nnn2 2 1 1 1 1 2 y1= + y p y2 ... yn 2 yn 1 y1 + + p 2 12 22 y y y ... y y y1 y1 1 y1y2 2 y... ynnn2 2 yy1 n n1 1 yy1 1 y1 1 n 1

) (

)

x2 ... xn 2 xn 1 x1 for A1(x1, y1), A12(x2, yx2), A3(x , y ), . . ., 3 3 x2 ... xn 2 xn 1 1 An-1(xn-1, yn-1) and 2 A 1yn(xn, yy n). ... y y = 1 our claim 2 The basis case 2for (whennn2 = 3) n 1 ... yn 2inducyn 1 was proved in case 1.y1Next,y2we assert the

tive hypothesis—that is, the statement holds ... xxnxxnn n22 x1 xxx1nn 11 x1xx1 x2 x2 ...(Note xn x1n 1 x1 x1 xx11 x1 xx1 22 xn x... x1 x1 polygon. ... xnthat x2n 2 our xx1x1n true for an (n–1)-sided 1n 1 2 n 2 n 1 1 1 11 1 11 1 1 1 = justification in=case = 2 admits polygons that are = + +p ++p 22 loss of generality, assume 2 2 2 2 2y y not convex.) Without y1 y1 y2 y2 ... ... yn y2n 2 yn y1n 1 y1 y1 yy11 y1 yy1 22 yn y ... yynynynn2n22 y1 yyyn1nn1 11 ... yy1y1n 1y1 1 1n 1 1 2 counterclockwise orientation of polygon A1A2… An–1 (which is equivalent counterclockwise orixx ...... xx xx x1xn x1 x1x1 x2 xn ... xnn 2 x1xn 1 x1x1 to x1 x1 xx1n 22 n n2 2 n n1 1 1 1 2 n 2 n 1 1 1 entation of A1A121An–1) so that we can drop the 1 == = + + absolute value bars. 22 20 2 y0y 2y 3y ... 2 y 4 y y1 0 yy y1y1 yy ... y y y ... y y y y y 22 n n2 2 true n n1 1 n 1 nn 1 n n 2 2 n 1 1 1 y1 1n 11 n 11 1 1 22 Now, supposing that the statement holds for such an (n–1)-sided polygon, we prove that it 2 x2 ... xn 2 xn 1 xn 5 x x11 x 3 ... 0 x 2 x1 x2n 1 5 xn x1 1 2 n 2 also holds true for 1 an n-sided polygon to establish 1 = = the inductive reasoning. That is, assume polygon 2 0 y1 0 y2 3 ... 2 yn 2 4 yn11 0 yn20 y y10 y 3 ... 2 y 4 A1A2 . . . An-1 has area y1n 1 0 yn y1 1 2 n 2 1 =1 1 2 2 2 5 1 3 0 2 2 5 5 1 3 0 2 2 5 x1 x2 ... xn 2 xn 1 x1 Using the Shoelace representation helps us see 1 = 10products 0 0 3 2 4 1 0 the 0 that 3 cancel 2 out.4 Using 1 the 0absolute 2 2 1 1 1 1 values gives a uniform representation for different y1 y= ... yn 2 yn 1 y1 = 2 2 2 2 orientations of vertices. 5 1 3 0 2 2 52 5 1 3 0 2 2 5 1 = 2

1 = 2

1 = 2

x=2 1 2 =1 2 y2

x1 y1

...

xn

2

xn

1

x1

...

yn

2

yn

1

y1

1 + p 2

x1= 1 xn 1 2 =1 2 y1 yn 1

x1

= x12 2

...

xn

2

xn

1

x1

y1

y2

...

yn

2

yn

1

y1

x1

x2

...

xn

2

xn

1

xn

x1

y1

y2

...

yn

2

yn

1

yn

y1

(a)

1 + 2

xn

x1

yn

y1

x=1 1 2

xn

1

xn

x1

y1

yn

1

yn

y1

xn

1

xn

x1

yn

1

yn

y1

xxn n1 1

xxn n

xx1 1

yyn n1 1

yyn n

yy1 1

xxn n1 1

xxn n

xx1 1

yyn n1 1

yyn n

yy1 1

xn

1

xn

x1

yn

1

yn

y1

(b)

0 nth 0vertex creates 3 2a triangle 4 which 1 enlarges 0 Fig. 6 The the (n–1)-sided polygon when the triangle’s orientation is counter1 clockwise (a) or reduces the polygon when the orientation is clockwise (b). 2

5

=1 2

1

3

0

2

2

5

Vol. 110, No. 8 • April 2017 | MATHEMATICS TEACHER 635 This content downloaded from 129.24.105.37 on Fri, 31 Mar 2017 17:43:20 UTC All use subject to http://about.jstor.org/terms

x2 ... xn 2 xn 1 x1 x1 1 x ... x xn 1 x 2 x1 x... ...yn 2 xn y xn y1 1 x1 xy1 2 1 y1 2 2 2 n 2 1 1 In this1 article, we examined the nShoelace formula 2 through y12 multiple y ...representations y 2 yn 1of they1vector cross y1 2 y2further ...n into ynthe yn 1 formula y1 2 Shoelace product. Delving x ... x x x xn 1 xn x1 x x1 1 2 n 2 n to 1 examine 1 provided us with the opportunity a path 1 1 = abstraction to interconnected x2 while ... engaging xn 2 inxan x1 + 2 p x1 x x x1 n 1 2 x1 x ... x x x1 y xy x1 x xy1 n 1 xn y1 n body of mathematical ideas, forms, skills, y1 2 ... yn 2 n y 1 y1 2 2n 1 n y 1 1 and1imagn1 1 n 1 n = + p 1 1We close our discussion by proposing ination. an 2 = 2 + p y1 2problem y ... secondary y yn 1 y y12 y y y exercise ... ...xn n2 2 y students. xn 1 y x1 1 y x1 y x2 2 yfor x1 y xn n1 1 y xn n y x1 1 y 1 2 n 2 n 1 1 1 n 1 n 1 1 1 = + x ... x x x x x x x x 2 n 1 1 2 1 2 1 APPLYING FORMULA x... ...yn 2 xnTO xn y1 1 x1 xy x1 xy1THE xy1 n 1 xn y1 n y y1 1 y1 2 2n 1 1 n1 2 n 2 1 n 1 n = + FIND THE 1 AREA OF A STAR 1 2 = 2 + A classroom of aystary yexercise ... to find y the area y y y y y ... xnn 22 y xnn 11 y x1n y x1 12 y n 1 y n x112 yx22 y ... yn 1 y1 1 2 be challenging. n 2 n 1 students 1 1 n 1 shaped decagon may Some 1 = partition x the polygon ... x into smaller xn 1 triangles x xor may 1 2 x1 x... ...yn 2 xn y xn 1yn xy x1 xy1 2 1 yrectangles 2 2 nvercreate to remove extra triangles. Ten 1 2 n 2 n 1 n 1 = 1 = deter some students from using the Shoe2 may tices y12 y2 ... yn 2 y yn y y1 but others y2 ... yn 2 n 1 ysymmetry yn 1 y1Fig. 7 A symmetric decagon can be partitioned into two lace formula, may recognize n 1 to simplify the calculations. The area of hexagon hexagons of equal areas. ABCDEF 0 0in figure 3 7 2can be 4 calculated 1 0as follows: 1 Vertex coordinates that contain zeros (as illus0 3 2 4 1 0 2 0 trated in the work above) reduce the work in 1 3 0 2 2 5 1 5 0 0 3 4 1 0 1 calculations. Students may also notice how the 2 5 2 1 3 0 2 2 5 orientation of the vertices relates to the sign of the 5 1 3 0 2 2 5 calculated value. A student who took a clockwise =1 2 orientation for the vertices in the Shoelace for1 = |(0 +10 + 0 + 4 + 8 + 5) – (0 + (–3) + (–6) + 0 + 2 + 0)| mula would see a negative value inside the abso2 = lute value instead. 2 = 1 |17 – (–7)| Students learn many ways to finding the area of a 2 triangle—the method involving base and height, the 1 = 12 2 =1 method involving lengths of two sides and the sine 2 value of the angle between them, Heron’s formula, and the Shoelace formula to name a few. Asking students to compare and contrast various methods as they work in different contexts can provide them with useful intuitions behind the formulas for area.

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REFERENCE Beyer, William. 1978. CRC Standard Mathematical Tables. Boca Raton, FL: CRC Press.

YOUNHEE LEE, [email protected], is a graduate student at The Pennsylvania State University. Her research focuses on relationships between collegiate mathematics and what is taught in school mathematics at the K–grade 12 level. WOONG LIM, [email protected], is an assistant professor of mathematics education at the University of New Mexico. His research interests include the development of algebraic thinking and the role of language in mathematics education.

636 MATHEMATICS TEACHER | Vol. 110, No. 8 • April 2017 This content downloaded from 129.24.105.37 on Fri, 31 Mar 2017 17:43:20 UTC All use subject to http://about.jstor.org/terms