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School and University Partnership for Educational Renewal in Mathematics

An NSF-funded Graduate STEM Fellows in K–12 Education Project

University of Hawai‘i Department of Mathematics

Folding Quadrilaterals From a Square

Abstract The purpose of this activity is to construct all regular quadrilaterals from a square by folding, then to demonstrate that the figures are correct using geometric properties and definitions.

Grade range 6th to 8th gradeStudent Prompt When you fold a square in half you get a rectangle. What

other quadrilaterals can you make from a square by folding?(Oral)

Equipment Part 1:Square sheets (15 x 15 cm)

Part 2:The figures obtained during the first partPaperPencil, eraser

Duration 3 x 45 minutesPlan Part 1:

Announce that you will work on describing quadrilaterals.

List the seven quadrilaterals (without specifying their geometric properties): square, kite, arrowhead, rhombus, parallelogram, rectangle, and trapezoid.

Distribute the instruction sheets and squares.

Allow the students to read the instructions and try to solve the given problem. At this stage it is not necessary to put students in groups - they should all try on their own.

After 10 minutes, if necessary, propose an initial class discussion to list all of the shapes they've come up with (this is to get rid of any shapes the students make that are not quads).

Class sharing of the part 1: list the figures obtained. (See solutions: part 1). Students explain how they got their figures and if there are any ways to make the same figures with fewer folds.

Part 2:

Tell students that they will try to prove that the figures

CEM Week of Math: Let the math unfold ! October 11 - 15 2010

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obtained in the previous class are correct.

Demonstrate the rectangle to the class. (See solutions: part 2)

Then have students work on the kite in groups. (establish groups of 2 or 3 students so they can compare their ideas.)

Class Discussion: Discuss the various proofs given by the students. (See solutions: part 2)

Next: Ask them to prove that the remaining figures are correct (do not distribute the rhombus -- that can be used for final evaluation). Distribute the figures so that several groups had the same figure. This will show that there may be multiple ways of solving the problem.

Sharing: Each group submits its proposal and the class discusses the validity of the proposals.

Finally, individually, have them prove that the rhombus is correct.

Preliminary analysis of the activity ( student's predictable steps, teacher response)

Result:You can get all the regular quadrilaterals (see solutions: part 1)

Proof (see solutions: part 2)

References to educational content, curricula, and teaching aids

Recognize, describe, and name surfaces according to their shape (internal symmetry, sides, angles, diagonals )Recognize and check if two straight lines are parallel or perpendicular (ruler and square-edge)Draw parallel or perpendicular lines(ruler and square-edge)Identify the lines of symmetry in a figure

Highlighted mathematical concepts

QuadrilateralsProperties of quadrilateralsAnglesTriangles and properties of similar triangles

PossibleDevelopment

● Write instructions to fold a shape ● What other geometric figures can you get by

folding a square sheet?


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STUDENT PROMT (It may be written on the blackboard rather than distributed to students)

When you fold a square in half you get a rectangle. What other quadrilaterals can you get from a square by folding?


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THEORETICAL ELEMENTS:

• Recall the definition of quadrilateralsA polygon is often described by an ordered sequence of vertices, such as ABCD. We only consider polygons whose edges do not intersect except at their ends. This avoids the butterfly shape:

NAME DEFINITION PROPERTIESAxes Diagonals Angles

Square • four congruent sides

• four right angles

• 4 • equal length• perpendicular• bisect each other

four 90º angles

Kite • two pairs of congruent adjacent sides

• convex

• 1 • perpendicular• one is bisected by

the other

twocongruent opposite angles

Arrowhead • two pairs of congruent

adjacent sides • not convex.

• 1 • the diagonals cross each other at right angles outside the body of the figure

one angle over 180° and two congruent opposite acute angles

Rhombus • four equal sides

• 2 • Perpendicular• intersects in the

middle

two pairs of congruent opposite angles

Parallelogram • two pairs of parallel

• 0 • One is bisected by the other

two pairs of equal opposite


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sides or

• two pairs of equal opposite sides

• convex

angles

Rectangle • two pairs of equal sides

• four right angles

• 2 • equal• intersects at their

midpoints

four 90º angles

Trapezoid • a pair of parallel sides

• 0 •

Isosceles Trapezoid

• a pair of parallel sides

• a pair of equal nonparallel sides

• 1 • equal Two pairs of equal consecutive angles

Rectangular Trapeziod

• 1 pair of parallel sides

• 2 right angles

• 0 • two angles

1) Definition relevant to anglesTwo angles are opposite if:

Two angles are related if:


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Two angles are complementary if their sum is 90º.

Two angles are supplementary if their sum is 180º.

A bisector cuts an angle in half


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2) Plotting Parallels

3) Equal TrianglesTwo triangles are similar if they don't overlap. So if:1) AB = A’B’ 4) α = α’2) AC = A’C’ 5) β = β’3) BC = B’C’ 6) γ = γ’

There are three tests of equality:Case 1: if we have 1), 2) and 3) then we have 4), 5) 6)In other words: SSS: side / side / sideCase 2:if we have 1), 2) and 4) then we have 3), 5) and 6)That SAS: side / angle / sideCase 3: if we have 1), 4) and 5) then we have 2), 3 ) and 6)In other words: ASA: angle / side / angleNote: since α + β + γ = 180 º, ASA is also AAS


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Solutions: End of Part 1 (creation of figures)Note: Only justified folds will be considered correct, so random folding doesn't count.

SQUARE

This is our starting shape.

RECTANGLE

1) The starting square is a rectangle.

OR

2)

Fold edge to edge


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KITE

1) The starting square is a kite.

OR

2)

Fold along the diagonal BD.Open back up to get the starting square.Fold CD to BDFold AD to BDThis leaves the kite LBMD


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Rhombus

1) The starting square is a rhombus.

OR

2)

Fold along the diagonal BDOpen back up to get the starting squareFold CD to BDFold AD to BDThis leaves the kite LBMD

Fold BL to BDFold BM to BDThis leaves the diamond HBTD


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PARALLELOGRAM

1) The starting square is a parallelogram.

OR

2)

Fold along the diagonal BDOpen to get the starting square.Fold CD to BDFold AB to BDThis leaves the parallelogram KBLD


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ANY TRAPEZOID, ISOSCELES OR RECTANGULAR, FROM A SQUARE

1) The starting square is a trapezoid. It is both isosceles and rectanglar.

OR

2) Any trapezoid :

Fold two opposite sides in any way.Fold any protruding flaps under the figure.We get XYCD

3) Rectangular trapezoid:

Fold one side in any way.Fold any protruding flaps under the figure.We get AXCD


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4) Isosceles trapezoid:

OBC is folded to AD.We get ATUD.

OAD is folded in any way over XY.

OOpen the figure and fold along XY and X'Y'.This gives X 'XY'Y.


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ANY TRAPEZOID, ISOSCELES OR RECTANGULAR, FROM A PARALLELOGRAM

1) The parallelogram is a trapezoid, neither isosceles nor rectangular.

OR

2) Any trapezoid :

Construct the parallelogramFold B onto DL

This leaves the trapezoid KXLD


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3) Rectangular trapezoidTake parallelogram KBLD.Fold D onto the line segment DL.We get XBLY.

4) Isosceles trapezoid:Take the parallelogram KBLD.

Fold D to LWe get K ' (the image of K)

Fold along K'L

We obtain KK'LD.


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ISOSCELES AND RECTANGULAR TRAPEZOID FROM A TRIANGLE

1) Isosceles trapezoid :

Mark the diagonals AC and BDOpen the paper back up to the squareNote: AC ∩ BD = M

Fold along the diagonal BDFold A along AMWe get the trapezoid YXBD


Construct an isosceles trapezoidFold B to BD on a line perpendicular to the segment XY

We get the trapezoid YXND


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ARROWHEAD

Fold along the diagonal BD

Fold B to D

Fold AB on to AM

We get the arrowhead AB''LD

Note: In the fourth step, any line through the apex A produces an arrowhead, possibly by folding any excess paper under the figure.


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SOLUTIONS : part 2 (proofs)

Note: We implicitly use the following result: An edge to edge fold defines the line bisecting the angle between the edges. This result is because the bisector is the locus of points equidistant from the rays defining the angle.

SQUARE

Initially assumed.

RECTANGLE

1) The beginning square.The square is a rectangle because it has 2 pairs of congruent sides and four right angles.

2)Properties we use:

- rectangles have 4 right angles

The fold is edge to edge.Half a straight angle is 90 degrees, so the angles of vertices M and N are 90 degrees each.The angle DAB = angle ABC = 90ºIt has 4 right angles, so it is a rectangle.


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KITE

1) A square has two pairs of congruent sides and it is convex. So it's a kite.


- on a kite, 2 pairs of sides are equal- angle/side/angle equality of triangles- definition of the bisector

If we open the figure we get:

DB bisects ADC so ADB = CDB DL and BM are the bisectors of ADB and BDC respectively, so ADL = LDB = BDM = MDC.By ASA (= ANGLE / SIDE / ANGLE), the triangles are equal.

We know that triangles ADL = LDA', and CDM = DMC' by construction.We must prove that these four triangles are identical.DC' = DA' because these are the sides of the original square.Angle LDA = angle MDC because these are the bisectors of the angle BDCangle MCD = angle FDA are right because they are the corners of the original square.DA = DC because they are the sides of the square, so by ASA (=ANGLE / SIDE / ANGLE) we obtain that the 2 triangles ALD = ACMMC'D =LA'D by folding one onto the other MC = LA because the triangles are similarBA = BC (sides of square)BM = BL


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RHOMBUS

1) The beginning square.The basic square is a rhombus because it has 4 congruent sides.

2) Proof 1Properties we use:

- a rhombus has 4 congruent sides - properties of complementary angles

If we open the figure we get:

We have four identical angles: they are double bisectorsFold the square back into a kite

Similarly we prove that:

The triangle BA'K= the triangle BC'FIt remains to show that the 4 sides are congruent.H is the intersection of DL and KB, the opposite angles theorem gives us that KHD = LHB

We know that angle LBH = angle HDK because we bisected the 90º angle of the original square twice.We know that DK = LBBy ASA we can say that the triangles HKD = LHB and therefore BH = DHSo we have 4 congruent sides and that is the definition of a rhombus.


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Proof 2 Properties we use:

- a rhombus has 4 congruent sides - properties of the diagonals of the square

We know that the 2 diagonals of a square are congruent and bisect at right angles (point O).So BO = ODangle HDO = angle HBO (bisectors)angle HOB = angle HOD = 90°by ASA ( = ANGLE/SIDE/ANGLE), triangle BHO = triangle DOHSo HB = DHSo we have a rhombus.


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PARALLELOGRAM

1) The starting square has two pairs of parallel sides.


- A parallelogram has 2 pairs of parallel sides- definition of bisectors- how to construct parallels lines

We know that BL is parallel to KD because they are the sides of the starting square.angle KBD = angle BDL because they are the bisectors of the diagonal of the square.

If we use our construction of parallel lines we can see that:

Two lines are parallel if they have the same interior angles

Angle BDL= angle KBD because it is the opposite angle of the angle corresponding to BDL.KB is parallel to DL


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ANY TRAPEZOID, ISOSCELES OR RECTANGULAR, FROM A SQUARE

1) The starting square is a trapezoid because it has a pair of parallel sides.

2) Trapezoid :Properties we use:

- trapezoids have one pair of parallel sides

By folding two opposite sides in any way, we obtain a trapezoid because it is sufficient that two sides are parallel and two sides of the square are parallel by definition.

3) Rectangular Trapezoid :Properties we use:

- trapezoids have one pair of parallel sides- a rectangular trapezoid has two right angles

If you fold one side in any way we obtain a trapezoid simply because it has a pair of parallel sides from the square. It is rectangular because it keeps two of the square's right angles.

4) Isosceles Trapezoid:Properties we use:- trapezoids have one pair of parallel sides- an isosceles trapezoid has two congruent sides

XX' is parallel to YY' because these are the sides of the original square.TX = TX 'and UY = UY' by construction.


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ANY TRAPEZOID, ISOSCELES OR RECTANGULAR, FROM A PARALLELOGRAM

1) The parallelogram has a pair of parallel sides.

2) Any trapezoid :Properties we use:- trapezoids have one pair of parallel sidesThe sides DL and KB are parallel, then the sides KX and DL are too.Therefore it is a trapezoid.

OR


- trapezoids have one pair of parallel sides

Start with a parallelogram

KX and DL are parallel as defined by the parallelogram.DB is parallel to DL by construction.So we have a trapezoid.

Note: this works for any fold through L such that the angle XLB is between 45 and 90 degrees.

3) Rectangular trapezoid :Properties we use:

- trapezoids have one pair of parallel sides- a rectangular trapezoid has two right angles

The sides YL and XB are parallel because they are the sides of the original parallelogram.It remains to show that XY is perpendicular to YL. This can be shown by construction: if a line can be folded onto itself, then it is perpendicular to the two parallel sides.It is therefore a trapezoid.


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4) Isosceles Trapezoid : Properties we use:

- trapezoids have one pair of parallel sides- an isosceles trapezoid has an axis of symmetry

The sides DL and KK' are parallel because they are the sides of the original parallelogram.

By folding D onto L, we construct the bisector m of the segment DL, which is also the bisector of the segment KK' (since DL and KK' are parallel). This implies that DKK'L has m as the axis of symmetry and is therefore an isosceles trapezoid


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ISOSCELES AND RECTANGULAR TRAPEZOID FROM A TRIANGLE

1) ISOSCELES TRAPEZOID :

Properties we use:- trapezoids have one pair of parallel sides- an isosceles trapezoid has two congruent sides- properties of the diagonals of a square

We must unfold the figure. We get:

We know that the diagonals of a square intersect at right angleswhere XY||BD because AM is perpendicular to BD by alternate interior angles. Itremains to show that XB = YDIf we show that the 4 triangles (AZY, AZX , XZM, YZM) are equal, then XB =YD andAX + XB =AY + YDX and Y are in the middle of AB and AD, respectively, since the diagonals AM and XY intersect at right angles in a rectangle.The sum of the angles of a triangle is 180ºZXM = 180 - (90 + 45) = 45Same for ZYMThen angle MYC = angle ZXM = 45Then ZM = ZX.So AXMY is a squareSame for XB and YDThis is therefore an isosceles trapezoid


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Properties we use:- trapezoids have one pair of parallel sides- a rectangular trapezoid has two right angles

We start with a parallelogram

If we fold perpendicular to KB at L, we get a right angle.


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ARROWHEAD

Proof 1:

Properties we use:- arrowheads have two pairs of equal sides- definition of bisectors- properties of the diagonals of a square

We know that AD = AB, because we start from a square.It remains to prove that DL = LB to show that we have an arrowhead.We must unfold the figure. We get:

The 4 angles are congruent because they are the bisectors of the diagonals of the original square.We know that the triangle ALM = triangle ABM So LM = BL'Thus we have an arrowhead


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Proof 2 :Properties we use:

- arrowheads have two pairs of equal sides- definition of bisectors- SAS equality of triangles

We know that AD = AB, as these are the sides of the original square.We must show that triangle ADL = triangle ABLWe know that the angles LAB and DAL are congruent because they were folded along bisectorsThus we know that DL = LB if we use SAS

Note: If the last fold is not along the bisector of the angle A, the proof that the figure is an arrowhead uses trigonometry and is therefore not suitable for the grade levels involved in this activity


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