Download 1 Folding Quadrilaterals From a Square Abstract The - SUPER-M

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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
Grade range
Student Prompt
Equipment
Duration
Plan
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.
6th to 8th grade
When you fold a square in half you get a rectangle. What
other quadrilaterals can you make from a square by
folding?(Oral)
Part 1:
Square sheets (15 x 15 cm)
Part 2:
The figures obtained during the first part
Paper
Pencil, eraser
3 x 45 minutes
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
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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)
References to
educational
content, curricula,
and teaching aids
Highlighted
mathematical
concepts
Possible
Development
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Result:You can get all the regular quadrilaterals (see
solutions: part 1)
Proof (see solutions: part 2)
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 squareedge)
Identify the lines of symmetry in a figure
Quadrilaterals
Properties of quadrilaterals
Angles
Triangles and properties of similar triangles
● 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 quadrilaterals
A 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
Square
Kite
Arrowhead
DEFINITION
•
four
congruent
sides
• four right
angles
• two pairs of
congruent
adjacent
sides
• convex
• two pairs of
congruent
adjacent sides
• not convex.
Axes
• 4
•
•
•
PROPERTIES
Diagonals
equal length
perpendicular
bisect each other
Angles
four 90º
angles
•
1
•
•
perpendicular
one is bisected by
the other
two
congruent
opposite
angles
•
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
two pairs of
congruent
opposite
angles
two pairs of
equal
opposite
Rhombus
•
four equal
sides
•
2
•
•
Perpendicular
intersects in the
middle
Parallelogram
•
two pairs of
parallel
•
0
•
One is bisected by
the other
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sides
angles
or
•
Rectangle
•
•
•
Trapezoid
•
Isosceles
Trapezoid
•
•
Rectangular
Trapeziod
•
•
two pairs of
equal
opposite
sides
convex
two pairs of
equal sides
four right
angles
a pair of
parallel
sides
a pair of
parallel
sides
a pair of
equal
nonparallel
sides
1 pair of
parallel
sides
2 right
angles
•
2
•
•
•
0
•
1
•
•
0
•
equal
intersects at their
midpoints
four 90º
angles
•
equal
Two pairs of
equal
consecutive
angles
two angles
1) Definition relevant to angles
Two 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 Triangles
Two 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 / side
Case 2:if we have 1), 2) and 4) then we have 3), 5) and 6)
That SAS: side / angle / side
Case 3: if we have 1), 4) and 5) then we have 2), 3 ) and 6)
In other words: ASA: angle / side / angle
Note: 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 BD
Fold AD to BD
This leaves the kite LBMD
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Rhombus
1) The starting square is a rhombus.
OR
2)
Fold along the diagonal BD
Open back up to get the starting
square
Fold CD to BD
Fold AD to BD
This leaves the kite LBMD
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Fold BL to BD
Fold BM to BD
This leaves the diamond HBTD
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PARALLELOGRAM
1) The starting square is a parallelogram.
OR
2)
Fold along the diagonal BD
Open to get the starting square.
Fold CD to BD
Fold AB to BD
This 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:
O
BC is folded to AD.
We get ATUD.
O
Open the figure and fold along XY and
X'Y'.
This gives X 'XY'Y.
O
AD is folded in any way over XY.
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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 parallelogram
Fold B onto DL
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This leaves the trapezoid KXLD
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3) Rectangular trapezoid
Take parallelogram KBLD.
Fold D onto the line segment DL.
We get XBLY.
4) Isosceles trapezoid:
Take the parallelogram KBLD.
Fold D to L
We get K ' (the image of K)
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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 BD
Open the paper back up to the square
Note: AC ∩ BD = M
Fold along the diagonal BD
Fold A along AM
We get the trapezoid YXBD
2) Rectangular trapezoid:
Construct an isosceles trapezoid
Fold B to BD on a line perpendicular to
the segment XY
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We get the trapezoid YXND
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ARROWHEAD
Fold B to D
Fold AB on to AM
Fold along the diagonal BD
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 = 90º
It 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.
2)
Properties we use:
- 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.
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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
BDC
angle 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 = ACM
MC'D =LA'D by folding one onto the
other
MC = LA because the triangles are
similar
BA = 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 1
Properties 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
bisectors
Fold the square back into a kite
Similarly we prove that:
The triangle BA'K= the triangle BC'F
It 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 = LB
By ASA we can say that the triangles HKD = LHB and
therefore BH = DH
So 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 = OD
angle HDO = angle HBO (bisectors)
angle HOB = angle HOD = 90°
by ASA ( = ANGLE/SIDE/ANGLE), triangle
BHO = triangle DOH
So HB = DH
So we have a rhombus.
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PARALLELOGRAM
1) The starting square has two pairs of parallel sides.
2)
Properties we use:
- 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 sides
The sides DL and KB are parallel, then the sides KX and DL are too.
Therefore it is a trapezoid.
OR
2)
Properties we use:
- 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 angles
where XY||BD because AM is perpendicular to BD by alternate interior angles. It
remains to show that XB = YD
If we show that the 4 triangles (AZY, AZX , XZM, YZM) are equal, then XB =YD and
AX + XB =AY + YD
X 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 180º
ZXM = 180 - (90 + 45) = 45
Same for ZYM
Then angle MYC = angle ZXM = 45
Then ZM = ZX.
So AXMY is a square
Same for XB and YD
This is therefore an isosceles trapezoid
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2) Rectangular trapezoid:
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 ABL
We know that the angles LAB and DAL are
congruent because they were folded along
bisectors
Thus 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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