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1 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 CEM Week of Math: Let the math unfold ! October 11 - 15 2010 2 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 CEM 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? Week of Math: Let the math unfold ! October 11 - 15 2010 3 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? CEM Week of Math: Let the math unfold ! October 11 - 15 2010 4 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 CEM Week of Math: Let the math unfold ! October 11 - 15 2010 5 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: CEM Week of Math: Let the math unfold ! October 11 - 15 2010 6 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 CEM Week of Math: Let the math unfold ! October 11 - 15 2010 7 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 CEM Week of Math: Let the math unfold ! October 11 - 15 2010 8 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 CEM Week of Math: Let the math unfold ! October 11 - 15 2010 9 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 CEM Week of Math: Let the math unfold ! October 11 - 15 2010 10 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 CEM Fold BL to BD Fold BM to BD This leaves the diamond HBTD Week of Math: Let the math unfold ! October 11 - 15 2010 11 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 CEM Week of Math: Let the math unfold ! October 11 - 15 2010 12 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 CEM Week of Math: Let the math unfold ! October 11 - 15 2010 13 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. CEM Week of Math: Let the math unfold ! October 11 - 15 2010 14 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 CEM This leaves the trapezoid KXLD Week of Math: Let the math unfold ! October 11 - 15 2010 15 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) CEM Fold along K'L We obtain KK'LD. Week of Math: Let the math unfold ! October 11 - 15 2010 16 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 CEM We get the trapezoid YXND Week of Math: Let the math unfold ! October 11 - 15 2010 17 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. CEM Week of Math: Let the math unfold ! October 11 - 15 2010 18 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. CEM Week of Math: Let the math unfold ! October 11 - 15 2010 19 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. CEM 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 Week of Math: Let the math unfold ! October 11 - 15 2010 20 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. CEM Week of Math: Let the math unfold ! October 11 - 15 2010 21 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. CEM Week of Math: Let the math unfold ! October 11 - 15 2010 22 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 CEM Week of Math: Let the math unfold ! October 11 - 15 2010 23 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. CEM Week of Math: Let the math unfold ! October 11 - 15 2010 24 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. CEM Week of Math: Let the math unfold ! October 11 - 15 2010 25 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 CEM Week of Math: Let the math unfold ! October 11 - 15 2010 26 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 CEM Week of Math: Let the math unfold ! October 11 - 15 2010 27 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. CEM Week of Math: Let the math unfold ! October 11 - 15 2010 28 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 CEM Week of Math: Let the math unfold ! October 11 - 15 2010 29 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 CEM Week of Math: Let the math unfold ! October 11 - 15 2010