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MEASUREMENT AND The Improving Mathematics Education in Schools (TIMES) Project GEOMETRY Module 21 RHOMBUSES, KITES AND TRAPEZIA A guide for teachers - Years 9–10 June 2011 YEARS 910 Rhombuses, kites and trapezia (Measurement and Geometry: Module 21) For teachers of Primary and Secondary Mathematics 510 Cover design, Layout design and Typesetting by Claire Ho The Improving Mathematics Education in Schools (TIMES) Project 2009‑2011 was funded by the Australian Government Department of Education, Employment and Workplace Relations. The views expressed here are those of the author and do not necessarily represent the views of the Australian Government Department of Education, Employment and Workplace Relations. © The University of Melbourne on behalf of the International Centre of Excellence for Education in Mathematics (ICE‑EM), the education division of the Australian Mathematical Sciences Institute (AMSI), 2010 (except where otherwise indicated). This work is licensed under the Creative Commons Attribution‑NonCommercial‑NoDerivs 3.0 Unported License. http://creativecommons.org/licenses/by‑nc‑nd/3.0/ MEASUREMENT AND The Improving Mathematics Education in Schools (TIMES) Project GEOMETRY Module 21 RHOMBUSES, KITES AND TRAPEZIA A guide for teachers - Years 9–10 June 2011 Peter Brown Michael Evans David Hunt Janine McIntosh Bill Pender Jacqui Ramagge YEARS 910 {4} A guide for teachers RHOMBUSES, KITES AND TRAPEZIA ASSUMED KNOWLEDGE The material in this module is a continuation of the module, Parallelograms and Rectangles, which is assumed knowledge for the present module. Thus the present module assumes: • Confidence in writing logical argument in geometry written as a sequence of steps, each justified by a reason. • Ruler‑and‑compasses constructions. • The four standard congruence tests and their application to: - proving properties of and tests for isosceles and equilateral triangles, - proving properties of and tests for parallelograms and rectangles. • Informal experience with rhombuses, kites, squares and trapezia. MOTIVATION Logical argument, precise definitions and clear proofs are essential if one is to understand mathematics. These analytic skills can be transferred to many areas in commerce, engineering, science and medicine but most of us first meet them in high school mathematics. Apart from some number theory results such as the existence of an infinite number of primes and the Fundamental Theorem of Arithmetic, most of the theorems students meet are in geometry starting with Pythagoras’ theorem. Many of the key methods of proof such as proof by contradiction and the difference between a theorem and its converse arise in elementary geometry. As in the module, Parallelograms and Rectangles, this module first stresses precise definitions of each special quadrilateral, then develops some of its properties, and then reverses the process, examining whether these properties can be used as tests for that particular special quadrilateral. We have seen that a test for a special quadrilateral is usually the converse of a property. For example, a typical property–test pair from the previous module is the pair of converse statements: The Improving Mathematics Education in Schools (TIMES) Project {5} • If a quadrilateral is a parallelogram, then its diagonals bisect each other. • If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. Congruence is again the basis of most arguments concerning rhombuses, squares, kites and trapezia, because the diagonals dissect each figure into triangles. A number of the theorems proved in this module rely on one or more of the previous theorems in the module. This means that the reader must understand a whole ‘sequence of theorems’ to achieve some results. This is typical of more advanced mathematics. In addition, two other matters are covered in these notes. • The reflection and rotation symmetries of triangles and special quadrilaterals are identified and related to congruence. • The tests for the kite also allow several important standard constructions to be explained very simply as constructions of a kite. {6} A guide for teachers CONTENT SYMMETRIES OF TRIANGLES, PARALLELOGRAMS AND RECTANGLES We begin by relating the reflection and rotation symmetries of isosceles triangles, parallelograms and rectangles to the results that we proved in the previous module, Rectangles and Parallelograms. The axis of symmetry of an isosceles triangle In the module, Congruence, congruence was used to prove A that the base angles of an isosceles triangle are equal. To a a prove that B = C in the diagram opposite, we constructed the angle‑bisector AM of the apex A, then used the SAS congruence test to prove that B M C ABM ACM This congruence result, however, establishes much more than the equality of the base angles. It also establishes that the angle bisector AM is the perpendicular bisector of the base BC. Moreover, this fact means that AM is an axis of symmetry of the isosceles triangle. These basic facts of about isosceles triangles will be used later in this module and in the module, Circle Geometry: Theorem In an isosceles triangle, the following four lines coincide: A a a • The angle bisector of the apex angle. • The line joining the apex and the midpoint of the base. • The line through the apex perpendicular to the base. B M C • The perpendicular bisector of the base. This line is an axis of symmetry of the isosceles triangle. It has, as a consequence, the interesting property that the centroid, the incentre, the circumcentre and the orthocentre of ABC all tie on the line AM. In general, they are four different points. See the module, Construction for details of this. Extension – Some further tests for a triangle to be isosceles The theorem above suggests three possible tests for a triangle to be isosceles. The first two are easy to prove, but the third is rather difficult because simple congruence cannot be used in this ‘non‑included angle’ situation. The Improving Mathematics Education in Schools (TIMES) Project {7} EXERCISE 1 Use congruence to prove that ABC is isosceles with AB = AC if: a the perpendicular bisector of BC passes through A, or b the line through A perpendicular to BC bisects A, or c the angle bisector of A passes through the midpoint M of BC. [Hint: For part c, let BAM = CAM = , and let C = . Suppose by way of contradiction that AC < AB. Choose P on the interval AB so that AP = AC, and join PM. The symmetries of an equilateral triangle An equilateral triangle is an isosceles triangle in three different ways, so the three vertex angle bisectors form three axes of symmetry meeting each other at 60°. In an equilateral triangle, each vertex angle bisector is the perpendicular bisector of the opposite side – we proved in the previous module that in any triangle, these three perpendicular bisectors are concurrent. They meet at a point which is the centre of a circle through all three vertices. The point is called the circumcentre and the circle is called the circumcircle of the triangle. An equilateral triangle is also congruent to itself in two other orientations: A ABC BCA CAB (SSS), corresponding to the fact that it has rotation symmetry of order 3. O The centre of this rotation symmetry is the circumcentre O described above, because the vertices are equidistant from it. B C Other triangles do not have reflection or rotation symmetry In a non‑trivial rotation symmetry, one side of a triangle is mapped to a second side, and the second side mapped to the third side, so the triangle must be equilateral. In a reflection symmetry, two sides are swapped, so the triangle must be isosceles. Thus a triangle that is not isosceles has neither reflection nor rotation symmetry. Such a triangle is called scalene. {8} A guide for teachers Rotation symmetry of a parallelogram Since the diagonals of a parallelogram bisect each other, a A D parallelogram has rotation symmetry of order 2 about the intersection of its diagonals. Joining the diagonal M AC of a parallelogram ABCD produces two congruent triangles, B C ABC CDA (AAS); Reflection symmetry of a rectangle A rectangle is a parallelogram, so it has rotationsymmetry of order 2 about the intersection of its diagonals. This is even clearer in a rectangle than in a general parallelogram because the diagonals have equal length, so their intersection is the circumcentre of the circumcircle passing through all four vertices. The line through the midpoints of two opposite sides of a rectangle dissects the rectangle into two rectangles that are congruent to each other, and are in fact reflections of each other in the constructed line. There are two such lines in a rectangle, so a rectangle has two axes of symmetry meeting right angles. It may seem obvious to the eye that the intersection of these two axes of symmetry is the circumcentre of the rectangle, which is intersection of the two diagonals. This is illustrated in the diagram to the right, but it needs to be proven. EXERCISE 2 A D Use the diagram to the right to prove that the line through the midpoints of opposite sides of a P Q rectangle bisects each diagonal. M B C The Improving Mathematics Education in Schools (TIMES) Project {9} Axes of symmetry of triangles, parallelograms and rectangles • An isosceles triangle has an axis of symmetry – this line is the bisector of the apex angle, it is the altitude from the vertex to the base, and it is the line joining the apex to the midpoint of the base. • An equilateral triangle has three axes of symmetry, which are concurrent in the circumcentre of the circumcircle through its three vertices. It also has rotation symmetry of order three about its circumcentre. • A triangle that is not isosceles has no axes of symmetry and no rotation symmetry. • A parallelogram has rotation symmetry of order two about the intersection of its diagonals.
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