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Applying Angle Theorems
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Homework
  • Improve Individual Solutions to the Assessment Task (10 minutes)
      • Return students' work on Four Pentagons along with a fresh copy of the assessment task
      • If you did not write questions on students' solutions, display them on the board.
      • Ask students to read through their responses, bearing in mind what they have learned during this lesson.
        • Look at your original responses, and think about what you have learned in this lesson.
        • Using what you have learned, try to improve your work.
  • Solutions: Four Pentagons
    • We give examples of some approaches taken by students in trials. There are other methods that lead to correctly reasoned solutions.

      1. The measure of angle AEJ is 144°.

        Explanation 1:

        The sum of the measures of the interior angles of an n-gon is 180°(n − 2).

        For a pentagon this is 180° × 3 = 540°.

        The pentagons are regular, so all their interior angles are congruent.

        Each interior angle of a regular pentagon is 540° ÷ 5 = 108°.

        The sum of the angles forming a straight line is 180°.

        Each exterior angle of a regular pentagon is 180° − 108° = 72°.

        Angle AEJ is twice the exterior angle of the pentagon = 2 × 72° = 144°

        Explanation 2:

        Angle AEF is an exterior angle of a regular pentagon, as is angle FEJ.

        The sum of the exterior angles of a polygon is 360°.

        There are five congruent exterior angles in a regular pentagon, each of measure 360 ÷ 5 = 72°.

        So AEJ = AEF + FEJ = 72 + 72 = 144°.

      2. The measure of angle EJF is 36°.

        Explanation 1:

        Since consecutive angles of a parallelogram are supplementary, angle EJF = 180° − 144 = 36°.

        Explanation 2:

        The sum of the interior angles of the quadrilateral AFJE is 360°.

        Since the four pentagons are regular and congruent, sides AF, AE, EJ, JF are equal in length.

        So AEJF is a rhombus.

        Opposite angles in a rhombus are congruent.

        From part 1, angle AEJ = 144° = AFJ.

        The sum of the angles in a quadrilateral is 360°.

        360 − 2 × 144 = 72°.

        FJE = FAE = ½ × 72 = 36°.

      3. The measure of angle KJM is 108°.

        Explanation:

        The sum of the measures of the four angles around the point J is 360°.

        The measure of each of the interior angles of a regular pentagon is 108°, and angle EJF is 36° from Question 2.

        Angle KJM = 360° − (36° + 2 × 108°) = 108°.

        From this diagram, we can see that regular pentagons and rhombuses together form a semi-regular tessellation that can be used, for example, as a floor or wall tiling.

  • Solutions: The Pentagon Problem
    • Each method gives a way of calculating the measure of angle x, 75°. Each method uses different definitions and angle properties in the explanation.

      1. Annabel's method

        In trials, some students did not understand the need to justify the assumption that the line "down the middle of the pentagon" bisects the 130° angle at the base of the pentagon.

        The construction line divides AC into segments of equal length.

        So AB = BC.

        AF = CD is given.

        Angle BAF is congruent to angle BCD.

        So by SAS, triangles ABF and BCD are congruent.

        Triangle BFE is congruent to triangle BDE by SSS.

        So:

        angle FEB = angle BED = 130 = 65°
        2

        To show that the two quadrilaterals ABEF and BCDE are congruent:

        The sides are all congruent as BA= BC, AF = CD, FE = DE, and BE is common to both quadrilaterals.

        The angle between sides AB and AF is congruent to the angle between sides BC and CD.

        The angle between sides AF and FE is congruent to the angle between sides and DE.

        So the quadrilaterals are congruent.

        The figure is therefore symmetrical. So angle ABE = angle CBE = 90°.

        Since the sum of the angles in a quadrilateral is 360°, x = 360 − (90 + 130 + 65) = 75°.

      2. Carlos's method

        In trials, some students made the false assumption that all the exterior angles are congruent.

        The sum of an interior and an exterior angle is 180°.

        Three of the angles of the pentagon are known; all three are 130°.

        The exterior angle for each of these interior angles is 180 − 130 = 50°.

        The sum of the exterior angles of a polygon is 360°.

        360 − 3 × 50 = 360 − 150 = 210°.

        This is the sum of the two missing exterior angles.

        The two missing interior angles are congruent.

        x = 180 − ½ × 210 = 180 − 105 = 75°.

      3. Brian's method

        The pentagon is divided into a quadrilateral and a triangle.

        In trials, some students did not understand the need to justify the claim that the quadrilateral is a trapezoid, and others did not understand the need to show that both triangle and trapezoid are isosceles.

        The triangle is isosceles because it has two congruent sides. So the angles marked a are congruent. So the angles marked b = x - a are also congruent to each other.

        The quadrilateral is an isosceles trapezoid because the two slant sides are congruent and meet the horizontal side at congruent angles.

        It follows that the base is parallel to the top, and angles marked b are also congruent.

        The angles in a triangle sum to 180°.

        2a = 180 − 130 = 50.

        a = 25°.

        The angles in a quadrilateral sum to 360°.

        2b = 360 − 2 × 130 = 100

        b = 50°.

        Alternatively, since the top and base of the trapezoid are parallel, the angles b and 130° are supplementary, and b = 180 − 130 = 50°.

      4. Diane's method

        Some students in trials, perhaps relying on the appearance of the diagram, assumed the three triangles were all isosceles.

        Diane shows the pentagon divided into three triangles. The sum of the angles in any triangle is 180°. The sum of the angles in the pentagon is thus 180 × 3 = 540°.

        The three known angles are all 130°.

        The two unknown angles are congruent. 2x = 540 − 3 × 130 = 150°

        x = 75°.

        The outer triangles are not isosceles.

  • Analysis of Sample Responses to Discuss
    • Erasmus used Annabel's method

      Erasmus does not justify the claim that the perpendicular bisector of the horizontal side divides the 130° into two equal parts. He could do this by showing that the pentagon is symmetrical so that the bisector of the vertical side passes through the opposite vertex.

      He also needs to explain that the perpendicular bisector then divides the pentagon into two congruent quadrilaterals. Then he can apply the property that the sum of the angles in a quadrilateral sum to 360°.

      His calculation method is correct but he did not finish his working out.

      Erasmus's use of Annabel's method gives the correct measure of x = 75°.


      Tomas used Carlos's method

      Tomas makes a false assumption that all the exterior angles are congruent.

      He did not notice that the pentagon is not regular. The exterior angles are all congruent only when the polygon is regular.

      Tomas should calculate the size of the exterior angles for each of the known 130° interior angles first.

      The angles on a line sum to 180°, so there are three exterior angles of 50°.

      360 − 3 × 50 = 360 − 150 = 210°.

      So, the two missing exterior angles are congruent and sum to 210°. Each is 210 ÷ 2 = 105°.

      Then, since the angles on a line sum to 180°, x + 105 = 180. So x = 75°


      Katerina used Brian's method

      Katerina is correct that a trapezoid and triangle are formed by the horizontal line, but she does not fully explain her reasoning. It is not clear that the quadrilateral is a trapezoid, or that the trapezoid is isosceles.

      She needs to show the base of the quadrilateral is parallel to the top to show that the quadrilateral is a trapezoid.

      The horizontal side has at each end the same angle. The slant sides are the same length. So the line joining the ends of those slant sides is parallel to the top (trapezoid).

      The trapezoid is isosceles because the slant sides are equal in length and joined to the top by congruent angles (symmetry). So both base angles can be labeled b.

      She is correct that the triangle is isosceles because it has two congruent sides. So the two unknown angles in the triangle are congruent and can both be labeled a.

      Katerina made a numerical error in stating a = 50°.

      The angles in a triangle sum to 180°.

      2a = 180 − 130 = 50°

      She had forgotten to divide by two.

      Katerina's next piece of reasoning is faulty.

      It is not true that the consecutive angles in every quadrilateral sum to 180°. For example, it is not true that any two consecutive angles in a trapezoid always sum to180°.

      In a trapezoid, the angles formed by a transversal crossing the parallel sides forms a pair of supplementary angles.

      Supplementary angles sum to 180°.

      So b = 180 − 130 = 50°

      Katerina also needs to finish her solution by finding x = a + b = 25 + 50 = 75°


      Megan uses Diane's method

      Diana divided the pentagon into three triangles to calculate the measure of x. There is not enough detail to specify a method.

      Megan uses faulty reasoning with Diane's trisection.

      She makes a false assumption that the triangles are all isosceles.

      Megan would need to give reasons to support the assertion that the triangles are isosceles, and there are none beyond surface appearance since they are not!

      Diane's trisection method can lead to a correct solution. The sum of the angles in a triangle is 180°.

      So the total angle sum of the pentagon is 3 × 180 = 540°.

      This could be provide using the formula for the sum of the angles in a polygon with n sides, 180(n − 2).

      The interior angles sum is 540 and there are three known angles of 130°.

      So 2x = 540 − 3 × 130, and x = 75°.

      In Q4, it is not expected that students will show that Megan's assumption is false. However, we supply a solution in case you want to work on this with students.

      Assuming that the triangles are isosceles leads to a contradiction, showing that the assumption is false. (Proof by contradiction.)

      Megan assumes the three triangles formed are all isosceles triangles with two congruent base angles of 65°. Suppose she is correct.

      Each has base line of equal length, the base angles of equal measure, two sides of equal length, the apex angles must also be congruent to each other, and the triangles are thus congruent.

      Each apex angle would be 130 ÷ 3 = 43⅓°.

      Since the triangles are isosceles, and the angles in a triangle sum to 180°, the two base angles are (180 − 43⅓ ) ÷ 2 = 68⅓°.

      x cannot be both 68⅓° and 65°. The assumption leads to a contradiction, and must be false.

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