Answers and Solutions

1. 1. Square
      
      \(AC=BD\)

      Trapezoid
      

      Rectangle

      

      \(IK=JL\)

      Rhombus
      

      Parallelogram
      

   2. 1. Squares, rectangles, rhombuses, and parallelograms have diagonals that always bisect each other.
      2. Squares and rhombuses have diagonals that are always perpendicular to each other.
      3. Squares and rectangles have diagonals that are always equal in length.
2. The diagonals of a square and a rhombus are similar in that they always bisect each other and they are perpendicular to each other.  However, the diagonals of a square are always equal in length, while the diagonals of a rhombus are not usually equal in length. 
3. A parallelogram has diagonals that always bisect each other while a trapezoid does not have to have diagonals that bisect each other. We could determine if the two diagonals bisect each other by measuring the relevant line segments.  If they do bisect each other, we would most accurately name the shape a parallelogram.  If the diagonals do not bisect each other, we would most accurately name the shape a trapezoid.
4. A rectangle has diagonals that bisect each other and are equal in length.  If Yukiko notices that the diagonals are also perpendicular to each other, then the diagonals have all three properties of the diagonals of a square. Therefore, Yukiko could most accurately classify the shape as a square.
5. The dimensions of the wall are given as \(3.6\) m by \(2.4\) m.  Take some time to convince yourself that a wall created with wood cut to these dimensions must be a parallelogram and therefore have two pairs of parallel sides.  Lilliana and her Mom want to make sure the wall is rectangular.  A rectangle has diagonals that are equal in length, while a parallelogram that is not a rectangle, does not have diagonals that are equal in length. Lilliana and her Mom can confirm that the wall is rectangular by using the measuring tape to make sure that the diagonals of the wall are equal in length.
6. The new shape, \(CDEF\), has four sides that are all \(4\) cm in length.  This means the new shape must either be a rhombus or a square.  If we draw in the diagonals and compare their lengths, we would notice that the diagonals of \(CDEF\) are both the same length.  This means \(CDEF\) is a square.
   
7. 1. The mystery quadrilateral could be a trapezoid, a parallelogram, or a rhombus since the diagonals on these quadrilaterals do not have to be equal in length.
   2. The mystery quadrilateral could now be a trapezoid or a parallelogram since the diagonals on these quadrilaterals do not have to be perpendicular to each other.
   3. A parallelogram would have diagonals that bisect each other.  Therefore, Shara's mystery quadrilateral must be a trapezoid since the diagonals of a trapezoid do not have to bisect each other.
8. 1. If you have learned about the properties of circles, then you will have seen that there is always a circle that can be drawn through a set of three points, as long as the points do not lie on a straight line.  (We use perpendicular bisectors to show this.)  This means that there is always a circle that passes through the three vertices of a triangle, and so every triangle is cyclic.  
      In fact, there is actually exactly one circle that passes through a set of three points.  So to create a quadrilateral that is not cyclic, first draw a triangle and the circle that passes through the three vertices of the triangle, and then select a fourth point (to make a quadrilateral) so that this fourth point does not lie on the circle.  
   2. Consider the following rectangle and parallelogram, each having side lengths \(2\), \(1\), \(2\), \(1\):
      Rectangle
      

      Parallelogram
      
      These two quadrilaterals have the same respective side lengths, but the rectangle is cyclic and the parallelogram is not. Can you explain why? (You may need think about circles to help answer this.) This means that we need more than just the side lengths to determine whether a quadrilateral is cyclic.
   3. 1. Since this is a rectangle, we know that both diagonals are the same length and so the diagonals are both \(5\) units long. Since \(4 \times 4 + 3 \times 3 = 25\) and \(5 \times 5 = 25\), the equation in Ptolemy's Theorem is satisfied and therefore, the rectangle is cyclic.
         
      2. Since \(67 \times 26 + 16 \times 50 = 2542\) and \(64 \times 39 = 2496\), which are not equal, the quadrilateral is not cyclic.  Since the two numbers are reasonably close, we might expect this quadrilateral to be "almost" cyclic.  See if you agree from the picture.
         
      3. Since \(2.5 \times 1.5 + 0.7 \times 1.5 = 3.75 + 1.05 = 4.8\) and \(2.4 \times 2 = 4.8\), the quadrilateral is cyclic.
         
   4. Recall that the two diagonals in a square are equal in length. Let \(a\) be the length of the diagonals of the square. We want to show that \(a \) must be \(\sqrt{2}\).
      
      Since the unit square is a cyclic quadrilateral, its side lengths and diagonal lengths must satisfy the relationship in Ptolemy's Theorem. Since the side lengths are all equal to \(1\), this relationship becomes:\[1 \times 1 + 1\times 1 = a \times a\]

      This equation simplifies to

      \[2 = a\times a\]Since \(a\) is a length, it must be a positive number. The only positive number that when multiplied by itself gives an answer of \(2\) is the number \(\sqrt{2}\). When we substitute \(a=\sqrt{2}\) into the right-hand side of the equation we get \(\sqrt{2} \times \sqrt{2} = 2\).  Thus, using Ptolemy's Theorem, we have shown that the diagonal of a unit square must be \(\sqrt{2}\).