Olive & Oppong: Transforming Mathematics with GSP 4, page 18
Chapter 2: Exploring Quadrilaterals
In this chapter we extend our investigation of dynamically constructed figures in GSP to
closed figures constructed from 4 line segments. These are known as quadrilaterals.
Activity 2.1: Starting with four free points
Open a new sketch and place 4 free points on the screen. Select all 4 points and choose
Segment from the Construct menu. What kind of figure did you create? Depending on how you
arranged your 4 points you could have very different looking figures from those of your
classmates. Look around at other screens and compare the different figures. What is common to
all of the figures? What is different? Move one or more of your points to create different kinds
of quadrilaterals. Begin to think about how you might classify these different quadrilaterals. For
instance, how might you classify the 3 figures shown in Figure 2.1?
Figure 2.1: Three Different Quadrilaterals
Activity 2.2: Starting with three free points
Open a new sketch and place 3 free points on the screen. Find a way to construct a fourth
point that is dependent in some way on these 3 free points. Construct a quadrilateral using your
3 free points and your new constructed point. In what ways is you quadrilateral constrained?
Which special quadrilaterals can you construct? Think about how you might classify all the
quadrilaterals you can form with this construction. The 2 quadrilaterals in Figure 2.2 were
constructed using different constructions but both started with 3 free points (M, N and O in
Figure 2.2 a and P, Q and R in Figure 2.2 b).
Olive & Oppong: Transforming Mathematics with GSP 4, page 19
M’
P
S
N
M
O
Figure 2.2 a
Q
R
Figure 2.2 b
Activity 2.3: Starting with two free points
This time start with only 2 free points in your sketch and construct a quadrilateral.
Remember to construct the other 2 points so that they are dependent upon the first 2 free points.
Which special quadrilaterals can you construct? How does moving either of your 2 free points
change your quadrilateral? Can you make more than one kind of special quadrilateral from your
construction? The quadrilaterals in Figure 2.3 were constructed from the 2 labeled points using
different construction methods.
W
T
U
X
Figure 2.3: Quadrilaterals Starting from 2 Free Points
Activity 2.4: Starting with the diagonals of a quadrilateral
Start with 2 arbitrary segments that intersect, construct the quadrilateral for which these
segments are the diagonals – investigate the relations between the diagonals that create each of
the special quadrilaterals you discovered in your explorations above.
In a new sketch, construct your diagonals so that they remain perpendicular to one
another but have no other constraint (i.e. they do not have to be congruent; they do not have to
Olive & Oppong: Transforming Mathematics with GSP 4, page 20
bisect each other). Investigate the properties of all of the quadrilaterals you can form using your
perpendicular diagonals. Can you find any “special” properties?
This time start with diagonals that remain congruent but have no other constraint.
Investigate any special properties of these quadrilaterals.
Start with diagonals that bisect each other but have no other constraint. Investigate any
special properties of these quadrilaterals.
Investigating midpoint quadrilaterals
Assignment 2.1: For each of the quadrilaterals formed by the special diagonal relations
(perpendicular, congruent, or bisectors) construct the midpoint quadrilateral – that is the
quadrilateral formed by connecting the midpoints of the 4 sides of the original quadrilateral (see
Figure 2.4). In what ways are these midpoint quadrilaterals determined by the relation between
the diagonals of the original quadrilateral?
Figure 2.4: A Quadrilateral Formed from Congruent Diagonals with Its Midpoint Quadrilateral
Construct the midpoint quadrilateral for a general quadrilateral (starting with 4 free
points). Is there anything special about this midpoint quadrilateral?
Provide a rationale (an explanation or informal proof) for the special qualities of the
midpoint quadrilaterals in each of the above cases. [Hint: Look for mid-segments of triangles.]
Olive & Oppong: Transforming Mathematics with GSP 4, page 21
Classifying quadrilaterals
By now you should have several ways in which you could classify all of your
quadrilaterals. Make a classification based on relations among the sides of a quadrilateral. Make
another classification based on relations between the diagonals of a quadrilateral. Could you
classify quadrilaterals based on properties of their mid-point quadrilaterals? Where do the
“special” quadrilaterals fit in each of these different classifications? What other properties need
to be considered to determine if a quadrilateral is a rectangle?
Quadrilaterals on a circle: another class of quadrilaterals
Start by drawing a circle using the Circle tool. Place 4 arbitrary points on the circle and
use these 4 points to construct a quadrilateral. This figure is called a cyclic quadrilateral.
Investigate the special properties of cyclic quadrilaterals. Make sure you move your points
around the circle to make as many different cyclic quadrilaterals as possible. Find out which (if
any) of your “special” quadrilaterals are cyclic. Test your conjectures by constructing the circle
to pass through all 4 vertices of the quadrilateral. This circle is called the circumscribed circle of
the quadrilateral.
Yet another class of quadrilaterals can be formed from those quadrilaterals for which a
circle can be inscribed inside the quadrilateral so that each side of the quadrilateral is tangent to
the circle. The easiest way to construct such quadrilaterals is to start with the inscribe circle,
place four points on the circle and construct the four tangents at these points. Figure 2.5 shows
both a cyclic quadrilateral (CDEF) and a quadrilateral (MNOP) with an inscribed circle. Find at
least 2 properties unique to these special quadrilaterals. If a quadrilateral is cyclic can a circle
also be inscribed inside it? If a quadrilateral has an inscribed circle is it also cyclic?
Olive & Oppong: Transforming Mathematics with GSP 4, page 22
M
D
N
E
C
F
P
O
Figure 2.5: A Cyclic Quadrilateral and a Quadrilateral with an Inscribed Circle.
Investigating the Symmetry of Special Quadrilaterals
Look at the 2 quadrilaterals in Figure 2.2. Could you create one of them by constructing
half of it and then reflecting that half about a mirror line? Which 3 points could you start with
and which segment would become your mirror line? Any figure that could be constructed in this
way is said to have “line symmetry.” This simply means that you can draw a mirror line
somewhere across the figure in such a way that reflecting the figure about that line would not
change the figure in any way (its reflected image would lie directly over the pre-image). Some
figures may have more than one line of symmetry. This means that you could reflect the figure
about two or more different lines and still not change the figure in any way. Do any of the
quadrilaterals in any of the figures in this chapter have more than one line of symmetry? If so,
construct all the lines of symmetry for those figures. Use the sketchpad Mark Mirror and Reflect
actions to check your conjectures concerning lines of symmetry for the various quadrilaterals.
[GSP Notes: You can double click on any straight object to mark it as a mirror for reflection.
Once a mirror has been marked, select the objects you want to reflect then go to the Transform
menu and choose Reflect.]
Another form of symmetry is called rotational symmetry. A figure has rotational
symmetry if you can rotate it about some point by less than 360 degrees (a full revolution) and its
Olive & Oppong: Transforming Mathematics with GSP 4, page 23
rotational image lies directly over the pre-image. In other words, the rotation does not change the
appearance of the figure in any way. Do any of the quadrilaterals you have constructed so far
have rotational symmetry? If so, determine the center of rotation and the smallest angle of
rotation that will produce an identical image of the quadrilateral, lying directly over its preimage. You can test your conjecture with GSP by constructing the center of rotation, marking it
as a center of rotation by double clicking on the point, and then selecting all objects in your
quadrilateral and choosing Rotate from the Transform menu. You will need to enter the angle of
rotation (positive degrees for counter-clockwise rotation).
Can a quadrilateral have line symmetry without rotational symmetry? Can a quadrilateral
have rotational symmetry without line symmetry? If so, provide examples. Make a classification
of quadrilaterals based on symmetry.
Look for examples of both types of symmetry in nature. Does a human face have line
symmetry? In Figure 2.6 the face on the right was constructed by reflecting the left half of the
natural face on the left. Is the real face symmetric?
Figure 2.6: Face of a Maori Warrior
Olive & Oppong: Transforming Mathematics with GSP 4, page 24
Constructing Similar Quadrilaterals
We found that for constructing similar triangles, all we had to do was to duplicate two of
the angles of the triangle on a new segment. The intersection of the rays of the two angles would
intersect in the third vertex of the similar triangle. We could then simply change the length of our
new segment to form many triangles similar to the original triangle. Can we do the same for
quadrilaterals? Do the angles of a quadrilateral fully determine its shape? To test this conjecture,
create a quadrilateral using 4 free points. Draw a new segment. Starting with this segment
duplicate the angles at each end of one side of your quadrilateral. This side will correspond to the
new segment in your similar quadrilateral. For example, in Figure 2.7 segment EF corresponds to
side AD of quadrilateral ABCD, angle FEF’ is congruent to angle DAB, and angle EFE’ is
congruent to angle ADC.
B
C
A
D
F’
E’
E
F
Figure 2.7: Duplicating the Vertex Angles of One Side of a Quadrilateral on a New Segment.
[GSP Note: I used Rotation by Marked Angle to duplicate angles DAB and ADC on segment EF.
First select points D, A and B (in that order) and choose Mark Angle from the Transform menu;
then double click on point E to mark it as Center of Rotation. Select point F and choose Rotate
from the Transform menu. F will be rotated about E to create point F’. Select the Ray tool,
Olive & Oppong: Transforming Mathematics with GSP 4, page 25
select points E and F’ in that order and construct the ray EF’. Use similar steps to rotate point E
about point F by angle ADC and construct ray FE’.]
Unfortunately, the intersection of the two rays in Figure 2.7 (if it exists) does not give
you another vertex of the similar quadrilateral. Regarding the quadrilateral ABCD as composed
of 2 triangles could help you complete the construction of the similar quadrilateral and also help
you “prove” that the shape of a quadrilateral is completely determined by its 4 angles (see Figure
2.8 for the completed construction.
B
C
A
D
F’
E’
E
F
Figure 2.8: Construction of a Similar Quadrilateral.
You can also use the Dilation tool to shrink or enlarge any figure you have constructed in
GSP. You need to select a center of dilation by double clicking on any point in your sketch, then
select the Dilation tool from the toolbar by holding the mouse button down over the Arrow tool.
The Dilation tool is the furthest to the right of the arrow select tool options. Use this special
selection tool to select the sides of your quadrilateral and then move these sides towards or away
from the center of dilation. Your quadrilateral should change its size but not its shape.
Olive & Oppong: Transforming Mathematics with GSP 4, page 26
Assignment 2.2: Investigate properties of similar quadrilaterals. Measure the ratios of the
lengths of corresponding sides. Measure the areas of the similar quadrilaterals. How does the
ratio of the areas relate to the ratio of corresponding side lengths? Will these properties
generalize to similar polygons with more than 4 sides?
Constructing Congruent Quadrilaterals
With triangles, 3 sides completely determine a triangle (SSS congruency). Will 4 sides
completely determine a quadrilateral? To test this conjecture, try using circle by center and
radius to duplicate each side of a free quadrilateral starting with a point on a new line. For
example, Figure 2.9 appears to show a possible construction for a duplicate of quadrilateral
ABCD (FIJH).
B
C
A
J
D
I
F
H
Figure 2.9: Attempt to Duplicate a Quadrilateral by Duplicating Each Side Length.
Figure 2.10, however, clearly indicates that the 2 quadrilaterals are not necessarily congruent,
even though they have the same side lengths!
Olive & Oppong: Transforming Mathematics with GSP 4, page 27
B
C
A
J
D
I
F
H
Figure 2.10: Non-congruent Quadrilaterals with the Same Side Lengths.
Assignment 2.3: Determine the minimum set of properties to construct a quadrilateral
congruent to a given quadrilateral. Use your minimum set to duplicate a quadrilateral in GSP.
Construction Problems for Special Quadrilaterals
Activity 2.5: Attempt at least one construction from each of the following sets, first with
straight-edge and compass, then with GSP. Each given length (side, diagonal or altitude) should
be a separate, free segment in GSP. Each given angle should be a set of 3 free points in GSP.
Check that your construction works for all possible (or reasonable) cases by varying the free
segments and free points that define your given elements. Create GSP custom tools for each of
your constructions. The “givens” in each custom tool should correspond to the givens in each
construction plus one or two free points on which to build your quadrilateral. Use comments in
the script-view of your tool to let the user know which given GSP objects define which givens in
the construction. Exchange your custom tools with other members of your class.
A. Construct a square, given the length of the diagonal
B. Construct a rhombus, given
1) one side and one angle
2) one angle and a diagonal
Olive & Oppong: Transforming Mathematics with GSP 4, page 28
3) the altitude and one diagonal
C. Construct a parallelogram, given
1) one side, one angle, and one diagonal
2) two adjacent sides and an altitude
3) one angle, one side, and the altitude on that side
D. Construct an isosceles trapezoid, given
1) the bases and one angle
2) the median, altitude, and one of the bases
3) one base, the diagonal, and the angle included by them
Reflection. Look back at your constructions and ponder the following questions:
1) Which was easier – straight-edge and compass, or GSP for solving these construction
problems?
2) What properties of the special quadrilaterals did you make use of in solving these
problems?
3) Did manipulating the constructions in GSP help you to find errors in your
construction?
4) Did some constructions only work for special relations among your givens (e.g. as
long as the diagonal was longer than the side for the parallelogram)?
Investigating Golden Quadrilaterals
The investigations that follow will focus on four sided figures in which at least one pair of
opposite sides are parallel. Among these there are four sided figures that have both pairs of their
opposite sides in the same ratio. Special cases are the square, rhombus, rectangle, and
parallelogram in which the opposite sides are in the ratio of one to one. Some of these figures
have their adjacent sides in a special ratio. An interesting case of this is the Golden Rectangle
with adjacent sides in the Golden Ratio.
The Golden Rectangle
The Golden Rectangle has the ratio of its adjacent sides as the Golden Ratio. The Golden
Rectangle can be constructed using straight edge and compass. The construction is based on the
Olive & Oppong: Transforming Mathematics with GSP 4, page 29
fact that, the Golden Ratio can be expressed as
1+ 5
2 , which is approximately 1.62. If a square
of side 2 units is constructed, then the length from a vertex to the midpoint of the opposite side
will be 5 . Locate the midpoint of a side of a square and extend this side in either direction.
Using the midpoint as center, draw an arc of radius equal to the length from the midpoint to an
opposite vertex, to cut the extended side of the square. Constructing a perpendicular line at the
point where the arc cuts the extended side of the square to meet the opposite extended side of the
square can complete the construction. The resultant rectangle has one side of length 2, and the
other 1+ 5 . The Golden Rectangle has a long history and many excellent references are
available (see Huntley, 1970; Garland, 1987; and Vajda, 1989). It is a centerpiece in the Walt
Disney classic film Donald in Mathmagic Land (1959).
Are there other quadrilaterals with adjacent sides or even opposite sides in the Golden
Ratio? The Golden Rectangle when constructed with straight edge and compass is rigid. The
dynamic features in Geometer's Sketchpad, however, permit us to transform the rectangle into a
parallelogram, by moving three sides without moving the base (by changing the angle between
base and side), and yet maintain the lengths and hence the ratio of its sides. The size of the whole
figure can also be varied to show that the ratio does not depend on the size of the parallelogram.
We will be able to generalize the notion of Golden Rectangle to the parallelogram.
The Golden Parallelogram
A dynamic Golden Parallelogram can be constructed by first constructing a Golden
Rectangle CMSH (see figure below) and then drawing a circle with center C and passing through
H. Next, place an object (point P) on the circle. Through P, construct PR parallel and equal to
CM. As P is moved along the circle, the parallelogram CMRP will always have the ratio of its
adjacent sides equal the Golden Ratio. We can use traditional methods or the dynamic features
in the Geometer's Sketchpad to do this construction. A parallelogram with this property could be
called a Golden Parallelogram. Due to the dynamic nature of the Geometer's Sketchpad, the
distances and ratios shown besides the figure can be set to change accordingly on the computer
monitor as the point P moves along the circle in the plane. None of the lengths or ratios of the
Olive & Oppong: Transforming Mathematics with GSP 4, page 30
sides, however, will change as long as P moves along the circle. But the lengths of the diagonals
will change as P moves along the circle. In the figure CH=CP=MS=MR=2.79 cm. Also given
are the corresponding measures for other sides and lengths, and the ratios of adjacent sides, and
lengths of diagonals. You can investigate the range of these changes and associated ratios.
Perhaps another Golden Ratio will turn up!
H
S
R
P
C
M
Dista nce(H t oS) = 3.23 cm
Dista nce(H t oC) = 2.00 c m
(H t o S)/(H t o C) = 1 .62
Dista nce(R t oP ) = 3 .23 c m
Dista nce(P t oC ) = 2 .00 c m
(R to P)/( P t oC) = 1.62
Dista nce(R t oC ) = 4 .71 c m
Dista nce(M t o P) = 2 .59 c m
(R to C)/( M t oP) = 1.82
Figure 2.11: Constructing a Golden Parallelogram
The Golden Parallelogram assumes all properties of a parallelogram. At this stage you can
explore the range of cases of the Golden Parallelogram. One way to do this is to investigate the
ratios of the diagonals as the side PR is moved around. Recall that the longer side of the Golden
Rectangle with shorter side of 2 units is {1+ 5 }. We can move PR close to CM such that P is
closest to M. At that position, the length of the diagonal PM will be very close to CMCP={1+ 5 }-2 or 1.24. The diagonal CR in that position will be close to CM+MR={1+ 5 }+2
CR
{1+ 5}+2
or 5.24. The ratio of the diagonals PM will then be
= 2+ 5 or 4.24. We can also
{1+ 5}-2
move PR close to CM such that P is farthest from M. At that position, the length of the diagonal
PM will be very close to CM+CP={1+ 5 }+2 or 5.24. The diagonal CR in that position will be
CR
{1+ 5}-2
close to CM-MR={1+ 5 }-2 or 1.24. The ratio of the diagonals PM will then be
={1+ 5}+2
2+ 5 or 0.24. The least of these ratios is -2+ 5 and the maximum is 2+ 5 . Somewhere in
between these extreme values is the ratio 1, where the Golden Parallelogram will be the Golden
Rectangle. Move PR to see if it is possible that the diagonals CR and PM are also in the Golden
Ratio.
Olive & Oppong: Transforming Mathematics with GSP 4, page 31
H
S
R
P
C
X
M
Distance(H to S) = 3 .23 cm
Distance(H to C) = 2.00 cm
(H to S)/(H to C) = 1 .62
Distance(R to P) = 3 .23 cm
Distance(P to C) = 2 .00 cm
(R to P)/(P to C) = 1.62
Distance(R to C) = 4.57 cm
Distance(M to P) = 2 .83 cm
(R to C)/(M to P) = 1 .62
Distance(P to X) = 2.0 0
Figure 2.12: Golden Parallelogram
This parallelogram CMRP is a Golden Parallelogram with diagonals in the Golden Ratio.
Methods for the construction of the Golden Parallelogram using traditional methods can be
developed, by first showing that CP=PX. You could consider a general position of the point P
on a unit circle such that the angle PCM is 0- . Let the Golden Ratio be 0/ . The coordinates of
the points C, P, R, and M are (0,0), (cos0- , sin0- ), (cos0- +0/ , sin0- ), and (0/ ,0) respectively.
Finding the lengths of the diagonals in terms of 0- , we have by Pythagoras theorem CR =
CR
2+20/cos0-+0/ and PM = 2-20/cos0-+0/ . The ratio of the diagonals
will then be
PM
2+20/cos0-+0/
. We leave it to the reader to verify that this ratio is equal to 0/ when 0- is π/3
2-20/cos0-+0/
radians (or 60º). The sketch of this graph showing the relationship between the ratio (y) of the
diagonals and 0- is shown in Figure 2. 13.
Olive & Oppong: Transforming Mathematics with GSP 4, page 32
Figure 2.13: Ratio of lengths of diagonals
From the graph the least of these ratios is approximately 0.2 and the maximum is
approximately 4.2 and these occur when 0- is 0 and π radians respectively.
It is also possible to construct a geometric proof for the 60º-120º golden parallelogram
having golden diagonals using similar parallelograms (construct a line through X parallel to MR
in Figure 2.12 to produce a golden parallelogram similar to PCMR) and the fact that PXMR will
be an isosceles trapezoid (thus diagonals PM and XR are congruent).
Up to this point, we have only considered parallelograms. We will next explore four-sided
figures with only a single pair of parallel sides, namely trapezoids.
Golden Trapezoids
Construct a trapezoid in which the non-parallel opposite sides are in the ratio of one to one.
The trapezoids for which this is true will all be isosceles trapezoids. Now construct a trapezoid
with opposite sides in the ratio 1 to 2. Two cases are possible, the parallel sides in the ratio 1 to
2, or the non-parallel sides in the ratio 1 to 2.
In the first case, the trapezoid can be constructed as follows. Draw a line segment AB.
Locate the midpoint of AB. At A, draw AD. Through the midpoint of AB, construct a line
parallel to AD. At D, construct a parallel line to AB. The intersection of this parallel line and
the line parallel to AD is the point C. A special case of this trapezoid will be an isosceles
trapezoid as illustrated below.
D
C
AB/DC=2
A
B
Figure 2.14
We can construct the second trapezoid as follows. Draw line segments AB and BC. Through C,
construct a line parallel to AB. Using the point A as center and half of the length of BC as
radius, construct an arc to cut the line through C parallel to AB at D. Does this construction
always work? Investigate the dependence of the construction on angle B.
Olive & Oppong: Transforming Mathematics with GSP 4, page 33
D
C
BC/AD=2
A
B
Figure 2.15
The construction becomes more interesting if we want both pairs of opposite sides to be in
the same ratio. For example, construct a trapezoid ABCD in which AB/DC=2 and BC/AD=2. In
general, one should ask whether there are any trapezoids with the parallel sides and the nonparallel sides in the same ratio. We could even make a conjecture that there exist such figures
for each given ratio. How could we consider our conjecture? We could begin by investigating
the case of a triangle EAB, noting that when a line DC parallel to the base AB is constructed, we
have a trapezoid ABCD whose non-parallel sides are in the same ratio as the sides of the triangle.
E
D
C
EB/EA=2
A
B
Figure 2.16
The point E can be moved around until CB/DA=2. A feature of Geometer' Sketchpad is
that the calculated and changing ratios can be instantly displayed as we move point E in the
plane! In Figure 2.16 above, this will be equivalent to moving E until EB/EA=2. As shown in
the figure, we can construct an object (point D) on AE such that the point D can move freely
along AE. At D, the line DC is constructed parallel to AB. The line DC can then be moved by
moving D along AE until AB is twice DC. The trapezoid ABCD has the ratio of both pairs of
opposite sides the same. What will be the limitations on the ratios that we use? In Figure 2.16
above, we can move the point E close to B. The ratio EB/AE will get closer to zero as E moves
closer to B. Instead, if E is moved close to A, the ratio will increase and may reach infinity.
This implies that, at some time, the ratio will be the Golden Ratio.
Olive & Oppong: Transforming Mathematics with GSP 4, page 34
We can move the point E until we have a special case in which the ratio of both pairs of the
opposite sides of the trapezoid equal to the Golden Ratio. Do the diagonals of this special
trapezoid always bisect each other in the Golden Ratio? Further investigation by measurement
indicates this is the case. Can we use traditional methods to construct this special Trapezoid? By
using the lengths of the sides of a Golden Rectangle, we can construct a triangle whose sides are
in the Golden Ratio.
Activity 2.6: Complete the construction of this Golden Trapezoid. Justify each step of the
construction.
Let us now consider trapezoids in which two pairs of adjacent sides are in the same ratio.
The trapezoid AECD in Figure 2.17 satisfies this condition. It can be constructed by
constructing a parallelogram ABCD with D on the circumference of a circle of center A. The
side AB is extended to F and the point E moves freely on AF. Drag E along AB until the ratio
AE/EC is the same as the ratio DC/AD. If however, the parallelogram ABCD is constructed to
be Golden, then the ratio DC/AD will be the Golden Ratio. In this special case, if we drag E
along AF such that the ratio AE/EC is also Golden, then the trapezoid AECD is a Golden
Trapezoid whose adjacent sides are in the Golden ratio.
D
Distance(D to C) = 3.59 cm
Distance(A to D) = 3.00 cm
(D to C)/(A to D) = 1.20
Distance(E to A) = 9.4 5 cm
Distance(C to E) = 7 .86 cm
(E to A)/(C to E) = 1.20
C
A
B
E
F
Figure 2.17
By now, we must be wondering whether there is a special case when both pairs of the
opposite sides, as well as two pairs of the adjacent sides of a trapezoid are in the Golden Ratio.
We can construct a Golden Parallelogram ABCD such that D moves freely on circle with center
A, and a point E moves freely on AB extended. Move D and E to positions such that in
Olive & Oppong: Transforming Mathematics with GSP 4, page 35
trapezoid AECD, DC/AD=EA/CE=EA/DC=CE/AD= the Golden Ratio. A trapezoid with these
features is called a Super Golden Trapezoid.
D
Distance(D to C) = 4.87 cm
Distance(A to D) = 3.00 cm
(D to C)/(A to D) = 1.62
Distance(E to A) = 7.8 7 cm
Distance(C to E) = 4 .87 cm
(E to A)/(C to E) = 1.62
(E to A)/(D to C) = 1.6 2
(C to E)/(A to D) = 1.6 2
C
J
A
B
E
F
Distance(C to J) = 1.8 5 cm
Distance(J to A) = 2 .99 cm
(J to A)/(C to J) = 1.62
Distance(J to E) = 5.72 cm
Distance(J to D) = 3.5 4 cm
(J to E)/(J to D) = 1.6 2
Figure 2.18
Activity 2.7: What are the properties of Super Golden Trapezoids? Investigate the
relationship between the sides.
We may make conjectures that DC=CE and AD+DC=AE. Following are justifications for
these two conjectures:
1) Since AE/DC=AE/CE=ø, DC=CE.
2) If AD=1 unit, then DC=ø (since ø =DC/AD) and AE=ø2 because AE=ø•(DC).
3) Recall that the Golden Ratio is such that ø /1=1/(ø –1). This implies that 1+ø=ø2.
4) Since AD=1, DC= ø and AE= ø2, AD+DC=AE.
Traditional methods for the construction of the Super Golden Trapezoid can now be developed.
Construct a Golden Parallelogram with AB as one of its sides. Construct point E given that
AE=1+ø. Construct a circle with center at E and radius equal to ø. Construct a circle with center
at B and radius equal to 1. The point of intersection of these two circles determines the point C.
Construct a line through C parallel to AE and intersecting the circle with center at A and radius
1. One of the points of intersection determines the point D. Find the one that satisfies the given
conditions.
Olive & Oppong: Transforming Mathematics with GSP 4, page 36
Some of the figures we have constructed here represent a new class of Golden figures.
Whereas the Golden Rectangle is symmetric, a Golden Parallelogram is not necessarily
symmetric, and the Super Golden Trapezoid is certainly not symmetric. Investigate the Super
Golden Trapezoid further by examining the triangles formed by its diagonals.
The Golden Ratio can be found in music and poetry, art and architecture, and in science
and technology. Since a Super Golden Trapezoid has more Golden proportions than the Golden
Rectangle, this could be the beginning of new investigations of even more areas associated with
the Golden Ratio.
Reflection
We began our exploration of quadrilaterals by starting with four free points in Sketchpad.
We added constraints on these initial conditions that lead to special kinds of quadrilaterals.
Reflect back on this dynamic approach to introducing a new class of geometric figures to
students. Did you discover properties of different kinds of quadrilaterals that you had not known
before beginning this chapter? Could students gain a more robust understanding of the relations
among different classes of quadrilaterals from this approach as opposed to introducing each class
based on definitions (e.g. a parallelogram is a quadrilateral with opposite sides parallel)?
References
Disney (?) (1959). Donald in Mathmagic Land (animated film). Los Angeles, CA: Walt Disney
Productions.
Garland, T. H. (1987). Fascinating Fibonaccis: Mystery and Magic of Numbers. Dale Seymour
Publications, Paloalto.
Huntley, H. E. (1970). The Divine Proportion: A Study in Mathematical Beauty. Dover
Publications, New York.
Markowsky, G. (1992), Misconceptions about the Golden Ratio, The College Mathematics
Journal 23, 2-19.