|
Introduction: After the concepts have been introduced,
we proceed to more advanced work in the study of equivalences,
and the study of the relative value between geometric figures.
For this work the constructive triangles are used.
Materials:
Three wooden boxes: One triangular, two hexagonal
The triangular box contains ten pieces; the fundamental piece
is the right-angled scalene triangle, 1/2 of the equilateral
triangle.
The larger hexagonal box contains eleven pieces; the fundamental
piece is the obtuse-angled isosceles triangle, 1/3 of the equilateral
triangle.
The smaller hexagonal box contains eighteen pieces; the fundamental
piece is the smaller equilateral triangle, 1/4 of the large equilateral
triangle of the first box.
Other Materials: Colored paper, scissors, glue
Right-angled ruler, graph paper, compass
Inset of the equilateral trapezoid from
"Additional insets"
Note: These materials are presented
at two different levels, the first of which can be done in the
Children's House as a sensorial exercise: One box is presented
at a time. The child empties its contents, sorts the pieces by
color and shape and unites the pieces along their black lines.
The figures formed may be named at this level. Prior to the second
level presentation in the elementary, make the box available
to the children so that they may repeat the sensorial exercise
and reacquaint themselves with the material.
[top]
A. Triangular box
Presentation: The child empties the box, sorts
the pieces and unites them as usual. They all make equilateral
triangles. superimpose each equilateral triangle formed of several
pieces on the grey unit triangle. Determine that the unit has
been divided into 2, 3, 4 equal pieces and assign the fraction
values to each piece. Determine how the unit triangle was divided
in each case.
Isolate the two right-angled scalene triangles. As the child
did with the blue triangles of the first series, he tries to
form all of the figures possible with these two triangles. Excluding
the original equilateral triangle, there are five: rectangle,
two different parallelograms, obtuse-angled isosceles triangle,
and a kite. The child names the figures as he makes them.
Determine equivalence between each figure and the unit triangle.
superimpose the two triangles on the unit triangle to show congruency,
and to recall their value. Any other figure made with these two
halves has the same value as the unit, and is equivalent. Show
that the figures are equivalent among themselves.
rectangle = equilateral triangle
> therefore: rectangle = kite
kite = equilateral triangle
As a parallel exercise, the
children may trace the triangles onto colored paper, cut them
out, construct the figures using a compass and ruler.
Next, we draw the child's attention to the relationships that
exist between the lines of the equilateral triangle and those
of the five figures. Then we examine the relationship between
the lines of the five figures.
Recall the nomenclature of the equilateral triangle: side, base,
height (shown by superimposing one-half) and semi-base. Be careful
to position the parallelograms so that the height is represented
by a side of the triangles which form it. Identify the properties
of the kite.
Rectangle: base = 1/2 the base of the equilateral triangle
height = the height of the triangle
Parallelogram: base = the height of the triangle
height = 1/2 the base
side = side
Kite: longer side = height
shorter side = 1/2 the base
long diagonal = side
Note: In identifying the nomenclature
of this figure, the child showed where the shorter diagonal would
be. The characteristic of this figure is that one diagonal is
the perpendicular bisector of the other. However, there is no
relationship to be made with the short diagonal. Later the child
may discover that this short diagonal is equal to the side of
T2.
Obtuse angled isosceles triangle:
base = twice the height
height = 1/2 the base
equal sides = sides
In examining the relationship
between the lines of the various figures, paper figures must
be constructed for each figure in order to leave the triangle
free for use as a measuring instrument. Be sure that the child
understands that the number of comparisons to be made will decrease
with each figure: the first is compared with four others, the
second with three, the third with two and the fourth with only
one, the last figure.
Arrange the figures so that the key figure is isolated above
the row of other figures. Place the triangle first on the key
figure and name a line, then find the corresponding line on the
figure below. Identify all of the relationships between the key
figure and the first figure, then go on to do the same with the
other figures in the row. After the possibilities of the first
key figure have been exhausted, one of the figures below becomes
the key figure, and so on.
[top]
B. Large hexagonal box
Presentation: Empty the contents of the box. Sorting
the pieces by their shape, the child observes that all but one
of the pieces are congruent obtuse-angled isosceles triangles.
Sorting them by color and uniting them as always, the child identifies
the four figures: hexagon, triangle, rhombus, parallelogram.
Isolate the figures in yellow. Turn the three triangles of the
hexagon over as if on hinges so that they are superimposed wrong
side up on the unit triangle. Thus the three outside triangles
are congruent to the unit triangle, and the hexagon is equal
to twice the equilateral formed by the three yellow triangles
nearby.
Bring back the other pairs of triangles and determine the value
of each figure in relation to the unit triangle. Invite the child
to experiment to try to find another figure - the arrowhead.
determine its value. conclude: since the rhombus = 2/3 unit triangle,
and the arrowhead = 2/3 unit triangle, then, rhombus = the parallelogram
= the arrowhead.
Invite the child to experiment with three obtuse- angled isosceles
triangles to make all of the possible figures. There are four:
two concave pentagons, and obtuse-angled trapezoid and an isosceles
trapezoid. The value of each of these figures is 3/3, thus they
are equivalent to the unit triangle and equivalent to each other.
Again the child forms the various figures by tracing, cutting,
pasting or drawing.
As before, the lines of the various figures are compared first
with the guide triangle, then among themselves. the same is done
with figures formed of three triangles.
Form the hexagon again, observing that the inscribed figure is
1/2 the circumscribing figure. The three black lines of the hexagon
are special diagonals (recall how diagonals are formed). These
connect a vertex with the first non-successive vertex it meets.
At this point we examine the equivalence between one figure (hexagon)
and the sum of several congruent figures (rhombi). Isolate the
hexagon and the triangle (the figures formed by the yellow pieces).
Open the hatch of the hexagon and remove the inscribed triangle,
replacing it with the triangle made of three parts, and reclose
the hatch. Notice that the hexagon still has the same value,
though it was made of four pieces and is now made of six. Divide
the hexagon into three rhombi to show that this one figure is
equivalent to the sum of these three congruent figures.
As usual, examine the relationship between the lines of the hexagon
and the equilateral triangle; the lines of the hexagon and those
of the rhombus; the lines of the rhombus and the equilateral
triangle (the diagonals are the only lines considered).
Note: This first hexagon will
be called H1. The unit triangle is congruent to that of the first
box and therefore is called T1. Here we have found that T1 =
1/2 H1 and T1 is inscribed in H1.
[top]
C. Smaller hexagonal box
Presentation: At first the large yellow equilateral
triangle and the six obtuse-angled isosceles triangles are removed
from the box. The remaining triangles are all small equilateral
triangles: six grey, three green, two red.
The child begins as usual. He names the figures he has formed:
hexagon, trapezoid, rhombus. Then values are assigned to the
three figures. Superimpose all of the triangles to show congruency.
Reconstruct each figure and count the number of congruent triangles
in each. The trapezoid has 1/2 as many pieces as the hexagon.
Separate the hexagon to show two trapezoids. The rhombus has
1/3 as many pieces as the hexagon. Separate the hexagon into
three rhombi. comparing the trapezoid to the rhombus, we see
that the rhombus is 2/3 of the trapezoid, or the trapezoid is
3/2 of the rhombus.
Examine the relationship between the lines of the three figures.
Note this time that the diagonals of the hexagon connect opposite
vertices. classify the trapezoid. It is an isosceles trapezoid,
but it is more than isosceles. Since it is made up of three equilateral
triangles, it has an extraordinary characteristic. The longer
base is equal to the equal legs. Present the inset and the label
and add it to the other insets. Just as the equilateral triangle
is an isosceles triangle plus, the equilateral trapezoid is an
isosceles trapezoid plus. show also that bilateral symmetry exists
in the hexagon and rhombus.
At this point the large yellow equilateral triangle and the six
red obtuse-angled triangles are returned. When the child joins
the triangles, three rhombi are formed. The triangles are stacked
up to prove that they are congruent. Thus the three rhombi are
congruent.
Put the three rhombi together to form a hexagon, thus the hexagon
is formed of six equal triangles. Open the hatch of the hexagon
and take out the three red triangles, replacing them with the
yellow equilateral triangle. Observe that the equilateral triangle
is inscribed in the hexagon. Superimpose the red triangles (that
were just removed) on the yellow triangle to show that the equilateral
triangle is made up of three red triangles. Since the hexagon
is made up of six red triangles, the triangle is 1/2 of the hexagon.
Note: This hexagon will be
called H2 and the large yellow equilateral triangle is called
T2. Therefore T2 = 1/2 H2 and T2 is inscribed in H2. The smaller
equilateral triangles which are congruent to the small equilateral
triangles of the first box, and therefore have the value of 1/4
T1, will be called T3.
[top]
D. Relationship between
T1 and T2
Presentation: With the pieces of the small hexagonal
box, form various figure: hexagon, trapezoid, rhombus, and equilateral
triangle and hexagon. Superimpose the grey hexagon on the red
and yellow hexagon to demonstrate congruency.
Recall the value relationships already established. Since the
equilateral triangle is equal to 1/2 the red hexagon, then it
is also equal to 1/2 of the grey hexagon - or a trapezoid.
Bring from the triangular box the grey unit triangle (T1) and
the four small red equilateral triangles (T3).
We know that the yellow triangle T2 is equivalent to the trapezoid
which is 1/2 of the grey hexagon.
One of the triangles of the trapezoid is congruent to one of
the small red equilateral triangles (T3). Superimpose them. Thus,
one red triangle is 1/3 of the trapezoid.
The red triangle as we know is 1/4 of the large grey unit triangle
(T1). Four of these red triangles are equivalent to the grey
unit triangle.
Because the yellow triangle (T2) is made up of three of the small
red triangles (T3), then we can say that the yellow triangle
(T2) is made up of 3/4 (T3) of the grey unit triangle (T2 = 3/4
T1).
[top]
E. Difference and ratios
between similar figures
Materials: From the triangular box : T1-grey
equilateral, and T3- red equilateral
From the large hexagonal box: T1-yellow equilateral, three yellow
obtuse-angled triangles having the black line on the hypotenuse
From the small hexagonal box: T2-yellow equilateral, six small
grey equilaterals
Presentation: Identify the two triangles by the
symbol names T1 and T2. Construct the hexagons and identify them
H1 and H2.
If the child has not already
discovered it, lead him to the conclusion that T1 - T2 = 1/4
T1. since T1 has the value of 4/4ths, and T2 has the value of
3/4ths, T1 - T2 is the same as saying 4/4 - 3/4 which equals
1/4. Set up the equation using the materials and card for the
signs. T2 = T1 - T3.
This means that the small red equilateral triangle has the same
value as the grey portion that is left showing when T2 is superimposed
on T1 concentrically, or so that one vertex coincides.
Knowing that H1 is the double of T1 and that H2 is the double
of T2, we can conclude that their difference would be the double
of 1/4 of T1, that is 2/4.
2T1 - 2T2 = 2 (1/4 T1) = 2/4
2(4/4) - 2 (3/4) = 8/4 - 6/4 = 2/4
Using T1 as the unit, assign
relative values to the hexagons: H1 = 8/4 and H2 = 6/4 (T1 =
4/4). The large hexagon is 8/4 of the grey triangle; the small
hexagon is 6/4 of the grey triangle. Therefore, the small hexagon
is 3/4 of the large hexagon. Examine also the inverse of each
of these statements as they are also true.
Since the ratio between the small triangle and the large triangle
is 3:4, then the same relationship exists between their doubles,
the small hexagon and the large hexagon ... 3:4.
Superimpose H2 on H1 concentrically with the sides parallel.
the frame is the difference between the two hexagons, and, therefore,
has the value of 2 1/4 equilateral triangles. Thus the difference
between the hexagons is a rhombus. This is the arithmetical way
of showing their difference.
[top]
F. Sensorial demonstration
of the difference between H1 and H2
Materials:
Triangle fraction insets: whole, 3/3
Constructive triangles: H1, H2 and two red T3
Presentation: verify the congruency between the
triangle whole inset and the small red equilateral triangle,
by superimposing the inset on the wooden triangle. In the same
way, we can see that this metal inset is equal to 1/6 of the
grey trapezoid.
Superimpose H2 on H1 so that the vertices of the grey hexagon
coincide with the midpoints of the sides of the yellow hexagon.
Place the 1/3 metal inset pieces over the yellow portion that
remains visible. With this material we can cover only three places.
However these three obtuse-angled triangles equal one small red
equilateral triangle. Move the three thirds to the three remaining
vacant spots; these three triangles correspond to the second
small red equilateral triangle.
[top]
G. Equivalence between two rhombi
Material:
From the small hexagonal box: two red obtuse-angled and two red
equilateral triangles
From the triangular box: grey equilateral, green right- angled
2/2 triangle fraction insets (optional)
Presentation: Join each pair of red triangles along
the black lines to form two rhombi. Superimpose one rhombus on
the other to show congruency. Demonstrate that each rhombus was
divided into two equal pieces. Since each triangle has the value
of 1/2 of the same rhombus, all of the triangles are equivalent.
Demonstrate this equivalence sensorially by superimposing the
two metal inset pieces on first one red triangle, and then in
another arrangement on the other.
We know that the red equilateral triangle has the value of 1/4
T1. Since the red obtuse-angled triangle is equivalent, it must
also have the value of 1/4. Therefore, 1/4 + 1/4 = 1/2 T1. Superimpose
the green right-angled triangle as a proof.
Following this line, the grey triangle can be constructed with
two rhombi. Also this 1/2 is the difference between the two hexagons
in the form of a rhombus or this right-angled scalene triangle.
[top]
H. Equivalence between the
trapezoid and T2
Materials:
From the triangular box: grey equilateral triangle
From the small hexagonal box: yellow equilateral, three green
equilateral triangles
Presentation: Unite the three green triangles to
form a trapezoid. Knowing that each of these small triangles
is congruent to the red triangle having the value of 1/4 T1,
we can say that the trapezoid has the value of 3/4 T1. The trapezoid
can be superimposed on the grey equilateral triangle to show
that 1/4 is lacking.
We've already seen that the red obtuse-angled triangle was equivalent
to the small red equilateral triangle having the value of 1/4
T1. So each green triangle would also be equivalent to a red
obtuse-angled triangle. These obtuse-angled triangles were each
1/3 of T2; T2 is composed of three obtuse-angled triangles having
the value of 1/4 T1. Therefore T2 is equal to 3/4 T1. Having
the same value of 3/4 T1, the trapezoid and T2 are equivalent.
[top]
I. Ratio between circumscribing
and inscribed figures
Materials:
From triangular box: grey equilateral, four red equilateral triangles
From large hexagonal box: yellow equilateral, three yellow obtuse-angled
From small hexagonal box: six grey equilateral triangles
Triangle fraction insets 4/4, square fraction insets 4/4 (diagonal)
Presentation: Superimpose the four small red equilateral
triangles on the grey equilateral triangle. Remove the three
triangles at the vertices, leaving two equilateral triangles.
An equilateral triangle inscribed in another equilateral triangle
is 1/4 of it. The ratio is 1:4. Demonstrate the same experience
using the metal insets. We can construct a chart to examine all
regular polygons.
Number of sides Ratio
3 1/4
4 2/4
5 ? (between 2/4 and 3/4
6 3/4
etc... etc...
circle 4/4
Use the metal inset of the
square. Recall the value of the triangular pieces. Put two pieces
aside. With the remaining two form an inscribed square. The inscribed
square is 2/4 of the circumscribing square.
We don't have any material to examine the pentagon. For the hexagon,
form H1 and H2 and superimpose them to recall the ratio of 3:4.
We have no more materials to examine the others. When will the
ratio be 4/4? When there is no difference between the circumscribing
and inscribing figures, that is when the figures are circles.
We can conclude that the ratio for the pentagon will be somewhere
between 2/4 and 3/4, and for the septagon and all the others,
the ratio will be between 3/4 and 4/4.
[top]
|
