**Concepts in Action**

**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.

**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.

**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.

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.

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).

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.

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.

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.

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.

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.

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