Note: The triangle, the smallest figure in reality is the constructor of all other figures in reality. The tetrahedron, the smallest solid in reality constructs all of the other solids in space.
Two yellow equilateral triangles
Two yellow, two green right-angled isosceles triangles
Two yellow, two green, two gray right-angled scalene triangles
One red smaller right-angled scalene triangle
On red obtuse-angled scalene triangle
(Each triangle has a black line along one side)
Box 2 –
Two blue equilateral triangles
Two blue right-angled isosceles triangles
Two blue right-angled scalene triangles
One blue obtuse-angled scalene triangle
One blue right-angled scalene triangle (corresponds to the red triangle from Box 1)
Box 3 –
Twelve blue right-angled scalene triangles with no black lines. The angles measure 30, 60, and 90 degrees.
Presentation: Invite the child to remove the triangles from the box, and then sort the triangles according to shape. Having done this, ask the child to separate each pile according to color, resulting in various piles of triangles having both shape and color in common. Isolate the two red ones to be used later.
The teacher takes the pile of two equilaterals and separates them in such a way that the two black lines are facing each other. Watch, these black lines are like a magnet. Slide the two triangles together so that the black lines meet. Invite the child to do the same, leaving the joined triangles in place.
Identify the figures that have been constructed: a yellow rhombus, a green square, and a gray rectangle. The teacher identifies the other three figures as common parallelograms (parallelogram: Greek parallelogramium <parallelos, parallel, and grame, figure>). therefore a parallelogram is a plane figure having parallel sides. By simultaneously running two fingers along two parallel sides, the teacher gives a sensorial impression of parallel. We also call them common parallelograms to differentiate them from the square, rectangle, and rhombus which could also be considered parallelograms. The child names each figure as they are indicated by the teacher.
SECOND BOX – CONSTRUCTION OF QUADRILATERALS
Presentation: Invite the child to sort the triangles by shape. As before, set aside the two small triangles which correspond to the red ones of the first box.
Isolate the two equilateral triangles and invite the child to form all of the possible quadrilaterals. try as he might, he can only form one. The child identifies it as the rhombus.
Leaving the rhombus intact, the teacher takes the two right-angled isosceles triangles and forms the possible figures. The child identifies the figures as they are made. there are two: the square and the common parallelogram.
The child may see two different parallelograms. Trace one on a sheet of paper. Form the other and superimpose it. The second parallelogram doesn’t fit inside the contours of the first. Trace the second parallelogram and cut out the two figures. By placing the cut-outs back to back, we can see that one is the mirror image of he other, therefore they are the same parallelogram.
The child is invited to form the possible quadrilaterals with the two right-angled scalene triangles. The child identifies the three:
rectangle, common parallelogram, a different parallelogram.
One by one, isolate each type of triangle, ask the child to classify the triangle according to its sides, and ask, “Of how many different lengths are the sides of this triangle”? Conclude that with a triangle whose sides have all one measure, we can form only one figure – the rhombus. With a triangle whose sides have two different measures, we can form to figures – square and common parallelogram. With a triangle whose sides have three different measures, we can form three figures – rectangle and two parallelograms.
Direct Aim: To give the relationship between the number of different lengths of the sides and the number of figures which can be possibly constructed.
SECOND BOX – HEADS & TAILS
Presentation: Every plane figure, like a coin, has two sides. One side is called obverse or heads; it is the side which has a face or the principle design. When you turn it over, you have the reverse side or tails. All of these figures have an obverse (blue) and a reverse (natural wood or white) side.
Isolate the two equilateral triangles. Invite the child to form as many figures as possible. As before, he can make only one. Suggest that he tries with one obverse and one reverse side. It won’t help. There is only one figure he can make.
With the two isosceles triangles he makes the two possible quadrilaterals with the obverse sides. Invite the child to make a triangle. The child classifies the triangle: isosceles. By turning one triangle to its reverse side, the child can make no new figures.
With the two scalene triangles, the child tries to form all possible quadrilaterals first with the two obverse sides (yielding the same three figures as before) and then with one reverse side. The child is able to form a new figure: a kite (or a deltoid, having the form of the Greek capital letter Delta ).
Invite the child to make triangles, first with obverse sides (yielding none) and then with one reverse side. The child classifies the triangles he makes: acute-angled isosceles and obtuse-angled isosceles triangles.
Presentation: With the two small red triangles, invite the child to unite the triangles along the black lines and identify the figure obtained – a trapezoid.
Using the two corresponding blue triangles, invite the child to form the figure which he already knows and to identify it – a trapezoid. Continue making other quadrilaterals using only the obverse sides. The teacher identifies the figure obtained. since it has four sides we can call it a quadrilateral. It is a concave quadrilateral – a boomerang (it may also be called a re-entrant).
Invite the child to turn over one triangle to form any other figures, quadrilaterals or triangles. The quadrilateral is called a common quadrilateral. The triangle is an obtuse-angled isosceles triangle. (Note: This triangle has great importance in the later study of the area of a trapezoid.) Recall the figures formed by these triangles; there are four.
Age: After 6 years
Aim: Exploration of the triangle as the constructor of triangles and quadrilaterals.
Presentation: Isolate one triangle. Ask the child to identify each of the angles, and the biggest and smallest angles. This angle which is neither smallest nor biggest we can call the medium angle. The child names each angle: smallest angle, medium angle, biggest angle.
First star: Let’s unite all the triangles by their smallest angle. The teacher positions a few and allows the child to continue. How many points does this star have? Twelve. with all of the triangles at our disposal, we can make only one star with twelve points.
Second star: Let’s unite all the triangles by the medium angles. How many points does this star have? Six. Try to make another star with the triangles that are left. With all the triangles at our disposal, we can make two stars with six points.
Third star: Let’s unite all of the triangles by the largest angle. How many points does this star have? Four. This symbol is very famous; it is the star of Saint Brigid, the patron saint of Ireland. Try to make another star like this. With all the triangles at our disposal, we can make three stars with four points.
Aim: Use of the triangle as a constructor to indirectly demonstrate the following:
30o x 12 triangles = 360o 60o x 6 triangles = 360o 90o x 4 = 360o
360o / 30o = 12 tris. 360o / 60o = 6 tris. 360o / 90o = 4 tris.
360o / 12 triangles 360o / 6 = 60o 360o / 4 = 90o
1. Construct the first star. Notice that the triangles meet at a point in the center. We must divide this star into two equal parts, leaving six triangles on one side and six on the other. Many possibilities exist; simply choose on and slide the triangles away to leave a gap.
We want to make the point at the top of one side meet the point on the top of the other. Slide one half along and then towards the other to make the two points meet at the top. We see that they have met at the bottom also, and where there was a point in the center there is now a line segment.
Again divide the figure in half, this time along the other side of the triangle which was displaced before. Separate the two halves to leave a gap. Identify the two points at the top and bottom which should meet. Slide one half into position. We see that a quadrilateral (a rhombus) has been created at the center.
Continue in the same manner, identifying the figure formed at the center each time: equilateral hexagon, equilateral octagon, equilateral decagon, equilateral and equiangular, therefore regular dodecagon. This is the first diaphragm. It is like the diaphragm of a camera. Bring one in to demonstrate.
2. Construct the second star. As before, divide into two equal parts. Slide one side so that the vertices of the extreme angles meet. Note the change from a point to a line segment. Continue naming each of the figures made, ending with the equilateral and equiangular, therefore – regular hexagon.
3. Construct the third star. Divide as before and slide one half. In this only the point, line segment and square are formed in the center. This is the third diaphragm.
Note: This third diaphragm will serve as a point of reference for two algebraic demonstrations of the Pythagorean theorem.
4. The children draw, cut, and paste the stars and diaphragms.
5. Older children may solve for the areas of the diaphragms and their internal figures, and find the relationship between them.
6. Constructing the second or third star, the child forms other figures by fitting in the angles.
7. Encourage further explorations using these triangles.
Direct Aim: Exploration of shapes using triangles.
Indirect Aim: Preparation for the sum of exterior and interior angles.