Geometric games
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''Polyhedrons and developments'' – this is a new set of teaching geometrical models, which is intended for demonstration of various polyhedrons and their developments. The set is arranged so that allows from each model of a polyhedron easily to obtain all types of its developments, and this in its turn gives the chance visually and logically to master inobtaining of the developments from polyhedrons and vice versa – from developments of polyhedrons.
''Polyhedrons and development''complement the geometric transformable models “Nane”. “Nane” models allow receiving numerous transformations of geometrical models, but with their help it is impossible to show the developments of stereometric bodies. The set of the models ''Polyhedrons and developments'' fills this gap and allows to demonstrate the geometric bodies. The models of a new set are arranged so that it is possible to place in them “Nane” models and combining both models, to show different opportunities of geometrical figures and bodies. Two main details of the components of a new set of models are the triangle and the square, which are connectedto each other with the help of a special lock. The set consists of ten triangles and ten squares,with the help ofwhich it is possible to obtain:
Receiving different models of the triangular pyramid, for example, is as follows:
Method I.
We take four triangles, we connect so that to obtain a triangular pyramid.
Method II.
At first we display the development/cut-out shape of this model on a table, and then we receive the required model.
The originality of models is that from each one it is possible to receive all developments/cut-out shapes of this model. By means of these models it is possible to make and solve interesting tasks. Forexample:
Task 1.
Choose from the figures those, which are considered as the developments/cut-out shapes of a cube.
Task 2.
To designate the tops of a triangular pyramid ABCD with the letters A, B, C and D, so that the triangular pyramid (octahedron) is turned out.
Task 3.
The figure depicts all developments of the correct octahedron (octahedron) EABCDF. The tops of the octahedron are designated by letters E, A, B, C, D and F, so as to obtain an octahedron.
By means of this set the educational process can be organized with high efficiency. For this purpose at first the pupils are given a task to draw the development of any polyhedron, then to carry out an inspection of correctness of the drawing (developments) with application of models. The training provided in such a way allows to acquire practical skills of receiving spatial bodies, and visually to display this process. It is also possible to carry out competitions on speed of assemblage of different polyhedrons, suggesting pupils to collect, for example, an icosahedron (icosahedra) and to receive its development.
Using these models in the course of geometry teaching, (especially stereometry), the teacher can reach more effective result, than working without them. The set can be applied as the constructor, it is also useful for pupils of all classes, and the use of models in pre-school groups can be organized in the form of geometrical games.
Thanks to the transformable qualities, the modelsgenerate an interest in geometry, promote logical and constructive thinking, and develop spatial imagination.
The possibilities to get new polygons are almost unlimited, wherein you can see for yourself.
Ball |
Pyramid |
Set Nane |
Sweep |
geometric |
Geometric |
geometric |
Geometric |
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Before starting working with the models it is necessary to be aware that for manipulations and transformations it is needed to work consistently only with each side separately. For the implementation of each of the following transformations it is needed to close consistently each of the sides up to the stop.
Getting the geometric figures out of triangular pyramid model
To implement the conversion it is necessary to close consistently each side of a triangular pyramid ABCD up to the stop and get the correct triangular pyramid (tetrahedron).
1. Romb–is obtained from the original pyramid through BD side stretching/pulling
2. Quadrat- is obtained from therhomb through AC side stretching/pulling and DB shortening
3. Trapezoi–is obtained through stretching/pulling of the sides AB, as a result the diagonal AC and BD are stretched themselvesd
4. Parallelogram–isobtainedthroughstretching/pullingofthesidesDCandthediagonalAC. Here it is possible to demonstrate operation on vectors (addition and subtraction).
5. Rectangle–is obtained stretching/pulling BD and shortening AC
6. Triangle
а) Close each side of the model sequentially up to the stop - the starting point is obtained
б) All sides of the base of pyramid ABC should be stretched/pulled until the edges DA, DB, DC will fit the four centers lie in a planeof the triangle ABC
7. Test of equality of triangles
AD must be disconnected at point D, and the AC and BD have to be stretched/pulled until the top of D does not coincide with the vertex A.
Watch the video "A triangular pyramid" with 0:56 seconds
8. The characteristics of an isosceles triangleStarting position –is tetrahedron. In the beginning it is necessary to obtain a rhomb, and thenstretch/pull the rods CB, AB and DB.
9. Pyramid
The pyramid can be obtained from the triangle (point 6)
The top/vertex D starts to move away from the tops/vertexes A, B and C a triangular pyramid is being formed.
9.1 The correct triangular pyramid DА=DВ=DС и АВ=ВС=АС.9.2 The projection of the triangle DAC in the plane of ABC is obtained by stretching DA and DC until DB does not become perpendicular to the plane of ABC.
9.3 The theorem on three perpendiculars/verticals.
It is necessary to stretch/pull the BC and AC until the side of the CA will not be perpendicular to AB. Studying the theorem of three perpendiculars, the pupils are asked to convert the four triangles of the pyramid ABCD into the right triangles which is not that simple. When repeated attempts of pupils are unsuccessful, the teacher, being inferior in resourcefulness to pupils, but superior in knowledge, shows the miracle about three perpendiculars.
9.4
1.1 The planes (DCA) and (DBC) are perpendicular to the plane ABC. From this it follows that the line of intersection of the planes DC is perpendicular to the plane of ABC.
10. A solution of the problem
Task. Given: a pyramid with ribs/links a, b, c, which are mutually perpendicular.Findthevolumeof the pyramid.
To add 2 or more numbers hung them to each other to a mark on his left shoulder. To find the answer to the same figure on the right shoulder are suspending 10. We obtain the following:
a) the right shoulder heavier - remove the number 10, and hanging on the line numbers, we find that at which equilibrium is established
b) if the right shoulder easier - that among the 10 numbers are suspending in turn and find the one at which equilibrium is established.
Чтобы найти состав любого числа (меньше 10). Необходимо это число подвесить к любой точки на правом плече. Чтобы получить состав например 10, надо на левом плече к той же точки подвесить цифры в таком составе чтобы получить равновесие.
Комментарий: Важно отметить для равновесия необходимо, чтобы плечи левое и правое были одинаковы.
To find the composition of any number (less than 10). This number is necessary to suspend any point on the right shoulder. To obtain the composition of example 10, it is necessary on the left shoulder to the point of hanging figures in this format to get the balance.
Comment: It is important for balance requires that the left and right shoulders were the same.
Например: Для получения ответа 9х7, подвешиваем 9 к отметке 7 на правом плече. Чтобы получить ответ на левом плече сначала надо найти десятки.
Алгоритм нахождения ответа: «Число 5 подвешиваем к отметке 10 (левое плечо), т.к. число 5 меньше, то подвешиваем 7, т.к. число 7 больше, то подвешиваем 6, т.к. 6 меньше, значит ответ десяток 6. По такому же принципу находим единицы на шкале 1
Пример: 4х7
Алгоритм: Число 5 подвешиваем к отметке 10 (левое плечо). Т.к. 5 больше, то подвешиваем 3, т.к. 3 больше, то подвешиваем 1, т.к. 1 меньше, значит десятки 2. Далее находим единицы по такому же принципу.
For example: To answer 9h7, hung around 9 to 7 on the right shoulder. To get the answer on the left shoulder first need to find dozens.
Algorithm for finding an answer: "The number 5 is suspended from the 10 mark (left shoulder), as 5 number less then 7 are suspending since the number 7 is greater then 6 are suspending since 6 smaller mean response dozen 6. The same principle find units on the scale 1
Example: 4h7
Algorithm: The number 5 is suspended from the 10 mark (left shoulder). Because 5 more, are suspending 3, because 3 more, are suspending 1 since 1 less then ten 2. Next, find the unit in the same way.
Деление Для получения ответа на 56/8 подвешиваем 5 к отметке 10 а 6 к отметке 1 на левом плече. На правом плече к отметке 8 подвешиваем 5. Так как 5 меньше, то подвешиваем 7, получается правильный ответ.
Dividing To answer 56/8 are suspending 5 to around 10 to around 6 and 1 on the left shoulder. On the right shoulder to the mark 8 are suspending 5. Since 5 less then 7 are suspending, get the right answers.
Пример: 77/9.
На левом плече число 7 подвешиваем к отметке 10 на шкале и 8 к единице (получаем 78). Чтобы делить на 9 на правом плече шкалы к 9 подвешиваем число 5, т.к 5 меньше, то подвешиваем 7, т.к. 7 меньше, то подвешиваем 9, т.к. 9 больше, то подвешиваем 8, т.к. 8 меньше, то правильный ответ 8. Таким же алгоритмом находим остаток на шкале отметки 1.
Весы очень полезны для демонстрации законов сложения и умножения.
Example: 77/9.
On the left shoulder number 7 is suspended from the 10 mark on the scale and 8 to one (get 78). To divide by 9 on the right shoulder of the scale to 9 are suspending number 5, because 5 is less, are suspending 7, as 7 less then 9 are suspending since 9 more, 8 are suspending since 8 less, the correct answer 8. In the same algorithm finds balance in the scale 1 mark.
Scales are very useful for demonstrating the laws of addition and multiplication.
a) An interesting demonstration of the distribution law of addition and multiplication, for example, (3 + 9) x 7 = 7 x 3 + 7x 9. The figure 3 is suspended from the number 9 and both suspended to around 7, and the mark of 3 other arm are suspending figure 7, another figure 7 to 9 and get a balance.
b) demonstration of the law of inversion of terms and factors evident.
6x7 = 7x6, 6 + 7 = 7 + 6