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# Geometric textbook

### Instructions for working with transformable models on stereometry and planimetry '' Nane ''

1. Composition and purpose
2. A means for the models
3. Obtaining of the main basic models and their modifications
4. The structure of models

Composition and purpose

The educational geometric transformable manual "Nane" is a set of six geometric figures to study stereometry.

• Triangularpyramid
• Triangular prism
• Cube
• Cone
• Cylinder
• Six additional rods for different constructions and sections.

''Нанэ'' is intended for increase of learning efficiency of the school course geometry and for development of spatial thinking of pupils. It is reached by visual demonstration of various standard and non-standard geometrical figures. The game character which is the cornerstone of the models functioning allows considerably increasing the efficiency of their use.

A means for the models

Each of the sides of each pattern consists of three rods forming the edge and entering each other on the principle of telescopic antenna. Quality material combined with this device allows the ribs with sufficient ease and clarity to make the necessary manipulations to change the size of the parties. Places docking edges are at the top and bottom geometric bodies connected by the original ring that, in fact, with an unlimited degree of freedom allows you to rotate the rods by changing the lengths of the sides. Such moving parts can be obtained from a model of different new models and figures, both standard and non-standard. It should also be noted that the removable nature of the node (vertex) parts allows changing the number of sides at each vertex of the original figure, to open the way to search for new geometric figures.

Obtaining the main models and their modifications

Triangularpyramid From this model it is necessary to obtain all types of a triangular pyramid. 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.

SB⊥(ABC)

AC⊥CS follows AC⊥BC и AC⊥BC follows AC⊥CS With this type of model it is possible to solve numerous tasks. From this model it is possible to receive all types of quadrangles and a triangle for the construction of which it is necessary to make the following: all sides of the basis of a pyramid A, B and C should be pulled, and the edges SA, SB and SC should be decreased in length until they all fit in the plane of the triangle ABC. This model allows receiving of all types of the quadrangular pyramid and pentagon. We remind that the transformations are carried out with the help of change of each of the sides separately. Of particular interest is the case when the basement of ABDC quadrate and the lateral edge are perpendicular to the plane of the basement. Then according to the theorem of three perpendiculars, we receive: AC⊥CD from what it follows AC⊥SC.  It is possible to receive a triangular pyramid from a quadrangular pyramid as follows. All sides of a triangle ABD we extend up to the end, then we hold the top C and shortening SC gradually we put it inside (i.e. top of C) of the triangular pyramid ABDS until SC becomes ⊥ (ABD) and C ∈ (ABD). There is also the second option of transformation. The top of C ∈ (SAD). As a result we obtain a triangular pyramid ABDS (Fig. 4) in which it is possible to construct the height BC with the help of an auxiliary rod. In the same way it is possible to obtain a triangular pyramid from the quadrangular pyramid. Shortening AD, we do so that the sides of DB and DC made one line to the line BC. It is clear that turning this pyramid, different types of triangular pyramids are created, which are used in the solutions of tasks. For implementation of each of the following transformations it is necessary to close consistently each of the sides of the model up to the stop. The quadrangular pyramid turns into a regular pentagon with its diagonals.

Triangularprism We get this model from the box; it is in a set separately from all other models. We close all sides. Then we pull A1BA1C, and we obtain a regular triangular prism. Opening the AB, AC and BC we obtain the truncated pyramid. Further, extending BB1 until the B1A1, A, B1, C1, C become straight lines, we obtain a triangular pyramid ABCB1, where different sections of a triangular pyramid are visible. This model is of interest that the simple turning from side to side generates more and more new pyramids with various sections. We hold from top B1 and pull in order to obtain a pentagonal pyramid. All other sides are B1A, B1B, B1C1, B1C1 and B1A1. We increase the sides of the quadrangle ABCA1, turning it into quadrate. We compress the C1B1 till the end and move upwards and inwards the top C1 and bring it up to the osculation with CA. As a result we obtain a tetragonal pyramid with the height B1C1. For receiving a pentagonal pyramid again we close all sides. The triangular prism turns into the regular hexagon, and you can obtain it yourself To obtain an octahedron from this model (regular octahedron), we do the following. Strongly holding from the top C and twisting CA1 midstream completely separating from the top C. In the same way we separate CA1 from A1. Then CA1 is strengthened in tops A and C1. Closing all sides, we obtain an octahedron.

Cylinder.   For creation of this model at first we fix those rods, which have no metal pins. We fix them against each other.
By means of separately taken circle and the diameter thereof is possible to construct an inscribed triangle, quadrangle.

Creation of sections The auxiliary rods are included in the kitfor sections construction, which allow to obtain any sections. It is easy to be convinced, if we notice that the rods can be fixed in any combinations. In a triangular pyramid by means of auxiliary rods, as shown in figure, we demonstrate that section which represents a quadrangle. By the way, it is possible to construct with their help the height, the median, the bisector, etc. Auxiliary rods can also serve for replacement of the broken rods models.

## купить телефон android Для сложения 2-х и более чисел подвешиваем их друг к другу к какой-либо отметке на левом плече. Чтобы найти ответ, к той же цифре на правом плече подвешиваем 10. Получаем следующее:
а) правое плечо тяжелее – снимаем число 10 и, подвешивая по очереди цифры, находим ту, при которой устанавливается равновесие
б) если правое плечо легче – то к числу 10 подвешиваем по очереди цифры и находим ту, при которой устанавливается равновесие.

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.

### 2. Состав числа. The composition of the number. 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.

### 3. Умножение. Multiplication. 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.

### 4. Вычитание и деление выполняется как противоположные действия сложению и умножению. Subtraction and division is performed as opposing the operations of addition and multiplication. 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.

### 5. Деление с остатком. Division with remainder 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.

### 6. Распределительный закон сложения и умножения. Distributive law of addition and multiplication. а) Интересна демонстрация распределительного закона сложения и умножения, например, (3 + 9 ) х 7 = 7 х 3 + 7х 9. Цифру 3 подвешиваем к цифре 9 и обе подвешиваем к отметке 7, а к отметке 3 другого плеча подвешиваем цифру 7, другую цифру 7 к 9 и получаем равновесие.

б) демонстрация закона о перестановке мест слагаемых и множителей очевидна.
6x7=7x6, 6+7=7+6

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