# How to cut a square into 7 triangles

Table of Contents:

## Extracurricular activity: “Cutting problems

Please note that in accordance with Federal Law N 273-FZ “On Education in the Russian Federation” in organizations carrying out educational activities, training and education of students with disabilities is organized both together with other students, and in separate classes or groups.

“Relevance of establishing school reconciliation/mediation services in educational organizations”

Certificate and discount for each participant

### Description of the presentation by individual slides:

Pentamino game, cutting and folding shapes.

Pentaminoes 12 pieces, each consisting of five identical squares, and the squares are adjacent to each other only by their sides. “PENTA”. “FIVE.” (from Greek)

Pentamino A game consisting of folding different shapes from a given set Invented by the American mathematician S.Golombom in the 1950s

Cover Chessboard 88 with all the pieces by cutting a square in the middle 22.

Put all 12 pentamino pieces in a rectangle 6 10. How many options are possible? Build 2 rectangles 5 6.

Cut the shapes shown in the figure into two equal parts.

Cut square 44 without a corner cell into 3 equal parts?

Cut a 2×1 rectangle into 7 pieces that can be used to make a square.

Divide a 4×4 square into two equal parts in four different ways so that the cutting line runs along the sides of the cells.

Cut each of the shapes into three equal parts. (Only the sides of the squares may be cut. The pieces must be equal not only in area but also in shape.)

Divide the rectangle into 4 equal parts in size and shape so that each part contains one circle.

Divide the rectangle into 4 equal parts in size and shape so that each part contains one circle.

Divide the rectangle into six equal parts so that each part has a cell with a dot in it.

Divide the square into 4 equal parts so that in each of them was on 2 coins. At the same time in any of the parts of the coins should not be in the neighboring cells.

Cut a figure consisting of three squares into 4 equal parts.

Four countries are shaped like triangles. How are the countries located relative to each other if each country has common borders with three other countries? 1 2 3 4

Cut the figure into 4 equal parts similar to the original. Students have difficulty solving this problem. Students divide the given figure into small **triangles** at random Divide by 6, but 6 by 4 is not divisible. Which number is also divisible by 4?? We thought that each “middle” triangle should be divided into 4 more parts, so we have 24 triangles, which we divide into 4 equal parts, consisting of six **triangles**, which are similar to the original.

Cut the figure into 4 equal parts, similar to the original.

With two straight lines “cut” the horseshoe into 6 parts. Counting: 1 2 3 4 5 6

How to cut a square into 7 squares so that 3 of the squares are equal and the other 4 are also equal?

Place 10 points on 5 segments so that each segment has 4 points (the segments do not belong to the same line) Answers

A local land dealer got an unusually shaped piece of land by chance (he expected to sell it for a profit in installments). But each of the eight buyers he found wanted a plot no worse than his neighbor’s. Where should the merchant set up dividing hedges so that there are 8 identical plots? Answer

A square consists of 16 identical cells, 4 of which are shaded. Cut the square into four equal parts so that each of them has only one painted cell. A cell may occupy any place in each part.

Spread 4 coins of one denomination and 4 coins of another on the 16 squares, so that no two identical coins can be found in any row. horizontal, vertical or diagonal.

How, without taking the pencil off the paper, to cross a figure with three straight lines, so that they cross all the squares one by one?

Cut the cross into 4 equal pieces so that you can make a square the same height and width as the height and width of the cross

Divide a cross into 4 equal parts so that you can make a square with a cross inside of it Hint Practice Task:

A successful real estate salesman bought 16 acres of land. He decided to divide the area into 16 one-acre lots for dwellings so that all the lots were the same size and the same shape as the main lot. How to do this? Answer In 1821, a collection of D.Jackson’s “Mental Entertainment for Winter Evenings,” which included a quatrain riddle:. (text on slide) IMPORTANT: Skip to the next slide. reasoning by clicking on the margin of the slide. Button. go to the slide with the correct answer.

Cut the figure into 2 equal parts Answer (3).

Cut the figure into 2 equal parts Answer.

### Methods of Teaching Mathematics in Basic and Secondary Schools in the Context of the 21st Century State Educational Standard

## APP for Math in Grade 2 by Rudnitskaya and Yudacheva, Part 1 Page 107-116

The guests placed cutlery: spoons, knives, forks. How many cutlery do you see in the picture?? How did the Wolf solve the problem and how did the Hare? Which one of them did the task faster?? Why? Answer: The wolf solved it this way: 1. put the cutlery on the first napkin with the cutlery on the second napkin. 2. the sum of the cutlery on two napkins added up to the cutlery on the third napkin. 3. The sum of the cutlery on three napkins added up with the cutlery on the fourth napkin. 4. added the sum of the cutlery on four napkins to the cutlery on the fifth napkin. 3 3 = 6 cutlery. on two napkins 6 3 = 9 cutlery. on three napkins 9 3 = 12 flatware. on four napkins 12 3 = 15 cutlery. on five napkins Hare multiplied the number of cutlery on one napkin by the number of napkins. 3 • 5 = 15 cutlery. on five napkins So the Hare did the task faster than the wolf.

Calculate. 3 • 2 3 • 7 3 • 9 3 • 6 3 • 4 3 • 5 Answer: 3 • 2 = 6 3 • 6 = 18 3 • 7 = 21 3 • 4 = 12 3 • 9 = 27 3 • 5 = 15

Compare the multiplication results, using a calculator. 3 • 8 и 8 • 3 3 • 6 и 6 • 3 3 • 9 и 9 • 3 Conclude. Answer: 3 • 8 = 24 8 • 3 =24 3 • 9 = 27 9 • 3 = 27 3 • 6 = 18 6 • 3 = 18 We can conclude that changing the places of the multipliers does not change the product.

Multiply 3 by 0, multiply 0 by 3, 3 by 1, 1 by 3. Compare the results of multiplication. Conclude. Answer: Answer: 3 • 0 = 0 0 • 3 = 0 1 • 3 = 3 3 • 1 = 3 Multiplying any number by 0 or 0 by any number produces 0. When you multiply any number by 1 or 1 by any number, you will get the number you multiplied.

Using the multiplication table for number 3, make and write a multiplication table for number 3. Answer: 3 • 1 = 3 3 • 2 = 6 3 • 3 = 9 3 • 4 = 12 3 • 5 = 15 3 • 6 = 18 3 • 7 = 21 3 • 8 = 24 3 • 9 = 27 3 • 10 = 30

State the result of multiplication. 4 • 3 5 • 3 8 • 3 2 • 3 6 • 3 9 • 3 7 • 3 3 • 3 Answer: 4 • 3 = 12 6 • 3 = 18 5 • 3 = 15 9 • 3 = 27 8 • 3 = 24 7 • 3 = 21 2 • 3 = 6 3 • 3 = 9

How many liters of juice are in nine three-liter jars?? Answer: 9 • 3 = 27 liters. juice in nine three-liter jars

8.The wire was cut into three pieces, six meters each. What was the length of the wire? Answer: 3 • 6 = 18 м. wire length

A bag of coffee costs 3 rubles. What is the cost of nine bags of coffee

Planted 3 rows of currant bushes, 8 bushes per row, and 18 gooseberries. How many bushes were planted? Answer: 1) 3 • 8 = 24 bushes. currants 2) 24 18 = 42 bushes. total

There are four children at each of the three tables in the school cafeteria. Five of them had lunch and left. How many children are left? Answer: 1) 3 • 4 = 12 children. total 2) 12. 5 = 7 children. remaining

into how many squares is each quadrilateral divided?? Count in different ways. Answer: Yellow quadrilateral 1 way: 1 1 1 1 1 1 1 1 1 1 = 12 squares 2 way: 3 3 3 3 = 12 squares 3 way: 4 4 4 = 12 squares 4 way: 3 • 4 = 12 squares Method 5: 4 • 3 = 12 squares

Blue quadrilateral 1 way: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 = 27 squares 2 way: 3 3 3 3 3 3 3 3 3 = 27 squares 3 way: 9 9 9 = 27 squares 4 way: 3 • 9 = 27 squares 5 way: 9 • 3 = 27 squares

Using the multiplication table for 3, divide. 21 : 3 12 : 3 27 : 3 24 : 3 18 : 3 9 : 3 3 3 : 3 15 : 3 Answer: 21 : 3 = 7 18 : 3 = 6 12 : 3 = 4 9 : 3 = 3 27 : 3 = 9 3 : 3 = 1 24 : 3 = 8 15 : 3 = 5

Do the steps. 2 • 3 3 • 9 7 • 2 2 • 9 12 : 2 27 : 3 2 : 2 18 : 2 Answer: 2 • 3 = 6 12 : 2 = 6 3 • 9 = 27 27 : 3 = 9 7 • 2 = 14 2 : 2 = 1 2 • 9 = 18 18 : 2 = 9

What is the result if you multiply 0 by 3 and divide the result by 3? Answer: 0 • 3 = 0 0 : 3 = 0

24 athletes are lined up in 3 rows equally. How many athletes are there in one row of? Answer: 24 : 3 = 8. in one row

A 27-meter long ribbon was cut into three equal pieces. How many meters of ribbon are in each piece? Answer: 27 : 3 = 9 meters. of the tape in each piece of

On each window sill in the apartment were placed 3 pots with houseplants. How many windows are there in the apartment if there are 12 pots? Answer: 12 : 3 = 4 windows. in an apartment

Petya can arrange his tin soldiers in 3 rows of 4 soldiers or in 2 rows of. soldiers. What number should be written instead of dots? Answer: 3 • 4 = 12 soldiers. a total of 12 : 2 = 6 soldiers. 2 rows Instead of dots the number should be written 6.

3 schoolchildren made 4 spatulas each in 2 days for the kindergarten. How many spades in total did the schoolchildren? Carefully read the text of the problem. Is there an extra given? Which is ? Answer: the excess given: in 2 days 3 • 4 = 12 aspen mushrooms. of the total number of plums made by the students

Masha found 27 mushrooms. One third of them are aspenberries. How many mountain aspen mushrooms did Masha find?? Answer: 27 : 3 = 9 aspen mushrooms. Masha found

Draw one segment of length 6 cm and another segment such that it is a third of its length. Answer: 6 : 3 = 2 cm. length of the second segment

We collected 18 cups of currants: a third of this number are blackcurrant cups, 9 cups of white currants and the rest are red currant cups. How many cups of red currants were picked? Answer: 1) 18 : 3 = 6 glasses. with black currants 2) (18. 6). 9 = 12. 9 = 3 glasses. with red currants

On the plate there are plums. Olya took 5 plums. This is a third of all plums. How many plums were on the plate? Answer: 5 • 3 = 15 plums. were on the plate

Compute. (19 2) : 3 (18 : 2). 9 (5 • 2) 46 (96. 87) • 2 (15 : 3) • 2 (3 • 4) : 2 Answer: (19 2) : 3 = 21 : 3 = 7 (5 • 2) 46 = 10 46 = 56 (15 : 3) • 2 = 5 • 2 = 10 (18 : 2). 9 = 9. 9 = 0 (96. 87) • 2 = 9 • 2 = 18 (3 • 4) : 2 = 12 : 2 = 6

What part of each figure is painted? Answer: 1) one half 2) one third 3) one third

If there are different ways of measuring how many squares each shape is divided into, do the following. Answer: Green shape 1 way: 1 1 1 1 1 1 1 1 1 1 1 = 13 squares 2 way: (3 3 3) (2 2) = 9 4 = 13 squares 3 way: 3 (5 5) = 3 10 = 13 squares 4 way: (3 • 3) (2 • 2) = 9 4 = 13 squares 5 way: 3 (5 • 2) = 3 10 = 13 squares

Blue shape 1 way: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 = 17 squares 2 way: (2 4 4) (4 3) = 10 7 = 17 squares 3 way: (4 4 4) (5 4) = 8 9 = 17 squares 4 way: (2 3) (4 • 3) = 5 12 = 17 squares 5 way: (4 • 3) 5 = 12 5 = 17 squares 6 way: (4 • 4) 1 = 17 squares

Draw a heptagon and color it. Answer:

Draw two circles: the length of the radius of one circle is 6 cm and the length of the other is 2 cm shorter. Consider two cases: 1) the centers of the circles do not coincide; 2) the centers of the circles are the same point. Answer: 6. 2 = 4 cm. radius of the second circle 1)

Masha embroidered two flowers on the handkerchief and folded the handkerchief in half. Are the florets symmetrical with respect to the braids of the folding line? Answer: If you bend the handkerchief along the scythe line of the fold, the flowers should match. a) are not symmetrical b) are symmetrical

If there are errors, correct them. 5 • 2 = 10 7 • 2 = 18 16 : 2 = 7 8 • 2 = 14 2 • 0 = 0 18 : 2 = 6 6 • 2 = 16 3 • 1 = 1 2 : 2 = 1 2 • 2 = 4 4 • 2 = 8 0 : 2 = 0 Answer: 8 • 2 = 16 7 • 2 = 14 16 : 2 = 8 6 • 2 = 12 3 • 1 = 3 18 : 2 = 9

Calculate. (6 4) : 2 (5 4) • 2 (2 • 3) • 3 (9. 5) • 2 (16. 8) : 2 (24 : 3) : 2 (10. 8) • 0 (5 5) : 2 0 : (9 • 3) Answer: (6 4) : 2 = 10 : 2 = 5 (9). 5) • 2 = 4 • 2 = 8 (10. 8) • 0 = 2 • 0 = 0 (5 4) • 2 = 9 • 2 = 18 (16. 8) : 2 = 8 : 2 = 4 (5 5) : 2 = 10 : 2 = 5 (2 • 3) • 3 = 6 • 3 = 18 (24 : 3) : 2 = 8 : 2 = 4 0 : (9 • 3) = 0 : 27 = 0

Find the sum and difference of the numbers: 6 and 4, 60 and 40, 30 and 5, 70 and 0. Answer: 6 4 = 10 6. 4 = 2 60 40 = 100 60. 40 = 20 30 5 = 35 30. 5 = 25 70 0 = 70 70. 0 =70

How many units must be added to each of the numbers: 15, 12, 18, 16, 19, to get 20? Answer: To find out how many units you must add to get 20, you must subtract each of the numbers from 20: 15, 12, 18, 16, 19. 20. 15 = 5 5 15 = 20 20. 12 = 8 8 12 = 20 20. 18 = 2 2 18 = 20 20. 16 = 4 4 16 = 20 20. 19 = 1 1 19 = 20

Name half of each number: 16, 8, 4, 2. Answer: 16 : 2 = 8 8 : 2 = 4 4 : 2 = 2 2 : 2 = 1

What can Dasha buy having 50? Answer: having 50, Dasha can buy: a bear, 15 30 = 45. the value of the book and the disk

When half the twine has been cut off, 6 m remain in the skein. How many meters of twine were in the skein?? What is the right answer: 3 m, 12 m. Answer: 1) 6 • 2 = 12 meters. the length of all the twine

The length of the segment is 1 dm 8 cm. What is the length of half of this segment?? Answer: 1 dm 8 cm = 18 cm 18 : 2 = 9 cm. the length of half of this segment

Which number does each machine add or subtract?? Answer: 20. 15 = 5 15 5 = 20 The machine added the number 5

30 = 10 40. 10 = 30 The machine subtracted the number 10

The department store sold 12 carpets, which was 7 more than they sold palaces. How many palases were sold? Change the text of the problem so that instead of the word more we use the word less. Solve the problems. Answer: 12. 7 = 5 palas. sold

The department store sold 12 rugs, which was 7 less than they sold palases. How many palases were sold? 12 7 = 19 palases. sold

The distance between neighboring strokes of a pole is 1 dm. How deep is the creek at this location?? Answer: The distance between adjacent strokes of the pole is 1 dm, and the water is at the 3. So the depth of the stream is 3 dm.

Compare the examples in each column. How are they similar and how are they different?? Make a guess about the values of these expressions: are they equal?? 72. (24 26) (48 49). 39 (72. 24). 26 48 (49. 39) Check your guess: do the math. Answer: In both columns in the first and second examples, the numbers are the same, but the order of operations is different because the brackets are in different places. The answer would be the same. 72. (24 26) = 72. 50 = 22 (48 49). 39 = 97. 39 = 58 (72. 24). 26 = 48. 26 = 22 48 (49. 39) = 48 10 = 58

1) To number 8, from the right side write the digit 0. How much greater is the resulting number than number 8? 2) To number 5, add the digit 9 on your left. By how much is the number 5 less than the resulting number? Answer: 1) 80 80. 8 = 72 80 is greater than 8 by 72

Calculate the perimeter of a triangle if one side is 7 cm long, the second is 13 cm long, and the third side is 3 cm shorter than the second. Answer: 1) 13. 3 = 10 cm. length of the third side 2) 7 13 10 = 30 cm. the perimeter of the triangle

Look at the drawing. Write down the notation of points: lying on a circle; belonging to a circle; lying outside a circle; not belonging to a circle. Answer: circle. is a closed line. A circle. is a figure that is bounded by a circle. Points lying on the circle: A, K. Circle points: A, C, K, O. The points lying outside the circle: C, D, M. Points that are not on the circle: D, M.

Name all the **triangles** you see in the drawing. Name some other shapes. Answer: **Triangles**: ABC, ABD, CBD, KBM, KBO, MBO. Quadrilaterals: AKMC, AKOD, CMOD.

Petya went to the forest to pick mushrooms. Petya left home at 12 o’clock. He returned home at 17 h. How many hours did Petya walk for mushrooms? How many hours it took Petya to go there and back, if it took him 4 hours in the forest? Answer: 1) 17.12 = 5 hours. Petya went mushroom picking 2) 5. 4 = 1 hour. it took to go there and back

Copy the **square** onto chequered paper and cut it out. Cut the resulting figure into two parts and make a triangle out of them. Answer: For this purpose cut a square on a diagonal, you will receive two small **triangles**. And make one big triangle out of them.

## Geometric problems (for cutting)

We pay your attention that according to the Federal law N 273-FZ “About education in the Russian Federation” in the organizations which are carrying out educational activity, training and education of students with disabilities is organized both together with other students, and in separate classes or groups.

“Relevance of creating school reconciliation/mediation services in educational organizations”

A certificate and a tuition discount for each participant

Exercise: Geometric problems (for cutting)

Development of students’ creative skills

Development of attention, memory, independent and teamwork skills

development of mental imagination, ingenuity, and “wit”

Today’s geometric (cutting) problems will be related to one seemingly simple geometric figure.

The main merit of the square was its use as a convenient unit of area. Indeed, squares are very convenient to mask flat areas, and say, circles do not do this without holes and overlaps. Often mathematicians say “squaring” instead of “finding the area”.

Thus, the problem of finding the area of a circle is called the problem of the square of a circle. The square is the protagonist of the Pythagoras theorem.

Cut a square piece of paper into 20 equal triangles and make 5 equal squares.

A cross made of five squares needs to be cut into such parts from which a single square could be made.

A square contains 16 cells. Divide a square into two equal parts so that the cutting line runs along the sides of the squares.

Cut a 7×7 square into five pieces and rearrange them so that you have three squares: 2×2, 3×3, and 6×6.

Cut a square into 4 pieces of equal shape and size so that each piece contains exactly one shaded square.

To divide a square into smaller squares of the same area is very simple: just draw a grid of equally spaced straight lines parallel to the sides of the square. The number of squares obtained will be a **square**, yes, yes! This is why the product of two identical numbers was called a square. Is it possible to cut a square into several squares that do not have identical squares??

This question remained unsolved for a long time. Many even prominent mathematicians believed that such a dissection was impossible. But in 1939 a subdivision of the square into 55 different squares was constructed. In 1940, two ways were found to divide the square into 28 different squares, followed by 26 squares, and in 1948, a division into 24 different squares was obtained. In 1978 a partition of 21 different squares was found and it was proved that a partition into a smaller number of different squares could no longer be found.

## Distant Olympiad

Divide two opposite sides of the square into 3 cm segments. Connecting the ends of the resulting segments, we obtain 1005 identical rectangles. Draw a diagonal in each rectangle to obtain 2010 identical triangles.

The smallest four-digit number is 1000, divide 1002 by 6 without the remainder, and we get 167, taking into account the remainder of 5, then we get:

Divide 1007 by 6 and you get 167 and the remainder is 5

- On the bookshelf stand the volumes in the following order: 1,2,6,10,3,8,4,7,9 and Arrange them in order from first to tenth, but you can only take two adjacent volumes and put them together in another place (without separating them). Complete the problem by rearranging only 3 pairs.

Pour water into a 9-liter bucket and pour it into a 5-liter bucket, then only 4 liters remain in the 9-liter bucket. Next, the five-liter bucket needs to be emptied, 4 liters poured into it. Take the 9-liter bucket from the lake again and pour into the five-liter bucket until it is full (you can only pour out 1 liter). This leaves 8 liters in the 9-liter bucket.

- Write down 19 sevens in a row: 77777. Put arithmetic signs “” or “-” between some of the digits, so that the result will be the number 2009.

Begin to break down the 19 sevens into amounts:

777777 is 222, and we need 287

that’s 11 six times 6 = 66 sevens

22266=288 and we have to add 287, so one seven must be subtracted

- The inner chambers of the sultan consist of 100 identical
**square**rooms arranged in a 10 × 10 square. If two rooms share a wall, there must be exactly one door. How many doors are in the palace?

All horizontal interior walls (top view) will divide by 10 from left to right and 9 from top to bottom, for a total of 90, and vice versa, also 90.

## An Intelligence Task: Where did the extra empty square come from?

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This is not really an optical illusion, but an interesting task. The squares of the painted-in areas are of course equal (32 cells), but what is visually seen as triangles 135 is not in fact so and has different areas (S135 = 32.5 cells). That is, the mistake, disguised in the problem condition, is that the initial figure is named a triangle (in fact, it is a concave 4-corner). This is clearly visible in the figure-scheme below. the “hypotenuse” of the upper and lower figures pass through different points: (8,3) at the top and (5,2) at the bottom. The secret is in the properties of blue and red triangles. This is easy to check by calculations.

The ratios of the lengths of the corresponding sides of the blue and red triangles are not equal to each other (2/3 and 5/8), so these triangles are not similar and therefore have different angles at the corresponding vertices. Let us call the first figure, which is a concave quadrilateral, and the second figure, which is a concave octagon, a pseudo-triangle. If the lower sides of these pseudotriangles are parallel, then the hypotenuses in both pseudotriangles 135 are actually broken lines (the upper figure creates an inward bend and the lower figure creates an outward bend). If we overlay the upper and lower figures 135 on each other then a parallelogram is formed between their “hypotenuse” and contains the “extra” area. In the figure-scheme this parallelogram is given in correct proportions.

The acute angle in this parallelogram is arcctg46 0°118.2. To such an angle the minute hand moves in 12.45 seconds in an efficient watch. Exactly on such magnitude the obtuse angle in the considered parallelogram differs from the unfolded one. Visually, such a negligible difference is invisible, but it is well visible in the animation.

According to Martin Gardner, this problem was invented by amateur New York illusionist Paul Curry in 1953. However, the principle behind it was already known in the 1860s. You may notice that the side lengths of the figures in this problem (2, 3, 5, 8, 13) are consecutive Fibonacci numbers.

## How to make a square out of 3 triangles

Quick patchwork techniques greatly simplify and accelerate the work of the masters. Over the years, needlewomen have come up with a huge number of different ways to simplify their work. About the most popular and we will tell.

At the Anglo-American school of quilting all measurements are made in inches (2,5 cm). While a square is a whole block unit, quilting **triangles** are divided into half square **triangles** (half square triangles), and quarter square **triangles** (quarter square triangles). The distinction is in the direction of the grain. Therefore, some triangles are cut out two triangles into squares, while others are cut out four triangles.

When using the quick triangle sewing method, it is important to correctly calculate the size of the workpiece and the allowance.

Half **square** triangles: Cut a **square** of the same size as the side of the triangle 2 allowance for a 6mm seam allowance on the diagonal seam allowance 8.4mm. The size of the seam allowance does not depend on the size of the triangles and squares, it depends on the size of the seam allowance. If you are used to using a different allowance: 7 mm or 5 mm. then the value of the extra seam allowance is calculated by the formula : (allowance x allowance). Calculate the square root of the result and divide it by 2. (Rationale: since according to the Pythagoras formula the square of the hypotenuse is equal to the sum of the squares of the cathetuses, and in this case the hypotenuse is two allowances, then the cathetus is an extra allowance, we take the hypotenuse, take the **square** root of it and divide by two ).

Example: You need 2 parts of two triangles square with a size of 5x5cm. You will use a 6 mm ( 0,6 cm ) allowance to cut two 7 x 7 cm ( 5 0,6 0,6 0,84 ) squares. Fold them right sides together. For convenience, you can mark the diagonal along which you will be stitching with a marker or chalk. Sew two stitches along the diagonal (6mm from the diagonal). Make a diagonal cut, between the stitches. Unfold them and press out the seam allowance.

### The basics on a Speed square

If you need more than two composite squares. You can not cut the fabric into squares, but use a strip width of several squares or a large **square** (see the diagram below).

Square of four triangles: Cut a square of the same size as the hypotenuse (long) of the triangle 2 seam allowances of 6 mm 2 diagonal seam allowances of 8.4 mm. Example: You need 2 parts of four triangles square 10 x 10 cm (and an allowance of 0.6 cm). (see the diagram below. You will need to cut two squares of size 10 (0.6 2) (0.84 2) = 12.9 cm. т.е. 12.9 x 12.9 cm. Fold the squares face-to-face, mark the diagonals and scythe the seam lines. Sew scythe stitches along one diagonal. Cut the squares in half along this diagonal. Unfold details and iron out. Fold the resulting squares face-to-face again, combining the scythe line seams. Sew scythe seams along the second diagonal. Cut along this diagonal squares in half. Unfold the parts and iron out. Create two squares of 4 triangles each.

Quick **triangles** half square. Quick method of assembling triangles into squares.

Layout diagram of the square for this method.

Cut out squares of two colors. Fold them right sides together. Sew two stitches along the diagonal ( 6mm away from the diagonal).

Make a cut along the diagonal. Unfold the pieces and fold out the seam allowances

Square of two triangles a few squares in a row.

Quarter square triangles: quarter-**square** Note that here the grain of the thread runs along the diagonal of the future square (along the hypotenuse of the triangle)

Cut out the square. Stitch along the sides of the square marked in red. Cut the square diagonally. Open and seal parts.

Make squares with a valley thread along the diagonal of the square.

In the same pattern, stitch along the blue. Or along the green lines, and cutting the square along the diagonals, you will get composite triangles of quarter-squares with a valley on the short side (cathetus) of the triangle.

Example of assembling a 3/3 square blocks of half-squares and quarter-squares.

Squares of 4 triangles (quarter-squares).

Cut out squares, fold, mark the braid fishing line seams and the braid fishing line cut. Stitch. Cut.

Re-stitch seam allowance. Fold them face-to-face, matching the scythe line seams. Mark the scythe line seams and scythe line the cut. Stitch. Cut.

We got two squares with bows (a square of quarter-squares).

I want to introduce another flat geometric constructor, however, which can be calledPythagoras-2.

A 10×10 cm square is cut as shown, resulting in 9 geometric shapes: 4 large triangles, 2 small triangles, one middle triangle, a square and a rectangle.

Before working with the patterns, the children do several tasks with specific shapes.

Assignment 1. Take 2 large triangles and a square. Make: a rectangle, a triangle, and 2 different quadrilaterals, one of which is a trapezoid.

Task 2. Take 2 small triangles and the middle one. Make: a square, a triangle, a rectangle, and 2 different quadrilaterals, one of which is a trapezium.

Task 3. Take 2 small triangles, a medium triangle, and a large triangle. Make: a square, a triangle, a rectangle, and 2 different quadrangles, one of which is a trapezoid.

Next, the children build different patterns, gradually moving on to unbroken patterns.

The following constructor is also author’s work and by analogy with Tangram I call itTregram, because it is obtained by cutting equilateral triangle. Playing exercises with such a construction set can be carried out for filling in and composing planar images from sets of geometric shapes. An equilateral triangle made of cardboard (side length 20 cm, each divided into 5 equal parts of 4 cm) is cut into 10 pieces, as shown in the figure.

The result is 4 small triangles, 2 rhombuses, a trapezoid, a parallelogram, a large triangle, and a hexagon.

__At the first stage__ children get acquainted with all the parts of the constructor, making them out of triangles and other small shapes:

By attaching two triangles to each other, the children make a rhombus. 2. By adding another triangle to a rhombus, children get a trapezoid, which can also be made with three triangles. 3. By adding another triangle to a trapezoid, the children make a parallelogram. This same figure can also be made up of other small shapes. 4. Then the children draw their own conclusions about which pieces they can be made of by putting the smaller pieces on the larger ones.

__On the second stage__ Children fill the inner space of the figures-silhouettes on the sheets, using all the parts of the construction set.

__At the third stage__ children compose flat images from exploded patterns with a gradual transition to partially exploded patterns.

__In the fourth stage__ children model the images according to their own ideas.

Similar to Nikitin’s cubes, I made another flat constructor. This set consists of 15 5×5 cm squares:

8 squares are painted half diagonally;

__At the first stage__ For our game tasks we use only 4 squares, painted half of them, and make all the images by the model from them only.

__In the second stage__ children make images of 9 squares using the entire set.

Goal. Teach children to make geometric figures out of a certain number of sticks, using the technique of fitting one figure taken as the basis with another.

Material: The children have counting sticks on their desks, the board, and chalk for this and the next class.

Course of work. 1. The teacher asks the children to count five sticks, check and put them in front of each other. Then says, “Tell me how many sticks it would take to make a triangle with each side equal to one stick. How many sticks would it take to make two such triangles?? You only have 5 sticks, but you need to make two equal triangles from them. Think about how to do this and make it.”.

After most of the children complete the task, the teacher asks them to tell how they should make two equal triangles out of five sticks. Draws the children’s attention to the fact that the task may be done in different ways. They should draw different ways of doing it. When explaining, use the expression “attached to one triangle the other triangle from below” (left, etc.).д.), and in explaining the solution of the problem use also the expression “attached to one triangle the other triangle using only 2 sticks”.

Make 2 equal squares out of 7 sticks (the teacher clarifies beforehand what geometric figure can be made out of 4 sticks). Gives the task: count 7 sticks and think how to make two equal squares on the table.

After completing the task, consider different ways of fitting one square to another, the teacher sketches them on the board.

Analysis questions: “How did you make 2 equal squares out of 7 sticks? What did I do first, what did I do next? How many sticks make one square?? Of how many sticks did he add the second **square** to it? How many sticks it took to make 2 equal squares?”

Goal. Make shapes by fitting in. See and show the new, resulting figure; use the expression: “added to one figure with another”, to think about practical actions.

Workflow. The teacher invites the children to remember what kind of shapes they made using the fitting technique. Informs them what they will be doing today to learn how to make new, more complex shapes. Gives tasks:

Count 7 sticks and think how they can make 3 equal **triangles**.

After completing the task, the teacher invites all the children to make 3 **triangles** in a row so that a new quadrilateral figure emerges This solution the children sketch with chalk on the board. The teacher asks them to show 3 separate triangles, a quadrilateral and a triangle (2 figures), a quadrilateral.

* Figure. 2 Making shapes from triangles*

Make 4 equal **triangles** from 9 sticks. Think about how to do it, tell them, then do the task.

After that, the teacher invites the children to draw the shapes they made with chalk on the board and tell them about the sequence of the task.

Analysis questions: “How do you make 4 equal triangles from 9 sticks?? Which of the triangles formed the first? What the resulting shapes are and how many?”

The teacher, clarifying the children’s answers, says: “You can start with any triangle and then add others to it on the right or left, above or below.

Goal. Exercise children to independently find ways to make shapes based on thinking about the solution beforehand.

How to do it. The kindergarten teacher asks the children questions: “How many sticks can be used to make a square, each side of which is equal to one stick? 2 squares? (from 8 and 7). How will you make 2 squares out of 7 sticks?”

Count 10 sticks and make 3 equal squares. Think about how to make a **square** and tell.

As the teacher calls several children to sketch their figures on the board and tell them the sequence of making them. Invites all the children to make a figure from three equal squares arranged in a row, horizontally. Draws one on the board and says, “Look at the board. Here is a picture of how this problem can be solved in different ways. You can add another **square** to one and then a third. (Showing.) And you can make a rectangle out of 8 sticks, then divide it into 3 equal squares with 2 sticks. (Shows.) Then asks questions: “What shapes are made and how many sticks are needed to make a square?? How many rectangles did we get?? Find them and show them.”.

Make a square and 2 equal **triangles** with 5 sticks. Tell first and then make up.

When doing this task, children tend to make the mistake of making 2 triangles in the learned way of fitting, resulting in a quadrilateral. Therefore the teacher draws the children’s attention to the condition of the problem, the need to make a square, and offers leading questions: “How many sticks do we need to make a **square**?? Since you have sticks? Is it possible to make up by fitting 1 triangle to another? How to make? Which figure should we start with?” After completing the task, children explain how they did: you have to make a square and divide it with 1 stick into 2 equal triangles.

Goal. Exercise children in the ability to express a presumptive solution, guessing.

Activity. 1. Make a square and 4 triangles from 9 sticks. Think and say how to make. (Several children make suggestions.)

If the children have difficulty, the teacher advises: “Recall how you made a square and two triangles out of five sticks. Think about and guess how to do the task. Whoever is the first to solve the problem will draw the resulting figure on the board.”.

After completing and sketching the answer, the teacher asks all the children to make the same figures for themselves

* Figs. 3 Making shapes out of triangles*

Analyze questions: “What geometric shapes are made?? How many **triangles**, squares, quadrilaterals? How did they make?? How is it more convenient, faster to compose?”

Make two squares out of 10 sticks, a small one and a big one.

Make five triangles from nine sticks.

If necessary, during the second and third tasks, the tutor gives leading questions and advice: “First think, then make up. Do not repeat mistakes, look for a new solution. Does the problem say about the size of the triangles? These are problems for ingenuity, you have to think, guess how to solve the problem.

So, in the initial period of teaching 5-year-old children to solve simple wit problems, they independently, mostly by practicing with sticks, look for a way of solving. In order to develop their ability to plan the course of thought, it is necessary to offer children to express preliminary reasoning or to combine it with practical probes, to explain the way and the way of solution.

Several kinds of solutions for the first group of problems are possible. Once children have learned the way of fitting shapes given the commonality of sides, it is very easy and quick for them to give 2-3 solutions. Each figure differs from the previous one in its spatial position. At the same time children master a way to build given figures by dividing the resulting geometric figure into several (a quadrilateral or **square** into 2 triangles, a rectangle into 3 squares).

For 5-6-year-old children, the more complicated reconstructing figure problems should begin with those in which a certain number of sticks must be removed to change the figure, and the simplest reconstructing sticks.

The process of children searching for solutions to problems of the second and third groups is much more complicated than that of the first group. For this purpose it is necessary to remember and comprehend character of transformation and result (what figures should turn out and how many) and constantly during search of the decision to correlate it with the expected or already carried out changes. In the process of solving it, a visual and thinking analysis of the problem is needed, as well as the ability to imagine possible changes in the figure.

So, in the process of solving tasks children should master such thinking operations of task analysis, as a result of which one can imagine various transformations, check them, then, discarding incorrect ones, look for and try new ways of solving them. Teaching should be aimed at forming in children the ability to think through the moves mentally, to solve the task completely or partly in their mind, and to limit practical probes.

In what order should we offer the second and third groups of the second and third group of problems on ingenuity to 5-6 year-old children??

* Fig. 4*

* Fig. 5*

* Fig. 6*

* Fig. 7*

* Figure. 8*

* Fig. 9*

* Fig. 10*

* Fig. 11*

* Fig. 12*

* Fig. 13*

These and other analogous knowledge problems are characterized by the fact that the transformation needed for solving them leads to a change in the number of squares the given figure is made of (problems 2, 5, etc.).), changing their size (problems 6, 7), and modifying shapes, e.g. transforming squares into rectangles in problem 1.

During the lessons, in order to guide the search activities of children, teachers use different techniques to foster a positive attitude towards a long and persistent search, but at the same time, the rapid response, abandoning the developed way of searching. Children’s interest is supported by a desire to achieve success, which requires active work of the mind.

## Olympiad, logical, and fun problems in mathematics. Cutting tasks

Here is a selection of entertaining and developing geometric cutting problems for math tutors and teachers of different extracurricular activities and circles. The tutor’s goal in using such tasks in his classes is not only to interest students in interesting and effective combinations of cells and shapes, but also to form a sense of lines, angles, and shapes. The problem set is mainly aimed at children in grades 4-6, although its use with even high school students is not excluded. Exercises require students to have a high and gregarious concentration and are great for developing and training visual memory. Recommended for mathematics tutors involved in preparing students for entrance exams for mathematics schools and classes that place special demands on the level of independent thinking and creativity. The level of problems corresponds to the level of the entrance exams in the Lyceum of the second school (the second mathematical school), the Faculty of Mechanics and Mathematics of Moscow State University, the Kurchatov school, etc.

Math Tutor’s Note: Some of the solutions to the problems, which you can view by clicking on the corresponding index, show only one possible example of cutting. I quite admit that you can get some other correct combination do not be afraid of this. Check your solution carefully and if it satisfies the condition you may continue with the next problem.

1) Try to cut the figure shown in the figure into 3 equal parts:

Math Tutor Hint: The little figures look a lot like the letter T View Math Tutor Solution

2) Now cut this figure into 4 equal shaped pieces:

Math tutor tip: It is easy to guess that the smaller pieces will have 3 squares, but there are not many pieces with 3 squares. There are only two kinds: a corner and a rectangle 1×3. View the math tutor’s solution:

3) Cut this figure into 5 pieces of equal shape:

Math Tutor Hint: Find the number of cells that make up each of these shapes. These pieces look like the letter G. View math tutor solution

4) Now you need to cut a figure of ten cells into 4 unequal rectangles (or squares).

Directions from the math tutor: Pick a rectangle and then try to fit three more squares into the remaining squares. If it doesn’t work, change the first rectangle and try again. View Math Tutor Solutions

5) The task is more complicated: you need to cut the figure into 4 different shapes (not necessarily into rectangles).

Math Tutor Hint: Draw all kinds of differently shaped shapes separately first (there will be more than four) and repeat the method of trying variations as in the previous problem. View math tutor solution:

6) Cut this figure into 5 shapes of four differently shaped cells so that only one green cell is shaded in each.

Math Tutor Hint: Try starting cutting from the top edge of a given shape and you will immediately understand how to proceed. View the math tutor solution:

7) Based on the previous problem. Find how many differently shaped shapes there are that have exactly four cells? Figures can be twisted and turned, but they cannot be lifted from the surface on which they lie. That is, the two figures given would not be considered equal because they cannot be made from each other by rotation.

Math Tutor Hint: Study the solution to the previous problem and try to imagine the different positions of these shapes as they rotate. It is not difficult to guess that the answer in our problem is the number 5 or more. (In fact, even more than six.). There are a total of 7 types of described shapes. View Math Tutor Solution

8) Cut a square of 16 squares into 4 equally shaped pieces so that each of the four pieces has exactly one green cell.

Math Tutor Hint: The appearance of the little figures is not a square or a rectangle, or even a corner of four cells. So what kind of shapes should we try to cut? View Math Tutor’s Solution

9) Cut the drawn figure into two parts so that it is possible to make a square from the received parts.

Math tutor tip: There are a total of 16 cells in the shape, so the square will be sized 4×4. And somehow fill in the box in the middle as well. How to do this? Maybe by some kind of shift? Then since the length of the rectangle is equal to an odd number of squares, the cut should not be made by a vertical cut, but by a broken line. So that the top part is cut from one side of the middle cell, and the bottom part from the other. View Math Tutor Solutions

10) Cut a rectangle of size 4×9 into two pieces so that you can make a square out of them.

Math Tutor Hint: There are a total of 36 cells in a rectangle. So the square will be a size 6×6. Since the long side consists of nine squares, three of them must be cut off. How does this cut go next?? View Math Tutor Solution

11) The five-cell cross shown in the picture needs to be cut (one can cut the cells themselves) into such pieces that a square can be made of.

Math tutor tip: It is clear that no matter how we cut along the lines of the squares we won’t get a square, since there are only 5 squares. This task is the only one in which it is allowed to cut outside the cells. However, it would still be good to leave them as a reference point. For example, it is worth noting that we somehow need to remove the recesses we have, namely, in the inner corners of our cross. How would this be done? For example, by cutting some protruding triangles from the outer corners of the cross. See Math Tutor’s Solution: Comment: Cut as shown in the picture and put the blue triangles into the empty areas shown by the purple triangles.

Alexander Kolpakov. Tutor in Mathematics Moscow, Strogino.

Cool site! Thank you for the most interesting problems on the entire Internet with answers!

## Generalize to arbitrary triangles

It is easy to generalize the above to the case of an arbitrary triangle by drawing three families of parallel lines (in each family, the lines are parallel to one side and divide each of the other two sides by *n* equal parts). Now it is easy to understand how to divide any triangle into *n* like it, where *n* 5. The partition into 6 triangles similar to the original triangle is obtained if we make a drawing similar to Fig. 2, *а*, and erase the extra mowing line (Fig. 3, *а*). Partitioning into 8 similar ones (Fig. 3, *б*) is obtained from Fig. 2, *б*, and t. д., for any even *n*, greater than 5. If *n.* is odd, then after erasing it is necessary to take one more step: divide the “upper” triangle by the middle lines into four equal. In figure 3, *в* is shown such a partition into 11 triangles.

But not every triangle can be dissected into 2, 3 or 5 triangles similar to the original triangle.

a) Let an arbitrary triangle be given *ABC*. Let’s draw a middle line *MN* parallel to the side of *AB*, and in the resulting triangle *CMN* Let us omit the height *CD*. In addition, let’s lower on the line *MN* perpendiculars *AK* и *BL*. Then it is easy to see that ∆*AKM* = ∆*CDM* и ∆*BLN* = ∆*CDN* as right triangles in which the corresponding pair of sides and pair of angles are equal.

Hence the method of dissecting this triangle and then rearranging the pieces. Exactly, let us make cuts along the segments *MN* и *CD*. After that, we rearrange the triangles *CDM* и *CDN* in the place of the triangles *AKM* и *BLN* respectively, as shown in Fig. 2. We have obtained a rectangle *AKLB*, as required in the problem.

Note that this method will not work if one of the angles *CAB* or *CBA*. obtuse. This is because in this case the altitude *CD* is not inside the triangle *CMN*. But this is not too terrible: if we draw the middle line parallel to the longest side of the original triangle, we will drop the height from the obtuse angle in the cut-off triangle, and it will necessarily lie inside the triangle.

b) Let the rectangle *ABCD*, whose sides *AD* и *AB* are equal to *a* и *b* respectively, whereby *a* *b*. Then the area of the square we want to end up with should be equal to *ab*. Hence, the length of the side of the square is √*ab*, which is less than *AD*, but greater than *AB*.

Let’s construct a square *APQR*, is equal to the sought one, so that the point *B* lies on the segment *AP*, and the point *R*. on the segment *AD*. Let *PD* intersects the segments *BC* и *QR* at points *M* и *N* respectively. Then it is easy to see that the triangles *PBM*, *PAD* и *NRD* are similar, and furthermore, *BP* = (√*ab*. *b*) и *RD* = (*a*. √*ab*). So,

Therefore, ∆*PBM* = ∆*NRD* on the two sides and the angle between them. It is also not difficult to deduce from here the equalities *PQ* = *MC* и *NQ* = *CD*, which means that ∆*PQN* = ∆*MCD* also for the two sides and the angle between them.

From all the above reasoning, the method of dissection. Exactly, first we adjourn on the sides of *AD* и *BC* segments *AR* и *CM*, whose lengths are equal to √*ab* (on how to draw segments of the form √*ab*, see. the problem “Right polygons”. box in the “Solution” section). Next, reconstruct the perpendicular to the segment *AD* at the point *R*. Now all that remains is to cut the triangles *MCD* и *NRD * and rearrange them as shown in Fig. 3.

Note that in order to use this method, it is required that the point *M* is inside the segment *BK* (otherwise not the whole triangle *NRD* is contained inside the rectangle *ABCD*). That is, it is necessary that

If this condition is not satisfied, then we must first make this rectangle wider and less long. To do this, simply cut it in half and rearrange the pieces as shown in Fig. 4. It is clear that after carrying out this operation, the ratio of the larger side to the smaller side will decrease by a factor of four. And so, by doing it a sufficiently large number of times, we end up with a rectangle to which the dissection with Fig. 3.

c) Consider the two given squares *ABCD* и *DPQR*, to each other so that they intersect on the side of *CD* of the smaller square and had a common vertex *D*. Let us assume that *PD* = *a* и *AB* = *b*, and, as we have already noted, *a* *b*. Then on the side of *DR* of the larger square can be considered such a point *M*, that *MR* = *AB*. According to the Pythagoras theorem.

Let the lines passing through the points *B* и *Q* parallel to the straight lines *MQ* и *BM* respectively intersect at *N*. Then the quadrilateral *BMQN* is a parallelogram and since all its sides are equal it is a rhombus. But ∆*BAM* = ∆*MRQ* on three sides, from which it follows (given that the angles of *BAM* и *MRQ* lines) that. Thus, *BMQN*. square. And since its area is equal (*a* 2 *b* 2 ), this is exactly the **square** we need to get.

In order to proceed to the cut, it remains to note that ∆*BAM* = ∆*MRQ* = ∆*BCN* = ∆*NPQ*. After this, what needs to be done becomes obvious: You need to cut the triangles *BAM* и *MRQ* and rearrange them as shown in Fig. 5.