What does the sum of the circumferences of the 4 small squares have to do with the circumference of

The sum of the circumferences of the four small squares is 2 times the circumference of the large square.

Hypothesis: A large square is a combination of four small and small squares of the same kind.

So let's start with a small square side length of a.

then the side length of the large square is 2a.

The sum of the perimeters of the small square is 4a*4=16a.

The sum of the circumferences of a large square is 2a*4=8a.

That is, the sum of the circumferences of the four small squares is twice the circumference of the large square.

Determination theorem1. Diagonal.

The equal diamond shape is a square.

2. A diamond with a right angle is a square.

3. Rectangles with diagonal lines perpendicular to each other are squares.

4. A group of rectangles with equal adjacent sides is a square.

5. A group of parallelograms with equal adjacent sides and one angle being a right angle.

It's a square. 6. A parallelogram with diagonal lines perpendicular to each other and equal to each other is a square.

7. Quadrilaterals with equal diagonals and perpendicular bisects are squares.

8. A group of quadrilaterals with equal adjacent sides and three right angles is a square.

9. A quadrilateral that is both a diamond and a rectangle is a square.

Hypothesis. The big square is made up of four small and identical small squares.

Then let's set the side length of the small square to a

then the side length of the large square is 2a.

The sum of the perimeters of the small square is 4a*4=16a.

The sum of the circumferences of a large square is 2a*4=8a.

That is, the sum of the circumferences of the four small squares is twice the circumference of the large square.

The sum of the circumferences of the 4 small squares is 2 times the circumference of the large square.

The solution is as follows: Assuming that the side length of the small square is 1, then the sum of the perimeters of the 4 small squares = (1 4) 4 = 16.

When placed into a large square, its side length = 1 2 = 2, and its circumference = 2 4 = 8;

Because 16 8 = 2, the sum of the circumferences of the 4 small squares is 2 times the circumference of the large square.

Each small square has two sides that coincide together and is no longer counted as the side length of the newly arranged large square.

The sum of the circumferences of the four small squares is twice the circumference of the large square.

The small ones are much larger than the big ones and twice as big.

Let the side length be small x

The small one is 16x in total

Big 8x

Four small squares make up a large square, then their circumference is twice the relationship, and four small are twice the size of the large.

Hypothesis. The big square is made up of four small and identical small squares.

Then let's set the side length of the small square to a

then the side length of the large square is 2a.

The sum of the perimeters of the small square is 4a*4=16a.

The sum of the circumferences of a large square is 2a*4=8a.

That is, the sum of the circumferences of the four small squares is twice the circumference of the large square.

The sum of the circumferences of the four small squares is 2 times the circumference of the large square.

Hypothesis. The big square is made up of four small and identical small squares.

Then let's set the side length of the small square to a

then the side length of the large square is 2a.

The sum of the perimeters of the small square is 4a*4=16a.

The sum of the circumferences of a large square is 2a*4=8a.

That is, the sum of the circumferences of the four small squares is twice the circumference of the large square.

Perimeter formulaCircle: c = d = 2 r (d is the diameter, r is the radius, ) triangle. The circumference of c = a+b+c (abc is the three sides of a triangle) quadrilateral.

c = a + b + c + d (abcd is the length of the sides of the quadrilateral) rectangle: c = 2 (a + b) (a is long, b is wide) square: c = 4a (a is the length of the side of the square).

Polygon. c = sum of all edge lengths.

Circumference of the fan: c = 2r + n r 180 (n = central angle.

Angle) =2r+kr (k=radians.

First of all, you have to tell the relationship between the big square and the small square, otherwise there is no answer to this question.

I assume that the big squares are made up of four small and identical small squares.

So let's start with a small square side length of a.

then the side length of the large square is 2a.

The sum of the perimeters of the small square is 4a*4=16a.

The sum of the circumferences of a large square is 2a*4=8a.

Answer: The sum of the circumferences of the four small squares is 2 times the circumference of the large square.

Summary. The circumference of 1 small square = half the circumference of a large square. Circumference of 4 small squares = Circumference of 2 large squares.

A large square is divided into four identical small squares, what does the sum of the circumferences of the four small squares have to do with the circumference of the large square? Can you explain it.

The circumference of 1 small square = half the circumference of a large square. Circumference of 4 small squares = Circumference of 2 large squares.

Look at the **, dear. The circumference of a large square is 8, and the circumference of a small square is 16.

So: the circumference of 1 small square = half the circumference of a large square. Circumference of 4 small squares = Circumference of 2 large squares.

The circumference of a rectangle is longer than the circumference of a square" is incorrect and lacks prerequisites, and the circumference of the square is the shortest, and if it is a pentagonal with the same area, it is a regular pentagon.

The perimeter is the shortest.

A rectangle, also called a rectangle, is a flat figure that is a parallelogram with a right angle at one angle.

A rectangle is also defined as a parallelogram with all four corners at right angles. A square is a special rectangle with four sides of equal length. Rectangle diagonal.

The square of the length is the sum of the two sides of the long flat front and the closed square.

30 cm. The analysis process is as follows:

1) The circumference of the large square is 24 cm, the circumference of the small square is 12 cm, and the sum of the circumference of the large square and the small square is 12 + 24 = 36 cm.

2) When the two are put together, the overlapping part is 2 times the side length of the small square, and the total circumference is subtracted by 2 times the side length of the small square.

3) The circumference of the small square is 12 cm, and the side length of the small square = 12 4 = 3 cm.

4) Therefore, the circumference of the piece together = 36-3 2 = 30 cm.

What is the circumference of a large square and a small square of a positive liquid shed?

If only one point touches when the big square and a small square are put together, then the total circumference is the sum of the perimeter of the large square and the perimeter of the small square, which is the maximum; If all the sides of the small square are contained within the sides of the large square, then the total perimeter is the sum of the perimeter of the large square and the perimeter of the small square, minus twice the length of the sides of the small square, which is the minimum. So the figure is put together, and the circumference of Kai buried dust is between the maximum and minimum values.

<> total circumference of the 4 small square-shaped bridges is twice the circumference of the large square.

First of all, you have to tell the relationship between the big square chain auspicious bridge shape and the small square, otherwise there is no answer to this question.

I assume that the big squares are made up of four small and identical small squares.

So I'll start with a small banquet square with a side length of a

then the side length of the large square is 2a.

The sum of the perimeters of the small square is 4a*4=16a.

The sum of the circumferences of a large square is 2a*4=8a.

Answer: The sum of the circumference of the four small squares is twice the circumference of the shed of the large square.