Problems about the area of the square and much more

Such an amazing and familiar square. It is symmetric about its center and axes drawn along the diagonals and through the centers of the sides. And to search for the area of a square or its volume does not make much effort. Especially if the length of his side is known.

A few words about the figure and its properties

The first two properties are related to the definition. All sides of the figure are equal to each other. After all, the square is the right quadrilateral. And he necessarily all sides are equal and the angles have the same value, namely - 90 degrees. This is the second property.

The third is related to the length of the diagonals. They are also equal to each other. And they intersect at right angles and at the points of the middle.

A formula in which only the side length is used

First about the designation. For the length of the side, it is customary to choose the letter "a". Then the square of the square is calculated by the formula: S = a ² .

It is easily obtained from the one known for the rectangle. In it, the length and width are multiplied. For a square, these two elements are equal. Therefore, the square of this one quantity appears in the formula.

The formula in which the length of the diagonal appears

It is the hypotenuse in the triangle, the legs of which are the legs of the figure. Therefore, we can use the formula of the Pythagorean theorem and derive an equality in which the side is expressed through the diagonal.

Carrying out such simple transformations, we obtain that the square of the square through the diagonal is calculated by the following formula:

S = d 2/2 . Here the letter d denotes the diagonal of the square.

Formula around the perimeter

In this situation, it is necessary to express the side through the perimeter and substitute it in the area formula. Since there are four sides of the figure, the perimeter will have to be divided by 4. This will be the value of the side, which can then be substituted into the initial one and the area of the square.

The formula in general form is as follows: S = (P / 4) ² .

Settlement tasks

No. 1. There is a square. The sum of its two sides is 12 cm. Calculate the area of the square and its perimeter.

Decision. Since the sum of the two sides is given, you need to know the length of one. Since they are the same, the known number must simply be divided into two. That is, the side of this figure is 6 cm.

Then its perimeter and area can be easily calculated from the above formulas. The first is 24 cm, and the second is 36 cm ² .

Answer. The perimeter of the square is 24 cm, and its area is 36 cm ² .

No. 2. Find out the area of the square with a perimeter of 32 mm.

Decision. It is enough to substitute the perimeter value into the above formula. Although you can first know the side of the square, and then its area.

In both cases, the actions will first be divided, and then raised to the power. Simple calculations lead to the fact that the area of the presented square is 64 mm ² .

Answer. The required area is 64 mm ² .

The side of the square is 4 dm. Dimensions of the rectangle: 2 and 6 dm. Which of the two figures has more area? How much?

Decision. Let the side of the square be denoted by the letter a ₁ , then the length and width of the rectangle a ₂ and ₂ . To determine the area of a square, the value of a _{1 is} to be squared, and the rectangle is multiplied by a ₂ and by ₂ . It is not difficult.

It turns out that the square of the square is 16 dm ² , and the rectangle is 12 dm ² . Obviously, the first figure is larger than the second. This is despite the fact that they are equal, that is, they have the same perimeter. To check, you can count the perimeters. At the square, the side should be multiplied by 4, it will be 16 dm. At the rectangle, fold the sides and multiply by 2. There will be the same number.

In the task it is still necessary to answer, on how many areas differ. To do this, a smaller number is subtracted from a larger number. The difference is 4 dm ² .

Answer. The areas are 16 dm ² and 12 dm ² . At the square it is more by 4 dm ² .

The problem of proof

Condition. The square is constructed on the leg of an isosceles right triangle . To its hypotenuse height is built on which another square is built. Prove that the area of the first is twice as large as the second.

Decision. We introduce the notation. Let the cathete be equal to a, and the height to the hypotenuse, x. The area of the first square is S ₁ , the second is S ₂ .

The square of the square constructed on the leg is easy to calculate. It turns out to be a ² . With the second value, everything is not so simple.

First, you need to know the length of the hypotenuse. For this, the formula of the Pythagorean theorem is useful. Simple transformations lead to the following expression: a√2.

Since the height in an isosceles triangle drawn to the base is also a median and a height, it divides a large triangle into two equal isosceles right triangles. Therefore, the height is half the hypotenuse. That is, x = (a√2) / 2. Hence it is easy to find out the area S ₂ . It is obtained as a 2/2.

Obviously, the recorded values differ exactly by a factor of two. And the second is a number of times smaller. Q.E.D.

An unusual puzzle - tangram

It is made from a square. It is necessary to cut it into various shapes according to certain rules. The total parts should be 7.

The rules assume that during the game all the resulting details will be used. Of these, you need to make other geometric shapes. For example, a rectangle, a trapezoid or a parallelogram.

But it's even more interesting when silhouettes of animals or objects are obtained from the pieces. And it turns out that the area of all derived figures is equal to that of the initial square.