Such an amazing and familiar square. It is symmetrical about its center and the axes drawn along the diagonals and through the centers of the sides. And to look for the square area or its volume is not at all difficult. Especially if the length of its 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, a square is a regular quadrangle. Moreover, he must have all sides 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 also turn out to be equal to each other. Moreover, they intersect at right angles and at midpoints.
Formula that uses only side length
First about the designation. For the length of the side, it is customary to choose the letter “a”. Then the square area is calculated by the formula: S = a 2 .
It is easily obtained from the one that is known for the rectangle. In it, the length and width are multiplied. In a square, these two elements are equal. Therefore, a square of this one quantity appears in the formula.
The formula in which the length of the diagonal appears
She is a hypotenuse in a triangle whose legs are the sides 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.
Having carried out such simple transformations, we find that the area 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.
Perimeter formula
In such a situation, it is necessary to express the side through the perimeter and substitute it in the area formula. Since the figure has four identical sides, 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 count the square area.
The formula in general terms looks like this: 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, it is necessary to find out 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 are easily calculated using 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 .
Number 2. Find out the area of the square with a perimeter equal to 32 mm.
Decision. Simply substitute the perimeter value in the above formula. Although you can first find out the side of the square, and only then its area.
In both cases, the actions will first be divided, and then exponentiation. 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 .
No. 3. The side of the square is 4 dm. Rectangle sizes: 2 and 6 dm. Which of these two figures has a larger area? How much?
Decision. Let the side of the square be marked with the letter a 1 , then the length and width of the rectangle a 2 and 2 . To determine the area of the square, the value of a 1 is supposed to be squared, and the rectangle is multiplied by a 2 and 2 . This is not difficult.
It turns out that the area of the square is 16 dm 2 , and the rectangle is 12 dm 2 . Obviously, the first figure is larger than the second. This despite the fact that they are equal, that is, they have the same perimeter. For verification, you can count the perimeters. At the square, the side needs to be multiplied by 4, we get 16 dm. Fold the sides of the rectangle and multiply by 2. It will be the same number.
In the problem, it is still necessary to answer how many areas differ. For this, the smaller is subtracted from the larger number. The difference is 4 dm 2 .
Answer. The areas are 16 dm 2 and 12 dm 2 . At the square, it is 4 dm 2 more .
Proof task
Condition. A square is built on the leg of an isosceles right triangle . To his hypotenuse, the height is built at 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 leg be equal to a, and the height drawn to the hypotenuse, x. The area of the first square is S 1 , the second is S 2 .
The area of a square built on a leg is easy to calculate. It turns out to be equal to 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 this expression: a√2.
Since the height in the isosceles triangle drawn to the base is also the median and the height, it divides the large triangle into two equal isosceles right-angled triangles. Therefore, the height is equal to half the hypotenuse. That is, x = (a√2) / 2. From here it is easy to know the area S 2 . It turns out equal to a 2/2.
Obviously, the recorded values differ exactly twice. And the second one is this number of times less. Q.E.D.
Unusual puzzle - tangram
It is made from a square. It must, according to certain rules, be cut into various figures. Total parts should be 7.
The rules assume that during the game all the resulting details will be used. From them you need to make other geometric shapes. For example, a rectangle, a trapezoid or a parallelogram.
But even more interesting is when silhouettes of animals or objects are made from pieces. Moreover, it turns out that the area of all derivatives is equal to that of the initial square.