How to find the diamond area?

How to find the diamond area? To give an answer, you first need to figure out what we think is a rhombus.

Firstly, it is a quadrangle.Secondly, it has all four equal sides. Thirdly, its diagonals at the intersection point are perpendicular. Fourthly, these diagonals are divided into equal parts by the point of intersection. Fifth, the same diagonals divide the corners of the diamond into two equal parts. Sixth, in total, the two corners that adjoin one side are the unfolded angle, that is, 180 degrees. And to put it simply, a diamond is a beveled square.

If we take a square whose sides are fastenedmovable, and it is easy to pull it for two opposing angles, then the square will lose its rectangularity and turn into a rhombus. Therefore, a rhombus with straight angles - this is the real square.

The first to introduce the concept of the diamond Hero and Papp of Alexandria, the mathematicians of ancient Greece. The word "rhombus" from Greek can be translated as "tambourine".

To find the area of ​​the diamond, it is worth considering that the rhombus is a parallelogram. And the area of ​​the parallelogram can be found by multiplying the base, that is, the side, and the height.

To prove this, it follows thatLower the perpendicular from the vertices of the upper corners of the rhombus. For example, given a QWER rhombus. From the vertices of the upper corners Q and W, the perpendiculars QT and WY are omitted. And the perpendicular QT will drop to the side RE, and the perpendicular WY will be on the extension of this side.

Thus, we have a new quadrilateral QWYT with parallel sides and right angles, which, based on the above, can be bravely called a rectangle.

The area of ​​this rectangle is multiplied by the side and height. Now we need to prove that the area of ​​the resulting rectangle in the area corresponds to the given condition of the rhombus.

Considering the results obtained with the additionalthe construction of triangles QYR and WET, we can say that they are equal in shape and hypotenuse. After all, the legs in triangles are drawn perpendiculars, which at the same time are also sides of the resulting rectangle. And the hypotenuse is the side of the rhombus.

The rhombus consists of the sum of the area of ​​the triangle QYR andtrapezoid QYEW. The resulting rectangle consists of the same trapezoid QYEW and the triangle WET, whose area is equal to the area of ​​the triangle QYR. Hence the conclusion suggests itself: the value of the square of the diamond QWER corresponds to the square of the rectangle QWYT.

Now it becomes clear how to find the area of ​​the diamond on the side and its height: they need to be multiplied.

You can find the area of ​​the diamond, knowing the angle of the diamond and the side. It is only necessary to know what the sine of the angle is equal to, and multiply it by the doubled side. You can find the sine by using the calculator or the Bradys table.

Sometimes, when talking about how to find the area of ​​a diamond, use the sine of the angle and the radius of the circle inscribed in it, which is necessarily the maximum.

However, most often calculate the area of ​​the diamond through the diagonal. From this formula it follows that the area is equal to the semi-product of the diagonals.

It is quite simple to prove this by considering twotriangle QWE and ERQ, which turned out when carrying out in a diamond one diagonal. These triangles are equal on three sides or on the base and two adjacent corners.

Having carried out in the rhombus the second diagonal, we getheight in these triangles, since the diagonals intersect at a point X at an angle of 90 degrees. The area of ​​the triangle QWE is equal to the product QE, which is one diagonal, on WX - half of the second diagonal, divided by two.

Now to the question of how to find the area of ​​the rhombus, the answerClear: the expression should be doubled. For the convenience of algebraic reduction of this expression, one diagonal can be denoted by the letter z, and the second by the letter u. We get:

2 (z Х 1 / 2u: 2) = z Х 1 / 2u, which just comes out - a semi-product of diagonals.