The method of coordinates, proposed in the 17th century by the French mathematicians R. Descartes (1596-1650) and P. Fermat (1601-1665), is a powerful apparatus for translating geometric concepts into algebraic language. This method is based on the concept of a coordinate system. We will consider the calculation of the area of \u200b\u200ba polygon by the coordinates of its vertices in a rectangular coordinate system.
Area of \u200b\u200ba triangle
Theorem 1... If is the area of \u200b\u200bthe triangle
then the equality
will be called the determinant of the area of \u200b\u200ba triangle.
Evidence. Let the vertices of the triangle be located in the first coordinate quarter. Two cases are possible.
Case 1... The direction (or, or) of the location of the vertices of the triangle coincides with the direction of movement of the end of the hour hand (Fig. 1.30).
Since the figure is a trapezoid.
Similarly, we find that
By performing algebraic transformations
we get that:
In equality (1.9) the determinant of the area, so there is a minus sign in front of the expression, since.
Let us show that. Indeed, here
(the area of \u200b\u200ba rectangle with base and height is greater than the sum of the areas of rectangles with bases and heights,; (Fig. 1.30), whence
Case 2... The indicated directions in case 1 are opposite to the direction of movement of the end of the hour hand (fig. 1.31)
since the figure is a trapezoid, and
where. Indeed, here
The theorem is proved when the vertices of the triangle are located in the first coordinate quarter.
Using the concept of a modulus, equalities (1.9) and (1.10) can be written as follows:
Remark 1... We have derived formula (1.8), considering the simplest arrangement of vertices shown in Figures 1.30 and 1.31; however, formula (1.8) is true for any location of the vertices.
Consider the case shown in Figure 1.32.
Therefore, by performing simple geometric transformations:
we get again that, where
Area of \u200b\u200ban n-gon
A polygon can be convex or non-convex; the vertex numbering order is considered negative if the vertices are numbered in the direction of the end of the clockwise movement. A polygon that does not have self-intersection of the sides will be called simple. For simple it is n-gon the following is true
Theorem 2... If is the area of \u200b\u200ba prime n-gon, where, then the equality
will be called the determinant of the area of \u200b\u200bthe prime n-gon.
Evidence. Two cases are possible.
Case 1. n-gon is convex. Let us prove formula (1.11) by the method of mathematical induction.
For it has already been proved (Theorem 1). Suppose it is valid for n-gon; let us prove that it remains valid for convex ( n+1) -gon.
Let's add one more vertex to the polygon (fig. 1.33).
Thus, the formula is valid for ( n+1) -gon, and hence the conditions of mathematical induction are satisfied, i.e., formula (1.11) for the case of a convex n-gon is proved.
Case 2. n-gon is nonconvex.
In any non-convex n-gon one can draw a diagonal lying inside it, and therefore the proof of Case 2 for a nonconvex n-gon is similar to the proof for the convex n-gon.
Remark 2... Expressions for are not easy to remember. Therefore, to calculate its values, it is convenient to write down the coordinates of the first, second, third, ..., n-th and again the first peaks n-gon and multiply according to the scheme:
The signs in column (1.12) must be arranged as indicated in the scheme (1.13).
Remark 3... When composing column (1.12) for a triangle, you can start at any vertex.
Remark 4... When composing column (1.12) for n-gon (), the sequence of writing out the coordinates of the vertices must be observed n-gon (it doesn't matter which vertex to start the traversal). Therefore, calculating the area n-gon should be started by building a "rough" drawing.
The triangle is one of the most common geometric shapes that we get to know already in elementary school. Every student faces the question of how to find the area of \u200b\u200ba triangle in geometry lessons. So, what features of finding the area of \u200b\u200ba given figure can be distinguished? In this article, we will look at the basic formulas required to complete such a task, and also analyze the types of triangles.
Types of triangles
You can find the area of \u200b\u200ba triangle in completely different ways, because geometry contains more than one type of shape containing three corners. These types include:
- Obtuse.
- Equilateral (correct).
- Right triangle.
- Isosceles.
Let's take a closer look at each of the existing types of triangles.
This geometric shape is considered the most common in solving geometric problems. When it becomes necessary to draw an arbitrary triangle, this option comes to the rescue.
In an acute-angled triangle, as the name implies, all angles are acute and add up to 180 °.
Such a triangle is also very common, but it is somewhat less common than an acute-angled one. For example, when solving triangles (that is, you know several of its sides and angles and you need to find the remaining elements), you sometimes need to determine whether the angle is obtuse or not. Cosine is a negative number.
In the value of one of the angles exceeds 90 °, so the remaining two angles can take small values \u200b\u200b(for example, 15 ° or even 3 °).
To find the area of \u200b\u200ba triangle of this type, you need to know some of the nuances, which we will talk about further.
Regular and isosceles triangles
A regular polygon is a figure that includes n corners, in which all sides and angles are equal. This is a regular triangle. Since the sum of all the angles of the triangle is 180 °, each of the three angles is 60 °.
A regular triangle, due to its property, is also called an equilateral figure.
It is also worth noting that only one circle can be inscribed into a regular triangle, and only one circle can be described around it, and their centers are located at one point.
In addition to the equilateral type, you can also distinguish an isosceles triangle, slightly different from it. In such a triangle, two sides and two angles are equal to each other, and the third side (to which equal angles are adjacent) is the base.
The figure shows an isosceles triangle DEF, the angles D and F of which are equal, and DF is the base.
Right triangle
The right-angled triangle is so named because one of its corners is straight, that is, it is equal to 90 °. The other two angles add up to 90 °.
The largest side of such a triangle, lying opposite an angle of 90 °, is the hypotenuse, while the other two sides are the legs. For this type of triangles, the Pythagorean theorem is applicable:
The sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.
The figure shows a right-angled triangle BAC with hypotenuse AC and legs AB and BC.
To find the area of \u200b\u200ba triangle with a right angle, you need to know the numerical values \u200b\u200bof its legs.
Let's move on to the formulas for finding the area of \u200b\u200bthis figure.
Basic formulas for finding the area
In geometry, two formulas can be distinguished that are suitable for finding the area of \u200b\u200bmost types of triangles, namely for acute-angled, obtuse, regular and isosceles triangles. Let's analyze each of them.
By side and height
This formula is universal for finding the area of \u200b\u200bthe figure we are considering. To do this, it is enough to know the length of the side and the length of the height drawn to it. The formula itself (half of the product of the base and the height) is as follows:
where A is the side of the given triangle and H is the height of the triangle.
For example, to find the area of \u200b\u200ban acute-angled triangle ACB, multiply its side AB by the height CD and divide the resulting value by two.
However, it is not always easy to find the area of \u200b\u200ba triangle this way. For example, to use this formula for an obtuse triangle, it is necessary to continue one of its sides and only then draw the height to it.
In practice, this formula is used more often than others.
On two sides and a corner
This formula, like the previous one, is suitable for most triangles and in its meaning is a consequence of the formula for finding the area by the side and height of the triangle. That is, the considered formula can be easily derived from the previous one. Its wording looks like this:
S \u003d ½ * sinO * A * B,
where A and B are the sides of the triangle and O is the angle between sides A and B.
Recall that the sine of an angle can be viewed in a special table named after the outstanding Soviet mathematician V.M. Bradis.
Now let's move on to other formulas that are suitable only for exceptional types of triangles.
Area of \u200b\u200ba right triangle
In addition to the universal formula, which includes the need to draw the height in a triangle, the area of \u200b\u200ba triangle containing a right angle can be found by its legs.
So, the area of \u200b\u200ba triangle containing a right angle is half the product of its legs, or:
where a and b are the legs of a right triangle.
Regular triangle
This type of geometric shapes differs in that its area can be found at the specified value of only one of its sides (since all sides of a regular triangle are equal). So, faced with the problem "find the area of \u200b\u200ba triangle when the sides are equal", you need to use the following formula:
S \u003d A 2 * √3 / 4,
where A is the side of an equilateral triangle.
Heron's formula
The last option for finding the area of \u200b\u200ba triangle is Heron's formula. In order to use it, you need to know the lengths of the three sides of the figure. Heron's formula looks like this:
S \u003d √p (p - a) (p - b) (p - c),
where a, b and c are the sides of this triangle.
Sometimes the problem is given: "the area of \u200b\u200ba regular triangle - find the length of its side." In this case, you need to use the formula already known to us for finding the area of \u200b\u200ba regular triangle and derive the value of the side (or its square) from it:
A 2 \u003d 4S / √3.
Exam tasks
In the problems of the GIA in mathematics, there are many formulas. In addition, it is often necessary to find the area of \u200b\u200ba triangle on checkered paper.
In this case, it is most convenient to draw the height to one of the sides of the figure, determine its length by cells and use the universal formula to find the area:
So, after studying the formulas presented in the article, you will not have problems finding the area of \u200b\u200ba triangle of any kind.