How can you find the area of a triangle. Calculating the area of a polygon from the coordinates of its vertices The area of a triangle from the coordinates of the vertices formula
The triangle is one of the most common geometric shapes, which we are already familiar with in elementary school. The question of how to find the area of a triangle is faced by every student in geometry lessons. So, what are the features of finding the area of \u200b\u200ba given figure can be distinguished? In this article, we will consider the basic formulas necessary to complete such a task, and also analyze the types of triangles.
Types of triangles
You can find the area of a triangle in completely different ways, because in geometry there is more than one type of figure containing three angles. These types include:
- obtuse.
- Equilateral (correct).
- Right triangle.
- Isosceles.
Let's take a closer look at each of the existing types of triangles.
Such a geometric figure 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 triangle, as the name implies, all angles are acute and add up to 180°.
Such a triangle is also very common, but 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 need to find the remaining elements), sometimes you 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 (for example, 15° or even 3°).
To find the area of a triangle of this type, you need to know some of the nuances, which we will talk about next.
Regular and isosceles triangles
A regular polygon is a figure that includes n angles, in which all sides and angles are equal. This is the right triangle. Since the sum of all the angles of a triangle is 180°, each of the three angles is 60°.
The right triangle, due to its property, is also called an equilateral figure.
It is also worth noting that only one circle can be inscribed in a regular triangle and only one circle can be circumscribed around it, and their centers are located at one point.
In addition to the equilateral type, one can also distinguish an isosceles triangle, which differs slightly from it. In such a triangle, two sides and two angles are equal to each other, and the third side (to which equal angles adjoin) 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
A right triangle is so named because one of its angles is a right angle, i.e. 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 of its 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 triangle BAC with hypotenuse AC and legs AB and BC.
To find the area of a triangle with a right angle, you need to know the numerical values of its legs.
Let's move on to the formulas for finding the area of \u200b\u200ba given figure.
Basic formulas for finding the area
In geometry, two formulas can be distinguished that are suitable for finding the area of most types of triangles, namely for acute-angled, obtuse-angled, regular and isosceles triangles. Let's analyze each of them.
By side and height
This formula is universal for finding the area of the 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 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 an acute-angled triangle ACB, you need to 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 a triangle in this way. For example, to use this formula for an obtuse-angled triangle, you need to continue one of its sides and only then draw a height to it.
In practice, this formula is used more often than others.
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 a triangle. That is, the formula under consideration can be easily deduced from the previous one. Its wording looks like this:
S = 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.
And now let's move on to other formulas that are suitable only for exceptional types of triangles.
Area of a right triangle
In addition to the universal formula, which includes the need to draw a height in a triangle, the area of \u200b\u200ba triangle containing a right angle can be found from its legs.
So, the area of a triangle containing a right angle is half the product of its legs, or:
where a and b are the legs of a right triangle.
right triangle
This type of geometric figures is distinguished by the fact that its area can be found with the specified value of only one of its sides (since all sides of a regular triangle are equal). So, having met with the task of “find the area of a triangle when the sides are equal”, you need to use the following formula:
S = A 2 *√3 / 4,
where A is the side of an equilateral triangle.
Heron's formula
The last option for finding the area of a 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 = √p (p - a) (p - b) (p - c),
where a, b and c are the sides of the given triangle.
Sometimes the task is given: "the area of \u200b\u200ba regular triangle is to find the length of its side." In this case, you need to use the formula already known to us for finding the area of a regular triangle and derive from it the value of the side (or its square):
A 2 \u003d 4S / √3.
Exam problems
There are many formulas in the tasks of the GIA in mathematics. In addition, quite often it is necessary to find the area of a 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 for finding the area:
So, after studying the formulas presented in the article, you will not have problems finding the area of a triangle of any kind.
The coordinate method, proposed in the 17th century by the French mathematicians R. Descartes (1596-1650) and P. Fermat (1601-1665), is a powerful tool that allows you to translate 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 a polygon by the coordinates of its vertices in a rectangular coordinate system.
Area of a triangle
Theorem 1. If is the area of the triangle
then the equality
will be called the determinant of the area of a triangle.
Proof. 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
Having performed algebraic transformations
we get that:
In equality (1.9) there is a determinant of the area, so there is a minus sign in front of the expression, since.
Let's show that. Indeed, here
(the area of a rectangle with a base and a height is greater than the sum of the areas of the 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 quadrant.
Using the concept of a module, equalities (1.9) and (1.10) can be written as follows:
Remark 1. We have derived formula (1.8) by considering the simplest arrangement of vertices shown in Figures 1.30 and 1.31; however, formula (1.8) is true for any arrangement of vertices.
Consider the case depicted in Figure 1.32.
Therefore, after performing simple geometric transformations:
we get again what, where
Area of an n-gon
The polygon can be convex or non-convex, the numbering order of the vertices is considered negative if the vertices are numbered in a clockwise direction. A polygon that does not have self-intersecting sides will be called simple. For a simple n-gon the following holds
Theorem 2. If is the area of a prime n-gon, where, then the equality
will be called the determinant of the area of a prime n-gon.
Proof. Two cases are possible.
Case 1. n-gon - convex. Let us prove formula (1.11) by the method of mathematical induction.
For it has already been proved (Theorem 1). We assume that it is valid for n-square; let us prove that it remains valid for a 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 - non-convex.
In any non-convex n-gon, one can draw a diagonal lying inside it, and therefore the proof of case 2 for a non-convex n-gon is similar to the proof for a convex n-gon.
Remark 2. Expressions for are not easy to remember. Therefore, to calculate its values, it is convenient to write in the column the coordinates of the first, second, third, ..., n th and again first vertices n-gon and carry out multiplication according to the scheme:
The signs in the column (1.12) must be arranged as indicated in the scheme (1.13).
Remark 3. When compiling column (1.12) for a triangle, you can start from any vertex.
Remark 4. When compiling column (1.12) for n-gon () you must follow the sequence of writing out the coordinates of the vertices n-gon (from which vertex to start the bypass is indifferent). So the area calculation n-gon should begin with the construction of a "rough" drawing.