The school curriculum provides for teaching children geometry from an early age. One of the most basic knowledge in this field is finding the area of ​​various shapes. In this article we will try to give all possible ways to obtain this value, from the simplest to the most complex.

The basis

The first formula that children learn at school involves finding the area of ​​a triangle through the length of its height and base. Height is a segment drawn from the vertex of the triangle at right angles to the opposite side, which will be the base. How to find the area of ​​a triangle using these quantities?

If V is the height and O is the base, then the area is S=V*O:2.

Another option for obtaining the desired value requires us to know the lengths of two sides, as well as the size of the angle between them. If we have L and M - the lengths of the sides, and Q - the angle between them, then you can get the area using the formula S=(L*M*sin(Q))/2.

Heron's formula

In addition to all the other answers to the question of how to calculate the area of ​​a triangle, there is a formula that allows us to obtain the value we need, knowing only the lengths of the sides. That is, if we know the lengths of all sides, then we do not need to draw the height and calculate its length. We can use the so-called Heron's formula.

If M, N, L are the lengths of the sides, then we can find the area of ​​the triangle as follows. P=(M+N+L)/2, then the value we need is S 2 =P*(P-M)*(P-L)*(P-N). In the end, all we have to do is calculate the root.

For a right triangle, Heron's formula is slightly simplified. If M, L are legs, then S=(P-M)*(P-L).

Circles

Another way to find the area of ​​a triangle is to use incircles and circumcircles. To get the value we need using an inscribed circle, we need to know its radius. Let's denote it "r". Then the formula by which we will carry out calculations will take the following form: S=r*P, where P is half of the sum of the lengths of all sides.

In a right triangle, this formula is slightly modified. Of course, you can use the one above, but it is better to use a different expression for calculations. S=E*W, where E and W are the lengths of the segments into which the hypotenuse is divided by the tangency point of the circle.

Speaking of the circumscribed circle, finding the area of ​​the triangle is also not difficult. By introducing the designation R as the radius of the circumscribed circle, you can obtain the following formula necessary to calculate the desired value: S= (M*N*L):(4*R). Where the first three quantities are the sides of the triangle.

Speaking about an equilateral triangle, through a number of simple mathematical transformations you can obtain slightly modified formulas:

S=(3 1/2 *M 2)/4;

S=(3*3 1/2 *R 2)/4;

S=3*3 1/2 *r 2 .

In any case, any formula that allows you to find the area of ​​a triangle can be changed in accordance with the data of the task. So all written expressions are not absolutes. When solving problems, reflect to find the most appropriate solution.

Coordinates

When studying coordinate axes, the tasks facing students become more complex. However, not so much as to panic. In order to find the area of ​​a triangle from the coordinates of the vertices, you can use the same, but slightly modified Heron's formula. For coordinates it takes the following form:

S=((x 2 -x 1) 2 *(y 2 -y 1) 2 *(z 2 -z 1) 2) 1/2.

However, no one forbids, using coordinates, calculating the lengths of the sides of a triangle and then, using the formulas written above, calculating the area. To convert coordinates to length, use the following formula:

l=((x 2 -x 1) 2 +(y 2 -y 1) 2) 1/2.

Notes

The article used standard notations for quantities that are used in most problems. In this case, the power "1/2" means that you need to extract the root of the entire expression under the brackets.

Be careful when choosing a formula. Some of them lose their relevance depending on the initial conditions. For example, the circumcircle formula. It is able to calculate the result for you in any case, but there may be a situation where a triangle with the given parameters may not exist at all.

If you are sitting at home and doing homework, then you can use an online calculator. Many sites provide the ability to calculate various quantities using given parameters, and it doesn’t matter which ones. You can simply enter the initial data into the fields, and the computer (website) will calculate the result for you. This way you can avoid mistakes made due to carelessness.

We hope our article answered all your questions regarding calculating the area of ​​a variety of triangles, and you will not have to look for additional information elsewhere. Good luck with your studies!

To determine the area of ​​a triangle, you can use different formulas. Of all the methods, the easiest and most frequently used is to multiply the height by the length of the base and then divide the result by two. However, this method is far from the only one. Below you can read how to find the area of ​​a triangle using different formulas.

Separately, we will look at ways to calculate the area of ​​specific types of triangles - rectangular, isosceles and equilateral. We accompany each formula with a short explanation that will help you understand its essence.

Universal methods for finding the area of ​​a triangle

The formulas below use special notation. We will decipher each of them:

• a, b, c – the lengths of the three sides of the figure we are considering;
• r is the radius of the circle that can be inscribed in our triangle;
• R is the radius of the circle that can be described around it;
• α is the magnitude of the angle formed by sides b and c;
• β is the magnitude of the angle between a and c;
• γ is the magnitude of the angle formed by sides a and b;
• h is the height of our triangle, lowered from angle α to side a;
• p – half the sum of sides a, b and c.

It is logically clear why you can find the area of ​​a triangle in this way. The triangle can easily be completed into a parallelogram, in which one side of the triangle will act as a diagonal. The area of ​​a parallelogram is found by multiplying the length of one of its sides by the value of the height drawn to it. The diagonal divides this conditional parallelogram into 2 identical triangles. Therefore, it is quite obvious that the area of ​​our original triangle must be equal to half the area of ​​this auxiliary parallelogram.

S=½ a b sin γ

According to this formula, the area of ​​a triangle is found by multiplying the lengths of its two sides, that is, a and b, by the sine of the angle formed by them. This formula is logically derived from the previous one. If we lower the height from angle β to side b, then, according to the properties of a right triangle, when we multiply the length of side a by the sine of angle γ, we obtain the height of the triangle, that is, h.

The area of ​​the figure in question is found by multiplying half the radius of the circle that can be inscribed in it by its perimeter. In other words, we find the product of the semi-perimeter and the radius of the mentioned circle.

S= a b c/4R

According to this formula, the value we need can be found by dividing the product of the sides of the figure by 4 radii of the circle described around it.

These formulas are universal, as they make it possible to determine the area of ​​any triangle (scalene, isosceles, equilateral, rectangular). This can be done using more complex calculations, which we will not dwell on in detail.

Areas of triangles with specific properties

How to find the area of ​​a right triangle? The peculiarity of this figure is that its two sides are simultaneously its heights. If a and b are legs, and c becomes the hypotenuse, then we find the area like this:

How to find the area of ​​an isosceles triangle? It has two sides with length a and one side with length b. Consequently, its area can be determined by dividing by 2 the product of the square of side a by the sine of angle γ.

How to find the area of ​​an equilateral triangle? In it, the length of all sides is equal to a, and the magnitude of all angles is α. Its height is equal to half the product of the length of side a and the square root of 3. To find the area of ​​a regular triangle, you need to multiply the square of side a by the square root of 3 and divide by 4.

Area of ​​a triangle - formulas and examples of problem solving

Below are formulas for finding the area of ​​an arbitrary triangle which are suitable for finding the area of ​​any triangle, regardless of its properties, angles or sizes. The formulas are presented in the form of a picture, with explanations for their application or justification for their correctness. Also, a separate figure shows the correspondence between the letter symbols in the formulas and the graphic symbols in the drawing.

Note . If the triangle has special properties (isosceles, rectangular, equilateral), you can use the formulas given below, as well as additional special formulas that are valid only for triangles with these properties:

• "Formula for the area of ​​an equilateral triangle"

Triangle area formulas

Explanations for formulas:
a, b, c- the lengths of the sides of the triangle whose area we want to find
r- radius of the circle inscribed in the triangle
R- radius of the circle circumscribed around the triangle
h- height of the triangle lowered to the side
p- semi-perimeter of a triangle, 1/2 the sum of its sides (perimeter)
α - angle opposite to side a of the triangle
β - angle opposite to side b of the triangle
γ - angle opposite to side c of the triangle
h a, h b , h c- height of the triangle lowered to sides a, b, c

Please note that the given notations correspond to the figure above, so that when solving a real geometry problem, it will be visually easier for you to substitute the correct values ​​in the right places in the formula.

• The area of ​​the triangle is half the product of the height of the triangle and the length of the side by which this height is lowered(Formula 1). The correctness of this formula can be understood logically. The height lowered to the base will split an arbitrary triangle into two rectangular ones. If you build each of them into a rectangle with dimensions b and h, then obviously the area of ​​these triangles will be equal to exactly half the area of ​​the rectangle (Spr = bh)
• The area of ​​the triangle is half the product of its two sides and the sine of the angle between them(Formula 2) (see an example of solving a problem using this formula below). Even though it seems different from the previous one, it can easily be transformed into it. If we lower the height from angle B to side b, it turns out that the product of side a and the sine of angle γ, according to the properties of the sine in a right triangle, is equal to the height of the triangle we drew, which gives us the previous formula
• The area of ​​an arbitrary triangle can be found through work half the radius of the circle inscribed in it by the sum of the lengths of all its sides(Formula 3), simply put, you need to multiply the semi-perimeter of the triangle by the radius of the inscribed circle (this is easier to remember)
• The area of ​​an arbitrary triangle can be found by dividing the product of all its sides by 4 radii of the circle circumscribed around it (Formula 4)
• Formula 5 is finding the area of ​​a triangle through the lengths of its sides and its semi-perimeter (half the sum of all its sides)
• Heron's formula(6) is a representation of the same formula without using the concept of semi-perimeter, only through the lengths of the sides
• The area of ​​an arbitrary triangle is equal to the product of the square of the side of the triangle and the sines of the angles adjacent to this side divided by the double sine of the angle opposite to this side (Formula 7)
• The area of ​​an arbitrary triangle can be found as the product of two squares of the circle circumscribed around it by the sines of each of its angles. (Formula 8)
• If the length of one side and the values ​​of two adjacent angles are known, then the area of ​​the triangle can be found as the square of this side divided by the double sum of the cotangents of these angles (Formula 9)
• If only the length of each of the heights of the triangle is known (Formula 10), then the area of ​​such a triangle is inversely proportional to the lengths of these heights, as according to Heron’s Formula
• Formula 11 allows you to calculate area of ​​a triangle based on the coordinates of its vertices, which are specified as (x;y) values ​​for each of the vertices. Please note that the resulting value must be taken modulo, since the coordinates of individual (or even all) vertices may be in the region of negative values

Note. The following are examples of solving geometry problems to find the area of ​​a triangle. If you need to solve a geometry problem that is not similar here, write about it in the forum. In solutions, instead of the "square root" symbol, the sqrt() function can be used, in which sqrt is the square root symbol, and the radical expression is indicated in parentheses.Sometimes for simple radical expressions the symbol can be used √

Task. Find the area given two sides and the angle between them

The sides of the triangle are 5 and 6 cm. The angle between them is 60 degrees. Find the area of ​​the triangle.

Solution.

To solve this problem, we use formula number two from the theoretical part of the lesson.
The area of ​​a triangle can be found through the lengths of two sides and the sine of the angle between them and will be equal to
S=1/2 ab sin γ

Since we have all the necessary data for the solution (according to the formula), we can only substitute the values ​​​​from the problem conditions into the formula:
S = 1/2 * 5 * 6 * sin 60

In the table of values ​​of trigonometric functions, we will find and substitute the value of sine 60 degrees into the expression. It will be equal to the root of three times two.
S = 15 √3 / 2

Answer: 7.5 √3 (depending on the teacher’s requirements, you can probably leave 15 √3/2)

Task. Find the area of ​​an equilateral triangle

Find the area of ​​an equilateral triangle with side 3 cm.

Solution .

The area of ​​a triangle can be found using Heron's formula:

S = 1/4 sqrt((a + b + c)(b + c - a)(a + c - b)(a + b -c))

Since a = b = c, the formula for the area of ​​an equilateral triangle takes the form:

S = √3 / 4 * a 2

S = √3 / 4 * 3 2

Answer: 9 √3 / 4.

Task. Change in area when changing the length of the sides

How many times will the area of ​​the triangle increase if the sides are increased by 4 times?

Solution.

Since the dimensions of the sides of the triangle are unknown to us, to solve the problem we will assume that the lengths of the sides are respectively equal to arbitrary numbers a, b, c. Then, in order to answer the question of the problem, we will find the area of ​​the given triangle, and then we will find the area of ​​the triangle whose sides are four times larger. The ratio of the areas of these triangles will give us the answer to the problem.

Below we provide a textual explanation of the solution to the problem step by step. However, at the very end, this same solution is presented in a more convenient graphical form. Those interested can immediately go down the solutions.

To solve, we use Heron’s formula (see above in the theoretical part of the lesson). It looks like this:

S = 1/4 sqrt((a + b + c)(b + c - a)(a + c - b)(a + b -c))
(see first line of picture below)

The lengths of the sides of an arbitrary triangle are specified by the variables a, b, c.
If the sides are increased by 4 times, then the area of ​​the new triangle c will be:

S 2 = 1/4 sqrt((4a + 4b + 4c)(4b + 4c - 4a)(4a + 4c - 4b)(4a + 4b -4c))
(see second line in the picture below)

As you can see, 4 is a common factor that can be taken out of brackets from all four expressions according to the general rules of mathematics.
Then

S 2 = 1/4 sqrt(4 * 4 * 4 * 4 (a + b + c)(b + c - a)(a + c - b)(a + b -c)) - on the third line of the picture
S 2 = 1/4 sqrt(256 (a + b + c)(b + c - a)(a + c - b)(a + b -c)) - fourth line

The square root of the number 256 is perfectly extracted, so let’s take it out from under the root
S 2 = 16 * 1/4 sqrt((a + b + c)(b + c - a)(a + c - b)(a + b -c))
S 2 = 4 sqrt((a + b + c)(b + c - a)(a + c - b)(a + b -c))
(see fifth line of the picture below)

To answer the question asked in the problem, we just need to divide the area of ​​the resulting triangle by the area of ​​the original one.
Let us determine the area ratios by dividing the expressions by each other and reducing the resulting fraction.

The triangle is a figure familiar to everyone. And this despite the rich variety of its forms. Rectangular, equilateral, acute, isosceles, obtuse. Each of them is different in some way. But for anyone you need to find out the area of ​​a triangle.

Formulas common to all triangles that use the lengths of sides or heights

The designations adopted in them: sides - a, b, c; heights on the corresponding sides on a, n in, n with.

1. The area of ​​a triangle is calculated as the product of ½, a side and the height subtracted from it. S = ½ * a * n a. The formulas for the other two sides should be written similarly.

2. Heron's formula, in which the semi-perimeter appears (it is usually denoted by the small letter p, in contrast to the full perimeter). The semi-perimeter must be calculated as follows: add up all the sides and divide them by 2. The formula for the semi-perimeter is: p = (a+b+c) / 2. Then the equality for the area of ​​the figure looks like this: S = √ (p * (p - a) * ( р - в) * (р - с)).

3. If you don’t want to use a semi-perimeter, then a formula that contains only the lengths of the sides will be useful: S = ¼ * √ ((a + b + c) * (b + c - a) * (a + c - c) * (a + b - c)). It is slightly longer than the previous one, but it will help out if you have forgotten how to find the semi-perimeter.

General formulas involving the angles of a triangle

Notations required to read the formulas: α, β, γ - angles. They lie opposite sides a, b, c, respectively.

1. According to it, half the product of two sides and the sine of the angle between them is equal to the area of ​​the triangle. That is: S = ½ a * b * sin γ. The formulas for the other two cases should be written in a similar way.

2. The area of ​​a triangle can be calculated from one side and three known angles. S = (a 2 * sin β * sin γ) / (2 sin α).

3. There is also a formula with one known side and two adjacent angles. It looks like this: S = c 2 / (2 (ctg α + ctg β)).

The last two formulas are not the simplest. It's quite difficult to remember them.

General formulas for situations where the radii of inscribed or circumscribed circles are known

Additional designations: r, R - radii. The first is used for the radius of the inscribed circle. The second is for the one described.

1. The first formula by which the area of ​​a triangle is calculated is related to the semi-perimeter. S = r * r. Another way to write it is: S = ½ r * (a + b + c).

2. In the second case, you will need to multiply all the sides of the triangle and divide them by quadruple the radius of the circumscribed circle. In literal expression it looks like this: S = (a * b * c) / (4R).

3. The third situation allows you to do without knowing the sides, but you will need the values ​​of all three angles. S = 2 R 2 * sin α * sin β * sin γ.

Special case: right triangle

This is the simplest situation, since only the length of both legs is required. They are designated by the Latin letters a and b. The area of ​​a right triangle is equal to half the area of ​​the rectangle added to it.

Mathematically it looks like this: S = ½ a * b. It is the easiest to remember. Because it looks like the formula for the area of ​​a rectangle, only a fraction appears, indicating half.

Special case: isosceles triangle

Since it has two equal sides, some formulas for its area look somewhat simplified. For example, Heron's formula, which calculates the area of ​​an isosceles triangle, takes the following form:

S = ½ in √((a + ½ in)*(a - ½ in)).

If you transform it, it will become shorter. In this case, Heron’s formula for an isosceles triangle is written as follows:

S = ¼ in √(4 * a 2 - b 2).

The area formula looks somewhat simpler than for an arbitrary triangle if the sides and the angle between them are known. S = ½ a 2 * sin β.

Special case: equilateral triangle

Usually in problems the side about it is known or it can be found out in some way. Then the formula for finding the area of ​​such a triangle is as follows:

S = (a 2 √3) / 4.

Problems to find the area if the triangle is depicted on checkered paper

The simplest situation is when a right triangle is drawn so that its legs coincide with the lines of the paper. Then you just need to count the number of cells that fit into the legs. Then multiply them and divide by two.

When the triangle is acute or obtuse, it needs to be drawn to a rectangle. Then the resulting figure will have 3 triangles. One is the one given in the problem. And the other two are auxiliary and rectangular. The areas of the last two need to be determined using the method described above. Then calculate the area of ​​the rectangle and subtract from it those calculated for the auxiliary ones. The area of ​​the triangle is determined.

The situation in which none of the sides of the triangle coincides with the lines of the paper turns out to be much more complicated. Then it needs to be inscribed in a rectangle so that the vertices of the original figure lie on its sides. In this case, there will be three auxiliary right triangles.

Example of a problem using Heron's formula

Condition. Some triangle has known sides. They are equal to 3, 5 and 6 cm. You need to find out its area.

Now you can calculate the area of ​​the triangle using the above formula. Under the square root is the product of four numbers: 7, 4, 2 and 1. That is, the area is √(4 * 14) = 2 √(14).

If greater accuracy is not required, then you can take the square root of 14. It is equal to 3.74. Then the area will be 7.48.

Answer. S = 2 √14 cm 2 or 7.48 cm 2.

Example problem with right triangle

Condition. One leg of a right triangle is 31 cm larger than the second. You need to find out their lengths if the area of ​​the triangle is 180 cm 2.
Solution. We will have to solve a system of two equations. The first is related to area. The second is with the ratio of the legs, which is given in the problem.
180 = ½ a * b;

a = b + 31.
First, the value of “a” must be substituted into the first equation. It turns out: 180 = ½ (in + 31) * in. It has only one unknown quantity, so it is easy to solve. After opening the brackets, the quadratic equation is obtained: 2 + 31 360 = 0. This gives two values ​​for "in": 9 and - 40. The second number is not suitable as an answer, since the length of the side of a triangle cannot be a negative value.

It remains to calculate the second leg: add 31 to the resulting number. It turns out 40. These are the quantities sought in the problem.

Answer. The legs of the triangle are 9 and 40 cm.

Problem of finding a side through the area, side and angle of a triangle

Condition. The area of ​​a certain triangle is 60 cm 2. It is necessary to calculate one of its sides if the second side is 15 cm and the angle between them is 30º.

Solution. Based on the accepted notation, the desired side is “a”, the known side is “b”, the given angle is “γ”. Then the area formula can be rewritten as follows:

60 = ½ a * 15 * sin 30º. Here the sine of 30 degrees is 0.5.

After transformations, “a” turns out to be equal to 60 / (0.5 * 0.5 * 15). That is 16.

Answer. The required side is 16 cm.

Problem about a square inscribed in a right triangle

Condition. The vertex of a square with a side of 24 cm coincides with the right angle of the triangle. The other two lie on the sides. The third belongs to the hypotenuse. The length of one of the legs is 42 cm. What is the area of ​​the right triangle?

Solution. Consider two right triangles. The first one is the one specified in the task. The second one is based on the known leg of the original triangle. They are similar because they have a common angle and are formed by parallel lines.

Then the ratios of their legs are equal. The legs of the smaller triangle are equal to 24 cm (side of the square) and 18 cm (given leg 42 cm subtract the side of the square 24 cm). The corresponding legs of a large triangle are 42 cm and x cm. It is this “x” that is needed in order to calculate the area of ​​the triangle.

18/42 = 24/x, that is, x = 24 * 42 / 18 = 56 (cm).

Then the area is equal to the product of 56 and 42 divided by two, that is, 1176 cm 2.

Answer. The required area is 1176 cm 2.

You can find over 10 formulas for calculating the area of ​​a triangle on the Internet. Many of them are used in problems with known sides and angles of the triangle. However, there are a number of complex examples where, according to the conditions of the assignment, only one side and angles of a triangle are known, or the radius of a circumscribed or inscribed circle and one more characteristic. In such cases, a simple formula cannot be applied.

The formulas given below will allow you to solve 95 percent of problems in which you need to find the area of ​​a triangle.
Let's move on to consider common area formulas.
Consider the triangle shown in the figure below

In the figure and below in the formulas the classical designations of all its characteristics are introduced
a,b,c – sides of the triangle,
R – radius of the circumscribed circle,
r – radius of the inscribed circle,
h[b],h[a],h[c] – heights drawn in accordance with sides a,b,c.
alpha, beta, hamma – angles near the vertices.

Basic formulas for the area of ​​a triangle

1. The area is equal to half the product of the side of the triangle and the height lowered to this side. In the language of formulas, this definition can be written as follows

Thus, if the side and height are known, then every student will find the area.
By the way, from this formula one can derive one useful relationship between heights

2. If we take into account that the height of a triangle through the adjacent side is expressed by the dependence

Then the first area formula is followed by the second ones of the same type

Look carefully at the formulas - they are easy to remember, since the work involves two sides and the angle between them. If we correctly designate the sides and angles of the triangle (as in the figure above), we will get two sides a, b and the angle is connected to the third With (hamma).

3. For the angles of a triangle, the relation is true

The dependence allows you to use the following formulas for the area of ​​a triangle in calculations:

Examples of this dependence are extremely rare, but you must remember that there is such a formula.

4. If the side and two adjacent angles are known, then the area is found by the formula

5. The formula for area in terms of side and cotangent of adjacent angles is as follows

By rearranging the indexes you can get dependencies for other parties.

6. The area formula below is used in problems when the vertices of a triangle are specified on the plane by coordinates. In this case, the area is equal to half the determinant taken modulo.

7. Heron's formula used in examples with known sides of a triangle.
First find the semi-perimeter of the triangle

And then determine the area using the formula

or

It is quite often used in the code of calculator programs.

8. If all the heights of the triangle are known, then the area is determined by the formula

It is difficult to calculate on a calculator, but in the MathCad, Mathematica, Maple packages the area is “time two”.

9. The following formulas use the known radii of inscribed and circumscribed circles.

In particular, if the radius and sides of the triangle, or its perimeter, are known, then the area is calculated according to the formula

10. In examples where the sides and the radius or diameter of the circumscribed circle are given, the area is found using the formula

11. The following formula determines the area of ​​a triangle in terms of the side and angles of the triangle.

And finally - special cases:
Area of ​​a right triangle with legs a and b equal to half their product

Formula for the area of ​​an equilateral (regular) triangle=

= one-fourth of the product of the square of the side and the root of three.