A triangle is three points that do not lie on the same straight line, and three line segments that connect them. Otherwise, a triangle is a polygon that has exactly three angles.

These three points are called the vertices of the triangle, and the segments are called the sides of the triangle. The sides of a triangle form three angles at the vertices of the triangle.

An isosceles triangle is one in which two sides are equal. These sides are called the sides, the third side is called the base. In an isosceles triangle, the angles at the base are equal.

An equilateral or right triangle is called, in which all three sides are equal. All angles of an equilateral triangle are also equal and equal to 60°.

The area of ​​an arbitrary triangle is calculated by the formulas: or

The area of ​​a right triangle is calculated by the formula:

The area of ​​a regular or equilateral triangle is calculated by the formulas: or or

Where a,b,c- sides of a triangle h- the height of the triangle, y- the angle between the sides, R- radius of the circumscribed circle, r is the radius of the inscribed circle.

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 dimensions. The formulas are presented in the form of a picture, here are explanations for the application or justification of their correctness. Also, in a separate figure, the correspondence of the letter symbols in the formulas and the graphic symbols in the drawing is shown.

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

• "Formulas 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- the radius of the circle inscribed in the triangle
R- the radius of the circumscribed circle around the triangle
h- the height of the triangle, lowered to the side
p- semiperimeter of a triangle, 1/2 the sum of its sides (perimeter)
α - the angle opposite side a of the triangle
β - the angle opposite side b of the triangle
γ - the angle opposite side c of the triangle
h a, h b , h c- the height of the triangle, lowered to the side a, b, c

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

• The area of ​​the triangle is half the product of the height of a triangle and the length of the side on 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 we complete each of them to 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). Despite the fact that 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 drawn by us, which will give us the previous formula
• The area of ​​an arbitrary triangle can be found through work half the radius of a circle inscribed in it by the sum of the lengths of all its sides(Formula 3), in other words, you need to multiply the half-perimeter of the triangle by the radius of the inscribed circle (it's easier to remember this way)
• 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 in terms of 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 a semiperimeter, 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 a circle circumscribed around it and the sines of each of its angles. (Formula 8)
• If the length of one side and the magnitude of the two angles adjacent to it are known, then the area of ​​\u200b\u200bthe 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 a triangle is known (Formula 10), then the area of ​​such a triangle is inversely proportional to the lengths of these heights, as by Heron's Formula
• Formula 11 allows you to calculate area of ​​a triangle according to the coordinates of its vertices, which are given 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 can be in the area of ​​negative values

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

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 ​​a triangle.

Decision.

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 statement into the formula:
S=1/2*5*6*sin60

In the table of values ​​\u200b\u200bof trigonometric functions, we find and substitute in the expression the value of the sine 60 degrees. It will be equal to the root of three by two.
S = 15 √3 / 2

Answer: 7.5 √3 (depending on the requirements of the teacher, it is probably possible to leave 15 √3/2)

Task. Find the area of ​​an equilateral triangle

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

Decision .

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 \u003d b \u003d c, the formula for the area of ​​​​an equilateral triangle will take the form:

S = √3 / 4 * a2

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 ​​a triangle increase if the sides are quadrupled?

Decision.

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 find the area of ​​this triangle, and then we find the area of ​​a triangle whose sides are four times larger. The ratio of the areas of these triangles will give us the answer to the problem.

Next, we give a textual explanation of the solution of the problem in steps. However, at the very end, the same solution is presented in a graphical form that is more convenient for perception. Those who wish can immediately drop down the solution.

To solve, we use the Heron 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 the first line of the picture below)

The lengths of the sides of an arbitrary triangle are given by the variables a, b, c.
If the sides are increased by 4 times, then the area of ​​\u200b\u200bthe new triangle c will be:

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

As you can see, 4 is a common factor that can be bracketed out of 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

From the number 256, the square root is perfectly extracted, so we will 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 the fifth line of the figure below)

To answer the question posed in the problem, it is enough for us to divide the area of ​​the resulting triangle by the area of ​​the original one.
We determine the area ratios by dividing the expressions into each other and reducing the resulting fraction.

Definition of a triangle

Triangle- This is a geometric figure that is formed as a result of the intersection of three segments, the ends of which do not lie on one straight line. Any triangle has three sides, three vertices and three angles.

Online calculator

Triangles are of various types. For example, there is an equilateral triangle (one in which all sides are equal), isosceles (two sides are equal in it) and right-angled (in which one of the angles is a right one, that is, equal to 90 degrees).

The area of ​​a triangle can be found in various ways, depending on which elements of the figure are known by the condition of the problem, whether it be angles, lengths, or, in general, the radii of the circles associated with the triangle. Consider each method separately with examples.

The formula for the area of ​​a triangle given its base and height

S = 1 2 ⋅ a ⋅ h S= \frac(1)(2)\cdot a\cdot hS=2 1 ​ ⋅ a ⋅h,

A a a- the base of the triangle;
h h h- the height of the triangle drawn to the given base a.

Example

Find the area of ​​a triangle if the length of its base is known, equal to 10 (cm) and the height drawn to this base, equal to 5 (cm).

Decision

A=10 a=10 a =1 0
h=5 h=5 h =5

Substitute in the formula for the area and get:
S = 1 2 ⋅ 10 ⋅ 5 = 25 S=\frac(1)(2)\cdot10\cdot 5=25S=2 1 ​ ⋅ 1 0 ⋅ 5 = 2 5 (see sq.)

Answer: 25 (see sq.)

The formula for the area of ​​a triangle given the lengths of all sides

S = p ⋅ (p − a) ⋅ (p − b) ⋅ (p − c) S= \sqrt(p\cdot(p-a)\cdot (p-b)\cdot (p-c))S=p ⋅ (p − a ) ⋅ (p − b ) ⋅ (p − c )​ ,

A , b , c a, b, c a, b, c- the length of the sides of the triangle;
pp p- half the sum of all sides of the triangle (that is, half the perimeter of the triangle):

P = 1 2 (a + b + c) p=\frac(1)(2)(a+b+c)p=2 1 ​ (a +b +c)

This formula is called Heron's formula.

Example

Find the area of ​​a triangle if the lengths of its three sides are known, equal to 3 (see), 4 (see), 5 (see).

Decision

A=3 a=3 a =3
b=4 b=4 b=4
c=5 c=5 c=5

Find half the perimeter pp p:

P = 1 2 (3 + 4 + 5) = 1 2 ⋅ 12 = 6 p=\frac(1)(2)(3+4+5)=\frac(1)(2)\cdot 12=6p=2 1 ​ (3 + 4 + 5 ) = 2 1 ​ ⋅ 1 2 = 6

Then, according to Heron's formula, the area of ​​a triangle is:

S = 6 ⋅ (6 − 3) ⋅ (6 − 4) ⋅ (6 − 5) = 36 = 6 S=\sqrt(6\cdot(6-3)\cdot(6-4)\cdot(6- 5))=\sqrt(36)=6S=6 ⋅ (6 − 3 ) ⋅ (6 − 4 ) ⋅ (6 − 5 ) ​ = 3 6 ​ = 6 (see sq.)

Answer: 6 (see sq.)

Formula for the area of ​​a triangle given one side and two angles

S = a 2 2 ⋅ sin ⁡ β sin ⁡ γ sin ⁡ (β + γ) S=\frac(a^2)(2)\cdot \frac(\sin(\beta)\sin(\gamma))( \sin(\beta+\gamma))S=2 a 2 ​ ⋅ sin(β+γ)sin β sin γ ​ ,

A a a- the length of the side of the triangle;
β , γ \beta, \gamma β , γ - angles adjacent to the side a a a.

Example

Given a side of a triangle equal to 10 (see) and two adjacent angles of 30 degrees. Find the area of ​​a triangle.

Decision

A=10 a=10 a =1 0
β = 3 0 ∘ \beta=30^(\circ)β = 3 0 ∘
γ = 3 0 ∘ \gamma=30^(\circ)γ = 3 0 ∘

According to the formula:

S = 1 0 2 2 ⋅ sin ⁡ 3 0 ∘ sin ⁡ 3 0 ∘ sin ⁡ (3 0 ∘ + 3 0 ∘) = 50 ⋅ 1 2 3 ≈ 14.4 S=\frac(10^2)(2)\cdot \frac(\sin(30^(\circ))\sin(30^(\circ)))(\sin(30^(\circ)+30^(\circ)))=50\cdot\frac( 1)(2\sqrt(3))\approx14.4S=2 1 0 2 ​ ⋅ sin(3 0 ∘ + 3 0 ∘ ) sin 3 0 ∘ sin 3 0 ∘ ​ = 5 0 ⋅ 2 3 ​ 1 ​ ≈ 1 4 . 4 (see sq.)

Answer: 14.4 (see sq.)

The formula for the area of ​​a triangle given three sides and the radius of the circumscribed circle

S = a ⋅ b ⋅ c 4 R S=\frac(a\cdot b\cdot c)(4R)S=4 Ra ⋅ b ⋅ c​ ,

A , b , c a, b, c a, b, c- sides of a triangle
R R R is the radius of the circumscribed circle around the triangle.

Example

We take the numbers from our second problem and add a radius to them R R R circles. Let it be equal to 10 (see).

Decision

A=3 a=3 a =3
b=4 b=4 b=4
c=5 c=5 c=5
R=10 R=10 R=1 0

S = 3 ⋅ 4 ⋅ 5 4 ⋅ 10 = 60 40 = 1.5 S=\frac(3\cdot 4\cdot 5)(4\cdot 10)=\frac(60)(40)=1.5S=4 ⋅ 1 0 3 ⋅ 4 ⋅ 5 ​ = 4 0 6 0 ​ = 1 . 5 (see sq.)

Answer: 1.5 (cm.sq.)

The formula for the area of ​​a triangle given three sides and the radius of an inscribed circle

S = p ⋅ r S=p\cdot rS= p⋅ r,

ppp- half the perimeter of the triangle:

p = a + b + c 2 p=\frac(a+b+c)(2)p= 2 a+ b+ c​ ,

a, b, c a, b, ca, b, c- sides of a triangle
r rr is the radius of the circle inscribed in the triangle.

Example

Let the radius of the inscribed circle be equal to 2 (see). We take the lengths of the sides from the previous problem.

Decision

$a=3b=4\; b=4b=4c=5\; c=5c=5r=2\; r=2r=2$

$p=23+4+5=6$

S = 6 ⋅ 2 = 12 S=6\cdot 2=12S= 6 ⋅ 2 = 1 2 (see sq.)

Answer: 12 (see sq.)

Formula for the area of ​​a triangle given two sides and the angle between them

S = 1 2 ⋅ b ⋅ c ⋅ sin ⁡ (α) S=\frac(1)(2)\cdot b\cdot c\cdot\sin(\alpha)S= 2 1 ​ ⋅ b⋅ c⋅ sin(α ) ,

b, c b, cb, c- sides of a triangle

α\alphaα - angle between sides bbb and c cc.

Example

The sides of the triangle are 5 (see) and 6 (see), the angle between them is 30 degrees. Find the area of ​​a triangle.

Decision

$b=5c=6\; c=6c=6\alpha \; =\; 3\; 0\; \circ \; \backslash alpha=30^(\backslash circ)\alpha =30^{\circ }$

S = 1 2 ⋅ 5 ⋅ 6 ⋅ sin ⁡ (3 0 ∘) = 7.5 S=\frac(1)(2)\cdot 5\cdot 6\cdot\sin(30^(\circ))=7.5S= 2 1 ​ ⋅ 5 ⋅ 6 ⋅ sin(3 0 ∘ ) = 7 . 5 (see sq.)

Answer: 7.5 (see sq.)

A triangle is the simplest geometric figure, which consists of three sides and three vertices. Due to its simplicity, the triangle has been used since ancient times for various measurements, and today the figure can be useful for solving practical and everyday problems.

Triangle features

The figure has been used for calculations since ancient times, for example, surveyors and astronomers operate with the properties of triangles to calculate areas and distances. Through the area of ​​this figure, it is easy to express the area of ​​any n-gon, and this property was used by ancient scientists to derive formulas for the areas of polygons. Constant work with triangles, especially with a right triangle, has become the basis for a whole section of mathematics - trigonometry.

triangle geometry

The properties of the geometric figure have been studied since ancient times: the earliest information about the triangle was found in Egyptian papyri 4000 years old. Then the figure was studied in ancient Greece and the greatest contribution to the geometry of the triangle was made by Euclid, Pythagoras and Heron. The study of the triangle never stopped, and in the 18th century Leonhard Euler introduced the concept of the figure's orthocenter and Euler's circle. At the turn of the 19th and 20th centuries, when it seemed that absolutely everything was known about the triangle, Frank Morley formulated the angle trisectrix theorem, and Vaclav Sierpinski proposed the fractal triangle.

There are several types of flat triangles familiar to us from the school geometry course:

• acute-angled - all corners of the figure are sharp;
• obtuse - the figure has one obtuse angle (greater than 90 degrees);
• rectangular - the figure contains one right angle equal to 90 degrees;
• isosceles - a triangle with two equal sides;
• equilateral - a triangle with all equal sides.
• In real life, there are all kinds of triangles, and in some cases we may need to calculate the area of ​​a geometric figure.

Area of ​​a triangle

Area is an estimate of how much of the plane the figure bounds. The area of ​​a triangle can be found in six ways, using sides, height, angles, the radius of an inscribed or circumscribed circle, as well as using Heron's formula or calculating a double integral over the lines bounding the plane. The simplest formula for calculating the area of ​​a triangle is:

where a is the side of the triangle, h is its height.

However, in practice it is not always convenient for us to find the height of a geometric figure. The algorithm of our calculator allows you to calculate the area, knowing:

• three sides;
• two sides and the angle between them;
• one side and two corners.

To determine the area in terms of three sides, we use Heron's formula:

S = sqrt (p × (p-a) × (p-b) × (p-c)),

where p is the half-perimeter of the triangle.

The calculation of the area on two sides and an angle is made according to the classical formula:

S = a × b × sin(alfa),

where alfa is the angle between sides a and b.

To determine the area through one side and two corners we use the relation that:

a / sin(alfa) = b / sin(beta) = c / sin(gamma)

Using a simple proportion, we determine the length of the second side, after which we calculate the area using the formula S = a × b × sin(alfa). This algorithm is fully automated and you only need to enter the given variables and get the result. Let's look at a couple of examples.

Real life examples

paving slabs

Let's say you want to pave the floor with triangular tiles, and to determine the amount of material needed, you should find out the area of ​​​​one tile and the area of ​​\u200b\u200bthe floor. Let it be necessary to process 6 square meters of a surface using a tile whose dimensions are a = 20 cm, b = 21 cm, c = 29 cm. Obviously, the calculator uses Heron's formula to calculate the area of ​​a triangle and will give the result:

Thus, the area of ​​\u200b\u200bone tile element will be 0.021 square meters, and you will need 6 / 0.021 \u003d 285 triangles to improve the floor. The numbers 20, 21 and 29 make up the Pythagorean triple - numbers that satisfy . And that's right, our calculator also calculated all the angles of the triangle, and the gamma angle is exactly 90 degrees.

school task

In a school problem, you need to find the area of ​​a triangle, knowing that the side a \u003d 5 cm, and the angles alpha and beta of the wound are 30 and 50 degrees, respectively. To solve this problem manually, we would first find the value of the side b using the ratio of the sides and the sines of the opposite angles, and then determine the area using the simple formula S = a × b × sin(alfa). Let's save time, enter the data in the calculator form and get an instant answer

When using a calculator, it is important to correctly specify the angles and sides, otherwise the result will be incorrect.

Conclusion

The triangle is a unique figure that occurs both in real life and in abstract calculations. Use our online calculator to find the area of ​​triangles of any kind.

Area of ​​a triangle. In many geometry problems related to the calculation of areas, formulas for the area of ​​a triangle are used. There are several of them, here we will consider the main ones.To list these formulas would be too simple and useless. We will analyze the origin of the main formulas, those that are used most often.

Before you familiarize yourself with the derivation of formulas, be sure to look at the article on.After studying the material, you can easily restore the formulas in memory (if they suddenly "fly out" at the right time for you).

First formula

The diagonal of a parallelogram divides it into two triangles of equal area:

Therefore, the area of ​​the triangle will be equal to half the area of ​​the parallelogram:

Triangle area formula

* That is, if we know any side of the triangle and the height lowered to this side, then we can always calculate the area of ​​\u200b\u200bthis triangle.

Formula Two

As already stated in the article on the area of ​​a parallelogram, the formula has the form:

The area of ​​a triangle is half its area, so:

*That is, if any two sides in a triangle and the angle between them are known, we can always calculate the area of ​​such a triangle.

Heron's formula (third)

This formula is difficult to derive and you do not need it. Look how beautiful she is, we can say that she is remembered.

*If three sides of a triangle are given, then using this formula we can always calculate its area.

Formula Four

where ris the radius of the inscribed circle

*If three sides of a triangle and the radius of the circle inscribed in it are known, then we can always find the area of ​​this triangle.

Formula five

where Ris the radius of the circumscribed circle.

*If three sides of a triangle and the radius of the circumscribed circle are known, then we can always find the area of ​​such a triangle.

The question arises: if three sides of a triangle are known, then is it not easier to find its area using Heron's formula!

Yes, it is easier, but not always, sometimes it becomes difficult. It has to do with root extraction. In addition, these formulas are very convenient to use in problems where the area of ​​a triangle is given, its sides are given and it is required to find the radius of an inscribed or circumscribed circle. Such tasks are included in the exam.

Let's take a look at the formula:

It is a special case of the formula for the area of ​​a polygon in which a circle is inscribed:

Consider it on the example of a pentagon:

We connect the center of the circle with the vertices of this pentagon and drop perpendiculars from the center to its sides. We get five triangles, with the dropped perpendiculars being the radii of the inscribed circle:

The area of ​​the pentagon is:

Now it is clear that if we are talking about a triangle, then this formula takes the form:

Formula six