In geometry, there are often problems related to the sides of triangles. For example, it is often necessary to find the side of a triangle if the other two are known.

Triangles are isosceles, equilateral and equilateral. From all the variety, for the first example, we will choose a rectangular one (in such a triangle, one of the angles is 90 °, the sides adjacent to it are called the legs, and the third is the hypotenuse).

Quick article navigation

The length of the sides of a right triangle

The solution of the problem follows from the theorem of the great mathematician Pythagoras. It says that the sum of the squares of the legs of a right triangle is equal to the square of its hypotenuse: a²+b²=c²

• Find the square of the leg length a;
• Find the square of the leg b;
• We put them together;
• From the result obtained, we extract the root of the second degree.

Example: a=4, b=3, c=?

• a²=4²=16;
• b²=3²=9;
• 16+9=25;
• √25=5. That is, the length of the hypotenuse of this triangle is 5.

If the triangle has no right angle, then the lengths of the two sides are not enough. This requires a third parameter: it can be an angle, height, area of ​​a triangle, radius of a circle inscribed in it, etc.

If the perimeter is known

In this case, the task is even easier. The perimeter (P) is the sum of all sides of the triangle: P=a+b+c. Thus, by solving a simple mathematical equation, we get the result.

Example: P=18, a=7, b=6, c=?

1) We solve the equation, transferring all known parameters to one side of the equal sign:

2) Substitute values ​​instead of them and calculate the third side:

c=18-7-6=5, total: the third side of the triangle is 5.

If the angle is known

To calculate the third side of a triangle given the angle and the other two sides, the solution is reduced to calculating the trigonometric equation. Knowing the relationship of the sides of the triangle and the sine of the angle, it is easy to calculate the third side. To do this, you need to square both sides and add their results together. Then subtract from the resulting product of the sides, multiplied by the cosine of the angle: C=√(a²+b²-a*b*cosα)

If the area is known

In this case, one formula is not enough.

1) First, we calculate sin γ by expressing it from the formula for the area of ​​a triangle:

sin γ= 2S/(a*b)

2) Using the following formula, we calculate the cosine of the same angle:

sin² α + cos² α=1

cos α=√(1 - sin² α)=√(1- (2S/(a*b))²)

3) And again we use the sine theorem:

C=√((a²+b²)-a*b*cosα)

C=√((a²+b²)-a*b*√(1- (S/(a*b))²))

Substituting the values ​​of the variables into this equation, we obtain the answer to the problem.

The first are segments that are adjacent to the right angle, and the hypotenuse is the longest part of the figure and is opposite the 90 degree angle. A Pythagorean triangle is one whose sides are equal natural numbers; their lengths in this case are called the "Pythagorean triple".

egyptian triangle

In order for the current generation to learn geometry in the form in which it is taught at school now, it has been developed for several centuries. The fundamental point is the Pythagorean theorem. The sides of a rectangle are known to the whole world) are 3, 4, 5.

Few people are not familiar with the phrase " Pythagorean pants equal in all directions." However, in fact, the theorem sounds like this: c 2 (the square of the hypotenuse) \u003d a 2 + b 2 (the sum of the squares of the legs).

Among mathematicians, a triangle with sides 3, 4, 5 (cm, m, etc.) is called "Egyptian". It is interesting that which is inscribed in the figure is equal to one. The name arose around the 5th century BC, when Greek philosophers traveled to Egypt.

When building the pyramids, architects and surveyors used the ratio 3:4:5. Such structures turned out to be proportional, pleasant to look at and spacious, and also rarely collapsed.

In order to build a right angle, the builders used a rope on which 12 knots were tied. In this case, the probability of constructing a right-angled triangle increased to 95%.

Signs of equality of figures

• acute angle in right triangle and the large side, which are equal to the same elements in the second triangle, is an indisputable sign of the equality of the figures. Taking into account the sum of the angles, it is easy to prove that the second acute angles are also equal. Thus, the triangles are identical in the second criterion.
• When two figures are superimposed on each other, we rotate them in such a way that, when combined, they become one isosceles triangle. According to its property, the sides, or rather, the hypotenuses, are equal, as well as the angles at the base, which means that these figures are the same.

By the first sign, it is very easy to prove that the triangles are really equal, the main thing is that the two smaller sides (i.e., the legs) are equal to each other.

The triangles will be the same according to the II sign, the essence of which is the equality of the leg and the acute angle.

Right angle triangle properties

The height, which was lowered from a right angle, divides the figure into two equal parts.

The sides of a right triangle and its median are easy to recognize by the rule: the median, which is lowered to the hypotenuse, is equal to half of it. can be found both by Heron's formula and by the statement that it is equal to half the product of the legs.

In a right triangle, the properties of angles of 30 o, 45 o and 60 o apply.

• At an angle that is 30 °, it should be remembered that the opposite leg will be equal to 1/2 of the largest side.
• If the angle is 45 o, then the second sharp corner also 45 o. This suggests that the triangle is isosceles, and its legs are the same.
• The property of an angle of 60 degrees is that the third angle has a measure of 30 degrees.

The area is easy to find by one of three formulas:

1. through the height and the side on which it descends;
2. according to Heron's formula;
3. along the sides and the angle between them.

The sides of a right triangle, or rather the legs, converge with two heights. In order to find the third, it is necessary to consider the resulting triangle, and then, using the Pythagorean theorem, calculate the required length. In addition to this formula, there is also the ratio of twice the area and the length of the hypotenuse. The most common expression among students is the first, as it requires less calculations.

Theorems that apply to a right triangle

The geometry of a right triangle includes the use of theorems such as:

Enter Known Triangle Data

Side a

Side b

side c

Angle A in degrees

Angle B in degrees

Angle C in degrees

Median per side a

Median per side b

Median per side c

Height per side a

Height per side b

Height per c side

Vertex A coordinates

X Y

Vertex B coordinates

X Y

Vertex C coordinates

X Y

Area of ​​triangle S

Semiperimeter of triangle sides p

We present you a calculator that allows you to calculate all possible.

I would like to draw your attention to the fact that this is a generic bot. It calculates all the parameters of an arbitrary triangle, with arbitrarily given parameters. You will not find such a bot anywhere.

Do you know the side and the two heights? Or two sides and a median? Or is the bisector two angles and the base of a triangle?

For any request, we can get the correct calculation of the parameters of the triangle.

You do not need to look for formulas and do the calculation yourself. Everything has already been done for you.

Create a request and get an accurate answer.

An arbitrary triangle is shown. We will immediately make a reservation how and what is indicated, so that in the future there will be no confusion and errors in the calculations.

Sides opposite to any angle are also called only a small letter. That is, opposite the angle A lies the side of the triangle a, the side c is opposed to the angle C.

ma is the medina falling on side a, respectively, there are also medians mb and mc falling on the corresponding sides.

lb is the bisector falling on side b, respectively, there are also bisectors la and lc falling on the corresponding sides.

hb is the height falling on side b, respectively, there are also heights ha and hc falling on the corresponding sides.

And secondly, remember that a triangle is a figure in which there is fundamental rule:

The sum of any (!) two sides must be greater thanthird.

So don't be surprised if you get an error P For such given data, the triangle does not exist. when trying to calculate the parameters of a triangle with sides 3, 3 and 7.

Syntax

For XMPP client enablers, the request is like this treug<список параметров>

For site users, everything is done on this page.

List of parameters - parameters that are known, separated by a semicolon

the parameter is written as parameter=value

For example, if side a is known with a value of 10, then we write a = 10

Moreover, the values ​​can be not only in the form of a real number, but also, for example, as the result of some kind of expression

And here is the list of parameters that can appear in the calculations.

side a

Side b

side c

Semiperimeter p

Angle A

Angle B

Angle C

Area of ​​triangle S

Height ha per side a

Height hb per side b

Height hc per side c

Median ma per side a

Median mb per side b

Median mc per side c

Vertex coordinates (xa,ya) (xb,yb) (xc,yc)

Examples

write treug a=8;C=70;ha=2

Triangle parameters by given parameters

Side a = 8

Side b = 2.1283555449519

Side c = 7.5420719851515

Semiperimeter p = 8.8352137650517

Angle A = 2.1882518638666 in degrees 125.37759631119

Angle B = 2.873202966917 in degrees 164.62240368881

Angle C = 1.221730476396 in 70 degrees

Triangle area S = 8

Height ha per side a = 2

Height hb per side b = 7.5175409662872

Height hc per side c = 2.1214329472723

Median ma per side a = 3.8348889915443

Median mb per side b = 7.7012304590352

Median mc per side c = 4.4770789813853

That's all, all the parameters of the triangle.

The question is why we named the party a, but not in or With? This does not affect the decision. The main thing is to withstand the condition about which I have already said " Sides opposite to any corner are called the same, only with a small letter." And then draw a triangle in your mind, and apply to the question asked.

could be taken instead a in, but then the included angle will not be FROM a BUT well, the height will be hb. The result if you check will be the same.

For example, like this (xa,ya) =3.4 (xb,yb) =-6.14 (xc,yc)=-6,-3

writing a request treug xa=3;ya=4;xb=-6;yb=14;xc=-6;yc=-3

and we get

Triangle parameters by given parameters

Side a = 17

Side b = 11.401754250991

Side c = 13.453624047073

Semiperimeter p = 20.927689149032

Angle A = 1.4990243938603 in degrees 85.887771155351

Angle B = 0.73281510178655 in degrees 41.987212495819

Angle C = 0.90975315794426 in degrees 52.125016348905

Triangle area S = 76.5

Height ha per side a = 9

Height hb per side b = 13.418987695398

Height hc per side c = 11.372400437582

Median ma per side a = 9.1241437954466

Median mb per side b = 14.230249470757

Median mc per side c = 12.816005617976

Good luck with your calculations!

Definition of a triangle

Triangle- this is geometric figure, which 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).

Solution

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).

Solution

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.

Solution

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).

Solution

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.

Solution

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

$p=23+4+5=6$

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.

Solution

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

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.)

ANDREY PROKIP: “MY LOVE IS RUSSIAN ECOLOGY. YOU SHOULD INVEST IN IT!”
On September 4-5, the ecological forum "Climatic shape of cities" was held. The initiator of the organization of the event is the C40 organization, which was founded in 2005 by the UN. The main task of the form and cities is to control climate change cities.
As practice has shown, unlike social events and "meetings in nightclubs", there were few deputies and public personalities. Among those who did reveal concerns environmental situation was Prokip Adrey Zinovievich. He took an active part in all plenary sessions, together with the Special Representative of the President Russian Federation on climate issues Ruslan Edelgeriev, Deputy mayor of Moscow on housing and communal services Petr Biryukov, as well as foreign representatives - the mayor Italian city Savona - Hilario Caprioglio. The participants presented their projects and also discussed strategies to keep the rise in global temperatures, and also proposed practical solutions sustainable urban development.
ANDREY PROKIP ABOUT SHASHLIKS, DEPUTY AND GREEN CONSTRUCTION
Of particular interest to the Russian side was the speech of the speakers, among whom were European architects, scientists and the mayor of Savona. The topic of the speech was the TOP direction - "green construction". As Andrei Prokip himself stated, “it is important to correctly redistribute resources, as well as take into account the standards of European construction for such a metropolis as Moscow. It is necessary that Russia at the federal level take a course towards “green financing”, especially since it is economically feasible and, as practice shows, profitable.” He also expressed concern about the deterioration of the health of Russians in connection with environmental disasters and non-compliance with environmental standards for waste disposal by large and small industrial enterprises". He also confirmed his fears thanks to the speech of Francesco Zambon, WHO European Bureau of Health Investment Professor.
With characteristic humor, Andrey turned to famous people who were invited to the forum, but never showed up, with a call “to remember nature, not only when they want barbecue or go fishing. After all, it is on the benevolence of nature that the health of the whole people depends, which, unfortunately, includes them.
In addition to passionate speeches about Andrei Zinovievich's new "mistress-nature" and the importance of taking responsibility for environment a significant event of the forum was the plenary session on the topic "How to educate a new generation." The forum participants were unanimous in their opinion that it is necessary to educate not only children, but also the adult generation. It is very important to bring up responsibility to nature in everyday behavior, as well as in business.
A special project “Learning to live in a civilized way” will be launched for Moscow. This is an educational project for all segments of the population and age categories. But no matter how wonderful the theory and good intentions are, the saying “until the roasted rooster pecks, the fool will not cross himself” is still relevant for Russia.
According to Timothy Netter, a famous theater director, art can change everything. In one of his speeches, he spoke about how the idea of ​​preserving nature should be presented in theater and cinema, and how important it is to educate people through art to be responsible for what will happen to us and nature tomorrow.
The attention of rentv operators and Andrei Prokirp was attracted by students Russian universities, presenting a project on environmentally friendly technology for the production of containers that are resistant to moisture and temperature. This is very actual problem, since laws are being passed around the world against plastic containers, which, by the way, decompose for more than 30 years, pollute the soil and cause the death of animals.
It is inspiring that Moscow is one of the 94 cities participating in the C40 organization and for the third time the forum has been held, which every year attracts the attention of more and more famous personalities and citizens.