Find the area of ​​a triangle given three sides online. How to find the area of ​​a triangle. Triangle formulas

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, a separate figure shows the correspondence of 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 below, as well as additionally special formulas that are valid 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 notation given corresponds to the figure above, so that when solving a real problem in geometry, it would 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 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, instead of the symbol " Square root" the sqrt() function can be used, 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.

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

In the table of values trigonometric functions 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.

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 \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?

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 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.

Instruction

Parties and corners are considered basic elements A. A triangle is completely defined by any of the following basic elements: either three sides, or one side and two angles, or two sides and an angle between them. For existence triangle defined by three sides a, b, c, it is necessary and sufficient that the inequalities, called inequalities triangle:
a+b > c
a+c > b
b+c > a.

For building triangle on three sides a, b, c, it is necessary from the point C of the segment CB=a how to draw a circle of radius b with a compass. Then, in a similar way, draw a circle from point B with a radius equal to the side c. Their intersection point A is the third vertex of the desired triangle ABC, where AB=c, CB=a, CA=b - sides triangle. The problem has , if the sides a, b, c, satisfy the inequalities triangle specified in step 1.

The area of ​​S constructed in this way triangle ABC with known parties a, b, c, is calculated by Heron's formula:
S=v(p(p-a)(p-b)(p-c)),
where a, b, c are sides triangle, p is the semiperimeter.
p = (a+b+c)/2

If the triangle is equilateral, that is, all its sides are equal (a=b=c). Area triangle calculated by the formula:
S=(a^2 v3)/4

If the triangle is right-angled, that is, one of its angles is 90 °, and the sides forming it are legs, the third side is the hypotenuse. In this case square equals the product of the legs divided by two.
S=ab/2

To find square triangle, you can use one of the many formulas. Choose the formula depending on what data is already known.

You will need

  • knowledge of formulas for finding the area of ​​a triangle

Instruction

If you know the value of one of the sides and the value of the height lowered to this side from the opposite corner, then you can find the area using the following: S = a*h/2, where S is the area of ​​the triangle, a is one of the sides of the triangle, and h - height, to side a.

There is a known way to determine the area of ​​a triangle if three of its sides are known. She is Heron's formula. To simplify its recording, an intermediate value is introduced - a semi-perimeter: p \u003d (a + b + c) / 2, where a, b, c - . Then Heron's formula is as follows: S = (p(p-a)(p-b)(p-c))^1, ^ exponentiation.

Suppose you know one of the sides of a triangle and three angles. Then it is easy to find the area of ​​the triangle: S = a²sinα sinγ / (2sinβ), where β is the angle opposite side a, and α and γ are angles adjacent to the side.

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note

The most general formula that is suitable for all cases is Heron's formula.

Sources:

Tip 3: How to find the area of ​​a triangle given three sides

Finding the area of ​​a triangle is one of the most common tasks in school planimetry. Knowing the three sides of a triangle is enough to determine the area of ​​any triangle. In special cases and equilateral triangles, it is enough to know the lengths of two and one side, respectively.

You will need

  • side lengths of triangles, Heron's formula, cosine theorem

Instruction

Heron's formula for the area of ​​a triangle is as follows: S = sqrt(p(p-a)(p-b)(p-c)). If you paint the semiperimeter p, then you get: S = sqrt(((a+b+c)/2)((b+c-a)/2)((a+c-b)/2)((a+b-c)/2) ) = (sqrt((a+b+c)(a+b-c)(a+c-b)(b+c-a)))/4.

You can also derive a formula for the area of ​​a triangle from considerations, for example, by applying the cosine theorem.

By the law of cosines, AC^2 = (AB^2)+(BC^2)-2*AB*BC*cos(ABC). Using the introduced notation, these can also be in the form: b^2 = (a^2)+(c^2)-2a*c*cos(ABC). Hence, cos(ABC) = ((a^2)+(c^2)-(b^2))/(2*a*c)

The area of ​​a triangle is also found by the formula S = a*c*sin(ABC)/2 through two sides and the angle between them. The sine of angle ABC can be expressed in terms of it using the basic trigonometric identity: sin(ABC) = sqrt(1-((cos(ABC))^2) By substituting the sine into the formula for the area and painting it, we can arrive at the formula for the area of ​​the triangle ABC.

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For repairs, it may be necessary to measure square walls. It is easier to calculate the required amount of paint or wallpaper. For measurements, it is best to use a tape measure or centimeter tape. Measurements should be taken after walls have been aligned.

You will need

  • -roulette;
  • -ladder.

Instruction

To count square walls, you need to know the exact height of the ceilings, as well as measure the length along the floor. This is done as follows: take a centimeter, lay it over the plinth. Usually a centimeter is not enough for the entire length, so fix it in the corner, then unwind it to the maximum length. At this point, put a mark with a pencil, write down the result and carry out further measurement in the same way, starting from the last measurement point.

Standard ceilings in typical - 2 meters 80 centimeters, 3 meters and 3 meters 20 centimeters, depending on the house. If the house was built before the 50s, then most likely the actual height is slightly lower than indicated. If you are calculating square for repair work, then a small margin will not hurt - consider based on the standard. If you still need to know the real height - take measurements. The principle is similar to measuring length, but you will need a stepladder.

Multiply the resulting figures - this is square your walls. True, for painting work or for it is necessary to subtract square door and window openings. To do this, lay a centimeter along the opening. If we are talking about a door that you are going to change later, then carry out with the door frame removed, considering only square the opening itself. The window area is calculated along the perimeter of its frame. After square window and doorway calculated, subtract the result from the total area of ​​​​the room obtained.

Please note that measurements of the length and width of the room are carried out together, it is easier to fix a centimeter or tape measure and, accordingly, get a more accurate result. Take the same measurement several times to make sure the numbers you get are accurate.

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Finding the volume of a triangle is indeed a non-trivial task. The fact is that a triangle is a two-dimensional figure, i.e. it lies entirely in one plane, which means that it simply has no volume. Of course, you can't find something that doesn't exist. But let's not give up! We can make the following assumption - the volume of a two-dimensional figure, this is its area. We are looking for the area of ​​the triangle.

You will need

  • sheet of paper, pencil, ruler, calculator

Instruction

Draw on a sheet of paper with a ruler and pencil. By carefully examining the triangle, you can make sure that it really does not have, since it is drawn on a plane. Label the sides of the triangle: let one side be side "a", the other side "b", and the third side "c". Label the vertices of the triangle with the letters "A", "B" and "C".

Measure any side of the triangle with a ruler and write down the result. After that, restore the perpendicular to the measured side from the opposite vertex, such a perpendicular will be the height of the triangle. In the case shown in the figure, perpendicular "h" is restored to side "c" from vertex "A". Measure the resulting height with a ruler and record the result of the measurement.

It may happen that you find it difficult to restore the exact perpendicular. In this case, you should use a different formula. Measure all sides of the triangle with a ruler. After that, calculate the half-perimeter of the triangle "p" by adding the resulting lengths of the sides and dividing their sum in half. Having at your disposal the value of the semi-perimeter, you can use the Heron formula. To do this, you need to take the square root of the following: p(p-a)(p-b)(p-c).

You have obtained the desired area of ​​the triangle. The problem of finding the volume of a triangle has not been solved, but as mentioned above, the volume is not . You can find volume which is essentially a triangle in the 3D world. If we imagine that our original triangle has become a three-dimensional pyramid, then the volume of such a pyramid will be the product of the length of its base and the area of ​​\u200b\u200bthe triangle we received.

note

Calculations will be more accurate the more carefully you take measurements.

Sources:

  • All-to-All Calculator - Reference Portal
  • triangle volume in 2019

The three points that uniquely define a triangle in the Cartesian coordinate system are its vertices. Knowing their position relative to each of the coordinate axes, you can calculate any parameters of this flat figure, including and limited by its perimeter square. This can be done in several ways.

Instruction

Use Heron's formula to calculate area triangle. It involves the dimensions of the three sides of the figure, so start the calculations with. The length of each side must be equal to the root of the sum of the squares of the lengths of its projections on the coordinate axes. If we denote the coordinates A(X₁,Y₁,Z₁), B(X₂,Y₂,Z₂) and C(X₃,Y₃,Z₃), the lengths of their sides can be expressed as follows: AB = √((X₁-X₂)² + (Y₁ -Y₂)² + (Z₁-Z₂)²), BC = √((X₂-X₃)² + (Y₂-Y₃)² + (Z₂-Z₃)²), AC = √((X₁-X₃)² + (Y₁-Y₃)² + (Z₁-Z₃)²).

To simplify the calculations, enter an auxiliary variable - the semi-perimeter (P). From that this is half the sum of the lengths of all sides: P \u003d ½ * (AB + BC + AC) \u003d ½ * (√ ((X₁-X₂)² + (Y₁-Y₂)² + (Z₁-Z₂)²) + √ ((X₂-X₃)² + (Y₂-Y₃)² + (Z₂-Z₃)²) + √((X₁-X₃)² + (Y₁-Y₃)² + (Z₁-Z₃)²).

To find the area of ​​a triangle, you can use different formulas. Of all the methods, the easiest and most often used is the multiplication of the height by the length of the base, and then dividing 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 consider methods for calculating the area of ​​specific types of triangle - rectangular, isosceles and equilateral. We accompany each formula with a short explanation that will help you understand its essence.

Universal ways to find the area of ​​a triangle

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

  • a, b, c are the lengths of the three sides of the figure we are considering;
  • r is the radius of a circle that can be inscribed in our triangle;
  • R is the radius of the circle that can be described around it;
  • α - the value of the angle formed by the sides b and c;
  • β is the angle between a and c;
  • γ - the value of the angle formed by the sides a and b;
  • h is the height of our triangle, lowered from angle α to side a;
  • p is 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 is easily completed to 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 should 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 they form. 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 multiplying the length of side a by the sine of angle γ, we get the height of the triangle, that is, h.

The area of ​​the figure under consideration is found by multiplying half the radius of the circle, which can be inscribed in it, by its perimeter. In other words, we find the product of the semiperimeter 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 circumscribed around it.

These formulas are universal, as they make it possible to determine the area of ​​any triangle (scalene, isosceles, equilateral, right-angled). This can be done with the help of 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? A feature of this figure is that its two sides are simultaneously its heights. If a and b are legs, and c becomes the hypotenuse, then the area is found as follows:

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

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

The triangle is a well-known figure. And this, despite the rich variety of its forms. Rectangular, equilateral, acute, isosceles, obtuse. Each of them is somewhat different. But for any it is required to know the area of ​​the triangle.

Common formulas for all triangles that use the lengths of the sides or heights

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

1. The area of ​​a triangle is calculated as the product of ½, the side and the height lowered onto it. S = ½ * a * n a. Similarly, one should write formulas for the other two sides.

2. Heron's formula, in which the semi-perimeter appears (it is customary to denote it with a 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 semi-perimeter formula: p \u003d (a + b + c) / 2. Then the equality for the area of ​​\u200b\u200bthe figure looks like this: S \u003d √ (p * (p - a) * ( p - c) * (p - c)).

3. If you do not want to use a semi-perimeter, then such a formula will come in handy, in which only the lengths of the sides are present: S \u003d ¼ * √ ((a + b + c) * (b + c - a) * (a + c - c) * (a + b - c)). It is somewhat longer than the previous one, but it will help out if you forgot how to find the semi-perimeter.

General formulas in which the angles of a triangle appear

The notation that is 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 \u003d (a 2 * sin β * sin γ) / (2 sin α).

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

The last two formulas are not the simplest. It's pretty hard to remember them.

General formulas for the situation when 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. In another way, it can be written as follows: S \u003d 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 the quadruple radius of the circumscribed circle. In literal terms, it looks like this: S \u003d (a * b * c) / (4R).

3. The third situation allows you to do without knowing the sides, but you need the values ​​of all three angles. S \u003d 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 denoted 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. She is the easiest to remember. Because it looks like the formula for the area of ​​a rectangle, only a fraction appears, denoting half.

Special case: isosceles triangle

Since its two sides are equal, 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 convert 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 \u003d ½ a 2 * sin β.

Special case: equilateral triangle

Usually, in problems about him, the side is known or can be somehow recognized. Then the formula for finding the area of ​​such a triangle is as follows:

S = (a 2 √3) / 4.

Tasks for finding the area if the triangle is depicted on checkered paper

The simplest situation is when a right-angled 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 must be drawn to a rectangle. Then in the resulting figure there will be 3 triangles. One is the one given in the task. And the other two are auxiliary and rectangular. The areas of the last two must be determined by 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.

Much more difficult is the situation in which none of the sides of the triangle coincides with the lines of the paper. Then it must 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.

An example of a problem on Heron's formula

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

Now you can calculate the area of ​​a 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 you do not need more precision, then you can take the square root of 14. It is 3.74. Then the area will be equal to 7.48.

Answer. S \u003d 2 √14 cm 2 or 7.48 cm 2.

An example of a problem with a right triangle

Condition. One leg of a right-angled triangle is 31 cm longer than the second. It is required to find out their lengths if the area of ​​the triangle is 180 cm 2.
Solution. You have to solve a system of two equations. The first has to do with area. The second is with the ratio of the legs, which is given in the problem.
180 \u003d ½ a * b;

a \u003d b + 31.
First, the value of "a" must be substituted into the first equation. It turns out: 180 \u003d ½ (in + 31) * in. It has only one unknown quantity, so it is easy to solve. After opening the parentheses, we get quadratic equation: in 2 + 31 in - 360 = 0. It gives two values ​​for "in": 9 and - 40. The second number is not suitable as an answer, since the length of the side of the 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.

The task of finding the side through the area, side and angle of a triangle

Condition. The area of ​​some triangle is 60 cm2. 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 designations, the desired side "a", the known "b", predetermined angle"γ". Then the area formula can be rewritten as follows:

60 \u003d 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 desired side is 16 cm.

The problem of 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 legs. The third belongs to the hypotenuse. The length of one of the legs is 42 cm. What is the area of ​​a right triangle?

Solution. Consider two right triangle. The first one is 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 24 cm (side of the square) and 18 cm (given leg 42 cm minus the side of the square 24 cm). The corresponding legs of the 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 \u003d 24 / x, that is, x \u003d 24 * 42 / 18 \u003d 56 (cm).

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

Answer. The desired area is 1176 cm 2.

The concept of area

The concept of the area of ​​any geometric figure, in particular a triangle, will be associated with such a figure as a square. For a unit area of ​​any geometric figure, we will take the area of ​​a square, the side of which is equal to one. For completeness, we recall two basic properties for the concept of areas of geometric shapes.

Property 1: If geometric figures are equal, their areas are also equal.

Property 2: Any figure can be divided into several figures. Moreover, the area of ​​the original figure is equal to the sum of the values ​​​​of the areas of all the figures that make it up.

Consider an example.

Example 1

It is obvious that one of the sides of the triangle is the diagonal of the rectangle , which has one side of length $5$ (since $5$ cells) and the other $6$ (since $6$ cells). Therefore, the area of ​​this triangle will be equal to half of such a rectangle. The area of ​​the rectangle is

Then the area of ​​the triangle is

Answer: $15$.

Next, consider several methods for finding the areas of triangles, namely using the height and base, using the Heron formula and the area of ​​an equilateral triangle.

How to find the area of ​​a triangle using the height and base

Theorem 1

The area of ​​a triangle can be found as half the product of the length of a side times the height drawn to that side.

Mathematically it looks like this

$S=\frac(1)(2)αh$

where $a$ is the length of the side, $h$ is the height drawn to it.

Proof.

Consider triangle $ABC$ where $AC=α$. The height $BH$ is drawn to this side and equals $h$. Let's build it up to the square $AXYC$ as in Figure 2.

The area of ​​rectangle $AXBH$ is $h\cdot AH$, and that of rectangle $HBYC$ is $h\cdot HC$. Then

$S_ABH=\frac(1)(2)h\cdot AH$, $S_CBH=\frac(1)(2)h\cdot HC$

Therefore, the desired area of ​​the triangle, according to property 2, is equal to

$S=S_ABH+S_CBH=\frac(1)(2)h\cdot AH+\frac(1)(2)h\cdot HC=\frac(1)(2)h\cdot (AH+HC)=\ frac(1)(2)αh$

The theorem has been proven.

Example 2

Find the area of ​​the triangle in the figure below, if the cell has an area equal to one

The base of this triangle is $9$ (since $9$ is $9$ cells). The height is also $9$. Then, by Theorem 1, we obtain

$S=\frac(1)(2)\cdot 9\cdot 9=40.5$

Answer: $40.5$.

Heron's formula

Theorem 2

If we are given three sides of a triangle $α$, $β$ and $γ$, then its area can be found as follows

$S=\sqrt(ρ(ρ-α)(ρ-β)(ρ-γ))$

here $ρ$ means the half-perimeter of this triangle.

Proof.

Consider the following figure:

By the Pythagorean theorem, from the triangle $ABH$ we obtain

From the triangle $CBH$, by the Pythagorean theorem, we have

$h^2=α^2-(β-x)^2$

$h^2=α^2-β^2+2βx-x^2$

From these two relations we obtain the equality

$γ^2-x^2=α^2-β^2+2βx-x^2$

$x=\frac(γ^2-α^2+β^2)(2β)$

$h^2=γ^2-(\frac(γ^2-α^2+β^2)(2β))^2$

$h^2=\frac((α^2-(γ-β)^2)((γ+β)^2-α^2))(4β^2)$

$h^2=\frac((α-γ+β)(α+γ-β)(γ+β-α)(γ+β+α))(4β^2)$

Since $ρ=\frac(α+β+γ)(2)$, then $α+β+γ=2ρ$, hence

$h^2=\frac(2ρ(2ρ-2γ)(2ρ-2β)(2ρ-2α))(4β^2)$

$h^2=\frac(4ρ(ρ-α)(ρ-β)(ρ-γ))(β^2 )$

$h=\sqrt(\frac(4ρ(ρ-α)(ρ-β)(ρ-γ))(β^2))$

$h=\frac(2)(β)\sqrt(ρ(ρ-α)(ρ-β)(ρ-γ))$

By Theorem 1, we get

$S=\frac(1)(2) βh=\frac(β)(2)\cdot \frac(2)(β) \sqrt(ρ(ρ-α)(ρ-β)(ρ-γ) )=\sqrt(ρ(ρ-α)(ρ-β)(ρ-γ))$

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