Ability to think mathematically one of the noblest human abilities.
Irish playwright Bernard Shaw
Heron's formula
In school mathematics, Heron's formula is very popular, the use of which allows you to calculate the area of \u200b\u200ba triangle along its three sides. At the same time, few students know that there is a similar formula for calculating the area of quadrilaterals inscribed in a circle. Such a formula is called Brahmagupta's formula. There is also a little-known formula for calculating the area of a triangle from its three heights, the derivation of which follows from Heron's formula.
Calculating the area of triangles
Let in a triangle sides, and . Then the following theorem (Heron's formula) is valid.
Theorem 1.
Where .
Proof. When deriving formula (1), we will use the well-known geometries tric formulas
, (2)
. (3)
From formulas (2) and (3) we obtain and . Since then
. (4)
If we designate then formula (1) follows from equality (4). The theorem has been proven.
Consider now the question of calculating the area of a triangle given that , that its three heights are known, And .
Theorem 2. The area is calculated by the formula
. (5)
Proof. Since , and , then
In this case, from formula (1) we obtain
or
Formula (5) follows from this. The theorem has been proven.
Calculating the area of quadrilaterals
Consider a generalization of Heron's formula for the case of calculating the area of quadrilaterals. However, it should immediately be noted that such a generalization is possible only for quadrilaterals that are inscribed in a circle.
Let the quadrilateral has sides , , and .
If is a quadrilateral, inscribed in a circle, then Theorem 3 (Brahmagupta's formula) holds.
Theorem 3. Square calculated by the formula
Where .
Proof. Draw a diagonal in a quadrilateral and get two triangles and . If we apply the cosine theorem to these triangles, which is equivalent to formula (3), then we can write
Since the quadrilateral is inscribed in a circle, the sum of its opposite angles is , i.e. .
Because or then from (7) we get
Or
. (8)
Since , then . However, and therefore
Since , then formulas (8) and (9) imply
If we put , then from here we obtain formula (6). The theorem has been proven.
If the inscribed quadrilateralis at the same time described, then formula (6) is greatly simplified.
Theorem 4. The area of a quadrilateral, inscribed in one circle and described around the other, is calculated by the formula
. (10)
Proof. Since a circle is inscribed in a quadrilateral, the equalities
In this case , , , and formula (6) is easily converted into formula (10). The theorem has been proven.
Let's move on to the consideration of examples of geometry problems, the solution of which is carried out on the basis of the application of the proven theorems.
Examples of problem solving
Example 1. Find area, if .
Solution. Since here , according to Theorem 1 we obtain
Answer: .
Note, if the sides of the triangletake on irrational meanings, then the calculation of its areaby using formula (1), usually , is ineffective. In this case, it is expedient to apply formulas (2) and (3) directly.
Example 2 Find the area if , and .
Solution. Taking into account formulas (2) and (3), we obtain
Since , then or .
Answer: .
Example 3 Find the area if , and .
Solution. Because the ,
then it follows from Theorem 2 that .
Answer: .
Example 4 The triangle has sides , and . Find and , where are the radii of the circumscribed and inscribed circles, respectively.
Solution. Let's calculate the area first. Since , then from formula (1) we obtain .
It is known that and . That's why .
Example 5 Find the area of a quadrilateral inscribed in a circle if , , and .
Solution. It follows from the conditions of the example that . Then, according to Theorem 3, we obtain .
Example 6 Find the area of a quadrilateral inscribed in a circle with sides , , and .
Solution. Since and , then the equality holds in the quadrilateral. However, it is known that the existence of such an equality is a necessary and sufficient condition for a circle to be inscribed in a given quadrilateral. In this regard, formula (10) can be used to calculate the area, from which follows .
For independent and high-quality preparation for entrance tests in the field of solving problems of school geometry, you can effectively use textbooks, listed in the list of recommended readings.
1. Gotman E.G. Problems in planimetry and methods for their solution. – M.: Enlightenment, 1996. – 240 p.
2. Kulagin E.D. , Fedin S.N. Triangle geometry in problems. - M .: KD "Librocom" / URSS, 2009. - 208 p.
3. Collection of problems in mathematics for applicants to higher educational institutions / Ed. M.I. Scanavi. - M .: World and Education, 2013. - 608 p.
4. Suprun V.P. Mathematics for high school students: additional sections of the school curriculum. – M.: Lenand / URSS, 2014. - 216 p.
Do you have any questions?
To get help from a tutor -.
blog.site, with full or partial copying of the material, a link to the source is required.
Gerona formula Hero formula
expresses the area s triangle in terms of the lengths of its three sides A, b And With and semiperimeter R = (A + b + With)/2: . Named after Heron of Alexandria.
HERONA FORMULAHERON FORMULA, expresses the area S triangle in terms of the lengths of its three sides a, b And c and semiperimeter P = (a + b + c)/2
Named after Heron of Alexandria.
encyclopedic Dictionary. 2009 .
See what the "Geron formula" is in other dictionaries:
Expresses the area S of a triangle in terms of the lengths of its three sides a, b and c and the semiperimeter P = (a + b + c) / 2 Named after Heron of Alexandria ... Big Encyclopedic Dictionary
Formula expressing the area of a triangle in terms of its three sides. Namely, if a, b, c are the lengths of the sides of the triangle, and a S is its area, then the G. f. has the form: where p denotes the semiperimeter of the triangle G. f. ... ...
The formula expressing the area of a triangle in terms of its sides a, b, c: where Named after Heron (c. 1st century AD), A. B. Ivanov ... Mathematical Encyclopedia
Expresses the area 5 of a triangle in terms of the lengths of its three sides a, b and c and the semi-perimeter p \u003d (a + b + c) / 2: s \u003d square. root p(p a)(p b)(p c). Named after Heron of Alexandria... Natural science. encyclopedic Dictionary
- ... Wikipedia
Allows you to calculate the area of a triangle (S) on its sides a, b, c: where p is the semiperimeter of the triangle: . Proof where the angle is triangular ... Wikipedia
Expresses the area of a quadrilateral inscribed in a circle as a function of the lengths of its sides. If an inscribed quadrilateral has side lengths and a semiperimeter, then its area is ... Wikipedia
This article lacks links to sources of information. Information must be verifiable, otherwise it may be questioned and removed. You can edit this article to include links to authoritative sources. This mark ... ... Wikipedia
- (Heronus Alexandrinus) (years of birth and death unknown, probably 1st century), an ancient Greek scientist who worked in Alexandria. The author of works in which he systematically outlined the main achievements of the ancient world in the field of applied mechanics, V ... ... Great Soviet Encyclopedia
Alexandrian (Heronus Alexandrinus) (years of birth and death unknown, probably 1st century), an ancient Greek scientist who worked in Alexandria. The author of works in which he systematically outlined the main achievements of the ancient world in the field of ... ... Great Soviet Encyclopedia
This formula allows you to calculate the area of a triangle along its sides a, b and c:
S=√(p(p-a)(p-b)(p-s),where p is the half-perimeter of the triangle, i.e. p = (a + b + c)/2.
The formula is named after the ancient Greek mathematician Heron of Alexandria (around the 1st century). Heron considered triangles with integer sides, the areas of which are also integers. Such triangles are called Heronian. For example, these are triangles with sides 13, 14, 15 or 51, 52, 53.
There are analogues of Heron's formula for quadrilaterals. Due to the fact that the problem of constructing a quadrilateral along its sides a, b, c and d has more than one solution, in the general case, the area of a quadrilateral is not enough to know the lengths of the sides. You have to enter additional parameters or impose restrictions. For example, the area of an inscribed quadrilateral is found by the formula: S \u003d √ (p-a) (p-b) (p-c) (p-d)
If the quadrilateral is both inscribed and circumscribed at the same time, its area is by a simpler formula: S=√(abcd).
Heron of Alexandria - Greek mathematician and mechanic.
He was the first to invent automatic doors, an automatic puppet theatre, a vending machine, a rapid-fire self-loading crossbow, a steam turbine, automatic scenery, a device for measuring the length of roads (an ancient odometer), etc. He was the first to create programmable devices (a shaft with pins with a rope wound around it ).
He studied geometry, mechanics, hydrostatics, optics. Main works: Metrics, Pneumatics, Autotopoetics, Mechanics (the work has been preserved entirely in Arabic translation), Catoptrics (the science of mirrors; it has been preserved only in Latin translation), etc. land survey, actually based on the use of rectangular coordinates. Heron used the achievements of his predecessors: Euclid, Archimedes, Strato from Lampsak. Many of his books are irretrievably lost (the scrolls were kept in the Library of Alexandria).
In the treatise "Mechanics" Heron described five types of the simplest machines: lever, gate, wedge, screw and block.
In the treatise "Pneumatics" Heron described various siphons, ingeniously arranged vessels, automata, set in motion by compressed air or steam. This is an aeolipil, which was the first steam turbine - a ball rotated by the power of jets of water vapor; door opener, holy water vending machine, fire pump, water organ, mechanical puppet theatre.
The book "On the Diopter" describes a diopter - the simplest device used for geodetic work. Geron sets out in his treatise the rules of land surveying based on the use of rectangular coordinates.
In "Katoptrik" Heron justifies the straightness of light rays with an infinitely high speed of their propagation. Heron considers various types of mirrors, paying special attention to cylindrical mirrors.
Heron's "Metric" and the "Geometrics" and "Stereometrics" extracted from it are reference books on applied mathematics. Among the information contained in the "Metric" information:
Formulas for areas of regular polygons.
Volumes of regular polyhedra, pyramid, cone, truncated cone, torus, spherical segment.
Heron's formula for calculating the area of a triangle from the lengths of its sides (discovered by Archimedes).
Rules for the numerical solution of quadratic equations.
Algorithms for extracting square and cube roots.
Heron's book "Definitions" is an extensive collection of geometric definitions, for the most part coinciding with the definitions of Euclid's "Elements".
Ability to think mathematically one of the noblest human abilities.
Irish playwright Bernard Shaw
Heron's formula
In school mathematics, Heron's formula is very popular, the use of which allows you to calculate the area of \u200b\u200ba triangle along its three sides. At the same time, few students know that there is a similar formula for calculating the area of quadrilaterals inscribed in a circle. Such a formula is called Brahmagupta's formula. There is also a little-known formula for calculating the area of a triangle from its three heights, the derivation of which follows from Heron's formula.
Calculating the area of triangles
Let in a triangle sides, and . Then the following theorem (Heron's formula) is valid.
Theorem 1.
Where .
Proof. When deriving formula (1), we will use the well-known geometries tric formulas
, (2)
. (3)
From formulas (2) and (3) we obtain and . Since then
. (4)
If we designate then formula (1) follows from equality (4). The theorem has been proven.
Consider now the question of calculating the area of a triangle given that , that its three heights are known, And .
Theorem 2. The area is calculated by the formula
. (5)
Proof. Since , and , then
In this case, from formula (1) we obtain
or
Formula (5) follows from this. The theorem has been proven.
Calculating the area of quadrilaterals
Consider a generalization of Heron's formula for the case of calculating the area of quadrilaterals. However, it should immediately be noted that such a generalization is possible only for quadrilaterals that are inscribed in a circle.
Let the quadrilateral has sides , , and .
If is a quadrilateral, inscribed in a circle, then Theorem 3 (Brahmagupta's formula) holds.
Theorem 3. Square calculated by the formula
Where .
Proof. Draw a diagonal in a quadrilateral and get two triangles and . If we apply the cosine theorem to these triangles, which is equivalent to formula (3), then we can write
Since the quadrilateral is inscribed in a circle, the sum of its opposite angles is , i.e. .
Because or then from (7) we get
Or
. (8)
Since , then . However, and therefore
Since , then formulas (8) and (9) imply
If we put , then from here we obtain formula (6). The theorem has been proven.
If the inscribed quadrilateralis at the same time described, then formula (6) is greatly simplified.
Theorem 4. The area of a quadrilateral, inscribed in one circle and described around the other, is calculated by the formula
. (10)
Proof. Since a circle is inscribed in a quadrilateral, the equalities
In this case , , , and formula (6) is easily converted into formula (10). The theorem has been proven.
Let's move on to the consideration of examples of geometry problems, the solution of which is carried out on the basis of the application of the proven theorems.
Examples of problem solving
Example 1. Find area, if .
Solution. Since here , according to Theorem 1 we obtain
Answer: .
Note, if the sides of the triangletake on irrational meanings, then the calculation of its areaby using formula (1), usually , is ineffective. In this case, it is expedient to apply formulas (2) and (3) directly.
Example 2 Find the area if , and .
Solution. Taking into account formulas (2) and (3), we obtain
Since , then or .
Answer: .
Example 3 Find the area if , and .
Solution. Because the ,
then it follows from Theorem 2 that .
Answer: .
Example 4 The triangle has sides , and . Find and , where are the radii of the circumscribed and inscribed circles, respectively.
Solution. Let's calculate the area first. Since , then from formula (1) we obtain .
It is known that and . That's why .
Example 5 Find the area of a quadrilateral inscribed in a circle if , , and .
Solution. It follows from the conditions of the example that . Then, according to Theorem 3, we obtain .
Example 6 Find the area of a quadrilateral inscribed in a circle with sides , , and .
Solution. Since and , then the equality holds in the quadrilateral. However, it is known that the existence of such an equality is a necessary and sufficient condition for a circle to be inscribed in a given quadrilateral. In this regard, formula (10) can be used to calculate the area, from which follows .
For independent and high-quality preparation for entrance tests in the field of solving problems of school geometry, you can effectively use textbooks, listed in the list of recommended readings.
1. Gotman E.G. Problems in planimetry and methods for their solution. – M.: Enlightenment, 1996. – 240 p.
2. Kulagin E.D. , Fedin S.N. Triangle geometry in problems. - M .: KD "Librocom" / URSS, 2009. - 208 p.
3. Collection of problems in mathematics for applicants to higher educational institutions / Ed. M.I. Scanavi. - M .: World and Education, 2013. - 608 p.
4. Suprun V.P. Mathematics for high school students: additional sections of the school curriculum. – M.: Lenand / URSS, 2014. - 216 p.
Do you have any questions?
To get the help of a tutor - register.
site, with full or partial copying of the material, a link to the source is required.
Preliminary information
To begin with, we introduce information and notation that will be needed in what follows.
We will consider triangle $ABC$ with acute angles $A$ and $C$. Draw a height $BH$ in it. Let us introduce the following notation: $AB=c,\ BC=a,\ $$AC=b,\ AH=x,\ BH=h\ $ (Fig. 1).
Picture 1.
We introduce without proof the triangle area theorem.
Theorem 1
The area of a triangle is defined as half the product of the length of its side and the height drawn to it, that is
Heron's formula
We introduce and prove a theorem on finding the area of a triangle given three known sides. This formula is called Heron's formulas.
Theorem 2
Let us be given three sides of a triangle $a,\ b\ and\ c$. Then the area of this triangle is expressed as follows
where $p$ is the half-perimeter of the given triangle.
Proof.
We will use the notation introduced in Figure 1.
Consider triangle $ABH$. By the Pythagorean theorem, we get
It is obvious that $HC=AC-AH=b-x$
Consider the triangle $\CBH$. By the Pythagorean theorem, we get
\ \ \
Equate the values of the squared height from the two obtained relations
\ \ \
From the first equation we find the height
\ \ \ \ \ \
Since the semiperimeter is equal to $p=\frac(a+b+c)(2)$, i.e. $a+b+c=2p$, then
\ \ \ \
By Theorem 1, we get
The theorem has been proven.
Examples of tasks for using the Heron formula
Example 1
Find the area of a triangle if its sides are $3$ cm, $6$ cm and $7$ cm.
Solution.
Let us first find the semiperimeter of this triangle
By Theorem 2, we get
Answer:$4\sqrt(5)$.
