Ability to think mathematically – one of the noblest abilities of man.
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 a triangle based on 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. This formula is called Brahmagupta's formula. Also little known is the formula for calculating the area of a triangle from its three altitudes, 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 true.
Theorem 1.
Where .
Proof. When deriving formula (1), we will use known geomes tric formulas
, (2)
. (3)
From formulas (2) and (3) we obtain and . Since then
. (4)
If we denote then from equality (4) formula (1) follows. The theorem has been proven.
Let us now consider the question of calculating the area of a triangle given that , that its three heights are known, And .
Theorem 2. The area is calculated using the formula
. (5)
Proof. Since , and , then
In this case, from formula (1) we obtain
or
From this follows formula (5). The theorem has been proven.
Calculating the area of quadrilaterals
Let's consider a generalization of Heron's formula to 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) is true.
Theorem 3. Square calculated by the formula
Where .
Proof. Let's draw a diagonal in the 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 a quadrilateral is inscribed in a circle, the sum of its opposite angles is equal, i.e. .
Because or then from (7) we obtain
Or
. (8)
Since, then. However and therefore
Since , then from formulas (8) and (9) it follows
If you put then from here we obtain formula (6). The theorem has been proven.
If a cyclic quadrilateralis also described, then formula (6) is significantly simplified.
Theorem 4. The area of a quadrilateral inscribed in one circle and circumscribed around another is calculated by the formula
. (10)
Proof. Since a circle is inscribed in a quadrilateral, the equalities hold:
In this case, , , , and formula (6) is easily converted into formula (10). The theorem has been proven.
Let's move on to consider examples of geometry problems, the solution of which is carried out on the basis of the application of proven theorems.
Examples of problem solving
Example 1. Find area, if , and .
Solution. Since here, then according to Theorem 1 we obtain
Answer: .
Note, if the sides of a triangletake on irrational values, then calculating its areaby using formula (1) usually is ineffective. In this case, it is advisable to directly apply formulas (2) and (3).
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. Since,
then from Theorem 2 it follows that .
Answer: .
Example 4. A triangle has sides , and . Find and , where are the radii of the circumscribed and inscribed circles, respectively.
Solution. First, let's calculate the area. Since , then from formula (1) we obtain .
It is known that . Therefore and.
Example 5. Find the area of a quadrilateral inscribed in a circle if , , and .
Solution. From the example conditions it follows that . Then, according to Theorem 3, we obtain .
Example 6. Find the area of a quadrilateral inscribed in a circle whose sides are , , and .
Solution. Since and , the equality holds in the quadrilateral. However, it is known that the existence of such an equality is a necessary and sufficient condition for the fact that a circle can be inscribed in a given quadrilateral. In this regard, to calculate the area, you can use formula (10), from which it follows.
For independent and high-quality preparation for entrance examinations in the field of solving school geometry problems, you can effectively use textbooks, listed in the list of recommended literature.
1. Gotman E.G. Problems in planimetry and methods for solving them. – M.: Education, 1996. – 240 p.
2. Kulagin E.D. , Fedin S.N. Geometry of a triangle in problems. – M.: CD “Librocom” / URSS, 2009. – 208 p.
3. Collection of problems in mathematics for applicants to colleges / Ed. M.I. Scanavi. – M.: Peace 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.
Still have questions?
To get help from a tutor -.
blog.site, when copying material in full or in part, a link to the original source is required.
Heron's formula Heron's formula
expresses area s of a triangle through the lengths of its three sides A, b And With and semiperimeter r = (A + b + With)/2: . Named after Heron of Alexandria.
HERONA FORMULAHERONA FORMULA, expresses area S of a triangle through 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 "Heron's formula" is in other dictionaries:
Expresses the area S of a triangle through the lengths of its three sides a, b and c and the semi-perimeter P = (a + b + c)/2Named after Heron of Alexandria... Big Encyclopedic Dictionary
Formula expressing the area of a triangle through its three sides. Namely, if a, b, C are the length of the sides of a triangle, and S is its area, then the G. f. has the form: where p denotes the semi-perimeter of the triangle G. f.... ...
A formula expressing the area of a triangle through 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 through the lengths of its three sides a, b and c and the semi-perimeter p = (a + b + c)/2: s = 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) based on its sides a, b, c: where p is the semi-perimeter 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 semi-perimeter, then its area is ... Wikipedia
This article lacks links to sources of information. Information must be verifiable, otherwise it may be questioned and deleted. You can edit this article to add links to authoritative sources. This mark... ... Wikipedia
- (Heronus Alexandrinus) (years of birth and death unknown, probably 1st century), 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), 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 based on its sides a, b and c:
S=√(р(р-а)(р-b)(р-с),where p is the semi-perimeter of the triangle, i.e. p = (a + b + c)/2.
The formula is named after the ancient Greek mathematician Heron of Alexandria (circa 1st century). Heron considered triangles with integer sides whose areas are also integers. Such triangles are called Heronian triangles. 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 does not have a unique solution, to calculate the area of a quadrilateral in the general case, it is not enough just 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=√(р-а)(р-b)(р-с)(p-d)
If a quadrilateral is both inscribed and circumscribed at the same time, its area is using a simpler formula: S=√(abcd).
Heron of Alexandria - Greek mathematician and mechanic.
He was the first to invent automatic doors, an automatic puppet theater, a vending machine, a rapid-fire self-loading crossbow, a steam turbine, automatic decorations, 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, and optics. Main works: Metrics, Pneumatics, Automatopoetics, Mechanics (the work is preserved entirely in Arabic translation), Catoptrics (the science of mirrors; preserved only in Latin translation), etc. In 1814, Heron’s essay “On Diopter” was found, which sets out the rules land surveying, actually based on the use of rectangular coordinates. Heron used the achievements of his predecessors: Euclid, Archimedes, Strato of Lampsacus. Many of his books are irretrievably lost (the scrolls were kept in the Library of Alexandria).
In his treatise “Mechanics,” Heron described five types of simple machines: lever, gate, wedge, screw and block.
In his treatise “Pneumatics,” Heron described various siphons, cleverly designed vessels, and automata driven by compressed air or steam. This is an aeolipile, which was the first steam turbine - a ball rotated by the force of jets of water vapor; a machine for opening doors, a machine for selling “holy” water, a fire pump, a water organ, a mechanical puppet theater.
The book “About the Diopter” describes the diopter - the simplest device used for geodetic work. Heron sets out in his treatise the rules for land surveying, based on the use of rectangular coordinates.
In Catoptrics, Heron substantiates the straightness of light rays with an infinitely high speed of propagation. Heron considers various types of mirrors, paying particular attention to cylindrical mirrors.
Heron's "Metrics" and the "Geometrics" and "Stereometrics" extracted from it are reference books on applied mathematics. Among the information contained in Metrica:
Formulas for the 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 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 abilities of man.
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 a triangle based on 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. This formula is called Brahmagupta's formula. Also little known is the formula for calculating the area of a triangle from its three altitudes, 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 true.
Theorem 1.
Where .
Proof. When deriving formula (1), we will use known geomes tric formulas
, (2)
. (3)
From formulas (2) and (3) we obtain and . Since then
. (4)
If we denote then from equality (4) formula (1) follows. The theorem has been proven.
Let us now consider the question of calculating the area of a triangle given that , that its three heights are known, And .
Theorem 2. The area is calculated using the formula
. (5)
Proof. Since , and , then
In this case, from formula (1) we obtain
or
From this follows formula (5). The theorem has been proven.
Calculating the area of quadrilaterals
Let's consider a generalization of Heron's formula to 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) is true.
Theorem 3. Square calculated by the formula
Where .
Proof. Let's draw a diagonal in the 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 a quadrilateral is inscribed in a circle, the sum of its opposite angles is equal, i.e. .
Because or then from (7) we obtain
Or
. (8)
Since, then. However and therefore
Since , then from formulas (8) and (9) it follows
If you put then from here we obtain formula (6). The theorem has been proven.
If a cyclic quadrilateralis also described, then formula (6) is significantly simplified.
Theorem 4. The area of a quadrilateral inscribed in one circle and circumscribed around another is calculated by the formula
. (10)
Proof. Since a circle is inscribed in a quadrilateral, the equalities hold:
In this case, , , , and formula (6) is easily converted into formula (10). The theorem has been proven.
Let's move on to consider examples of geometry problems, the solution of which is carried out on the basis of the application of proven theorems.
Examples of problem solving
Example 1. Find area, if , and .
Solution. Since here, then according to Theorem 1 we obtain
Answer: .
Note, if the sides of a triangletake on irrational values, then calculating its areaby using formula (1) usually is ineffective. In this case, it is advisable to directly apply formulas (2) and (3).
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. Since,
then from Theorem 2 it follows that .
Answer: .
Example 4. A triangle has sides , and . Find and , where are the radii of the circumscribed and inscribed circles, respectively.
Solution. First, let's calculate the area. Since , then from formula (1) we obtain .
It is known that . Therefore and.
Example 5. Find the area of a quadrilateral inscribed in a circle if , , and .
Solution. From the example conditions it follows that . Then, according to Theorem 3, we obtain .
Example 6. Find the area of a quadrilateral inscribed in a circle whose sides are , , and .
Solution. Since and , the equality holds in the quadrilateral. However, it is known that the existence of such an equality is a necessary and sufficient condition for the fact that a circle can be inscribed in a given quadrilateral. In this regard, to calculate the area, you can use formula (10), from which it follows.
For independent and high-quality preparation for entrance examinations in the field of solving school geometry problems, you can effectively use textbooks, listed in the list of recommended literature.
1. Gotman E.G. Problems in planimetry and methods for solving them. – M.: Education, 1996. – 240 p.
2. Kulagin E.D. , Fedin S.N. Geometry of a triangle in problems. – M.: CD “Librocom” / URSS, 2009. – 208 p.
3. Collection of problems in mathematics for applicants to colleges / Ed. M.I. Scanavi. – M.: Peace 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.
Still have questions?
To get help from a tutor, register.
website, when copying material in full or in part, a link to the source is required.
Preliminary information
First, let's introduce the information and notation that we will need later.
We will consider a triangle $ABC$ with acute angles $A$ and $C$. Let us draw the height $BH$ in it. Let us introduce the following notation: $AB=c,\ BC=a,\ $$AC=b,\ AH=x,\ BH=h\ $(Fig. 1).
Figure 1.
Let us introduce, without proof, the theorem on the area of a triangle.
Theorem 1
The area of a triangle is defined as half the product of the length of its side and the altitude drawn to it, that is
Heron's formula
Let us introduce and prove a theorem about finding the area of a triangle from 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 semi-perimeter of the given triangle.
Proof.
We will use the notation introduced in Figure 1.
Consider triangle $ABH$. According to the Pythagorean theorem, we get
It is obvious that $HC=AC-AH=b-x$
Consider the triangle $\CBH$. According to the Pythagorean theorem, we get
\ \ \
Let us equate the values of the squared height from the two obtained ratios
\ \ \
From the first equality we find the height
\ \ \ \ \ \
Since the semi-perimeter is equal to $p=\frac(a+b+c)(2)$, that is, $a+b+c=2p$, then
\ \ \ \
By Theorem 1, we get
The theorem has been proven.
Examples of problems using Heron's 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 semi-perimeter of this triangle
By Theorem 2, we get
Answer:$4\sqrt(5)$.
