Let's find the triangle using Heron's formula. Area of a triangle. Calculating the area of quadrilaterals
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 has more than one 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, 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".
Lesson summary
Subject: "Heron's formula and other formulas for the area of a triangle."
Lesson type : a lesson in discovering new knowledge.
Class: 10.
Lesson objectives: during the lesson, ensure conscious repetition of formulas for calculating the area of a triangle, which are studied in school curriculum. Show the need to know Heron's formula II, the formula for the area of a triangle given in a rectangular coordinate system. Ensure conscious assimilation and application of these formulas when solving problems.
Tasks:
Educational: development logical thinking, ability to decide independently learning objectives; development curiositystudents, cognitive interest in the subject; development of creative thinking and mathematical speech of students;
Educational: nurturing interest in mathematics; creating conditions forformation of communication skills and strong-willed qualities personality.
Educational: deepening knowledgeth modulus of a real number; teach the ability to solve typical problems.
Universal learning activities:
Personal: respect for the individual and his dignity; stable cognitive interest; the ability to conduct dialogue on the basis of equal relations and mutual respect.
Regulatory: set goals for activities in the lesson; plan ways to achieve the goal; make decisions in a problem situation based on negotiations.
Cognitive: V master general techniques for solving problems, performing tasks and calculations; perform tasks based on the use of real number modulus properties.
Communicative: A adequately use speech to plan and regulate one’s activities; formulate your own opinion.
Technical support : computer, projector, interactive whiteboard.
Lesson structure
Motivational stage – 2 min.
Homework – 1 min.
The stage of updating knowledge on the proposed topic and carrying out the first trial action – 10 minutes.
Identifying difficulties: what is the complexity of the new material, what exactly creates the problem, searching for contradictions - 4 min.
Development of a project, a plan to resolve their existing difficulties, consideration of many options, search for the optimal solution - 2 min.
Implementation of the chosen plan to resolve the difficulty - 5 min.
Primary consolidation of new knowledge - 10 min.
Independent work and checking against the standard - 5 min.
Reflection, which includes reflection on learning activities, self-analysis, and reflection on feelings and emotions – 1 min.
During the classes.
Motivational stage.
Hello guys, have a seat. Today our lesson will follow the following plan: during the lesson we will study a new topic: “ Heron's formula and other formulas for the area of a triangle "; Let's repeat the formulas that you know; Let's learn how to apply these formulas when solving problems. So, let's get to work.
The stage of updating knowledge on the proposed topic and carrying out the first trial action.
Slide 1.
Write down the topic of the lesson. Before proceeding directly to the formulas, let's remember what formulas for calculating the area of a triangle do you know?
Slide 2.
Write these formulas.
What formulas do you know for calculating the area of a triangle?(students recall all the formulas they have learned)
Slide 3.
Area of a right triangle. S=ab. Write down the formula
Slide 4.
Area of any triangle. S= A . a = , = Write down the formula.
Slide 5. The area of a triangle based on two sides and the angle between them.
S=½·ab·sinα. Write down the formula.
Now we will study new formulas for finding area.
Slide 6.
The area of a triangle in terms of the radius of the inscribed circle. S= P r. Write down the formula.
Slide 7.
Area of a triangle in terms of R-radius of the circumcircle.
Write down the formula.
Slide 8.
Heron's formula.
Before we begin the proof, let's remember two theorems of geometry - the theorem of sines and the theorem of cosines.
1. , a=2R; b=2R; c=2R
2.,cosγ = .
Slide 9- 10
Proof of Heron's formula. Write down the formula.
Slide 11.
The formula for the area of a triangle based on three sides was discovered by Archimedes in the 3rd century BC. However, the corresponding work has not reached our days. This formula is contained in the “Metrics” of Heron of Alexandria (1st century AD) and is named after him. Heron was interested in triangles with integer sides whose areas are also integer. Such triangles are called Heronian triangles. The simplest Heronian triangle is the Egyptian triangle
Identifying the difficulty: what is the complexity of the new material, what exactly creates the problem, searching for a contradiction.
Slide 12.
Find the area of a triangle with the given sides: 4,6,8. Is there enough information to solve the problem? What formula can you use to solve this problem?
Development of a project, a plan to resolve their existing difficulties, consideration of many options, search for an optimal solution.
This problem can be solved using Heron's formula. First, you need to find the semi-perimeter of the triangle, and then substitute the resulting values into the formula.
Implementation of the chosen plan to resolve the difficulty.
Finding p
p=(13+14+15)/2=21
p- a=21-13=8
p-b=21-14=7
p-c=21-15=6
S = 21*8*7*6=84
Answer :84
Task No. 2
Find the sides of the triangleABC, if the area of the trianglesABO, BCO, ACO, where O is the center of the inscribed circle, equal to 17,65,80 dc 2 .
Solution: 
S=17+65+80=162 – add up the areas of the triangles. According to the formula
S ABO =1/2 AB* r, therefore 17=1/2AB* r; 65=1/2ВС* r; 80=1/2 A.C.* r
34/r=AB; 130/r=BC; 160/r=AC
Find p
p= (34+130+160)/2=162/ r
(r-a)=162-34=128 (r- c)=162-160=2
(R- b)=162-130=32
According to Heron's formulaS= 128/ r*2/ r*32/ r*162/ r=256*5184/ r 4 =1152/ r 2
Because S=162, thereforer = 1152/162=3128/18
Answer: AB=34/3128/18, BC=130/3128/18, AC=160/3128/18.
Primary consolidation of new knowledge.
№10(1)
Find the area of a triangle with the given sides:
№12
Independent work and testing against the standard.
№10.(2)
Homework . P.83, No. 10(3), No. 15
Reflection, which includes reflection on educational activities, introspection, and reflection on feelings and emotions.
What formulas did you repeat today?
What formulas did you learn just today?
Can be found by knowing the base and height. The whole simplicity of the diagram lies in the fact that the height divides the base a into two parts a 1 and a 2, and the triangle itself into two right triangles, the area of which is and. Then the area of the entire triangle will be the sum of the two indicated areas, and if we take one second of the height out of the bracket, then in the sum we get back the base:
A more difficult method for calculations is Heron’s formula, for which you need to know all three sides. For this formula, you first need to calculate the semi-perimeter of the triangle: 
Heron's formula itself implies the square root of the semi-perimeter, multiplied in turn by its difference on each side.
The following method, also relevant for any triangle, allows you to find the area of the triangle through two sides and the angle between them. The proof of this comes from the formula with height - we draw the height on any of the known sides and through the sine of the angle α we obtain that h=a⋅sinα. To calculate the area, multiply half the height by the second side.
Another way is to find the area of a triangle, knowing 2 angles and the side between them. The proof of this formula is quite simple and can be clearly seen from the diagram.
We lower the height from the vertex of the third angle to the known side and call the resulting segments x accordingly. From right triangles it is clear that the first segment x is equal to the product
Theorem. The area of a triangle is equal to half the product of its side and its altitude:
The proof is very simple. This triangle ABC(Fig. 1.15) let's build it up to a parallelogram ABDC. Triangles ABC And DCB are equal on three sides, so their areas are equal. So the area of the triangle ABC equal to half the area of the parallelogram ABDC, i.e.
But here the following question arises: why are the three possible half-products of the base and height for any triangle the same? This, however, is easy to prove from the similarity of rectangles with a common acute angle. Consider a triangle ABC(Fig. 1.16):
And therefore
However, in school textbooks That's not how it's done. On the contrary, the equality of the three half-products is established on the basis that all these half-products express the area of the triangle. Thus, the existence of a single function is implicitly exploited. But here comes a convenient and instructive opportunity to demonstrate an example mathematical modeling. Indeed, there is a physical reality behind the concept of area, but direct verification of the equality of three half-products shows the quality of the translation of this concept into the language of mathematics.
Using the above triangle area theorem, it is often convenient to compare the areas of two triangles. Below we present some obvious but important consequences from the theorem.
Corollary 1. If the vertex of a triangle is moved along a straight line parallel to its base, then its area does not change.
In Fig. 1.17 triangles ABC And ABD have a common ground AB and equal heights lowered onto this base, since a straight line A, which contains the vertices WITH And D parallel to the base AB, and therefore the areas of these triangles are equal.
Corollary 1 can be reformulated as follows.
Corollary 1?. Let a segment be given AB. Many points M such that the area of the triangle AMV equal to given value S, there are two lines parallel to the segment AB and those located at a distance from it (Fig. 1. 18)
Corollary 2. If one of the sides of a triangle adjacent to a given angle is increased by k times, then its area will also increase by k once.
In Fig. 1.19 triangles ABC And ABD have a common height BH, therefore the ratio of their areas is equal to the ratio of the bases
Important special cases follow from Corollary 2:
1. The median divides the triangle into two small parts.
2. Bisector of an angle of a triangle, enclosed between its sides A And b, divides it into two triangles, the areas of which are related as a : b.
Corollary 3. If two triangles have a common angle, then their areas are proportional to the product of the sides enclosing this angle.
This follows from the fact that (Fig. 1.19)
In particular, the following statement holds:
If two triangles are similar and the side of one of them is k times larger than the corresponding sides of the other, then its area is k 2 times the area of the second.
We derive Heron's formula for the area of a triangle in the following two ways. In the first we use the cosine theorem:
where a, b, c are the lengths of the sides of the triangle, r is the angle opposite to side c.
From (1.3) we find.
Noticing that
where is the semi-perimeter of the triangle, we get.
Loading...