Can be found by knowing the base and height. The whole simplicity of the scheme 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 \u200b\u200bwhich is obtained and. Then the area of the entire triangle will be the sum of the two indicated areas, and if we take one half of the height out of the bracket, then in total we get back the base:
A more difficult method for calculations is the Heron formula, for which you need to know all three sides. For this formula, you must first calculate the semiperimeter 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 follows from the formula with height - we draw the height to any of the known sides and through the sine of the angle α we get 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 given 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 top of the third corner to the known side and call the resulting segments x, respectively. It can be seen from right triangles that the first segment x is equal to the product
Lesson summary
Subject: "Heron's formula and other formulas for the area of a triangle".
Lesson type : a lesson in the discovery of new knowledge.
Class: 10.
Lesson Objectives: ensure during the lesson the conscious repetition of the formulas for calculating the area of a triangle, which are studied in the school curriculum. Show the need for knowledge of Heron's formula II, the formula for the area of a triangle given in a rectangular coordinate system. Ensure the conscious assimilation and application of these formulas in solving problems.
Tasks:
Developing: development of logical thinking, the ability to independently solve educational problems; curiosity developmentstudents, cognitive interest in the subject; development of creative thinking, mathematical speech of students;
Educational: fostering interest in mathematics; creating conditions forformation of communication skills and volitional qualities of a person.
Educational: deepening knowledgeth module of a real number; teach the ability to solve typical problems.
Universal learning activities:
Personal: respect for the individual and his dignity; sustainable cognitive interest; the ability to conduct a dialogue based on 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 on the basis of negotiations.
Cognitive: V get along with general methods of solving problems, performing tasks and calculations; perform tasks based on the use of properties of the modulus of a real number.
Communicative: A adequately use speech to plan and regulate their 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 the implementation of the first trial action - 10 minutes.
Identification of the difficulty: what is the complexity of the new material, what exactly creates the problem, search for a contradiction - 4 min.
Development of a project, a plan to overcome their difficulties, consideration of many options, search for the optimal solution - 2 min.
Implementation of the selected plan to resolve the difficulty - 5 min.
Primary consolidation of new knowledge - 10 min.
Independent work and verification according to the standard - 5 min.
Reflection includes reflection learning activities, and introspection, and reflection of feelings and emotions - 1 min.
During the classes.
motivational stage.
Hello guys, have a seat. Today our lesson will be held according to the following plan: during the lesson we will study a new topic: “ Heron's formula and other formulas for the area of a triangle »; repeat those formulas that you know; Let's learn how to apply these formulas to solve problems. So, let's get to work.
The stage of updating knowledge on the proposed topic and the implementation of 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 for calculating the area of a triangle do you know?(students remember all the formulas they studied)
Slide 3.
Area of a right triangle. S=ab. Write down the formula
slide 4.
The area of any triangle. S= A . a = , = Write down the formula.
Slide 5. The area of a triangle on two sides and the angle between them.
S=½ ab sinα. Write down the formula.
And now we will learn new formulas for finding the 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.
The area of a triangle in terms of the R-radius of the circumscribed circle.
Write down the formula.
slide 8.
Heron formula.
Before proceeding to the proof, we recall two theorems of geometry - this is the sine theorem and the cosine theorem.
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 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 "Metric" of Heron of Alexandria (I in 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.
Identification of difficulties: what is the complexity of the new material, what exactly creates the problem, the search for a contradiction.
slide 12.
Find the area of a triangle with given sides: 4,6,8. Is there enough information to solve the problem? What formula can be used to solve this problem?
Development of a project, a plan to get out of their difficulties, consideration of many options, search for the 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 obtained values into the formula.
Implementation of the selected 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 #2
Find the sides of the triangleABCif the area of the trianglesABO, BCO, ACO, where O is the center of the inscribed circle, are equal to 17.65.80 dc 2 .
Solution: 
S\u003d 17 + 65 + 80 \u003d 162 - add up the areas of the triangles. According to the formula
S ABO =1/2 AB* r, hence 17=1/2AB* r; 65=1/2BC* r; 80=1/2 AC* 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 given sides:
№12
Independent work and verification according to the standard.
№10.(2)
Homework . P.83, No. 10(3), No. 15
Reflection, which includes both the reflection of educational activity, and introspection, and the reflection of feelings and emotions.
What formulas did you repeat today?
What formulas did you learn just today?
Theorem. The area of a triangle is equal to half the product of its side and the height drawn to it:
The proof is very simple. This triangle ABC(Fig. 1.15) we will complete the 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 the 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, this is not done in school textbooks. On the contrary, the equality of three half-products is established on the basis of the fact that all these half-products express the area of a triangle. Thus, the existence of a single function is implicitly exploited. But here there is a convenient and instructive opportunity to demonstrate an example of mathematical modeling. Indeed, there is a physical reality behind the concepts of area, but a direct check of the equality of three semi-products shows the quality factor of the translation of this concept into the language of mathematics.
Using the triangle area theorem above, it is very often convenient to compare the areas of two triangles. We present below some obvious but important corollaries of 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.
On fig. 1.17 triangles ABC And ABD have a common ground AB and equal heights lowered on this base, since the straight line A, which contains the vertices WITH And D parallel to the base AB, so the areas of these triangles are equal.
Corollary 1 can be reformulated as follows.
Consequence 1?. Let a segment be given AB. Many points M such that the area of the triangle AMV equal to the given value S, there are two lines parallel to the segment AB and located at a distance from it (Fig. 1. 18)
Consequence 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.
On fig. 1.19 triangles ABC And ABD have a common height BH, so 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 equal parts.
2. The bisector of the angle of a triangle enclosed between its sides A And b, divides it into two triangles whose areas are related as a : b.
Corollary 3. If two triangles have a common angle, then their areas are related as the products of the sides containing this angle.
This follows from the fact that (Fig. 1.19)
In particular, the following assertion holds:
If two triangles are similar and the side of one of them is k times 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 side c.
From (1.3) we find.
Noticing that
where is the semiperimeter of the triangle, we get.
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:$4\sqrt(5)$.
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".
