The purpose of the lesson: give a definition of similar triangles, prove the theorem on the ratio of similar triangles.
Lesson objectives:
- Educational: students should know the definition of similar triangles, the theorem on the ratio of similar triangles, be able to apply them in solving problems, implement interdisciplinary connections with algebra and physics.
- Educational: to cultivate diligence, attentiveness, diligence, to cultivate a culture of behavior of students.
- Developing: development of students' attention, development of the ability to reason, think logically, draw conclusions, develop students' competent mathematical speech and thinking, develop skills of introspection and independence.
- Health saving: observance of sanitary and hygienic standards, change of activities in the lesson.
Equipment: computer, projector, didactic material: independent and control work in algebra and geometry for grade 8 A.P. Ershova, etc.
Lesson type: learning new material.
During the classes
I. Organizational moment(greeting, checking readiness for the lesson).
II. The topic of the lesson.
Teacher: In everyday life there are objects of the same shape, but different sizes.
Example: soccer and tennis balls.
In geometry, figures of the same shape are called similar: any two circles, any two squares.
Let us introduce the concept of similar triangles.
Definition: Two triangles are said to be similar if their angles are respectively equal and the sides of one triangle are proportional to the similar sides of the other.
Number k, equal to the ratio of similar sides of similar triangles is called the coefficient of similarity. ∆ABC ~ A 1 B 1 C 1
1. Orally: Are triangles similar? Why? (prepared drawing on the screen).
a) Triangle ABC and triangle A 1 B 1 C 1 if AB = 7, BC = 5, AC = 4, ∠A = 46˚, ∠C = 84˚, ∠A 1 = 46˚, ∠B 1 = 50 ˚, A 1 B 1 \u003d 10.5, B 1 C 1 \u003d 7.5, A 1 C 1 \u003d 6.
b) In one isosceles triangle, the angle at the apex is 24˚, and in the other isosceles triangle, the angle at the base is 78˚.
Guys! Recall the theorem on the ratio of the areas of triangles having an equal angle.
Theorem: If the angle of one triangle is equal to the angle of another triangle, then the areas of these triangles are related as the products of the sides containing equal angles.
2. Written work according to prepared drawings.
On-screen drawing:
a) Given: BN: NC = 1:2,
BM=7cm, AM=3cm,
S MBN \u003d 7 cm 2.
Find: S ABC
(Answer: 30 cm2.)
b) Given: AE = 2 cm,
S AEK \u003d 8 cm 2.
Find: S ABC
(Answer: 56 cm2.)
3. Let us prove the theorem on the ratio of the areas of similar triangles ( the student proves the theorem on the blackboard, the whole class helps).
Theorem: The ratio of two similar triangles is equal to the square of the similarity coefficient.
4. Actualization of knowledge.
Problem solving:
1. The areas of two similar triangles are 75 cm 2 and 300 cm 2. One of the sides of the second triangle is 9cm. Find the side of the first triangle that is similar to it. ( Answer: 4.5 cm.)
2. Similar sides of similar triangles are 6 cm and 4 cm, and the sum of their areas is 78 cm 2. Find the areas of these triangles. ( Answer: 54 cm2 and 24 cm2.)
If there is time independent work educational character.
Option 1
Similar triangles have similar sides equal to 7 cm and 35 cm.
The area of the first triangle is 27 cm2.
Find the area of the second triangle. ( Answer: 675 cm2.)
Option 2
The areas of similar triangles are 17 cm 2 and 68 cm 2. The side of the first triangle is 8cm. Find the similar side of the second triangle. ( Answer: 4 cm)
5. Homework: geometry textbook 7-9 L.S. Atanasyan and others, pp. 57, 58, no. 545, 547.
6. Summing up the lesson.
Type of lesson: lesson of acquaintance with new material.The purpose of the lesson: To prove the property of the areas of similar triangles and show its practical significance in solving problems.
Lesson objectives:
teaching - to prove the property of the areas of similar triangles and show its practical significance in solving problems;
developing - to develop the ability to analyze and select arguments when solving a problem, the method of solving which is unknown;
educational - to cultivate interest in the subject through the content of the educational process and the creation of a situation of success, to cultivate the ability to work in a group.
The student has the following knowledge:
The unit of activity content that students need to learn:
During the classes.
1. Organizational moment.
2. Actualization of knowledge.
3. Dealing with a problem situation.
4. Summing up the lesson and recording homework, reflection.
Teaching methods: verbal, visual, problem-search.
Forms of training: frontal work, work in mini-groups, individual and independent work.
Technologies: task-oriented, information technologies, competency-based approach.
Equipment:
a computer, a projector for demonstrating a presentation, an interactive whiteboard, a document camera;
computer presentation in Microsoft PowerPoint;
reference summary;
During the classes
1. Organizational moment.
Today at the lesson we will work not in notebooks, but in supporting notes, which you will fill out for the duration of the entire lesson. Sign it. The assessment for the lesson will consist of two components: for the reference notes and for active work in the lesson.
2. Actualization of students' knowledge. Preparation for active educational and cognitive activity at the main stage of the lesson.
We continue to study the topic "similarity of triangles". So let's remember what we learned in the last lesson.
Theoretical workout. Test. In your reference notes, the first task has a test character. Answer the questions by choosing one of the suggested answers, where necessary, enter your answer.
Teacher: What is the ratio of two segments?
Answer: The ratio of two segments of two segments is the ratio of their lengths.
Teacher: In what case are the segmentsAB And CDproportional to segmentsA 1 B 1 and C 1 D 1
Answer: cuts AB And CDproportional to segmentsA 1 B 1 and C 1 D 1 if
your options. Fine. Don't forget to correct whoever is wrong.
Teacher: What is the definition of similar triangles? Refer to your reference abstract. You have three answers to this question. Choose the right one. Circle it.
So, please, which option did you choose _______
Answer: Two triangles are called similar if their angles are respectively equal and the sides of one triangle are proportional to the sides of the other triangle.
Well done! Correct whoever is wrong.
Teacher: What is the ratio of the areas of two triangles that have the same angle?
Answer: If the angle of one triangle is equal to the angle of another triangle, then the areas of these triangles are divided as the products of the sides containing equal angles.
Solution of problems according to ready-made drawings.Further, our warm-up will take place in the course of solving problems according to ready-made drawings. You also see these tasks in your reference notes.
Reflection. Let's clarify what knowledge and skills allowed us to solve these problems. What solution methods did we use (fixing the answers on the board).
Possible answers:
Definition of similar triangles;
Application of the definition of similar triangles in solving problems;
Theorem on the ratio of the areas of triangles having an equal angle;
And now I propose a method for solving several problems that resonates with the topic of the lesson, but they are more related to geography.
situation of success.
The first task is in front of you. We are working on this issue on our own. The first one who succeeds will show his solution at the blackboard, and someone will demonstrate his solution through a document camera, so we write beautifully and accurately.
Answer: the sides of the Bermuda Triangle are 2000 km, 1840 km, 2220 km. The length of the border is 6060 km.
Reflection.
Possible answer: Similar triangles have similar sides that are proportional.
situation of success.
We figured out the dimensions of the Bermuda Triangle. Well, now let's find out the measurements of the flower bed. Flipping the base notes. Second task. We solve this problem by working in pairs. We check in a similar way, but only the result will be the first pair that has completed the task.
Answer: the sides of a triangular flower bed are 10m and 11m 20 cm.
So, let's check in. Does everyone agree? Who decides in a different way?
Reflection.
What course of action did you use to solve this problem? Record in your master note.
Possible answer:
similar triangles have corresponding angles equal;
The areas of triangles with equal angles are the products of the sides containing equal angles.
Failure situation.
5. Learning new material.
When solving the third task, students are faced with a problem. They fail to solve the problem, because in their opinion the condition of the problem is not complete enough or they receive an unreasonable answer.
Students have not encountered this type of problem before, so there was a failure in solving the problem.
Reflection.
What method did you try to solve?
Why didn't you solve the last equation?
Pupils: We cannot find the area of a triangle if only the area of a similar triangle and the coefficient of similarity are known.
Thus, the purpose of our lesson find the area of a triangle if only the area of a similar triangle and the coefficient of similarity are known.
Let's reformulate the problem in geometric language. Let's solve it, and then return to this problem.
Conclusion: The ratio of the areas of similar triangles is equal to the square of the similarity coefficient.
Well, now let's go back to problem number 3 and solve it, based on a proven fact.
7. Summary of the lesson
What did you learn to do today?
Solve problems in which the coefficient of similarity and the area of one of the similar triangles are known.
What geometric property helped us in this?
The ratio of the areas of similar triangles is equal to the square of the similarity coefficient.
Homework.
P. 58 p.139 No. 546, 548
Creative task.
Find what is the ratio of the perimeters of two similar triangles (№547)
Goodbye.
Definition and properties of similar triangles
The numbers a 1 , a 2 , a 3 , ..., a n are called proportional to the numbers b 1 , b 2 , b 3 , ..., b n if the equality holds: a 1 / b 1 = a 2 / b 2 = a 3 / b 3 = ... = a n /b n = k, where k is a certain number, which is called the coefficient of proportionality.
Example. Numbers 6; 7.5 and 15 are proportional to -4; 5 and 10. The proportionality factor is -1.5 because
6/-4 = -7,5/5 = 15/-10 = -1,5.
The proportionality of numbers takes place if these numbers are related by proportion.
It is known that a proportion can be composed of at least four numbers, so the concept of proportionality is applicable to at least four numbers (one pair of numbers is proportional to another pair, or one triple of numbers is proportional to another triple, etc.).
Consider on rice. 1 two triangles ABC and A 1 B 1 C 1 with equal angles in pairs: A \u003d A 1, B \u003d B 1, C \u003d C 1.
Sides that are opposite equal pairs of angles of both triangles are called similar. Yes, on rice. 1 sides AB and A 1 B 1 , AC and A 1 C 1 , BC and B 1 C 1 , similar because they lie opposite the equal angles of triangles ABC and A 1 B 1 C 1 respectively.
Let's define similar triangles:
The two triangles are called similar, if their angles are pairwise equal, and similar sides are proportional.
The ratio of similar sides of similar triangles is called similarity coefficient.
Similar triangles are denoted as follows: Δ ABC ~ Δ A 1 B 1 C 1 .
So on rice. 2 we have: Δ ABC ~ Δ A 1 B 1 C 1
angles A \u003d A 1, B \u003d B 1, C \u003d C 1 and AB / A 1 B 1 \u003d BC / B 1 C 1 \u003d AC / A 1 C 1 \u003d k, where k is the similarity coefficient. From rice. 2 it can be seen that similar triangles have the same proportions, and they differ only in scale.
Note 1: Equal triangles are similar with a factor of 1.
Note 2: When designating similar triangles, their vertices should be ordered in such a way that the angles at them are equal in pairs. For example, for the triangles shown in Figure 2, it is incorrect to say that Δ ABC ~ Δ B 1 C 1 A 1. Observing the correct order of the vertices, it is convenient to write out the proportion connecting the similar sides of the triangles without referring to the drawing: the numerator and denominator of the corresponding ratios should contain pairs of vertices that occupy the same positions in the designation of similar triangles. For example, from the notation "Δ ABC ~ Δ KNL" it follows that the angles A = K, B = N, C = L, and AB / KN = BC / NL = AC / KL.
Note 3: The requirements listed in the definition of similar triangles are redundant. Triangle similarity criteria, which contain fewer requirements for similar triangles, will be proved a little later.
Let's formulate properties of similar triangles:
- The ratio of the corresponding linear elements of similar triangles is equal to the coefficient of their similarity. Such elements of similar triangles include those that are measured in units of length. This is, for example, the side of a triangle, perimeter, median. Angle or area are not such elements.
- The ratio of the areas of similar triangles is equal to the square of their similarity coefficient.
Let triangles ABC and A 1 B 1 C 1 be similar with coefficient k (Fig. 2).
Let us prove that S ABC /S A1 B1 C1 = k 2 .
Since the angles of similar triangles are pairwise equal, i.e. A \u003d A 1, and according to the theorem on the ratio of the areas of triangles with equal angles, we have:
S ABC /S A1 B1 C1 \u003d (AB AC) / (A 1 B 1 A 1 C 1) \u003d AB / A 1 B 1 AC / A 1 C 1.
Due to the similarity of triangles AB/A 1 B 1 = k and AC/A 1 C 1 = k,
so S ABC /S A1 B1 C1 = AB/A 1 B 1 AC/A 1 C 1 = k k = k 2 .
Note: The properties of similar triangles formulated above are also valid for arbitrary figures.
Signs of similarity of triangles
The requirements that are imposed on similar triangles by definition (this is the equality of angles and proportionality of sides) are redundant. You can also set the similarity of triangles by a smaller number of elements.
So, when solving problems, the first sign of similarity of triangles is most often used, stating that for the similarity of two triangles, it is enough that their angles are equal:
The first sign of the similarity of triangles
(on two angles): If two angles of one triangle are respectively equal to two angles of the second triangle, then these triangles are similar (Fig. 3).
Let triangles Δ ABC, Δ A 1 B 1 C 1 be given, in which the angles A = A 1 , B = B 1 . It is necessary to prove that Δ ABC ~ Δ A 1 B 1 C 1 .
Proof.
1) According to the theorem on the sum of angles of a triangle, we have:
angle C = 180° (angle A + angle B) = 180° (angle A 1 + angle B 1) = angle C 1 .
2) According to the theorem on the ratio of the areas of triangles having an equal angle,
S ABC /S A1 B1 C1 \u003d (AB AC) / (A 1 B 1 A 1 C 1) \u003d (AB BC) / (A 1 B 1 B 1 C 1) \u003d (AC BC) / (A 1 C 1 B 1 C 1).
3) From the equality (AB AC) / (A 1 B 1 A 1 C 1) = (AB BC) / (A 1 B 1 B 1 C 1) it follows that AC / A 1 C 1 = BC /B 1 C 1 .
4) From the equality (AB BC) / (A 1 B 1 B 1 C 1) = (AC BC) / (A 1 C 1 B 1 C 1) it follows that AB / A 1 B 1 = AC /A 1 C 1 .
Thus, for triangles ABC and A 1 B 1 C 1 DA \u003d DA 1, DB \u003d DB 1, DC \u003d DC 1, and AB / A 1 B 1 \u003d AC / A 1 C 1.
5) AB / A 1 B 1 \u003d AC / A 1 C 1 \u003d BC / B 1 C 1, that is, similar sides are proportional. So, Δ ABC ~ Δ A 1 B 1 C 1 by definition.
Theorem on proportional segments. Division of a segment in a given ratio
The proportional interval theorem is a generalization of the Thales theorem.
To use the Thales theorem, it is necessary that parallel lines intersecting two given lines cut off equal segments on one of them. The generalized Thales theorem states that if parallel lines intersect two given lines, then the segments cut off by them on one line are proportional to the segments cut off on the second line.
The theorem on proportional segments is proved similarly to the Thales theorem (only instead of the equality of triangles, their similarity is used here).
Theorem on proportional segments (generalized Thales theorem): Parallel lines intersecting two given lines cut off proportional segments on them.
Triangle median property
The first sign of the similarity of triangles allows us to prove the median property of a triangle:
Triangle median property: The medians of a triangle intersect at one point, and are divided by this point in a ratio of 2: 1, counting from the top (Fig. 4).
The point of intersection of the medians is called centroid triangle.
Let Δ ABC be given, for which AA 1 , BB 1 , CC 1 are medians, in addition, AA 1 ∩CC 1 = O. It is necessary to prove that BB 1 ∩ CC 1 = O and AO/OA 1 = BO/OB 1 \u003d CO / OS 1 \u003d 2.
Proof.
1) Draw the middle line A 1 C 1 . By the triangle midline theorem A 1 C 1 || AC, and A 1 C 1 = AC/2.
2) Triangles AOC and A 1 OC 1 are similar in two angles (angle AOC = angle A 1 OC 1 as vertical, angle OAC = angle OA 1 C 1 as internal cross lying at A 1 C 1 || AC and secant AA 1) , therefore, by the definition of similar triangles AO / A 1 O \u003d OS / OS 1 \u003d AC / A 1 C 1 \u003d 2.
3) Let BB 1 ∩CC 1 = O 1 . Similarly to points 1 and 2, it can be proved that BO / O 1 B 1 \u003d CO 1 / O 1 C \u003d 2. But since there is a single point O on the segment SS 1 that divides it in relation to CO: OS 1 \u003d 2: 1, then points O and O 1 coincide. This means that all the medians of the triangle intersect at one point, dividing each of them in a ratio of 2: 1, counting from the top.
In the course of geometry in the topic “area of polygons”, the fact is proved that the median divides an arbitrary triangle into two equal parts. In addition, when three medians of a triangle intersect, six triangles of equal area are formed.
Do you have any questions? Don't know how to solve triangle problems?
To get help from a tutor -.
The first lesson is free!
blog.site, with full or partial copying of the material, a link to the source is required.
Lesson 34 THEOREM. The ratio of the areas of two similar triangles is equal to the square of the similarity coefficient. where k is the similarity coefficient. The ratio of the perimeters of two similar triangles is equal to the similarity coefficient. V. A. S. R. M. K. Problem solving: No. 545, 549. Homework: p. 56-58, No. 544, 548.
slide 6 from the presentation "Geometry "Similar Triangles"". The size of the archive with the presentation is 232 KB.Geometry Grade 8
summary of other presentations"Definition of axial symmetry" - Symmetry in nature. Clue. Axes of symmetry. Draw a point. Building a point. Construction of a triangle. Building a segment. Peoples. Symmetry in poetry. Figures that do not have axial symmetry. Figures with two axes of symmetry. Rectangle. Symmetry. Straight. Plot points. Axial symmetry. Line segment. Axis of symmetry. Draw two lines. Points that lie on the same perpendicular. Proportionality.
"Finding the area of a parallelogram" - Find the area of a parallelogram. The area of a parallelogram. Height. Find the area of the square. Square area. Parallelogram heights. Find the area of the triangle. Signs of equality of right triangles. Find the area of the rectangle. Determining the height of a parallelogram. Base. Area of a triangle. Find the perimeter of the square. Area properties. oral exercises.
"Tasks for finding the area" - Lesson - an explanation of the new material, made in the form of a "Power point" presentation. Primary goal. "Area of a parallelogram". "Trapezoid Square". CHECKING THE LEARNED MATERIAL. Solve a problem. Workbook No. 42, repeat all the studied formulas. Derive formulas for the area of a rectangle, parallelogram, trapezoid, triangle. Expand and deepen ideas about measuring areas. Introduce the concept of area to students.
"Geometry "Similar Triangles"" - Two triangles are called similar. Proportionality of the sides of the angle. Sine, cosine and tangent values. The first sign of the similarity of triangles. Proportional segments in a right triangle. property of the bisector of a triangle. Mathematical dictation. Find the area of an isosceles right triangle. proportional cuts. Sine, cosine and tangent values for 30°, 45°, 60° angles.
"Rectangles" - Man. opposite sides. The side of the rectangle. The Tale of the Rectangle. sides of the rectangle. Rectangle in life. The perimeter of the rectangle. Rectangle. Diagonals. Paintings. Diagonal. Definition. The area of the rectangle.
""Square of the rectangle" Grade 8" - The area of the shaded square. The sides of each of the rectangles. ABCD and DSMK are squares. A parallelogram is drawn on side AB. Area units. Find the area of the square. The area of the rectangle. ABCD is a parallelogram. Area properties. Find the area of the quadrilateral. Areas of squares built on the sides of a rectangle. The floor of the room is rectangular in shape. The area of a square is equal to the square of its side.
Proportional segments
To introduce the concept of similarity, we first need to recall the concept of proportional segments. Recall also the definition of the ratio of two segments.
Definition 1
The ratio of two segments is the ratio of their lengths.
The concept of proportionality of the segments also takes place for a larger number of segments. Let, for example, $AB=2$, $CD=4$, $A_1B_1=1$, $C_1D_1=2$, $A_2B_2=4$, $C_2D_2=8$, then
That is, the segments $AB$, $A_1B_1$, $\ A_2B_2$ are proportional to the segments $CD$, $C_1D_1$, $C_2D_2$.
Similar triangles
To begin with, let us recall what the concept of similarity is in general.
Definition 3
Figures are said to be similar if they have the same shape but different sizes.
Let us now deal with the concept of similar triangles. Consider Figure 1.
Figure 1. Two triangles
Let these triangles have $\angle A=\angle A_1,\ \angle B=\angle B_1,\ \angle C=\angle C_1$. We introduce the following definition:
Definition 4
The sides of two triangles are called similar if they lie opposite the equal angles of these triangles.
In Figure 1, sides $AB$ and $A_1B_1$, $BC$ and $B_1C_1$, $AC$ and $A_1C_1$ are similar. We now introduce the definition of similar triangles.
Definition 5
Two triangles are called similar if the angles and all the angles of one triangle are respectively equal to the angles of the other and the triangle, and all similar sides of these triangles are proportional, that is
\[\angle A=\angle A_1,\ \angle B=\angle B_1,\ \angle C=\angle C_1,\] \[\frac(AB)(A_1B_1)=\frac(BC)((B_1C) _1)=\frac(AC)(A_1C_1)\]
Figure 1 shows similar triangles.
Designation: $ABC\sim A_1B_1C_1$
For the concept of similarity, there is also the concept of the coefficient of similarity.
Definition 6
The number $k$ equal to the ratio of similar sides of similar figures is called the coefficient of similarity of these figures.
Areas of Similar Triangles
Consider now the theorem on the ratio of the areas of similar triangles.
Theorem 1
The ratio of the areas of two similar triangles is equal to the square of the similarity coefficient, that is
\[\frac(S_(ABC))(S_(A_1B_1C_1))=k^2\]
Proof.
Consider two similar triangles and denote their areas as $S$ and $S_1$, respectively (Fig. 2).
Figure 2.
To prove this theorem, recall the following theorem:
Theorem 2
If the angle of one triangle is equal to the angle of the second triangle, then their areas are related as the products of the sides adjacent to this angle.
Since triangles $ABC$ and $A_1B_1C_1$ are similar, by definition $\angle A=\angle A_1$. Then, by Theorem 2, we get that
Since $\frac(AB)(A_1B_1)=\frac(AC)(A_1C_1)=k$, we get
The theorem has been proven.
Problems related to the concept of triangle similarity
Example 1
Given similar triangles $ABC$ and $A_1B_1C_1.$ The sides of the first triangle are $AB=2,\ BC=5,\ AC=6$. The coefficient of similarity of these triangles is $k=2$. Find the sides of the second triangle.
Solution.
This problem has two possible solutions.
Let $k=\frac(A_1B_1)(AB)=\frac((B_1C)_1)(BC)=\frac(A_1C_1)(AC)$.
Then $A_1B_1=kAB,\ (B_1C)_1=kBC,\ A_1C_1=kAC$.
Therefore, $A_1B_1=4,\ (B_1C)_1=10,\ A_1C_1=12$
Let $k=\frac(AB)(A_1B_1)=\frac(BC)((B_1C)_1)=\frac(AC)(A_1C_1)$
Then $A_1B_1=\frac(AB)(k),\ (B_1C)_1=\frac(BC)(k),\ A_1C_1=\frac(AC)(k)$.
Hence $A_1B_1=1,\ (B_1C)_1=2,5,\ \ A_1C_1=3$.
Example 2
Given similar triangles $ABC$ and $A_1B_1C_1.$ The side of the first triangle is $AB=2$, the corresponding side of the second triangle is $A_1B_1=6$. The height of the first triangle is $CH=4$. Find the area of the second triangle.
Solution.
Since triangles $ABC$ and $A_1B_1C_1$ are similar, $k=\frac(AB)(A_1B_1)=\frac(1)(3)$.
Find the area of the first triangle.
By Theorem 1, we have:
\[\frac(S_(ABC))(S_(A_1B_1C_1))=k^2\] \[\frac(4)(S_(A_1B_1C_1))=\frac(1)(9)\] \
