Ratio of areas and perimeters of similar triangles. Theorem on the ratio of areas of similar triangles
Lesson 34. Theorem on the ratio of areas of similar triangles. 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. Solving problems: 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 8th grade
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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 called the proportionality coefficient.
Example. Numbers 6; 7.5 and 15 are proportional to the numbers -4; 5 and 10. The proportionality coefficient is the number -1.5, since
6/-4 = -7,5/5 = 15/-10 = -1,5.
Proportionality of numbers takes place if these numbers are related by proportion.
It is known that a proportion can be made up 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.).
Let's look at rice. 1 two triangles ABC and A 1 B 1 C 1 with equal pairwise angles: A = A 1, B = B 1, C = C 1.
The 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 respectively equal angles of triangles ABC and A 1 B 1 C 1.
Let us define similar triangles:
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 = A 1, B = B 1, C = C 1 and AB/A 1 B 1 = BC/B 1 C 1 = AC/A 1 C 1 = k, where k is the similarity coefficient. From rice. 2 it is clear that similar triangles have the same proportions, and they differ only in scale.
Note 1: Equal triangles are similar by a factor of 1.
Note 2: When designating similar triangles, their vertices should be ordered so that their angles are pairwise equal. 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 similar sides of 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 angles A = K, B = N, C = L, and AB/KN = BC/NL = AC/KL.
Note 3: Those requirements that are listed in the definition of similar triangles are redundant. We will prove the similarity criteria for triangles that contain fewer requirements for similar triangles 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. These are, for example, the side of a triangle, the perimeter, the median. Angle or area do not apply to 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 equal in pairs, i.e. A = A 1, and according to the theorem on the ratio of the areas of triangles having each equal angle, we have:
S ABC /S A1 B1 C1 = (AB · AC) / (A 1 B 1 · A 1 C 1) = 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,
therefore 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 (these are equality of angles and proportionality of sides) are redundant. It is possible to establish the similarity of triangles using a smaller number of elements.
Thus, when solving problems, the first criterion of similarity of triangles is most often used, which states that for two triangles to be similar, the equality of their angles is sufficient:
The first sign of similarity of triangles
(by 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 the angles of a triangle, we have:
angle C = 180° (angle A + angle B) = 180° (angle A 1 + angle B 1) = angle C 1.
2) By the theorem on the ratio of the areas of triangles having equal angles,
S ABC /S A1 B1 C1 = (AB AC) / (A 1 B 1 A 1 C 1) = (AB BC) / (A 1 B 1 B 1 C 1) = (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, triangles ABC and A 1 B 1 C 1 DA = DA 1, DB = DB 1, DC = DC 1, and AB/A 1 B 1 = AC/A 1 C 1.
5) AB/A 1 B 1 = AC/A 1 C 1 = BC/B 1 C 1, that is, similar sides are proportional. This means that Δ ABC ~ Δ A 1 B 1 C 1 by definition.
Theorem on proportional segments. Dividing a segment in a given ratio
The proportional segment theorem is a generalization of Thales' theorem.
To use Thales' theorem, it is necessary that parallel lines intersecting two given lines cut off equal segments on one of them. The generalized theorem of Thales 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 proven similarly to Thales' theorem (only instead of the equality of triangles, their similarity is used here).
Theorem on proportional segments (generalized Thales' theorem): Parallel lines that intersect two given lines cut off proportional segments on them.
Property of medians of a triangle
The first criterion for the similarity of triangles allows us to prove the property of medians of a triangle:
Property of medians of a triangle: The medians of a triangle intersect at one point, and are divided by this point in a ratio of 2: 1, counting from the vertex (Fig. 4).
The intersection point 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 = VO/OB 1 = CO/OS 1 = 2.
Proof.
1) Draw the middle line A 1 C 1. By the theorem on the midline of a triangle 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 crosswise lying with A 1 C 1 || AC and secant AA 1) , therefore, by definition of similar triangles AO/A 1 O = OC/OS 1 = AC/A 1 C 1 = 2.
3) Let BB 1 ∩CC 1 = O 1 . Similar to points 1 and 2, it can be proven that VO/O 1 B 1 = CO 1 /O 1 C = 2. But since on the segment CC 1 there is a single point O dividing it in the ratio CO: OS 1 = 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 vertex.
In the geometry course, in the topic “area of polygons,” the fact is proven that the median divides an arbitrary triangle into two equal parts. In addition, when the three medians of a triangle intersect, six equal triangles are formed.
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Proportional segments
To introduce the concept of similarity, we first need to recall the concept of proportional segments. Let us also recall 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 segments also applies to more 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
Let us first remember what the concept of similarity generally represents.
Definition 3
Figures are called similar if they have the same shape but different sizes.
Let us now understand the concept of similar triangles. Consider Figure 1.
Figure 1. Two triangles
Let these triangles have $\angle A=\angle A_1,\ \angle B=\angle B_1,\ \angle C=\angle C_1$. Let us introduce the following definition:
Definition 4
The sides of two triangles are called similar if they lie opposite equal angles of these triangles.
In Figure 1, the sides $AB$ and $A_1B_1$, $BC$ and $B_1C_1$, $AC$ and $A_1C_1$ are similar. Let us now introduce the definition of similar triangles.
Definition 5
Two triangles are called similar if the angles of 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 similarity coefficient.
Definition 6
The number $k$ equal to the ratio of similar sides of similar figures is called the similarity coefficient of these figures.
Areas of similar triangles
Let us now consider 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.
Let's 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 product of the sides adjacent to this angle.
Since triangles $ABC$ and $A_1B_1C_1$ are similar, then, by definition, $\angle A=\angle A_1$. Then, by Theorem 2, we obtain 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 similarity coefficient 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 $A_1B_1=\frac(AB)(k),\ (B_1C)_1=\frac(BC)(k),\ A_1C_1=\frac(AC)(k)$.
Therefore, $A_1B_1=1,\ (B_1C)_1=2.5,\ \ A_1C_1=3$.
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, then $k=\frac(AB)(A_1B_1)=\frac(1)(3)$.
Let's find the area of the first triangle.
By Theorem 1, we have:
\[\frac(S_(ABC))(S_(A_1B_1C_1))=k^2\] \[\frac(4)(S_(A_1B_1C_1))=\frac(1)(9)\] \
The purpose of the lesson: give a definition of similar triangles, prove a theorem on the relation of similar triangles.
Lesson objectives:
- Educational: Students must know the definition of similar triangles, the theorem about the relationship of similar triangles, be able to apply them when solving problems, and implement interdisciplinary connections with algebra and physics.
- Educational: to cultivate hard work, attentiveness, diligence, and to cultivate a culture of student behavior.
- Educational: development of students' attention, development of the ability to reason, think logically, draw conclusions, development of students' competent mathematical speech and thinking, development of skills of self-analysis and independence.
- Health saving: compliance with sanitary and hygienic standards, changing types of activities in the lesson.
Equipment: computer, projector, teaching material: independent and test papers in algebra and geometry for grade 8 A.P. Ershova, etc.
Lesson type: learning new material.
During the classes
I. Organizing time (greeting, checking readiness for the lesson).
II. Lesson topic message.
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 called 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 similarity coefficient. ΔABC ~ A 1 B 1 C 1
1. Orally: Are the 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 = 10.5, B 1 C 1 = 7.5, A 1 C 1 = 6.
b) In one isosceles triangle, the vertex angle is 24˚, and in the other isosceles triangle, the base angle is 78˚.
Guys! Let us recall the theorem on the ratio of the areas of triangles having equal angles.
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 product of the sides enclosing equal angles.
2. Written work according to prepared drawings.
On screen drawing:
a) Given: BN: NC = 1:2,
BM = 7 cm, AM = 3 cm,
S MBN = 7 cm 2 .
Find: S ABC
(Answer: 30 cm 2.)
b) Given: AE = 2 cm,
S AEK = 8 cm 2 .
Find: S ABC
(Answer: 56 cm 2.)
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. Updating 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 9 cm. Find the side of the first triangle 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 cm 2 and 24 cm 2.)
If you have time independent work educational in nature.
Option 1
Similar triangles have similar sides equal to 7 cm and 35 cm.
The area of the first triangle is 27 cm 2.
Find the area of the second triangle. ( Answer: 675 cm 2.)
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 et al., paragraphs 57, 58, No. 545, 547.
6. Summing up the lesson.
Lesson type: lesson on introducing new material.Objective of the lesson: To prove the property of areas of similar triangles and show its practical significance in solving problems.
Lesson objectives:
teaching – to prove the property of 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 content educational process and creating a situation of success, developing the ability to work in a group.
The student has the following knowledge:
Unit of activity content that students need to learn:
During the classes.
1. Organizational moment.
2. Updating knowledge.
3. Working with a problematic 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 Technology, competency-based approach.
Equipment:
computer, projector for demonstrating presentations, interactive board, document camera;
computer presentation in Microsoft PowerPoint;
supporting summary;
During the classes
1. Organizational moment.
Today in the lesson we will work not in notebooks, but in reference notes, which you will fill out for the continuation of the entire lesson. Sign it. The grade for the lesson will consist of two components: for the supporting notes and for active work in the lesson.
2. Updating 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 studied in the last lesson.
Theoretical warm-up. Test. In your reference notes, the first task is of a test nature. Answer the questions by choosing one of the proposed answer options and enter your answer where necessary.
Teacher: What is the ratio of two segments called?
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 the segmentsA 1 B 1 and C 1 D 1
Answer: segments AB And CDproportional to the segmentsA 1 B 1 and C 1 D 1 if
Your options. Fine. Don't forget to correct anyone who has it wrong.
Teacher: Define similar triangles? Refer to your reference note. You have three options for answering 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 anyone who has it wrong.
Teacher: What is the ratio of the areas of two triangles that have equal angles?
Answer: If the angle of one triangle is equal to the angle of another triangle, then the areas of these triangles are related as the product of the sides enclosing equal angles.
Solving problems using ready-made drawings.Next, our warm-up will take place while solving problems using ready-made drawings. You can 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 (recording answers on the board).
Possible answers:
Determination of similar triangles;
Application of the definition of similar triangles in solving problems;
Theorem on the ratio of the areas of triangles having equal angles;
And now I propose a solution to several problems, which have something in common with the topic of the lesson, but they are more related to geography.
A situation of success.
The first task is before you. We are working on this problem ourselves. The first person to solve it will show his solution at the board, and someone else 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.
A situation of success.
We figured out the dimensions of the Bermuda Triangle. Well, now let’s find out the measurements of the flower bed. We turn over the supporting notes. Second task. We solve this problem by working in pairs. We check in a similar way, but only the result will be presented by the first couple to complete the task.
Answer: the sides of a triangular flower bed are 10m and 11m 20 cm.
So, let's check it out. Does everyone agree? Who decided in a different way?
Reflection.
What method of action did you use to solve this problem? Write it down in your reference note.
Possible answer:
similar triangles have equal corresponding angles;
The areas of triangles having equal angles are the product of the sides containing equal angles.
Failure situation.
5. Studying new material.
When solving the third problem, students are faced with a problem. They are unable to solve the problem because, in their opinion, the conditions of the problem are not complete enough or they receive an unfounded answer.
The students had 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 couldn't you solve the last equation?
Students: 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 into 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 return to problem No. 3 and solve it based on a proven fact.
7. Lesson summary
What new things did you learn to do today?
Solve problems in which the similarity coefficient and area of one of the similar triangles are known.
What geometric property helped us with 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 (No. 547)
Goodbye.