Paper 1: Junior high school mathematics teaching paper: the infiltration of classification thought in junior high school teaching
Implement quality education, cultivate qualified talents facing the new century, and make students have innovative consciousness and learn to learn through creation. Education should pay more attention to students' learning methods and strategies. Mathematician George. Paulia said: "The perfect way of thinking is like the North Star, through which many people find the right path." With the deepening of curriculum reform, in the process of changing from "exam-oriented education" to "quality education", the investigation of students not only examines basic knowledge and skills, but also pays more attention to the cultivation of exam ability. For example, the mathematical ideas and methods embodied in the process of learning and exploring the concepts, laws, properties, formulas, axioms and theorems of basic knowledge; Ask students to observe, compare, analyze, synthesize, abstract and summarize; I will explain my thoughts and opinions. So as to improve students' mathematics literacy and carry out mathematics education at the ideological level.
Mathematics learning is inseparable from thinking, and mathematics exploration needs to be realized through thinking. It is not only in line with the new curriculum standards, but also a starting point of mathematics quality education to gradually infiltrate mathematical thinking methods in junior high school mathematics teaching, cultivate thinking ability and form good mathematical thinking habits.
The idea of mathematical classification is to divide mathematical objects into several different categories according to their similarities and differences in essential attributes. It is not only an important mathematical thought, but also an important mathematical logic method.
The so-called mathematical classification discussion method is a mathematical method that divides mathematical objects into several categories and discusses them separately to solve problems. It is logical, comprehensive and exploratory to discuss mathematical problems related to thinking in categories, which can train people's thinking order and generality.
The idea of classified discussion runs through all the contents of middle school mathematics. The mathematical problems that need to be solved in the idea of classified discussion can be summarized as follows: ① Classify and define the mathematical concepts involved; ② Classify the mathematical theorems, formulas, operation properties and rules used; ③ There are many situations or possibilities for the conclusion of the solved mathematical problem; ④ There are parameter variables in mathematical problems, and the values of these parameter variables will lead to different results. The application of classified discussion can often simplify complex problems. The process of classification can cultivate the thoroughness and orderliness of students' thinking, and classified discussion can promote students' ability to study problems and explore laws.
Different from general mathematics knowledge, the idea of classification can be mastered through several classes of teaching. According to the age characteristics of students, students' understanding level and knowledge characteristics at various learning stages gradually penetrate and spiral, and constantly enrich their own connotations.
In teaching, students can form an active application of classification ideas through analogy, observation, analysis, synthesis, abstraction and generalization in the process of mathematics learning from the following aspects.
First, infiltrate the idea of classification and cultivate the consciousness of classification.
Every student has certain classification knowledge in daily life, such as the classification of people and stationery. We use this knowledge base of students to transfer the classification of life to mathematics, infiltrate the idea of mathematical classification into teaching, tap the opportunities provided by textbooks and seize the opportunities of infiltration. The classification of numbers, the meaning of absolute values and the nature of inequalities are all good opportunities to penetrate the idea of classification.
Integer,
mark
Positive rational number
zero
Negative rational number
After teaching the concepts of negative numbers and rational numbers, guide students to classify rational numbers in time, so that students can understand that rational numbers have different classification methods for different standards, such as:
rational number
Lay the foundation for the following classification discussion.
After knowing that the number A can represent any number, ask the students to classify the number A and get three categories: positive number, zero number and negative number.
When explaining the meaning of absolute value, guide students to get the following classification:
By knowing the absolute values of positive numbers, zero numbers and negative numbers, we can understand how to learn and understand mathematical concepts through classified discussion.
Another example is the comparison of two rational numbers, which are divided into positive numbers and positive numbers, positive numbers and zero, positive numbers and negative numbers, negative numbers and zero, negative numbers and negative numbers. The comparison between negative numbers and negative numbers is a new knowledge point, which highlights the focus of learning.
Combined with the teaching of the chapter "Rational Numbers", the idea of mathematical classification is repeatedly strengthened, so that students gradually form the consciousness of classification in mathematics learning. And when discussing classification, we can pay attention to some basic principles, such as the objects of classification are certain and the standards are unified, otherwise the objects are mixed and the standards are different, and there will be errors such as omission and repetition. If rational numbers are divided into positive numbers, negative numbers and integers, it is wrong to make different classification standards. After determining the objects and standards, we should also pay attention to distinguish the levels and not discuss them beyond the levels.
Second, learn classification methods to enhance the rigor of thinking.
When the idea of classification is permeated in teaching, students should understand that the so-called classification is to choose appropriate standards, divide the objects into several categories, and then answer the questions of each subcategory. Mastering a reasonable classification method has become the key to solving the problem.
Classification methods are usually as follows:
1, according to the concept of mathematical classification.
Some mathematical concepts are given by classification, and solving such problems is generally classified according to the classification form of concepts.
Example 1, Simplified solution:
This is classified according to the meaning of absolute value.
Example 2, comparison and error-prone, leading us to fail to notice that numbers can represent different kinds of numbers. Logarithms are classified and discussed, and the correct answer can be obtained:
> 0, = 0,
When learning the discriminant of the roots of quadratic equations, for deformation equations
To solve problems with bilateral square roots, it is necessary to classify the cases where the solution of the equation is greater than 0, equal to 0 and less than 0. The sign of this question determines whether it can be squared, which is the basis of classification. Thus, three cases of the roots of a quadratic equation with one variable are obtained.
Example 3. Solve the inequality about x: ax+3 > 2x+a.
The analysis is transformed into the form of (a-2) x >: A-3, and then the inequalities can be divided into a-2 > 0, a-2 = 0, and a-2 < 0, respectively.
When a-2 > 0, that is, A > 2, the solution of the inequality is X >；;
When a-2 = 0, that is, A = 2, the left side of inequality = 0 and the right side of inequality =- 1.
Because 0 1- 1, the solutions of inequalities are all real numbers.
When a-2 < 0, that is, a < 2, the solution of the inequality is X.
3, according to the characteristics of graphics or the relationship between them.
For example, triangles are classified by angle, including acute triangle, right triangle and obtuse triangle. According to the number of times a straight line intersects a circle, it can be divided into: separation, tangency and intersection.
For example, the angle between the height of one waist and the height of the other waist of an isosceles triangle is 30, and the length of the bottom is a, so the height of its waist is
Analysis: According to the characteristics of the figure, isosceles triangles can be divided into acute triangles and obtuse triangles according to the height CD. As shown in the figure, waist heights can be classified according to the different positions of points and lines in the geometric figure.
When proving the theorem of circle angle. Because the position of the center of the circle is on the edge, inside and outside the corner, it is discussed and proved in three different situations. First of all, it is proved that the center of the circle is on one side of the fillet, which is the easiest situation to solve. Then, by making the diameter of the vertex of the fillet, using the prior proof (the center of the circle is on one side of the fillet), the two situations that the center of the circle is inside the fillet and the center of the circle is outside the fillet are solved respectively. This is an idea and method of classified discussion embodied in the process of theorem proving. It is a method to solve problems one by one according to the different positions of points and lines in geometric figures. Textbooks have proved the theorem of chord tangent angle: the chord tangent angle is equal to the circumferential angle of the arc it encloses. It is also solved in three different situations: the center of the circle is on one side of the chord tangent angle, the inner side of the chord tangent angle and the outer side of the chord tangent angle.
Third, guide classified discussions and improve the ability to solve problems reasonably.
There are many theorems, rules, formulas and exercises in junior high school textbooks, which need to be discussed separately. When teaching these contents, students should constantly strengthen the awareness of classified discussion and make them aware of these problems. Only after classified discussion can the conclusion be complete and correct. If you don't discuss it in categories, it is easy to make mistakes. In problem-solving teaching, classified discussion also helps students to generalize and summarize regular things, thus strengthening the order and meticulousness of students' thinking.
Generally speaking, there are two kinds of problems solved by classified discussion ideas and methods:; One is to discuss and solve problems in algebraic expressions, functions or equations in different values according to different values of letters. The second is to discuss and solve the problems one by one according to the different positions of points and lines in geometric figures.
Example 4: The function y = (m-1) x2+(m-2) x-1(m is a real number) is known. If there is only one intersection point between the image of the function and the x axis, find the value of m.
Analysis: From the perspective of function classification, this paper studies and solves the problem in two cases: M- 1 = 0 and m-1/0.
Solution: When m = L, the function is a linear function Y =-x-1, and there is only one intersection with the X axis (-1, 0).
When m 1 1, the function is a quadratic function y = (m-1) x2+(m-2) x-1.
When △ = (m-2) 2+4 (m- 1) = 0, m=0.
The vertex (-1, 0) of the parabola y =-x2-2x- 1 is on the x axis.
Example 5, the function y = X6–X5+x4-x3+x2–x+1proves that the value of y is always positive.
Analysis: it is difficult to prove the conclusion by factorizing the expression of y, and it can be found that if the variable x is properly classified within the real number range, the problem will be solved.
Prove: (1) When x ≤0
∵ x5-x3-x ≥0, ∴ y≥ 1 holds;
0 is 2.
y = X6+(x4–X5)+(x2–x3)+(x– 1)
∵x4 & gt； x5，x2 & gtx3， 1 & gt； x
∴y & gt； 0 has been created;
(3) when x = 1, y = 1 > 0 holds;
(4) when x > 1
y =(X6–X5)+(x4–x3)+(x2–x)+ 1
∵X6 & gt； x5，x4 & gtx3，x2 & gtx
∴y & gt； 1 hold.
To sum up, y > 0 holds.
Example 6: It is known that △ABC is an equilateral triangle with a side length of 2, and △ACD is a right triangle with an angle of 30. △ABC and△△ ACD are combined into a convex quadrilateral ABCD. (1) Draw a quadrilateral ABCD(2) and find the area of the quadrilateral ABCD.
When analyzing the right triangle ACD with an angle of 30, we can study two cases: AC is the hypotenuse and AC is the right angle. As shown in figure 1, it is a quadrilateral ABCD composed of an equilateral triangle ABC with AC as the hypotenuse (the areas of quadrilateral ABCD calculated by DDAC = 30 and DDAC=60 are the same, so they belong to the same category). AC is a right-angled side, which can be divided into two different situations, as shown in Figure 2 and Figure 3. From figure 1, s quadrilateral ABCD =；; It can be calculated from Figure 2 that the S quadrilateral ABCD =；; It can be calculated as S quadrilateral ABCD=3.
As can be seen from the above examples, classified discussion can often make some complex problems extremely simple, and the ideas and steps for solving problems are very clear. On the other hand, in the process of discussion, students' interest in learning mathematics can be stimulated.
Make use of the existing teaching materials, try to help students master classified thinking methods in teaching, and pay attention to the comprehensive use of several thinking methods in combination with the study of other mathematical thinking methods, so as to provide students with enough materials and time to inspire students to think positively. I believe it will greatly improve students' understanding level and get twice the result with half the effort.
Paper 2: Junior high school mathematics teaching paper: Teaching students to solve difficult problems in junior high school mathematics exams.
It is the key to enable students to consolidate basic knowledge, have certain problem-solving skills, cultivate students' ability of analysis, synthesis and association, and cultivate students' intuitive thinking, so that students can quickly grasp the basic knowledge involved in math problems.
Keywords: problem-solving skills, association, grasping the essence of the problem
The annual junior high school math exam is generally divided into easy questions (basic questions), intermediate questions and difficult questions. In recent years, in junior high school mathematics exams, difficult questions generally account for more than a quarter of the total score of the whole paper, so it is difficult for students to get good grades in the exams without breaking through the difficult questions.
There are several problems in junior high school mathematics examination: 1, which requires a certain depth of thinking or strong skills. 2, the meaning of the question is novel or the idea of solving the problem is novel. 3. Explore or open math problems.
Different types of questions should have different teaching strategies. No matter what kind of problems are solved, students are required to have certain basic knowledge of mathematics and basic problem-solving skills (have a good understanding of mathematical concepts, theorems and formulas, and have some understanding of the proof of theorems and formulas; It is very skillful and quick to solve the basic problems of directly applying theorem formulas, so it is necessary to cultivate "double-base" students. Of course, the first stage of graduation review in grade three is "double-base" training, but if students have a deeper knowledge of mathematics and strengthen their basic skills, the review effect is still good.
Some teachers think that only the whole class needs to review the mild questions, and the difficult questions don't need to be reviewed. Those students with good intelligence, you do it without helping them review, and those students with poor intelligence, you waste time teaching them. In fact, students with certain mathematical knowledge and basic problem-solving skills may not be able to solve difficult problems. This is because starting from the basic knowledge of mathematics to get the answers to the difficult questions in the senior high school entrance examination, or the thinking depth is high-the students' lack of thinking depth or new ideas-the students have never been exposed to it. However, many experienced teachers' years of practice in the graduating class of Grade Three have proved that it is necessary to review difficult problems on special topics. As long as the review is good, it will greatly improve the ability of students above technical secondary school to solve problems. In this regard, we should train students' thinking ability and broaden their thinking in the second stage of review. Of course, this kind of training should also be aimed at students' "double basics" and math problems. This kind of training should pay attention to the choice of topics, not only for the examination, but also for the lack of students' thinking. Some training is necessary, but students should be given enough time to reflect and summarize their own problem-solving methods and ideas. Only by reflecting more and summarizing more can students improve their problem-solving ability. Teachers should pay attention to guidance and can't replace students' ideas with their own ideas, because everyone's methods of solving problems are not necessarily the same.
In the past, when reviewing in HKCEE, some teachers in the graduating class of Grade Three looked for simulation questions from all over the country and districts to train their students round by round. After the lecture and practice, both teachers and students worked hard, but the effect was not satisfactory. This is because this tactical review method of questioning fails to teach students in accordance with their aptitude, and the teacher's teaching is not targeted to students' knowledge, skills, thinking ability and mathematical problems. Students do not reflect the subjectivity of learning, nor do they have enough time to sum up and reflect. Therefore, students' problem-solving skills and thinking ability have not been really improved.
Some teachers think the examination questions are difficult, new and elusive. The special review of difficult questions is to practice the difficult questions in this year's exam and the simulated questions in various regions in that year. This topic-based topic review is also difficult to substantially improve students' ability to solve problems.
The proposition of the proposer's junior high school mathematics test questions aims to examine the degree of our junior high school graduates' mastery of the basic knowledge of junior high school mathematics, and the test questions are of course inseparable from the basic knowledge of junior high school. The so-called problem is just a few layers of yarn on the cage, which makes it difficult for us to see its true face clearly. Our teacher's task is to teach our students to uncover those mysterious veils and grasp their true colors. Cheng can win with three axes on the battlefield. Our students have mastered all the basic knowledge of junior high school mathematics and have certain problem-solving ability. As long as we guide and train students correctly, our students will surely win in the examination room.
The key point is that our review training can help students master knowledge and strengthen problem-solving skills. At the same time, our teachers' proper guidance, students' reflection and summary after training, and the independent construction of knowledge can help us grasp the essence of various mathematical problems-the connection with the basic knowledge of junior high school mathematics.
When reviewing difficult problems by classification, we should focus on cultivating students' ability to grasp the connection between mathematical difficult problems and basic knowledge, guide students to quickly and correctly analyze the thinking of solving problems, and cultivate students' intuitive thinking of solving problems. Problems should be classified first. Then carry out classification training. In class, students don't have to write down the problem-solving process in detail for each question, just one or two questions of one kind, and some only require students to write down the problem-solving ideas quickly and then write down the detailed problem-solving process after returning.
I think the problems in junior high school entrance examination can be divided into the following categories for special review:
The first category: problems closely related to one or two knowledge points:
Example 1 As shown in the figure, in ⊙O, C is the midpoint of arc AB, and D is any point on arc AC (not coincident with D C points A and C), then () A.
(A)AC+CB = AD+DB(B)AC+CB & lt； AD+DB
(C)AC+CB & gt； The relationship between AD+DB (D)AC+CB and AD+DB is uncertain.
Teaching guidance: What are the related knowledge of line segment size comparison? (The sum of any two sides of a triangle is greater than the third side or the big side is opposite to the big angle, etc. )
How to combine AC+CB and AD+DB into a triangle to compare their sizes?
Attached solution: take C as the center, CB as the radius, as the extension line of the arc intersection BD, and connect AE, CE and AB at point E.
∫ce = CB ∴∠ceb=∠cbe and ∠DAC=∠CBE.
∴∠CEB=∠CAD and CA=CE ∠CEA=∠CAE.
∴∠CEA-∠CEB=∠CAE-∠CAD
∴∠DEA=∠DAE
∴DE=DA
In △CEB，CE+CB >； BE is AC+CB & gt；; AD+DB。 So choose (c).
Comments: The key of this example teaching is to guide students to construct AC, CB, AD and DB on a triangle.
Example 2 is known: ⊙O 1 and ⊙O2 intersect at point A and point B. If PM cuts ⊙O 1 to m, PN cuts ⊙O2 to n, PM >: Try to point out the range of point P.
Teaching guidance: (1) Draw pictures first, try to judge and try to prove. (2) Look at several possible situations.
(3) Show the picture on the right and ask the students to point out the range of point P (point P is within ⊙O2 of straight line AB).
On one side, outside ≧O2), students point out the range of point P and ask them to
Prove (4) When students have difficulties in proving, they should give directions: If point P is on the straight line AB, what can be proved? (PM=PN), how to prove it?
(Using secant theorem: PM2=PA*PB, PN2=PA*PB, therefore, PM=PN) Now we can apply secant theorem to prove PM & gtPN?
(5) When students can't prove it, give a hint:
Connect PB, intersect ⊙O 1 at point C, intersect ⊙O2 at point D, and use the cutting line theorem.
(Proof: PM2=PC*PB, PN2=PD*PB, because PC >；; PD, so PC * PB & gtPD*PB, namely PM2 >; PN2, so PM & gtPN)
(6) Is there anything else? (Guide the students to find out the following two situations: Figure 2 and Figure 3. Ask the students to point out the range of point P and prove it.)
Comments: The key to this question is to guide students to prove it with secant theorem and discuss it in categories.
The key to this kind of difficult teaching is to guide students to stick to the knowledge points related to the topic until the problem is solved.
The second category: problems that combine multiple knowledge points or require certain problem-solving skills to solve.
The key to teaching this kind of difficult problems is to ask students to use analytical and comprehensive methods, some mathematical ideas and methods, and certain problem-solving skills to solve them.
Example 1 triangle ABC, point I is the center, straight lines BI, CI cross AC, AB is in d, e. Known ID=IE.
Verification: ∠ABC=∠BCA, or ∠ A = 60.
Teaching guidance: This topic should be analyzed from two directions: conditions and conclusions. From the conditional analysis, from ID=IE and I are the heart, it can be concluded that △AID and △AIE are two corresponding diagonals, and there are two possibilities: AD=AE or AD≠AE.
From this, the relationship between ∠ADI and ∠AEI can be deduced. From the conclusion analysis, we need to find out the relationship between ∠ABC and ∠ ACB ∠ ADI =1/2 ∠ ABC+∠ ACB ∠ AEI =1/2 ∠ AC.
Attached proof process: There are two possible situations for linking AI, in △AID and △AIE, and the size of AD and AE: AD=AE, or AD≠AE.
(1) if AD=AE, then △ aid △ AIE, where ∠ADI=∠AEI.
And ∠ Adi = 1/2∠ABC+∠ACB, ∠AEI= 1/2∠ACB+∠ABC.
So1/2 ∠ ABC+∠ ACB =1/2 ∠ ACB+∠ ABC.
That is ∠ABC=∠ACB.
(2) if AD≠AE, then AD >；; AE, intercept AE' = AE on AE, connect IE'. Then △ AIE △ AIE.
Therefore ∠ AE' I = ∠ AEI. IE' = IE = ID。
Therefore, △IDE' is an isosceles triangle,
There is ∠ e 'di = ∠ de 'i.
Because ∠ AE' i+∠ de' i = 180,
Therefore ∠ AEI+∠ AIE = 180.
Therefore, (1/2 ∠ ACB+∠ ABC)+(1/2 ∠ ABC+∠ ACB) =180.
Therefore ∠ ABC+∠ ACB = 120,
So ∠ a = 180- 120 = 60.
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Example 2: As shown in the figure, AB is the diameter ⊙O, AE bisects ∠BAF and intersects with ⊙O at point E, a straight line perpendicularly intersects with AF at point E, intersects with AF extension line at point D and intersects with AB extension line at point C. 。
(1) Verification: CD and ⊙O are tangent to point E. 。
(2) If CE*DE= 15/4 and AD=3, find the tangents of diameters ⊙O and ∠AED.
Teaching guidance: (1) certificate OE⊥CD.
(2) If the diameter ⊙O is required, the radius OE can be obtained first.
Because OE∑AD, OE/AD=CO/CA, AD=3, CO and CA are all related to BC and OB, AB (radius and diameter ⊙ O).
So, if you get BC, you can get OE. How to get BC? Can we use the condition of CE*DE= 15/4?
Let the students discuss.
Attached solution process: (1) omitted. (2) If point D is DG∑AC and AE.
If the extension line is at point G, connecting BE and OE, then ∠BAG=∠G, ∠C=∠EDG. ∵CD and⊙ O are tangent to point E,
∴∠BEC=∠BAG.
∴∠BEC=∠G. ∴△BEC∽△EGD.∴DE/CB=DG/CE.
∴CB*DG=DE*CE.
∠ Bao =∠ Da =∠G. ∴AD=DG=3. And ∵CE*DE= 15/4. ∴CB=5/4.
Let OE∑ad of (1) and ∴CO/CA=OE/AD be OE = x(x >；); 0), then CO=5/4+x=(5+4x)/4,
CA=5/4+2x=(5+8x)/4, ∴(5+4x)/(5+8x)=x/3. The arrangement is 8x2-7x- 15=0. The solution is x 1=- 1. CE2=CB*CA=25/4，∴de= 15/4* 1/ce=3/2. ∴ce=5/2
In Rt△ADE, tan∠AED=AD/DE=2.
The third kind of open and exploratory mathematical problems.
Whether it is an open or exploratory mathematical problem, the teaching focus is the key to teach students to grasp the problem.
Example 1 Please write an analytical formula of a quadratic function, in which an image only passes through two, three and four quadrants.
Teaching instructions: the image of quadratic function only passes through the second, third and fourth quadrants, but not the first quadrant, that is, x>0, Y; y < 0； What about 0? This is the core of the problem.
(answer: when a, b and c in the quadratic function y=ax2+bx+c are all negative, there must be x >；; 0，y
Example 2 is known: As shown in the figure, AB and AC are two chords of ⊙ O, AB=AC= 1,
∠ BAC = 120, p is any point on the optimal arc BC,
(1) verification: PA shares ∠BPC,
(2) If the length of PA is m, find the perimeter of quadrilateral PBAC,
(3) If point P moves on the optimal arc BC, is there a certain position P that makes S△PAC=2S△PAB? If yes, please prove it; If not, please explain why.
Teaching guidance: (2) Because AB=AC= 1 and PA=m, it can be proved that ∠ APB = ∠ APC = 30, so ∠ AOB = 60, so OA = OB = AB = 6544. So PB and PC have changed, but there are only two changes, and PB+PC should remain unchanged. Find PB+PC, and you can find the perimeter of the quadrilateral PBAC. The key to this problem is to combine PB and PC to find out. (3) The key to this problem is how to determine point P, which can be deduced from the area relationship between triangle PAC and triangle PAB. P
(emphasis: (1) omitted. (2) Extend PC to P', make CP'=BP, connect BC, find BC, prove △ PAB △ P' AC, get AP'=AP, prove △ ABC ∽△ APP', find PP', that is, Pb+PC, the proportional relationship of the corresponding sides. (3) Connect BC and PA at point G, BG/CG = BM/CN = s △ PAB/s △ PAC =1/2. Therefore, the intersection of point A and point G is the intersection of ray and ⊙O, which is the position of point P that meets the requirements of the topic. )
The fourth new question type (a question type that only appeared in junior high school entrance examinations all over the country in recent years)
No matter how new the examination questions are, they can't be separated from the basic knowledge of junior high school, so the key to solving such questions is to find the basic knowledge related to the questions from the meaning of the questions, and then use the basic knowledge related to them to find a solution to the problem through analysis, synthesis, comparison and association.
Example 1 As shown in Figure 1, the pentagon is a schematic diagram of a piece of land contracted by Uncle Zhang ten years ago. After years of reclamation, the wasteland has become a hexagonal ABCMNE as shown in figure 1, but the boundary path between the contracted land and the reclaimed wasteland (that is, the dotted line CDE in figure 1) still exists. Uncle Zhang wants to build a straight road at point E. After the straight road is repaired, it is necessary to keep the land area on the left side of the straight road as much as when contracting, and the land area on the right side as much as reclaiming wasteland. Please use relevant geometric knowledge to design the road construction scheme according to Uncle Zhang's requirements. (Not including the floor space of branch roads and straights)
(1) Write the design scheme and draw the corresponding figure in Figure 2;
(2) Explain the reasons for the scheme design.
Teaching guidance:
As shown in Figure 2, I tried to make E into a straight line EHF, crossing CD in H and CM in F. According to the meaning of the question, the area of EABCF = the area of =EABCD, and the area of EDCMN = the area of =EFMN (to meet the requirements of Uncle Zhang). That is, the area of triangle EHD = the area of triangle CHF. What are the conditions? (answer: connect EC, cross D is DF∨EC, cross CM at point F, and EF is where Uncle Zhang wants to build roads. )
Comments: This topic is practical application, and its related basic knowledge is some properties of trapezoid, as shown below.
In trapezoidal ABCD, ABCD, the area of triangular ADC = the area of triangular BCD. Subtract the area of triangular CDO to get the area of triangular ADO = the area of triangular BCO. Being able to associate this knowledge is the key to solve this problem.
The computer CPU chip is made of a material called "monocrystalline silicon", and the uncut monocrystalline silicon material is a thin wafer called "wafer". In order to produce a CPU chip, a large number of small silicon wafers with the length and width of 1cm are needed. If the wafer diameter is 10.05cm, can 66 small silicon wafers with the required size be cut from one wafer? Please explain your methods and reasons. (excluding cutting loss)
Teaching guidance: everyone will start to do this problem, but they must cut it in a certain order to get the correct answer.
Method: (1) First, arrange 10 small squares in a row.
As a long rectangle, this rectangle can just fit into a circle with a diameter of 10.05cm, as shown in the figure, rectangle ABCD.
∫AB = 1，BC= 10，
∴ Diagonal AC2 =102+12 =100+1=1
(2) Nine small squares can be placed above and below the rectangular ABCD.
In this way, the newly added two rows of small squares plus a part of ABCD can be regarded as a rectangle EFGH, with a length of 9 and a height of 3, and a diagonal EG2 = 92+32 = 8 1+9 = 90.
(3) Similarly, ∫82+52 = 64+25 = 89.
Therefore, eight small squares can be arranged above and below the rectangle EFGH, so there are now five layers of small squares.
(4) On the basis of the original, add another layer up and down, ***7 layers, and the height of the new rectangle can be regarded as 7, so the new two rows can be 7 but not 8.
∫72+72 = 49+49 = 98 & lt； 10.052
82+72 = 64+49 = 1 13 & gt； 10.052.
(5) On the basis of seven layers, add another layer up and down. The height of the new rectangle can be regarded as 9, and each row can be 4, but not 5. ∫42+92 = 16+8 1 = 97 & lt； 10.052 and 52+92 = 25+81=106 > 10.052.
At present, the total number of rows is 9, the height reaches 9, and there is a space of about 0.5cm above and below. Because the position of the rectangular ABCD cannot be adjusted, there is no room for 1 small squares.
Therefore, 10+2*9+2*8+2*7+2*4=66 (pieces).
Comments: The key to solve this problem is to arrange small squares in rows, using the knowledge that the diagonal of the inscribed rectangle of a circle is the diameter of the circle.