The first part of the ninth grade mathematics "circle" courseware
1. Understand the concept of rotation, its rotation center and rotation angle, the concept of rotation corresponding point and its application, and solve some practical problems.
2. By reviewing the related concepts and properties of translation and axial symmetry, starting from mathematics in life, we have experienced observation, produced concepts and applied them to solve some practical problems.
3. The basic nature of rotation.
focus
Concepts of rotation and corresponding points and their applications.
difficulty
Basic properties of rotation.
First, review the introduction.
(Student Activity) Please complete the following questions.
1. Translate the quadrilateral ABCD as shown in the figure, so that the point corresponding to point B is point D, and make the translated figure.
2. As shown in the figure, given the △ABC and the straight line L, please draw the symmetrical graph of △ABC about L △ A ′ B ′ C ′.
3. Is the circle an axisymmetric figure? What about the isosceles triangle? Can you point out anything else?
(Oral) Teacher's Comments Summary:
Some concepts and properties of (1) translation.
(2) How to draw a symmetrical figure about a straight line (axis of symmetry) and dictate some of its properties.
(3) What is an axisymmetric figure?
Second, explore new knowledge.
We have reviewed translation and other related contents before. Are there any other sports changes in our life? The answer is yes, so let's study it.
1. Please look at the big clock on the podium. What keeps turning? What point does the rotation revolve around? How many degrees has the hour hand turned from now until class is over? How many degrees did the minute hand turn? How many degrees did the second hand turn?
Teacher's comment: The hour hand, minute hand and second hand are constantly turning around the center of the clock. From now until the class is over, the hour hand has turned to _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
Look at my homemade toy, it looks like the wheel of a windmill. It can keep turning. How to change jobs to a new position? (Teacher's comment)
3. What are the characteristics of question 1, 2?
* * * The same feature is that if we regard the clock and windmill wheel as a figure, then these figures can rotate around a fixed point at a certain angle.
In this way, the graphic transformation that the graphic rotates by an angle around a certain point O is called rotation, the point O is called rotation center, and the rotation angle is called rotation angle.
If point P on the graph rotates into point P', then these two points are called corresponding points of this rotation.
Let us use these concepts to solve some problems.
Example 1 As shown in the figure, if the pointer of a clock is regarded as a triangle OAB and rotated clockwise around the O point, the △OEF is obtained. In the process of rotation:
(1) What is the center of rotation? What is the rotation angle?
(2) After the rotation, where do the points A and B move respectively?
Solution: (1) The center of rotation is O, ∠AOE and ∠BOF are both rotation angles.
(2) After rotation, point A and point B move to point E and point F respectively.
Independent investigation:
Please look at the cardboard in my hand. I dig a triangular hole in the cardboard, and then dig a point O as the center of rotation. I put the dug cardboard on the blackboard, first draw the dug triangular pattern (△ABC) on the blackboard, then rotate the cardboard around the rotation center O, and then draw the dug triangle (△ A ′ B ′ C ′) on the blackboard to remove the cardboard.
(Discuss in groups) Answer the following questions according to the pictures (one group recommends one person to take the stage)
1. What is the relationship between line segments OA and OA', OB and OB', OC and OC'?
2. What is the relationship between "AOA", "Bob" and "COC"?
3. What is the relationship between the shapes of 3.△ ABC and △ A ′ B ′ C ′?
Teacher's comments: 1. OA = OA', OB = OB', OC = OC', that is, the distances from the corresponding points to the rotation center are equal.
2.∠AOA'=∠ Bob '=∠COC', we call these three equal angles, that is, the angle between the corresponding point and the straight line connecting the rotation center, the rotation angle.
3.△ABC and △ A ′ B ′ C ′ are the same in shape and size, that is, congruence.
Based on the above experimental operations, the following conclusions are drawn:
(1) The distance from the corresponding point to the rotation center is equal;
(2) The included angle of the connecting line between the corresponding point and the rotation center is equal to the rotation angle;
(3) Graphic congruence before and after rotation.
Example 2 is shown in the figure. △ After △ABC rotates around point C, the corresponding point of vertex A is point D. Try to determine the position of vertex B and the corresponding point of the rotated triangle.
Analysis: When rotating around point C, the point corresponding to point A is point D, then the rotation angle is ∠ACD. According to the angle of the connecting line between the corresponding point and the rotation center is equal to the rotation angle, that is, ∠BCB'=∠ACD, and the distance between the corresponding point and the rotation center is equal, that is, CB = CB', the position of b' can be determined, as shown in the figure.
Solution: (1) connect the CD;
(2) Make ∠BCE with CB as one side, so that ∠ BCE = ∠ ACD;
(3) If CB'=CB is intercepted on the ray CE, then B' is the corresponding point of finding B;
(4) Connect DB', then △DB'C is the graph after △ABC rotates around point C. 。
Third, the class summary
(Student summary, teacher comment)
This lesson should master:
1. The distance from the corresponding point to the rotation center is equal;
2. The included angle of the connecting line between the corresponding point and the rotation center is equal to the rotation angle;
3. Graphic consistency before and after rotation and its application.
Fourth, homework
Exercise 4, 5, 6 on page 62-63 of the textbook.
The second part of the ninth grade mathematics "circle" courseware
1. Correctly understand what is central symmetry and symmetric center, and understand the properties and characteristics of central symmetric graphics.
2. According to the nature of central symmetry, we can make a symmetrical figure with central symmetry about a certain point.
focus
The concept and properties of central symmetry.
difficulty
Derivation and understanding of central symmetry.
Introduction to comments
Question: Make the pattern after the two figures in the figure below are rotated around the O point 180, and answer the following questions:
1. Do the two figures overlap after rotating 180 with O as the rotation center?
2. After the corresponding point rotates around O180, are these three points in a straight line?
Teacher's comment: After rotating around O180, we can find that the two patterns as shown in the figure overlap, that is, Figure A and Figure B overlap, and δ△OAB and δ△COD overlap.
In this way, rotate a graph around a certain point by 180, and if it can coincide with another graph, then the two graphs are said to be symmetrical or symmetrical about this point, which is called the center of symmetry.
The corresponding points in these two graphs are called symmetrical points about the center.
Explore new knowledge
(Teacher) Draw a triangle ABC on the blackboard and make two figures in two situations:
(1) is a symmetric graph with the vertex of △ABC as the center of symmetry;
(2) Make a symmetrical figure with a certain point o as the symmetrical center.
The first step is to draw △ABC.
Step 2, rotate point c (or point o) of △ABC 1 80 to draw △ a ′ b ′ c and △ a ′ b ′ c, as shown in Figure (1) and Figure (2).
From figure (1), it can be concluded that △ABC and △ a ′ b ′ c are congruent triangles;
Connect the symmetry points AA', BB' and CC' respectively. Point O is on these line segments, and O bisects these line segments.
Below, we will take Figure (2) as an example to prove these two conclusions.
It is proved that (1) in △ABC and △ A ′ B ′ C ′, OA = OA' ′, ob = ob' ′, ∠ AOB = ∠ A ′ OB ′, ∴△ AOB △ A ′ OB ′.
(2) Point A' is obtained after point A rotates180 around point O, that is, line segment OA is obtained when line segment OA rotates180 around point O, so point O is on line segment AA' and OA = OA', that is, point O is the midpoint of line segment AA'.
Similarly, the point O is also on the lines BB ′ and CC ′, and OB = OB' ′ and OC = OC' ′, that is, the point O is the midpoint of BB ′ and CC ′.
So, we get
1. For two graphs with symmetric centers, the line segments connected by symmetric points pass through the symmetric center and are equally divided by the symmetric center.
2. Two figures symmetrical about the center are congruent figures.
A detailed example
Example 1 As shown in the figure, △ABC and point O are known, draw △DEF, so that △DEF and △ABC are symmetrical about the center of point O. 。
Analysis: Central symmetry refers to rotation 180, and central symmetry about O point refers to rotation around O point 180, so we connect and extend AO, BO and CO to get the same line segment.
Solution: (1) connect AO, extend AO to D, make OD=OA, and then get the symmetrical point D of point A, as shown in the figure.
(2) Also draw the symmetrical points E and F of point B and point C. 。
(3) Connect DE, EF and FD in turn, and then △DEF is the required triangle.
Example 2 (student exercises, teacher's comments) As shown in the figure, given quadrilateral ABCD and point O, draw quadrilateral A ′ b ′ c ′ d ′ so that quadrilateral A ′ b ′ c ′ d ′ and quadrilateral ABCD are symmetrical about the center of point O (only traces are reserved, and writing is not required).
Class summary (student summary, teacher's comment)
This lesson should master:
Two basic properties of central symmetry;
1. For two graphs with symmetrical centers, the straight line connecting the corresponding points passes through the symmetrical center and is equally divided by the symmetrical center;
2. Two centrosymmetric graphs are congruent graphs and their applications.
work arrangement
Exercise on page 66 of the textbook
The third part of the ninth grade mathematics "circle" courseware
Understand the concept of central symmetric figure and the concept of symmetric center of central symmetric figure, and master the application of these two concepts.
Review the related concepts of two graphs about central symmetry, and use this knowledge to explore the related concepts and other applications of a graph being central symmetry.
focus
Related concepts of centrosymmetric graph and its application.
difficulty
Distinguish between two figures symmetrical about the center and one figure symmetrical about the center.
First, review the introduction.
1. (asked by the teacher) Answer: What are the properties of two figures with central symmetry?
(The teacher dictated): For two figures with symmetrical centers, the line segments connected by symmetrical points pass through the symmetrical center and are equally divided by the symmetrical center.
Two figures symmetrical about the center are congruent figures.
2. (Student activities) Drawing questions.
(1) Make a symmetrical figure of line segment AO about point O, as shown in the figure.
(2) Make a symmetrical figure of triangle AOB about point O, as shown in the figure.
Extend AO to make OC=AO, extend BO to make OD=BO, and connect CD, then △COD is what you want, as shown in the figure.
Second, explore new knowledge.
From another point of view, the above (1) problem is to rotate the line segment AB around its midpoint by 180. Because OA=OB, the line segment AB coincides with itself after rotating around its midpoint by 180.
In the above question (2), connecting AD and BC, the two figures which are just symmetrical about the center O become parallelograms, as shown in the figure.
AO = OC，BO=OD，∠AOB=∠COD
∴△AOB≌△COD
∴AB=CD
That is, ABCD rotates around the intersection o of its two diagonals180 and coincides with itself.
Therefore, if a graph rotates around a certain point like this 180, and if the rotated graph can coincide with the original graph, then this graph is called a central symmetric graph, and this point is its symmetric center.
(Student activities) Example 1 From the point of view that line segments and parallelograms are centrosymmetric figures, each student gives three figures, which are also centrosymmetric figures.
Teacher's comment: the characteristics of teachers asking and answering.
(Student Activities) Example 2 Please tell me what are the characteristics of the central symmetrical figure?
Teacher's comment: The central symmetrical figure has the characteristics of symmetry, beauty and stability.
Example 3 proves that any quadrilateral with a symmetrical center is a parallelogram as shown in the figure.
Analysis: The symmetrical center of the central symmetrical figure is the intersection of the connecting lines of the corresponding points and the midpoint of the line segments between the corresponding points. So we can get the diagonal bisection directly.
Prove that O is the symmetric center of quadrilateral ABCD as shown in the figure. According to the central symmetry property, the line segment AC, BD point O, and AO=CO, BO=DO, that is, the diagonal of the quadrilateral ABCD is bisected, so the quadrilateral ABCD is a parallelogram.
Third, class summary (student induction, teacher comments)
This lesson should master:
1. Related concepts of centrosymmetric graphs;
2. Solve related problems by using central symmetric graphics.
Fourth, homework
Exercise 8, 9, 10 on page 70 of the textbook.