GIÁO ÁN TOÁN SONG NGỮ HÌNH HỌC 7 CẢ NĂM - Chuẩn

CHAPTER I- PERPENDICULAR LINES PARALLEL LINES Date of teaching: PERIOD 1: VERTICAL ANGLES A OBJECTIVES Knowledge: Students know the properties of vertical angles Skill: Train skill doing exercises about: skill drawing figure Education: Education about carefully, precisely in learning for students B PREPARATIONS - Teacher: Straight ruler, protractor - Students: Straight ruler, protractor C PROCESS ORGANIZATION OF TEACHING I Organize 7C: II Check your homeworks III Teaching and learning new lesson TEACHER’S AND STUDENTS’ ACTIVITIES CONTENTS Activity What are Vertical Angles? Introduce the chapter I geometry Observe figure page 81: What are Vertical Angles? x T: two angles O1 and O3 are called vertical angles y T: Comment on the relationship of side of Ô1 and Ô3 S: answer T: What are Vertical Angles? S read definition S ?2, T comment 4O y’ x’ -Two angles O1 and O3 are called vertical angles -Vertical angles are two angles such that each side of this angles is an opposite ray of the side of that angles Activity Properties of Vertical Angles S ?3: Observe figure and a Measure angles Ô1 and Ô3 Compare their measurements b Measure angles Ô2 and Ô4 Compare their measurements c Predict results drawn from the question a) and b) Practice reaoning: Ô1 + Ô3 =? ; Ô2 + Ô3 = ? It follows that Ô1 = Ô3 We have the following property Properties of Vertical Angles Results a) Ô1 = Ô3 b) Ô2 = Ô4 c) Two vertical angles are congruent Two vertical angles are congruent IV Consolidation: Read text book about property Do exercises and in the textbook V Guide home: - Learn about the definition and property of vertical angles -Do exercises 3-10 in the textbook -Do exercise in the workbook Date of teaching: PERIOD 2: PRACTICE A OBJECTIVES Knowledge: Students know the properties of vertical angles and use exercises Skill: Train skill doing exercises about: skill drawing figure Education: Education about carefully, precisely in learning for students B PREPARATIONS - Teacher: Straight ruler, protractor - Students: Straight ruler, protractor C PROCESS ORGANIZATION OF TEACHING I Organize 7C: II Check your homeworks III Teaching and learning new lesson TEACHER’S AND STUDENTS’ CONTENTS ACTIVITIES Activity Review - Student 1: recalled the definition and property of vertical angles and draw to illustrate - Student 2: exercises in the textbook Activity Practice Exercise page 83 Exercise page 83 - Students read problem and how to draw + Method: - Draw angle xOy = 470 figure - Draw two opposite ray of Ox x y’ and Oy - Angle x’Oy’ is vertically 470 O opposite to angle xOy and y congruent 470 x’ We have: Ô1 = Ô3 = 470 (vertical angles) Ô1 + Ô2 = 1800 (adjacentsupplementary angles) Hence Ô2 = 1800 – 470 =1330 Ô4 = Ô2 = 1330 (vertical angles) Exercise page 83 S: Work in pair to finish the task in minutes How can you comment about exercise 7? z x’ y’ O y x z’ The students comment Pairs of congruent angles are : ⇒ Exercise page 83 Student work in groups and answer Worksheet to the student Exercise page 83 Students read problem y Teacher hints students to draw figure x’ T: Name two right angles not vertically opposite Student work in groups and answer A y’ Exercise 10 page 83 Student work in groups and answer T: How we fold the paper to show that two vertical angles are congruent? IV Consolidation: Read text book about property of vertical angles x V Guide home: - Learn about the definition and property of vertical angles -Do exercises 4-5 in the workbook Date of teaching: PERIOD 3: TWO PERPENDICULAR LINES A OBJECTIVES Knowledge: Students know the two perpendicular lines and denoted by ⊥ Students know the property : There is one and only one a’ passing through O and perpendicular to given line a Skill: Train skill doing exercises about: skill drawing figure Education: Education about carefully, precisely in learning for students B PREPARATIONS - Teacher: Straight ruler, protractor - Students: Straight ruler, protractor C PROCESS ORGANIZATION OF TEACHING I Organize 7C: II Check your homeworks III Teaching and learning new lesson TEACHER’S AND STUDENTS’ ACTIVITIES CONTENTS Activity What are two perpendicular lines? S ?1 and ?2 y Teacher hints students to practice reasoning: Using a linear pair of angles or two vertical angles T: What are two perpendicular lines? S read definition T: We have the definition x x’ O y’ When two lines xx’ and yy’ intersect so that one the angles formed is a right angle, the lines are called two perpendicular lines and denoted by xx’⊥yy’ Activity How to draw two perpendicular lines S ?3 and ?4 T comment T introduce some drawing ways are illustrated in figure and in textbook, page 85 T: We accept the following property *There is one and only one a’ passing through O and perpendicular to given line a Exercise 11 page 86: Fill in the blanks in the following statements: a) Two perpendicular lines are … b) Two perpendicular lines a and a’ are denoted by … c) Given a point A and a line d … line d’ passing through A and perpendicular to line d Activity Perpendicular bisector of a segment Look at figure 7, we recognize that: I is the midpoint of segment AB Line xy is perpendicular to the line AB at I We say: The line xy is the perpendicular bisector of the segment AB What is perpendicular bisector of a segment? S read definition When xy is the perpendicular bisector of the segment AB, it is also said that A is x A I B -The line perpendicular to a segment at its midpoint is called the perpendicular bisector of that segment the reflected image of B in line xy or B is the reflected imabe of A in line xy IV Consolidation: - Recall of two perpendicular and perpendicular bisector of a segment - S exercise 14 page 86 in the textbook d C I D V Guide home: - Learn about the definition and property of perpendicular and perpendicular bisector of a segment -Do exercises 12, 13, 14-18 in the textbook Preparing date: 10/9/2016 Teaching date: 17/9 PERIOD 4: PRACTICE A OBJECTIVES Knowledge: Students know explaining two perpendicular lines and use exercises Skill: Train skill draw two perpendicular lines and perpendicular bisector of a segment Education: Education about carefully, precisely in learning for students B PREPARATIONS - Teacher: Straight ruler, protractor - Students: Straight ruler, protractor C PROCESS ORGANIZATION OF TEACHING I Organize 7A: II Check your homeworks S1: What are two perpendicular lines and drawing illustrate? S2: What is perpendicular bisector of a segments and draw perpendicular bisector of segment AB=4cm? S come out to board T comment and give point III Teaching and learning new lesson TEACHER’S AND STUDENTS’ ACTIVITIES CONTENTS Exercise 15 page 86: S read problem S exercises 15 and give the conclusions + zt ⊥ xy at O + There are right angles: Exercise 18 page 87: T: Draw image in a way expressed in the following words: d2 C - Draw - Take any point A in xOy angle - Draw line d1 through A and y A O 450 B perpendicular to the ray Ox at B x d1 Draw line d2 through A and perpendicular to the ray Oy at C S come out to board and worksheet Exercise 19 page 87: Redeaw figure 11 and show clearly the drawing steps Observe figure 11 and answer S work in group d1 O B 600 C Exercise 20 page 87: S read problem and T: Draw in two cases: three points A, B, C are collinear and three point A, B, C are not collinear IV Consolidation: - Recall of two perpendicular and perpendicular bisector of a segment V Guide home: - Learn about the definition and property of perpendicular and perpendicular bisector of a segment -Do exercise in the workbook d2 A Làm tập 53, 54, 55 (SGK-Trang 80) Preparing date : 18/10 Teaching date: Period 64 PROPERTIES OF THREE PERPENDICULAR BISECTORS OF A TRIANGLE I AIMS: - Knowledge: Know the concept of a perpendicular bisector line of the triangle, a triangle has perpendiculars; Knowing how to use the straight-edge, online compasses to draw middle of the triangle; Understand the nature of the isosceles triangle, prove theorem 2, said the concept of the circle of a triangle - Skills: Train drawing skills bisector triangle; use the theorem to solve exercises - Attitude: Work seriously, responsibly II METHOD: Verbal, suggestive, stating and solving problems III PREPARE: GV: Side table HS: Textbook, group table IV PROCEDURE: 1.Organzation: 7A: Warm-up: 3.New lesson Teacher and students’activities Content 2.Properties of three perpendicular bisector of a triangle ?2 a)Theorem : Three perpendicular bisectors of a triangle are convergent at the same point, this point is equidistant to thre three vertical of that triangle Ba đường trung trực tam giác qua - Teachers demonstrate mentioned điểm, điểm cách cạnh tam giác direction: Because belong to perpendicular bisector of AB OB = OA Because O belong to perpendicular GT ∆ ABC, b is perpendicular bisector of AC visector of BC c is perpendicular bisector of AB, b and c cu • OC = OA OB = OC ⇒ O belong to KL O lies on the perpendicular bisector of BC perpendicular bisector of BC OA = OB = OC ⇒ Also from (1) OB = OC = OA b)Remark: ie three straightforward passing O is center of subcribe circle ∆ ABC point, points equidistant from the O tâm đường tròn ngoại tiếp ∆ ABC top of the triangle Consolidation - State properties of perpendicular bisector of a triangle - Do exercise 52 - Làm tập 52 (HD: xét tam giác) Homework - Exercise 53, 54, 55(textbook-P.80) Preparing date : 18/10 Teaching date: Period 65 PRACTICE I AIMS: - Knowledge: Consolidation properties of perpendiculars bisector in the triangle - Skills: Train skills straightforward draw triangle - Seeing the practical application of the central line of the nature of the segment; - Attitude: positive Train, accuracy and careful II METHOD: Verbal, suggestive, stating and solving problems III PREPARE: GV: Side table HS: Textbook, group table IV PROCEDURE: 1.Organzation: 7A: Warm-up: 3.New lesson Teacher and students’activities Content Exercise 52 * Request students to exercise 52 - Call student draw figure , record given, prove Guide sts demonstrate: ? Proven methods mentioned isosceles triangle - St : + M1: two equal sides + M2: congruent angles A B M C ∆ ABC, AM is medina and perpendicular b GT ∆ ABC isosceles at A KL Proof: Consider ∆ AMB, ∆ AMC have : ? proofs mentioned equal sides ⇒ ∆ AMB= ∆ AMC(sas) Teachers request students to read ⇒ AB = AC ⇒ ∆ ABC isosceles at A Figure 55 Exercise 55 ? Request what the problem Segment AB ⊥ AC;ID is perpendicular bisecto - T drawing figure 51 on the board GT perpendicular bisector of AC ? Said given, prove of the problem trung trực AB; KD trung trực AC - T suggestions: KL B, D, C collinear To prove B D, C collinear can St: To prove that B, D, C collinear we can prove prove how? Thẳng hàng ta chứng minh ? Calculate the angle BDA angle = 180o hay + = 180o A1 (T proven record on the table) St : Have D belong to perpendicular bisector of ? Similarly, consider the angle AD ⇒DA ( according to property of ADC angle A2 perpendicular bisector of line) ? Since then, compute the angle Có D thuộc trung trực AD ⇒ DA = DB BDC? (theo tính chất đường trung trực đoạn thẳng) ⇒∆DBA isosceles ⇒ = ⇒ = 180o - ( + ) = 180o - 2A1 - Similar =180o-2 =+=180o -2+180o-2 = 360o - 2( + )= 30 - 2.90o= 180o Consolidation * Request students to exercise 54 - Students read carefully requirements of the post - Teacher for each student to one part (if students not it,guide ) ? The center of the circle through the three vertices of the triangle in any position, it is the intersection of the public road - Students: delivery of midperpendiculars * Yêu cầu học sinh làm tập 54 - Học sinh đọc kĩ yêu cầu - Giáo viên cho học sinh làm phần (nếu học sinh không làm HD) ? Tâm đường tròn qua đỉnh tam giác vị trí nào, giao đường Homework - Do exercise 68, 69(workbook) Làm tập 68, 69 (SBT) Guide 68: AM is perpendicular bisector Preparing date : 18/10 Teaching date: 7C: 22/10; 7E,A:25/10 Period 66 properties three altitudes of triangles I AIMS: - Knowledge: Review concept, properties altitude of triangles; drawing altitude of triangle - Skills: Applying solve some problems - Attitude: Work seriously, responsibly II METHOD: Verbal, suggestive, stating and solving problems III PREPARE: GV: Side table HS: Textbook, group table IV PROCEDURE: 1.Organzation: 7A: Warm-up: 3.New lesson Teacher and students’activities - Request students to exercise 59 - Call students read carefully the first post, drawing recordgiven, Content Exercise 59 (textbook) prove ? SN ⊥ ML, SL is what line of ∆ LNM (The altitude of the triangle) ? So what's the point S triangle (Center) - Teachers guide students to find answers part b) · MSP =? ↑ ∆ SMP · SMP =? ↑ ∆ MQN · QNM - Request students to base on presentation solution - Request students to exercises 61 ? How to determine the center of the triangle L Q S 50° M P N ∆ LMN, MQ ⊥ NL, LP ⊥ ML a) NS ⊥ ML GT KL · b) With LNP = 50 Compute ; a) Since MQ ⊥ LN, LP ⊥ MN → S is or Orthocenter ∆ LMN → NS ⊥ ML b) Consider ∆ MQL have : + =900=500+ ⇒ =400 Consider ∆ MSP have: + =900=400+ ⇒ =500 · · * MSP + PSQ = 180 =500+ ⇒ =1300 Exercise 61 - Identifty intersection point of altitudes Xác định giao điểm đường cao A N - Call students on the board presents the a, b, class comments, supplemented, corrected M H B K C a HK, BN, CM are altitudes of ∆ BHC HK, BN, CM ba đường cao ∆ BHC Orthocenter of ∆ BHC is A Trực tâm ∆ BHC A b Orthocenter of ∆ AHC is B trực tâm ∆ AHC B Orthocenter of ∆ AHB is C Trực tâm ∆ AHB C Consolidation - Draw altitude Vẽ đường cao - Properties of altitude, altitude of triangle Tính chất đường cao, đường cao tam giác Homework - St part question Học sinh làm phần câu hỏi ôn tập - Do exercise 63, 64, 65(textbook) Làm tập 63, 64, 65 (SGK) - Next lesson review Tiết sau ôn tập Guide exercise 63 (P.87) A B D C E · a) We have ADC is exterior angle of ∆ · · ABD → ADC > BAD → (1) · We have again BDA is exterior angle of ∆ ADE → (2) From 1, → · · → AE > AD > AEB b) In ∆ ADE: ADC Preparing date : 18/10 Teaching date: Period 67 properties three altitudes of triangles I AIMS: - Knowledge: Review concept, properties altitude of triangles; drawing altitude of triangle - Skills: Applying solve some problems - Attitude: Work seriously, responsibly II METHOD: Verbal, suggestive, stating and solving problems III PREPARE: GV: Side table HS: Textbook, group table IV PROCEDURE: 1.Organzation: 7A: Warm-up: 3.New lesson Teacher andstudents’activities T put questions to the board textbook side-table Content 6.7 a) The centriod point of the triangle is the three median line, every peak median length Draw a triangle ABC and determine the centroid G of the triangle that that goes through the top Drawing : a) Trọng tâm tam giác điểm chung ba T given the assignment to monitor and đường trung tuyến, cách đỉnh độ dài guide the students draw pictures trung tuyến qua đỉnh Vẽ hình : T suggestions: a) Have a comment about the triangle MPQ and RPQ? T drawing altitude PH b) Similarly ratio than SRNQwith SMNQ like? Why c Compare SRNQ and SRPQ GV đưa đề lên hình hướng dẫn HS vẽ hình - T call st up table drawing: Draw angle xOy, grab a ∈ Ox; B ∈ Oy a) For how are the two sides of the angle xOy , the point M must xOy located? b) If OA = OB, how many points M satisfying the conditions of a sentence a? N A M G B C Properties of: - Three bisectors; Three perpendiculars bisector ; Three altitudes triangle Tính chất của: - Ba đường phân giác; Ba đường trung trực ; Ba đường cao tam giác Exercise 67 P.87 textbook GT ∆MNP is median of MR Q: is centroid Kl a) Compute SMPQ : SRPQ b) Compute SMNQ : SRNQ c) Compare SRPQand SRNQ ⇒ SQMN = SQNP = SQPM a) Triangle MPQ and RPQ have common point P, two side MQ and QR along lie on a straight line should share the atitude dropped from P to the line MR (altitude PH) There MQ = 2QR (property of centriod of S MPQ triangle) ⇒ S RPQ =2 S MNQ S RNQ =2 b) Similarly: Since the triangles has common altitude NK and MQ = 2QR c) SRPQ = SRNQ as two triangles has altitude QI and side NR = RP (gt) SQMN = SQNP = SQPM (= 2SRPQ = 2SRNQ) a) Tam giác MPQ RPQ có chung đỉnh P, hai cạnh MQ QR nằm đường thẳng nên có chung đường cao hạ từ P tới đường thẳng MR (đường cao PH) Có MQ = 2QR (tính chất trọng tâm tam S MPQ giác)⇒ S RPQ =2 S MNQ S RNQ =2 b) Tương tự: Vì hai tam giác có chung đường cao NK MQ = 2QR c) SRPQ = SRNQ hai tam giác có chung đường cao QI cạnh NR = RP (gt) SQMN = SQNP = SQPM (= 2SRPQ = 2SRNQ) Exercise 68 P.88 textbook St : To equidistant from the two sides of the angle, the point M located on the angle xOy - For equidistant two points A and B, the point M must lie on the perpendicular bisector line of the segment AB - Point M is the intersection of bisector rays xOy with perpendicular biscetor line of the segment AB b) If OA = OB, then the bisector Oz of xOy angle coincides with the perpendicular bisector line of the segment AB, so every point on Oz rays are satisfied the conditions in question a Cosolidation Exercise 91 P.34 workbook Homework Review chapter , learn concept , theorem , properties Ôn tập lý thuyết chương, học thuộc khái niệm, định lí, tính chất Trình bầy lại câu hỏi, tập ôn tập chương III SGK Do exercise 82, 84, 85 tr.33, 34 workbook Làm tập số 82, 84, 85 tr.33, 34 SBT Preparing date : 18/10 Teaching date: 7C: 22/10; 7E,A:25/10 Period 68 PRACTICE I AIMS: - Knowledge: Review and systematization of knowledge mainly about the case of the triangle are equal - Skills: To apply the knowledge learned to solve some exercises last study part geometry - Attitude: Train drawing skills, homework picture II METHOD: Verbal, suggestive, stating and solving problems III PREPARE: GV: Side table HS: Textbook, group table IV PROCEDURE: 1.Organzation: 7A: Warm-up: 3.New lesson REVIEW THE CONGRUENCES OF A TRIANGLE Exercise P.92 textbook st read exercise xOy = 90o DO = DA; CD ⊥ OA EO = EB; CE ⊥ OB Prove a) CE = OD b) CE ⊥ CD c) CA = CB d) CA // DE e) A, C, B collinear Given St present math exercise HS trình bày miệng toán a) ∆CED ∆ ODE have: ⇒ ∆CED=∆ODE(a.s.a) ⇒ CE = OD (corresponding side) b) ECD = DOE = 90o (corresponding side) ⇒ CE ⊥ CD c) ∆ CDA and ∆ DCE have: ⇒∆CDA = ∆DCE (s.a.s) T suggested sts to analyze the ⇒ CA = DE (corresponding side ) problem Then request students to present turn of all the questions GV gợi ý để HS phân tích toán Sau yêu cầu HS trình bày câu hỏi HOMEWORK Review definition Remaining exercise in textbook Date of teaching: PERIOD 69 CHAPTER III REVIEW A OBJECTIVES Knowledge: Students know to the review tables about knowledge of the chapter Skill: draw figure Education: carefully, precisely in learning for students B PREPARATIONS - Teacher: Straight ruler, protractor - Students: Straight ruler, protractor C PROCESS ORGANIZATION OF TEACHING I Organize 7c:………… II Check your homeworks Using the question in the chapter III Teaching and learning new lesson TEACHER’S AND STUDENTS’ CONTENTS ACTIVITIES Part : review table I Knowledge S: Do the groups ; AB > AC a) AB > AH; AC > AH b) If HB > HC then AB > AC c) If AB > AC then HB > HC DE + DF > EF; DE + EF > DF, T: Guide : Math the facts in the two columns to get the correct statements Part 2: Apply S: Read Exercise 63 Math the facts in the two columns to get the correct statements a - d' b - a' c - b' d - c' Math the facts in the two columns to get the correct statements: a - b' b - a' c - d' d - c' II Exercise Exercise 63 page 87 text book: S: write given and prove T: repeat theorem T: guide: ? is external angle of triangle a) We have is external angle ABD ? what is ABD triangle (1)(because ABD is isosceles B) but is external angle of (2) so, > T: Read proplem , from 1, ACE b) In ADE: AE > AD Exercise 65 - relations among the three sides of a triangle IV Consolidation: read to repeat property of three perpendicular bisector of a triangle V Guide home: - Exercise 64, 67, 68, 69 text book Date of teaching: PERIOD 70: YEAR REVIEW PART GEOMETRY A OBJECTIVES Knowledge: Students know to the review about knowledge of the year Skill: draw figure, knows proof figure, apply theorem to the Exercise Education: carefully, precisely in learning for students B PREPARATIONS - Teacher: Straight ruler, protractor - Students: Straight ruler, protractor C PROCESS ORGANIZATION OF TEACHING I Organize 7c:………… II Check your homeworks III Teaching and learning new lesson TEACHER’S AND STUDENTS’ ACTIVITIES Part 1: review about parallel T: what are two parallel sides ? S: Draw figure c a CONTENTS parallel Given Prove a // b ; .; b S: repeat theorem a b Â3+ … =1800 Given Two sides a, b Â3 or + … =1800 or Exercise 2,3 page 91 text book Prove a // b Oclit Exercise page 91 text book S: One haft Exercise Other student Exercise S: Do with groups a) We have a ⊥ MN (given); b ⊥ MN (given)⇒ a // b b) a // b (proof a) ⇒ + = 180o (from property) 50o+ =180o⇒ = 180o - 50o = 130o S: Exercise with groups Exercise page 91 text book: from O draws ray Ot // a // b Because a // Ot ⇒ = = 44o Because b // Ot ⇒ + = 180o Part 2: review about two triangles Exercise page 92 text book congruents y S: read problem B C E O ∆ CED = ∆ ODE (a.s.a) x ED common = (because CD//Oy) ⇒ ∆CED = ∆ODE (a.s.a) ⇒ CE = OD (two sides the same way) b) And = = 90o (two angles the CE⊥CD = A solution: a) Consider ∆CED and ∆ ODE we have: = (because EC//Ox) T: Guide S1: CE = OD S2: D = 900 same ) ∆CED = ∆ODE G/v gợi ý để học sinh chứng minh ⇒ CE ⊥ CD c) Consider ∆ CDA and ∆ DCE we have: CD common = = 90o DA = CE (= DO) ⇒ ∆CDA = ∆DCE (s.a.s) ⇒ CA = DE Proof the same way => CB = DE => CA = CB = DE d from ∆CDA = ∆DCE => = (two same angles ) => CA // DE e We have CA // DE the same way => CB // DE => A, C, B on a side IV Consolidation: read to repeat property which is known V Guide home: - Exercise 6,7,8,9 text book ... - Draw angle xOy = 470 figure - Draw two opposite ray of Ox x y’ and Oy - Angle x’Oy’ is vertically 470 O opposite to angle xOy and y congruent 470 x’ We have: Ô1 = Ô3 = 470 (vertical angles)... protractor - Students: Straight ruler, protractor C PROCESS ORGANIZATION OF TEACHING I Organize 7A: 7B 7C II Check your homeworks III Teaching and learning new lesson CONTENTS TEACHER’S AND STUDENTS’... square - Students: Straight ruler, set square C PROCESS ORGANIZATION OF TEACHING I Organize 7C: 7A: 7B II Check your homeworks Question: What are rules identify two parallel lines? And drawing