
- •Kazakh-American University Subject Mathematics Geometry Total 68 hours
- •Glossary
- •Kazakh-American University Subject Mathematics Geometry Total 68 hours
- •Glossary
- •Kazakh-American University Subject Mathematics Geometry Total 68 hours
- •Glossary
- •Kazakh-American University Subject Mathematics Geometry Total 68 hours
- •Glossary
- •Kazakh-American University Subject Mathematics Geometry Total 68 hours
- •Glossary
- •Kazakh-American University Subject Mathematics Geometry Total 68 hours
- •Glossary
- •Kazakh-American University Subject Mathematics Geometry Total 68 hours
- •Glossary
Glossary
English |
Russian |
Kazakh |
Quadrilateral |
Четырехугольник |
Төртбұрыш |
Vertex |
Вершина |
Төбесі |
Convex |
Выпуклый |
Дөңес |
Angle |
Угол |
Бұрыш |
Square |
Квадрат |
Квадрат |
Rhomb |
Ромб |
Ромб |
Home task Lesson 1 ex 97, 98 Lesson 2 ex 101, ex 102
Individual work of student: Collection for preparing to examination Group 15 (15A16 –15A20).
Office–hours Geometry 8grade Шыныбеков А.Н. 2011, ex 21, 28 ( p 921-33)
Literature: Geometry 8grade Becboev Zh, 2012, [21-33], Collection for preparing to examination Group15
Additional literature Tulebaeva S.K. Geometry . 8 grade. Collections of problems
200]
Notes
Kazakh-American University Subject Mathematics Geometry Total 68 hours
Theme of the lecture Phales Theorem 2 hours/ week
Lecturer Grachyova Svetlana Semenovna. Profile Economics
Grade 8 Academic year 2014-2015 2 term
Lesson 13,14
Statement of theorem
If parallel lines cross one side of angle and cut equal segments, then they cut equal segments on the other side too.
Proof
Lets A1, A2, A3 – points of crossing parallel lines with one side of angle. Lets denote B1, B2, B3 corresponding points of crossing this lines with another side of angle. Prove that if A1A2=A2A3 than respectively B1B2= B2B3. Lets draw through point B2 straight line of FE parallel to A1A2. According to the property of the parallelogram A1A2=FB2, A2A3=B2E. So if A1A2=A2A3 than FB2=B2E. Triangles B1B2F and B2B3E are equal according to the second test of equality. Really B2F=B2E, angles FB2B1=B3B2E as vertical angles, angles B2FB1 and B2EB3 are equal as interior alternate angles at parallel lines A1B1 and A3B3 and crossing line FE. From the equality of triangles B2B1F and B2B3F we can say that B1B2=B2B3. The theorem proved.
Task 1:
Divide segment into definite number of parts.
Glossary
English |
Russian |
Kazakh |
Quadrilateral |
Четырехугольник |
Төртбұрыш |
Vertex |
Вершина |
Төбесі |
Convex |
Выпуклый |
Дөңес |
Angle |
Угол |
Бұрыш |
Square |
Квадрат |
Квадрат |
Rhomb |
Ромб |
Ромб |
Home task Lesson 1 task 128, 129 Lesson 1 task 132,133
Individual work of student: Collection for preparing to examination Group 15 (15B21 –15B22).
Office–hours Geometry 8grade Шыныбеков А.Н., 2012, ex 35 ( p 33-37)
Literature: Geometry 8grade Шыныбеков А.Н. 2012, [33-37], Collection for preparing to examination Group15
Additional literature Tulebaeva S.K. Geometry . 8 grade. Collections of problems
Notes