CBSE-i. Shiksha Kendra, 2, Community Centre, Preet Vihar,Delhi India

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1 CBSE-i CLASS X UNIT-11 Mathematics Similar Triangles (Core) Shiksha Kendra, 2, Community Centre, Preet Vihar,Delhi India

2 The CBSE-International is grateful for permission to reproduce and/or translate copyright material used in this publication. The acknowledgements have been included wherever appropriate and sources from where the material may be taken are duly mentioned. In case any thing has been missed out, the Board will be pleased to rectify the error at the earliest possible opportunity. All Rights of these documents are reserved. No part of this publication may be reproduced, printed or transmitted in any form without the prior permission of the CBSE-i. This material is meant for the use of schools who are a part of the CBSE-International only.

3 PREFACE The Curriculum initiated by Central Board of Secondary Education -International (CBSE-i) is a progressive step in making the educational content and methodology more sensitive and responsive to the global needs. It signifies the emergence of a fresh thought process in imparting a curriculum which would restore the independence of the learner to pursue the learning process in harmony with the existing personal, social and cultural ethos. The Central Board of Secondary Education has been providing support to the academic needs of the learners worldwide. It has about schools affiliated to it and over 158 schools situated in more than 23 countries. The Board has always been conscious of the varying needs of the learners in countries abroad and has been working towards contextualizing certain elements of the learning process to the physical, geographical, social and cultural environment in which they are engaged. The International Curriculum being designed by CBSE-i, has been visualized and developed with these requirements in view. The nucleus of the entire process of constructing the curricular structure is the learner. The objective of the curriculum is to nurture the independence of the learner, given the fact that every learner is unique. The learner has to understand, appreciate, protect and build on values, beliefs and traditional wisdom, make the necessary modifications, improvisations and additions wherever and whenever necessary. The recent scientific and technological advances have thrown open the gateways of knowledge at an astonishing pace. The speed and methods of assimilating knowledge have put forth many challenges to the educators, forcing them to rethink their approaches for knowledge processing by their learners. In this context, it has become imperative for them to incorporate those skills which will enable the young learners to become 'life long learners'. The ability to stay current, to upgrade skills with emerging technologies, to understand the nuances involved in change management and the relevant life skills have to be a part of the learning domains of the global learners. The CBSE-i curriculum has taken cognizance of these requirements. The CBSE-i aims to carry forward the basic strength of the Indian system of education while promoting critical and creative thinking skills, effective communication skills, interpersonal and collaborative skills along with information and media skills. There is an inbuilt flexibility in the curriculum, as it provides a foundation and an extension curriculum, in all subject areas to cater to the different pace of learners. The CBSE has introduced the CBSE-i curriculum in schools affiliated to CBSE at the international level in 2010 and is now introducing it to other affiliated schools who meet the requirements for introducing this curriculum. The focus of CBSE-i is to ensure that the learner is stress-free and committed to active learning. The learner would be evaluated on a continuous and comprehensive basis consequent to the mutual interactions between the teacher and the learner. There are some non-evaluative components in the curriculum which would be commented upon by the teachers and the school. The objective of this part or the core of the curriculum is to scaffold the learning experiences and to relate tacit knowledge with formal knowledge. This would involve trans-disciplinary linkages that would form the core of the learning process. Perspectives, SEWA (Social Empowerment through Work and Action), Life Skills and Research would be the constituents of this 'Core'. The Core skills are the most significant aspects of a learner's holistic growth and learning curve. The International Curriculum has been designed keeping in view the foundations of the National Curricular Framework (NCF 2005) NCERT and the experience gathered by the Board over the last seven decades in imparting effective learning to millions of learners, many of whom are now global citizens. The Board does not interpret this development as an alternative to other curricula existing at the international level, but as an exercise in providing the much needed Indian leadership for global education at the school level. The International Curriculum would evolve on its own, building on learning experiences inside the classroom over a period of time. The Board while addressing the issues of empowerment with the help of the schools' administering this system strongly recommends that practicing teachers become skillful learners on their own and also transfer their learning experiences to their peers through the interactive platforms provided by the Board. I profusely thank Shri G. Balasubramanian, former Director (Academics), CBSE, Ms. Abha Adams and her team and Dr. Sadhana Parashar, Head (Innovations and Research) CBSE along with other Education Officers involved in the development and implementation of this material. The CBSE-i website has already started enabling all stakeholders to participate in this initiative through the discussion forums provided on the portal. Any further suggestions are welcome. Vineet Joshi Chairman

4 ACKNOWLEDGEMENTS Advisory Conceptual Framework Shri Vineet Joshi, Chairman, CBSE Shri G. Balasubramanian, Former Director (Acad), CBSE Shri Shashi Bhushan, Director(Academic), CBSE Ms. Abha Adams, Consultant, Step-by-Step School, Noida Dr. Sadhana Parashar, Head (I & R),CBSE Ideators Ms. Aditi Misra Ms. Anuradha Sen Ms. Jaishree Srivastava Dr. Rajesh Hassija Ms. Amita Mishra Ms. Archana Sagar Dr. Kamla Menon Ms. Rupa Chakravarty Ms. Anita Sharma Ms. Geeta Varshney Dr. Meena Dhami Ms. Sarita Manuja Ms. Anita Makkar Ms. Guneet Ohri Ms. Neelima Sharma Ms. Himani Asija Dr. Anju Srivastava Dr. Indu Khetrapal Dr. N. K. Sehgal Dr. Uma Chaudhry Material Production Groups: Classes IX-X English : Ms. Sarita Manuja Ms. Renu Anand Ms. Gayatri Khanna Ms. P. Rajeshwary Ms. Neha Sharma Ms. Sarabjit Kaur Ms. Ruchika Sachdev Geography: Ms. Deepa Kapoor Ms. Bharti Dave Ms. Bhagirathi Ms. Archana Sagar Ms. Manjari Rattan English : Ms. Rachna Pandit Ms. Neha Sharma Ms. Sonia Jain Ms. Dipinder Kaur Ms. Sarita Ahuja Material Production Group: Classes I-V Dr. Indu Khetarpal Ms. Rupa Chakravarty Ms. Anita Makkar Ms. Nandita Mathur Ms. Vandana Kumar Ms. Anuradha Mathur Ms. Kalpana Mattoo Ms. Seema Chowdhary Ms. Anju Chauhan Ms. Savinder Kaur Rooprai Ms. Monika Thakur Ms. Ruba Chakarvarty Ms. Deepti Verma Ms. Seema Choudhary Mr. Bijo Thomas Ms. Mahua Bhattacharya Ms. Ritu Batra Ms. Kalyani Voleti Coordinators: Dr. Sadhana Parashar, Ms. Sugandh Sharma, Dr. Srijata Das, Dr. Rashmi Sethi, Head (I and R) E O (Com) E O (Maths) E O (Science) Shri R. P. Sharma, Consultant Ms. Ritu Narang, RO (Innovation) Ms. Sindhu Saxena, R O (Tech) Shri Al Hilal Ahmed, AEO Ms. Seema Lakra, S O Mathematics : Dr. K.P. Chinda Mr. J.C. Nijhawan Ms. Rashmi Kathuria Ms. Reemu Verma Dr. Ram Avtar Mr. Mahendra Shankar Political Science: Ms. Sharmila Bakshi Ms. Archana Soni Ms. Srilekha Science : Dr. Meena Dhami Mr. Saroj Kumar Ms. Rashmi Ramsinghaney Ms. Seema kapoor Ms. Priyanka Sen Dr. Kavita Khanna Ms. Keya Gupta Ms. Preeti Hans, Proof Reader Science : Ms. Charu Maini Ms. S. Anjum Ms. Meenambika Menon Ms. Novita Chopra Ms. Neeta Rastogi Ms. Pooja Sareen Economics: Ms. Mridula Pant Mr. Pankaj Bhanwani Ms. Ambica Gulati Material Production Groups: Classes VI-VIII Mathematics : Ms. Seema Rawat Ms. N. Vidya Ms. Mamta Goyal Ms. Chhavi Raheja Political Science: Ms. Kanu Chopra Ms. Shilpi Anand History : Ms. Jayshree Srivastava Ms. M. Bose Ms. A. Venkatachalam Ms. Smita Bhattacharya Geography: Ms. Suparna Sharma Ms. Leela Grewal History : Ms. Leeza Dutta Ms. Kalpana Pant

5 Content 1. Syllabus 1 2. Scope document 2 3. Teacher's Support Material 4 vteacher Note 5 vactivity Skill Matrix 9 vwarm Up W1 11 Classification of Triangles vwarm Up W2 11 Revising Congruency vpre -Content P1 12 Recognizing Similar Figures Based on its Dictionary Meaning vpre -Content P2 12 Corresponding Parts of Congruent Triangles vpre -Content P3 13 Recapitulation of Ratio and Proportion vcontent Worksheet CW1 13 Similar Figures vcontent Worksheet CW2 14 Basic Proportionality Theorem vcontent Worksheet CW3 15 Application of BPT vcontent Worksheet CW4 16 Converse of BPT vcontent Worksheet CW5 16 Pythagoras Theorem vcontent Worksheet CW6 17 Hands on Activity to Verify Pythagoras Theorem vcontent Worksheet CW7 17

6 Application of Pythagoras Theorem vcontent Worksheet CW8 18 ICT Activity on Pythagoras Theorem vcontent Worksheet CW9 18 vconverse of Pythagoras Theorem vcontent Worksheet CW10 19 Application of Similar Triangles vpost Content Worksheet PCW1 20 vpost Content Worksheet PCW2 20 vpost Content Worksheet PCW3 20 vpost Content Worksheet PCW4 20 vpost Content Worksheet PCW5 20 vpost Content Worksheet PCW6 20 vpost Content Worksheet PCW7 vassessment Plan 21 vstudy Material 22 vstudent Support Material 52 vsw1: Warm Up (W1) 53 Classification of Triangles vsw2: Warm Up (W2) 55 Revising Congruency vsw3: Pre Content (P1) 57 Recognizing Similar Figures Based on its Dictionary Meaning vsw4: Pre Content (P2) 59 Corresponding Parts of Congruent Triangles vsw5: Pre Content (P3) 61 Recapitulation of Ratio and Proportion vsw6: Content (CW1) 62 Similar Figures

7 vsw7: Content (CW2) 73 Basic Proportionality Theorem (BPT) vsw8: Content (CW3) 75 Application of BPT vsw9:content (CW4) 76 Converse of BPT vsw10: Content (CW5) 78 Pythagoras Theorem vsw11: Content (CW6) 85 Hands on Activity to Verify Pythagoras Theorem vsw12:content (CW7) 86 Application of Pythagoras Theorem vsw13:content (CW8) 89 ICT Activity on Pythagoras Theorem vsw14:content (CW9) 92 Converse of Pythagoras Theorem vsw15:content (CW10) 99 Application of Similar Triangles vsw16: Post Content (PCW1) 104 vsw17: Post Content (PCW2) 105 vsw 18: Post Content (PCW3) 106 vsw 19: Post Content (PCW4) 107 vsw 20: Post Content (PCW5) 109 vsw 21: Post Content (PCW6) 110 vsw 22: Post Content (PCW7) 113 vsuggested Videos & Extra Readings. 114

8 SYLLABUS UNIT 11 GEOMETRY SIMILAR TRIANGLES (CORE) Introduction to similarity Similar figures, similarity of two polygons, similarity of two triangles. Criteria of similarity of two polygons Two polygons are similar when their a) corresponding angles are same and. b) corresponding sides are in proportion Criteria of similarity AAA, SSS, SAS through exploration Application problems Basic Proportionality theorem Proof and applications Pythagoras theorem Proof and applications Converse of Pythagoras theorem Statement and simple applications 1

9 SCOPE DOCUMENT Key Concepts 1. Similarity of polygons 2. Similarity of triangles 3. Basic Proportionality Theorem 4. Pythagoras Theorem Learning Objective: 1. To understand the difference between similar and congruent figures. 2. To define similar triangles. 3. To state Basic Proportionality Theorem (Thales Theorem) 4. To verify BPT using explorations or models. 5. To apply BPT in geometrical problems. 6. To state converse of BPT. 7. To verify converse of BPT using explorations or models. 8. To apply converse of BPT in geometrical problems. 9. To state Pythagoras Theorem. 10. To verify Pythagoras Theorem using explorations or models. 11. To apply Pythagoras Theorem in geometrical problems. 12. To state converse of Pythagoras Theorem. 13. To verify converse of Pythagoras Theorem using explorations or models. 14. To apply converse of Pythagoras Theorem in geometrical problems. 15. To state and understand all the criterion of similarity- SSS, AA,SAS. 16. To verify all criterions of similarity using exploration or models. 17. To apply the similarity criterion in problems 2

10 Extension Activities: 1. Sierpinski triangle can be generated by continuously repeating equilateral triangles. Create sierpinski triangle in series and find the ratio of the original equilateral triangle to the tenth triangle in the series. 2. Extension of Pythagoras theorem: a) Generalize the Pythagoras theorem to mean that if similar figures are drawn on each side of the right triangle, the sum of the areas of two smaller figures equals the area of larger figure. Perspective: Fractal is a new branch of geometry which is based on concept of self similarity. It has been observed that the shape of a portion of the object, if magnified, look like original object. For example clouds, mountains, tress etc. Students can find more on fractals. SEWA: Similar triangles find their application in many real life situations. For example, consider the following problems: a) Campers walking along the south side of a river want to fell a tree tall enough so that they can walk on the tree to get across the river. How can they find the width of the river at its narrowest point without swimming across the river? b) What must be the distance between the camera and statue if statue of some height is to be photographed? Research: The Greek mathematician Thales used the knowledge of similar triangles to estimate the height of the Greek Pyramid? Research about his technique and prepare a presentation. b) Extend the result to similar polygons for four or more sides. 3

11 TEACHER S SUPPORT MATERIAL 4

12 TEACHER S NOTE The teaching of Mathematics should enhance the child s resources to think and reason, to visualize and handle abstractions, to formulate and solve problems. As per NCF 2005, the vision for school Mathematics includes: 1. Children learn to enjoy mathematics rather than fear it. 2. Children see mathematics as something to talk about, to communicate through, to discuss among them, to work together on. 3. Children pose and solve meaningful problems. 4. Children use abstractions to perceive relationships, to see structures, to reason out things, to argue the truth or falsity of statements. 5. Children understand the basic structure of Mathematics: Arithmetic, algebra, geometry and trigonometry, the basic content areas of school Mathematics, all offer a methodology for abstraction, structuration and generalisation. 6. Teachers engage every child in class with the conviction that everyone can learn mathematics. Students should be encouraged to solve problems through different methods like abstraction, quantification, analogy, case analysis, reduction to simpler situations, even guess-and-verify exercises during different stages of school. This will enrich the students and help them to understand that a problem can be approached by a variety of methods for solving it. School mathematics should also play an important role in developing the useful skill of estimation of quantities and approximating solutions. Development of visualisation and representations skills should be integral to Mathematics teaching. There is also a need to make connections between Mathematics and other subjects of study. When children learn to draw a graph, they should be encouraged to perceive the importance of graph in the teaching of Science, Social Science and other areas of study. Mathematics should help in developing the reasoning skills of students. Proof is a process which encourages systematic way of argumentation. The aim should be to develop arguments, to evaluate arguments, to make conjunctures and understand that there are various methods of reasoning. Students should be made to understand that mathematical communication is precise, employs unambiguous use of language and rigour in formulation. Children should be encouraged to appreciate its significance. At the secondary stage students begin to perceive the structure of Mathematics as a discipline. By this stage they should become familiar with the characteristics of Mathematical communications, various terms and concepts, the use of symbols, precision of language and systematic arguments in proving the proposition. At this stage a student should be able to integrate the many concepts and skills that he/she has learnt in solving problems. Unit on similar triangles focus on lots of exploration and geogebra activities in order to attain the following learning objectives: 5

13 1. To understand the difference between similar and congruent figures. 2. To define similar triangles. 3. To state Basic Proportionality Theorem (Thales Theorem) 4. To verify BPT using explorations or models. 5. To Prove BPT logically. 6. To apply BPT in geometrical problems. 7. To state converse of BPT. 8. To verify converse of BPT using explorations or models. 9. To apply converse of BPT in geometrical problems. 10. To state Pythagoras Theorem. 11. To verify Pythagoras Theorem using explorations or models. 12. To apply Pythagoras Theorem in geometrical problems. 13. To state converse of Pythagoras Theorem. 14. To verify converse of Pythagoras Theorem using explorations or models. 15. To apply converse of Pythagoras Theorem in geometrical problems. 16. To state and understand all the criterion of similarity- SSS, AA,SAS. 17. To verify all criterions of similarity using exploration or models. 18. To apply the similarity criterion in problems Concept of similarly in geometry is analogous to algebraic concept of ratio and proportion. It finds its application in making maps, scale drawings, enlargement of photos and indirect measurements of distance e. g. height of a tall building etc. To introduce the concept of similar figures teacher can use interesting computer applications. He/She can perform the copy-paste operation on screen for various figures and demonstrate that all figures identical in shape and size are congruent figures. Further teacher can increase of decrease the size of copied figure and show that every time he/she performs this operation, size changes but shape is retained. When shape is identical but size changes figures are known as similar figures. 6

14 Further the figures can be overlapped, inserted, rotated or flipped in order to refine the understanding of relation between congruency and similarity and to thrust upon following points: i. ii. All congruent figures are similar but all similar figures are not congruent. All similar figures hold reflexive, symmetric and transitive property. Reflexive Δ ABC ~Δ ABC Transitive Δ ABC ~ Δ DEF and Δ DEF ~ Δ GHI > Δ ABC ~ ΔGHI Symmetric Δ ABC ~ Δ DEF Also Δ DEF ~ Δ ABC Further, students can observe that when the figures are similar the corresponding sides are in proportion. It can be again shown on the screen that when the size is not increased or reduced proportionately the shape is changed. They can further be allowed to explore why some polygons like square, circle, regular hexagon etc. are always similar? Why all triangles are not similar? Why all equilateral triangles are similar but right triangles are not similar? Under what conditions triangles are similar? To find the answer to above questions students can be allowed to draw various figures to measure them to cut, to overlap the sides or angles etc. Teacher must facilitate the conditions which can help them to come up with the idea on their own thati. All squares are similar, because the corresponding angles are always same i.e. 90. ii. When all the corresponding angles are same triangles are similar. iii. Equilateral triangles have all angles of 60 for any dimension of triangles, so all equilateral triangles are similar. iv. In right triangles all angles are not always same so all right triangles are not similar. 7

15 All observations made till this stage leads to AA similarity criteria. With the help of Geo-Gebra activities, other criteria of similarity can be introduced. Students can also draw different similar triangles and measure the dimensions of required parts to verify the similarity criteria. Basic Proportionality Theorem can also be easily understood with the help of Geo-Gebra activities. Converse of the BPT can also be explained with the help of Geo-gebra activities. While giving the similarity criterions teacher should thrust again and again on the writing of corresponding parts of similar triangles correctly. To understand each theorem proof of every theorem is given in Study material, but for the students of (core) the proofs will not be asked in examination. Teacher should ensure the clarity of the theorems to students using lots of Geo-Gebra Activities and Hands on activities. Pythagoras Theorem and its applications are not new for students. They can also verify physically the significance of this theorem. For example, using the squared paper one can verify the Pythagoras Theorem. One can observe that ² 5² Lots of models are also used to verify the validity of Pythagoras Theorem. But how the Pythagoras Theorem can be proved? It is important to understand that physical verification of any statement does not establish its validity in all conditions. One can generalize a statement if it is true for all real numbers or for all dimensions. To accept Pythagoras Theorem as a standard result it is necessary to prove it for any right angled triangle. Proof of Pythagoras Theorem and its converse can be discussed in class in detail. Further using similar triangle problems based on application of Pythagoras Theorem can be taken up. 8

16 ACTIVITY SKILL MATRIX Type of Activity Warm UP(W1) Warm UP(W2) Pre-Content (P1) Pre-Content (P2) Pre-Content (P3) Content (CW 1) Content (CW 2) Content (CW 3) Content (CW 4) Content (CW 5) Name of Activity Classification of triangles Revising Congruency Skill to be developed Recognition and segregation. Recognizing similar figures based on its dictionary meaning Corresponding parts of congruent triangles Vocabulary building, Identifying relations, Application of knowledge. Recapitulation of ratio and proportion Similar figures Computational skills. Observation and Knowledge Observation and Analysis. Analytical thinking, Reflection, Concept Development Observation, ICT skill, drawing inferences. Basic Proportionality theorem Application of BPT Thinking skill, Analysis and synthesis of knowledge, Application. Converse of BPT Knowledge, Thinking skills, Problem solving. Pythagoras Observation, Analytical skills, Reasoning, theorem Drawing Inferences. Content (CW 6) Hands on activity Pythagoras theorem Thinking skill, Analysis and synthesis of knowledge, Application, ICT. Content (CW 7) Application of Pythagoras theorem ICT Activity on Pythagoras theorem Problem solving skills. Converse of Pythagoras theorem Observation, Analytical skills, Reasoning, Drawing Inferences. Content (CW 8) Content (CW 9) ICT skills and observation 9

17 Content (CW 10) Application of similar triangles Problem solving skills Post - Content (PCW 1) Post - Content (PCW 2) Post - Content (PCW 3) Assignment (BPT) Problem solving skills. Assignment (Similar triangles) Assignment based Pythagoras theorem and converse. Concept Check Problem solving skills. Crossword Knowledge based MCQ Knowledge application Hands on Application Post - Content (PCW 4) Post - Content (PCW 5) Post - Content (PCW 6) Post - Content (PCW 7) Conceptual knowledge. Knowledge and application. 10

18 ACTIVITY 1 WARM UP (W1) Classification of Triangles Specific objective: To motivate the learners to classify the triangles according to sides and angles. Description: Students has the knowledge of types of triangles according to sides and angles. This is a warm up activity to gear up the students for further learning. Execution: Distribute the worksheet (W1) in which various triangles are drawn. Students may be asked to work in pairs. They will classify the given triangles into two categories: (i) according to sides namely scalene, isosceles and equilateral and (ii) according to angles namely acute angled, right angled and obtuse angled. Parameters for assessment: Able to classify the triangles according to sides Able to classify the triangles according to angles Extra reading: ACTIVITY 2 WARM UP (W2) Revising Congruency Specific objective: To motivate the learners to revise of triangles. Description: Students has the knowledge of concept of congruent figures. This is a warm up activity to gear up the students for further learning. Execution: Distribute the worksheet (W2) in which pair of geometrical figures are drawn. Students may be asked to work in pairs. They will find from the given pairs, the congruent figures. Parameters for assessment: Has knowledge of concept of congruent figures Able to find the congruent figures. Extra reading: 11

19 ACTIVITY 3 PRE CONTENT (P1) Recognising Similar Figures Based on its Dictionary Meanings Specific objective: To enable students to recognise similar figures based on dictionary meaning. Description: Students has the knowledge of concept of congruent figures. Execution: Firstly ask the students to find meaning of the word similar from the dictionary. Distribute the worksheet (P1) in which pair of geometrical figures are drawn. Students may be asked to work in pairs. They will find from the given pairs, the similar figures. Parameters for assessment: Has understood the meaning of the word similarity. Able to find similar figures. Extra reading: ACTIVITY 4 PRE CONTENT (P2) Corresponding Parts of Congruent Triangles Specific objective: To test the understanding of meaning of corresponding sides and corresponding angles in two given congruent figures. Description: Students has the knowledge of concept of congruent figures. They know what the corresponding sides and corresponding angles are. Through this activity, their knowledge and understanding about the same concept will be tested. Execution: Draw two congruent figures on the board and ask a few students to label pair of corresponding sides. Ask few students to mark pair of corresponding angles. Distribute the worksheet (P2) in which pair of geometrical figures are drawn. Students may be asked to work in pairs. They will write pairs of corresponding sides and corresponding angles. Parameters for assessment: Can correctly mark pair of corresponding sides and corresponding angles in two congruent figures. Extra reading: 12

20 ACTIVITY 5 PRE CONTENT WORKSHEET (P3) Recapitulation of Ratio and Proportion Specific objective: To test the understanding of concept of proportionality. Description: Students has the knowledge of concept of proportion. In the applications of basic proportionality theorem, we need to use the previous knowledge of this concept. Through this activity, their knowledge and understanding about the same concept will be tested. Execution: In a specified time of (say 10 minutes) ask the students to solve the problems based on the concept of proportion. Parameters for assessment: Can correctly find an unknown when a:b :: c:d and value of any 3 out of a, b, c and d is given. Extra reading: ACTIVITY 6 CONTENT WORKSHEET (CW1) Similar Figures Specific objective: To explore the concept of similar figures. Description: This activity is based on exploring the concept of similarity. Students will be encouraged to use strategies like paper cutting, overlapping, measurement etc. to visualise the concept of similarity. They will learn, to enlarge a given shape using the concept of proportion. Execution: Firstly, ask the students to cut out pieces in task I given in (CW1). 13

21 Use pair and share strategy and have a discussion in the classroom. This will enhance the communication skills as well. In part II of (CW1), a reflection activity is given. Students are asked to brainstorm and think on statements like all circles are similar, all squares are similar etc. You may ask the students to draw the shapes and then reflect. In part III, students will be asked to draw any shape on a squared paper or a dotted sheet, and then enlarge it. It will enhance their drawing skills, thinking skills and application of Math concepts. After this, the students will be asked to work in group and take up task IV given in (CW1). Parameters for assessment: Able to find correct pairs of similar figures Able to reflect correctly on similarity of two given geometrical shapes Able to enlarge and make a similar figure Extra reading: ACTIVITY 7 CONTENT WORKSHEET (CW2) Basic Proportionality Theorem (BPT) Specific objective: To verify, the basic proportionality theorem by using a parallel line board activity Description: This is a hands on activity based on exploring and learning the basic proportionality theorem. Students will verify the fact that if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then the other two sides are divided in the same ratio. Execution: Ask the students to bring a parallel line board sheet. It is a sheet on which parallel lines are drawn. 14

22 Distribute the instruction sheet given in (CW2). Students will verify the result following the instructions given in the sheet. In part II of (CW2), an instruction worksheet on verifying BPT using GeoGebra is given. You may ask the students to work in a multimedia room and explore the result using this software. Parameters for assessment: Able to verify the basic proportionality theorem. Extra reading: ACTIVITY 8 CONTENT WORKSHEET (CW3) Application of BPT Specific objective: To learn to apply, the basic proportionality theorem. Description: This is a worksheet based on exploring and learning the application of basic proportionality theorem. Students will apply the fact that if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then the other two sides are divided in the same ratio ordinary font sum one goldness in solving the given problems. Execution: Distribute (CW3) and ask the students to solve the given problems using BPT. Use pair and share and discuss. Parameters for assessment: Able to apply the basic proportionality theorem. Extra reading: 15

23 ACTIVITY 9 CONTENT WORKSHEET (CW4) Converse of BPT Specific objective: To explore the converse of basic proportionality theorem. Description: This is a worksheet based on exploring and learning the converse of basic proportionality theorem. Execution: Students will be given an activity sheet (CW4). In task I, they will explore the fact that if a line divides any two sides of a triangle in the same ratio, the line must be parallel to the third side. Parameters for assessment: Able to find that converse of BPT holds true Able to apply the basic proportionality theorem and its converse in solving problems Extra reading: ACTIVITY 10 CONTENT WORKSHEET (CW5) Pythagoras Theorem Specific objective: To understand and apply Pythagoras theorem. Description: In this activity, students will recall Pythagorean triplets. They are already familiar with the statement of Pythagoras theorem. Now they will learn more about Pythagoras theorem. Execution: Teacher may write the triplets on black board. Students will work out to find a rule satisfying all the triplets. A class discussion should be taken up as an extension activity to verify whether multiplying a Pythagorean triplet with a number results in another Pythagorean triplet. Parameters for assessment: Can check whether a triplet is Pythagorean or not. Can state Pythagoras Theorem. Extra reading:

24 gorean.html /pythagoras.html ACTIVITY 11 CONTENT WORKSHEET (CW6) Hands on Activity to Verify Pythagoras Theorem Specific objective: To verify Pythagoras theorem by hands on activity. Description: CW6 contains instructions to perform hands on activity for verifying Pythagoras theorem using paper cutting and pasting. Execution: Students will be asked to bring materials in advance and perform hands on, in the class. Parameters for assessment: Able to verify Pythagoras theorem. Extra reading: ACTIVITY 12 CONTENT WORKSHEET (CW7) Application of Pythagoras Theorem Specific objective: To learn to apply Pythagoras theorem in solving problems Description: This is a worksheet based on problems on the application of Pythagoras theorem. Execution: Students will solve the given problems using Pythagoras theorem. Parameters for assessment: Able to apply Pythagoras theorem in solving problems Extra reading: 17

25 ACTIVITY 13 CONTENT WORKSHEET (CW8) ICT Activity on Pythagoras Theorem Specific objective: Verification of Pythagoras theorem using the software GeoGebra Description: Instruction sheet for working on GeoGebra is given in CW8. Execution: students will follow the steps given in the instruction sheet and verify the Pythagoras theorem. Parameters for assessment: Able to verify Pythagoras theorem Extra reading: ACTIVITY 14 CONTENT WORKHEET (CW9) Converse of Pythagoras Theorem Specific objective: To explore the converse of Pythagoras theorem and apply in solving problems. Description: In this worksheet (CW9), there is a brainstorming question, followed by self exploration on the converse of Pythagoras theorem. The instructions for verifying the converse of Pythagoras theorem using Geo Gebra are also given. Further problems are given based on the application of converse of Pythagoras theorem. Execution: Firstly, students will have a discussion on the given statement in CW9. They will be asked to explore the converse of Pythagoras theorem. Using the instruction sheet, further they will verify the result using GeoGebra software. After learning the theorem, students will be asked to solve the problems based on the theorem. Parameters for assessment: Able to verify that converse of Pythagoras theorem is true Able to apply the theorem in solving problems Extra reading: 18

26 ACTIVITY 15 CONTENT WORKSHEET (CW10) Application of Similar Triangles Specific objective: Exploring the criterions of similarity namely SSS, SAS and AA. Description: This is a self exploratory worksheet. Students will explore the following using triangular cut-outs: 1. If the corresponding sides of two triangles are proportional, then they are similar. This is called SSS Similarity. 2. If in two triangles, corresponding angles are equal, then the triangles are similar. This is called AA similarity. 3. If in two triangles one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar. This is called SAS Similarity. Execution: To begin with ask the students to make creative designs using similarity. One design is shown in the worksheet. Ask them to take triangle cut-outs and verify the similarity criterions. Proceeding further students will solve problems based on similarity of triangles. Parameters for assessment: Able to verify AAA, SSS and SAS Similarity in given triangles. Able to apply similarity criterions in solving problems. Extra reading: ACTIVITY 16 ACTIVITY 22- POST CONTENT (PCW1 TO PCW7) Specific objective: To enhance the understanding of concepts learnt. Description: PCW1 to PCW6 are designed for further practicing the concepts learnt in the classroom. Execution: Assessment Plan Assessment guidance plan for teachers 2.22 Assessment Plan Assessment guidance plan for teachers With each task in student support material a self assessment rubric is attached for students. Discuss with the students how each rubric can help them to keep in tune their own progress. These rubrics are meant to develop the learner as the self motivated learner. 19

27 SUGGESTIVE RUBRIC FOR FORMATIVE ASSESSMENT (EXEMPLARY) Parameter Mastered Understanding of BPT (Thales Theorem). Able to state Basic Proportionality Theorem Able to prove the theorem. Developing Needs motivation Able to state Able to state Basic BPT partially Proportionality correct. Able to draw the figure correctly, able to write the proof partially correct. Able to apply BPT Able to apply for the given BPT for the figure and able to given figure draw accurate but not able to figure for a given draw accurate problem based on figure for a BPT. given problem based on BPT. Needs personal attention Not Able to state BPT Able to draw the Not Able to figure according prove. to the statement but can t prove. Able to apply BPT for some problems with given figures but not able to draw accurate figure for a given problem based on BPT. Not able to apply BPT. From above rubric it is very clear that Learner requiring personal attention is poor in concepts and requires the training of basic concepts before moving further. Learner requiring motivation has basic concepts but face problem in calculations or in making decision about suitable substitution etc. He can be provided with remedial worksheets containing solutions, methods of given problems in the form of fill-ups. Learner who is developing is able to choose suitable method of solving the problem and is able to get the required answer too. May have the habit of doing things to the stage where the desired targets can be achieved, but avoid going into finer details or to work further to improve the results. Learner at this stage may not have any mathematical problem but may have attitudinal problem. Mathematics teachers can avail the occasion to bring positive attitudinal changes in students personality. Learner who has mastered has acquired all types of skills required to solve the problems based on BPT. 20

28 TEACHERS RUBRIC FOR SUMMATIVE ASSESSMENT OF THE UNIT Parameter Understanding similarity Able to differentiate between similar and the congruent figures. Not able to differentiate between similar and the congruent figures. Able to identify similar triangles. Not able to identify similar triangles. Able to identify the corresponding parts of similar triangles. Not able to identify the corresponding parts of similar triangles. Able to define the conditions of similarity. Not able to define the conditions of similarity. Able to state all similarity criterions SSS, AA, SAS. Not able to state all similarity criterions SSS, AA, SAS. Able to apply all similarity criterions SSS, AA, SAS in problems. Basic Proportionality theorem and Converse. Can state BPT and its Converse. Not able to apply all similarity criterions SSS, AA, SAS in problems. Cannot state BPT and its Converse. Can apply BPT or its converse appropriately in problems of geometry. Cannot apply BPT or its converse appropriately in problems of geometry. Theorem based on area of similar triangles. Pythagoras theorem and its converse. Can state the theorem. Cannot state the theorem. Cannot apply the theorem appropriately in problems of geometry. Cannot state Pythagoras Theorem and its Converse. Cannot apply Pythagoras Theorem or its converse appropriately in problems of geometry. Can apply the theorem appropriately in problems of geometry. Can state Pythagoras Theorem and its Converse. Can apply Pythagoras Theorem or its converse appropriately in problems of geometry. 21

29 STUDY MATERIAL 22

30 SIMILAR TRIANGLES Introduction You are already familiar with congruent figures. Recall that two figures are said to be congruent if they are of same shape and same size, i.e. one figure can be considered as a trace copy of the other. However, we come across many figures which are of the same shape but not necessarily of the same size. Such figures are called similar figures. In this chapter, we shall discuss such figure in general and triangles in particular. We shall also discuss various criteria for similarity of triangles and use them to arrive at a famous theorem related to a right angled triangle commonly known as Pythagoras Theorem. (1) Congruent and similar Figures, See the following figures: (i) (ii) Fig.1 (iii) They all are of same shape and same size. So, these squares are congruent Now see the following figures. (i) (ii) (iii) (iv) Fig. 2 23

31 They are of same shape but are of different sizes. We call these squares as similar figures. Now observe the following equilateral triangles (i) (ii) (iii) Fig.3 Fig.3 These are of same shape and same size and hence congruent equilateral triangles. Again observe the following three equilateral triangles (i) Fig. 4(ii) (iii) Fig.4 These are of same shape but different sizes. We call then similar equilateral triangles Now see the following circles Fig.5 24

32 These are of same shape and same size. Hence, congruent circles. Again observe the following circles. Fig. 6 These are of the same shape but not of same size. We call then similar circles. From the above discussion, it can be said that all congruent figures are similar but all similar figure need not be congruent Look at the following photograph (i) (ii) (iii) Fig.7 25

33 In (i) there are photographs of same monument Tajmahal. They are of the same shape. They are similar, Note that angles in each photograph ore of same measure. Also the length of pillars are proportional In (ii), triangle are also similar. By measurement it can be seen that corresponding angles of triangles are of the same measure and corresponding sides are proportional in any two triangles taken at a time. In (iii), quadrilaterals are also similar. Here again, corresponding angles are of same measure and sides are proportional In view of the above, we say that Two polygons are similar if their corresponding (i) angles are of same measure (equal) (ii) sides are proportional Example 1: State whether the following pairs of polygons are similar or not. Give reasons. (i) (ii) (ii) 26

34 (iv) Solution: (i) Not similar. Corresponding angles are equal but sides are not proportional. as 1 (ii) Similar. Corresponding angles are equal and sides are proportional (iii) Not similar. Corresponding sides are proportional but corresponding angles are not equal. (iv) Not Similar. Corresponding angles are equal but corresponding sides are not proportional. but (2) Similarity of triangles Recall that a triangle is also a polygon. Therefore, the conditions for similarity of two polygons shall also be valid for the similarity of two triangles. Thus, we can say Two Triangles are similar, if their (i) Corresponding angles are equal, i.e. of the same measure and (ii) Corresponding sides are proportional, i.e., they are in the same ratio. As in the case of congruence, the words corresponding vertices; corresponding angles and corresponding sides are of great significance: For example, if A B C and DEF are similar under a correspondence says A D, B C and E F, then the two triangles need not be similar under the correspondence A E, B D and C F. It may also be noted that symbol is used to represent is similar to Thus, if ABC and DEF are similar under the correspondence A D,B E and C F, then we write it as ABC DEF (read as triangle ABC is similar to DEF). In this case, side corresponding to AB is DE, side corresponding to BC is EF and side corresponding to CA is FD. Similarly A and D are corresponding angles, B and E are corresponding angles and C and F are corresponding angles. It will not be correct to write the similarity of the above triangles as ABC EDF. 27

35 From the above stated two conditions of similarity of two triangles, it appears that for knowing whether are not the two triangles are similar, we shall have to check the measures of all the six elements of the triangles. Can we not obtain some criteria for similarity of triangles involving lesser number of elements as we obtain for congruency of triangles? Answer to this question was provided by famous Greek mathematician. Thales ( BC). It is believed that he had used either a theorem (which is known as Basic Proportionality theorem or Thales theorem) or some results related to this theorem to prove that the ratio of the corresponding sides of two equiangular triangles is always the same. The review. Therefore, before going further, let us have some understanding about the Basic proportionality theorem: Basic Proportionality Theorem : If a line is drawn parallel to one sides of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio: It means that if in, DE BC, then (Fig. 8). This can be verified by drawing DE BC and then measuring AD, DB, AE and EC. (The theorem can be proved as shown below. Fig.8 Given: ABC in which a line drawn parallel to BC intersects AB at D and AC at E. To prove: Construction: Draw EM AB and DN AC. Also, join BE and CD. Proof: We have: (Recall that ar ADE) mean area of ADE. Fig.9 Also, area of a triangle base x altitude) (1) Note : Proof of all theorems are for purpose of understanding and not to be used for purpose of examination. 28

36 Similarly: - But ar( BDE) ar ( CDE) (2) (3) (Recall that triangles on the same base and between the same parallels are equal in area) So, [From (1) and (2)] Hence, Proved, We can obtain the following corollary from the above theorem: Corollary: If in ABC, a line parallel to BC intersects AB and AC at D and E respectively, then Proof: See fig.10. We have: (1) [From Basic proportionality theorem (BPT)] (Adding on both sides or or (2) Dividing (1) by (2), We get. Hence, Proved Fig. 10 The converse of the BPT is also true. It is as follows: Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. It means that if in ABC, D and E are points on sides AB and AC respectively, such that, then DE BC (Fig 11). Fig

37 This result can be verified by taking suitable points D and E on AB and AC respectively and then. Showing that D B or E C which will give It can be proved as follows: Given: ABC and a line at D and E such that To prove: intersecting AB and AC respectively BC, i.e., DE BC. Fig. 12 Proof: Let us assume that is not parallel to BC, i.e. DE is not parallel to BC. So, there must be some line parallel to BC, through D. Let this line intersects AC at F (Fig. 12). Thus, we have DF BC. Since DF BC, therefore by BPT, we have But it is given that So, Or, i.e., or, So, FC EC This is impossible. It is possible only when E and F coinside. i.e. DF and line. Hence, BC or DE BC. We now take some examples to illustrate the use of these results. 30 are the same

38 Example 2: In Fig. 13, if PQ AB, find QC. Solution: PQ AB (Given) So, (By BPT) Or, Or, QC cm 4.5 cm. Fig. 13 Example 3: In Fig 14, find whether, AB is parallel to DF or not Solution: From the figure, And Thus, So, AB is not parallel to DF (By converse of BPT) Example 4: In Fig.15, if AB RQ and AC RS, prove that Solution: In Fig. 14. PQR, we have: AB RQ (Corollary of BPT) (1) Again in PQR, We have: AC RS (corollary of BPT) (2) So, From (1) and (2), we have: 31 Fig.15

39 Example 5: In Fig 16, if and AQP ABC, prove that ABC is an isosceles triangle. Solution: (Given) So, PQ BC (Converse of BPT) Hence, and APQ AQP ABC (Corresponding Angles) ACB (corresponding angles) AQP Also it is given that (1) Fig. 16 ABC So, from (1), we have ACB ABC Therefore AB AC (Sides opposite to the equal angles of a triangle) i.e. ABC is an isosceles triangle. Example6: Prove that any line parallel to the parallel sides of a trapezium divides its non-parallel sides in the same ratio. Solution: Let ABCD is a trapezium in which line AB CD intersects the sides AD and BC at E and F respectively. We are to prove that To apply some knowledge of BPT, We must have a triangle So, let us form AC and let line Now, from intersect AC at G. ADC, we have: EG DC (why?) So, Again, from (1) CAB, we have: GF AB (why?) So, 32 Fig.17

40 Or, (2) Therefore, from (1) and (2), we have (3) Criteria for similarity of Two Triangles AAA Similarity: Draw any two triangles ABC and PQR, such that A P, B Q and C R (Fig. 18) Now, measure, AB, BC, CA, PQ, QR and RP. After this find You will observe that, and Fig. 18 In this way, you have verified that if angles of one triangle are equal to corresponding angles of the other triangle, then the corresponding sides of the two triangles are in the same ratio or proportional. In other words, if corresponding angles of two triangles are equal then their corresponding sides are proportional. Thus, we see that both the conditions of similarity of polygons are satisfied and hence the two triangles are similar. So, we may state the following criterion for similarity of two triangles. If corresponding angles of two triangles are equal then their corresponding sides are proportional and hence the two triangles are similar. This criterion is referred to as the AAA criterion for similarity of two triangles In fact, the above result can be proved as shown below: Given: Two triangles ABC and PQR such that To Prove: ABC and hence PQR 33 A P, B Q and C R.

41 Construction : Assuming that AB<PQ and AC<PR, take points S and T on PQ and PR respectively such that ABPS and ACPT. Join ST (Fig 19). Note: In case AB>PQ and AC>PR, we can take ASPQ and ATPR and If ABPQ, then and hence similar. Proof: (SAS congruence criterion) B S and So, C T (CPCT) Q (Given) B But Fig. 19 So, S Q Hence, ST QR (Since corresponding angles are equal) Therefore, So, (corollary of BPT) (Since AB PS and AC PT) (1) Similarly, by taking points on other pair of sides PQ and QR, it can be seen that (2) Hence, from (1) and (2), i.e, corresponding sides are proportional and hence, the triangles are similar, PQR. i.e. ABC SSS Similarity Now, draw two triangles ABC and PQR Such that 34 (Fig. 20)

42 A, B, C, P, Q and Q and C R. That is if the corresponding sides of the two triangles are proportional then the corresponding angles are equal. Thus, the two conditions of similarity of two polygons are satisfied and so the two triangles ABC and PQR are similar. Thus, We can say that: Now, measure angle R. You will see that A P, B Fig. 20 If Corresponding sides of two triangles are proportional, then their corresponding angles are equal and hence the two triangles are similar. This criterion is referred to as the SSS criterion for similarity of two triangles. In fact, the above result can be proved as shown below: Given: Two triangles ABC and PQR such that A P, B Q, and and hence ABC PQR. To Prove: C R Construction: Take Points S and T on PQ and PR respectively such that ABPS and ACPT. Join ST (Fig.21) Proof: So, Therefore, Fig. 21 (Given) (By construction) Hence, ST QR (converse of BPT) So, S Q and T R (corresponding angles) Therefore, PST PQR (AAA Similarity, (1) P is common) So, (Sides will be proportional) So, (Since AB PS by construction) 35 (2)

43 But it is since that So, from (2), Therefore, ST BC Hence, ABC So, therefore But PST (SSS criterion for congruence) A P, B S and S Q and So, we have C T T R [Already proved in (1) ] A P, B Q and C R i.e., corresponding angles are equal. Hence, PQR. Note: In view of the above two results, it can be stated that (i) If the corresponding angles of two triangles are equal, then the two triangles are similar (AAA similarity criterion). Further, if two angles of one triangle are respectively equal to the two angels of the other triangles, then their third angles will automatically, be equal. Therefore, we may say that if two angles of a triangle are equal to corresponding two angles of the other triangles then the triangles are similar (AA similarity criterion). (ii) If the corresponding sides of two triangles are proportional, then the triangles are similar (SSS criterion for similarity). SAS similarity Criterion. To understand this criterion, draw two triangles ABC and PQR such that and A Fig. 22 P (Fig. 22) Here, two pairs of sides are proportional and the angles included between these pairs of sides are equal. 36

44 Now measure the remaining angles and sides, of the triangle i.e, measure B, C, Q, R. Side BC and Side QR. You will see that B Q, C R and is the same as or Thus, two triangles are similar (By AAA as well as SSS criterion of similarity). So, may state the third criterion of similarity of two triangles as follows: If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the triangles are similar. This is referred to as the SAS similarity criterion for similarity of triangles. In fact, We can prove the above result as shown below: Given: ABC and PQR in which To prove: ABC A P and. PQR Construction: Take points S and T on PQ and PR respectively such that AB PS and ACPT. Join ST (Fig.23) (SAS congruence criterion) Proof: So, S and C T (CPCT) Also, Since Fig. 23 (1), therefore (By construction) So, ST QR (Converse of BPT) Hence S Q and T R (Corresponding angles) (2) So, B Q and C R (From (1) and (2) i.e., (AAA similarity) Let us now take some examples to illustrate the use of this criteria. 37

45 Example7: In Fig. 24, ABC is a right triangle, right angled at B and D is any point on side AB. If DE AC, prove that Solution: In A A (Common angle) B E (Each 90 ) So, (AA similarity criterion) Fig. 24 Example 8 : Examine each of the following pairs of triangles and state which of them are similar. If similar, state the criterion used by you for it. Also, write there pairs of similar triangles in symbolic notation: 38

46 Fig. 25 Solution: (i) Triangles are Similar. AAA similarity Criterion. Symbolic Notation: (ii) Triangles are similar. SSS similarity criterion. Symbolic Notation: (iii) Triangles are not similar, because (iv) Triangles are not similar, because (v) Triangles are similar, because F 30 and N 80 AAA similarity criterion. Symbolic Notation. (vi) Triangles are similar. SAS Similarity criterion Symbolic Notation: 39

47 Example9: In Fig 26, if AB CD. then show that Solution: (Given) So, A C and B D (Corresponding angles) Since A C (or B D), Therefore AB CD (Because A and C are alternate angles) Fig. 26 Example 10: ABCD is a trapezium in which AB CD and its diagonals meet each other at O. Prove that Solution: See Fig. 27 In, We have: A C (alternate angles, AB CD) and B D (alternate angles, AB CD) So, (AA similarity) Hence, Fig. 27 (corresponding sides are proportional) Example 11: In Fig. 28, AM and DN are respectively the medians of triangle ABC and DEF. Such that (i) (ii) ABM ABC, Show that DEN DEF Solution: (i) BM BC and EN EF Fig. 28 Since AM and DN are medians) It is given that So, (Since BM BC and EN EF) 40

48 Or, Hence, or (ii) (Proved in (i) above) So, B E (corresponding angles are equal) Now, is (1) we have. (Given) And B E (Proved is (1)) So, (SAS similarity criterion) Example 12: In Fig 29. AM and D N are respectively medians of Prove that Solution: Produce AM to P and DN to Q such that AM MP and DN NQ Now, (SAS congruence) Fig. 29 So, ABPC (CPCT) (1) congruence) Also, So, DE QF (CPCT) (2) Now, in (Given)(Since AP2AM and DQ2DN) and (From (1) and (2) So, So, (SSS similarity of criterion) Therefore, CAM FDN (Corresponding angles are equal) 41 (3)

49 And CPM FQN (Corresponding angles are equal) Also, Since (4) (Already proved) So, BAM CPM (CPCT) Further, Since (5) (Already proved) So, EDN FQN (CPCT) (6) Therefore, from (4), (5) end (6), we have: BAM EDN (7) Adding (3) and (7), We have CAM + BAM FDN + EDN i.e., BAC EDF Now, we have So, and BAC EDF (SAS Criterion) (4) Area of Similar Triangles You have learnt that in two similar triangles, their corresponding sides are in the same ratio. What can we say about the ratio of areas of two similar triangles? You also know that area is measure in square units. Do we expect that the ratio of two similar triangles has something to do with the ratio of sides of these triangles? Answer to this question is provided by the following Theorem. Theorem: The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. Given: ABC and DEF (Fig 30) Such that ABC and DEF are similar. To Prove: Fig

50 Construction : Draw altitudes AM and DN (Fig.30) Proof: ar ( ABC) and ar Thus (1) Also ABM ~ DEN (By AA similarity criterion) Hence (2) Also From (As ABC~ DEF) (3) (1), (2) and (3) Thus Thus, we have proved the theorem. Let us consider some examples illustrating the use of the theorem above. Example13: Two triangles ABC and PQR are similar to each other, in which AB12 cm and PQ8 cm. Find the ratio of areas of ABC and PQR. Solutions: ABC ~ PQR So, Thus Area of Example 14: Area of two triangles ABC and DEF are 242 cm2 and 162cm2 respectively. If ABC ~ EDF and AB22cm, find the length of the side of DEF corresponding to side AB. Solution: Since, ABC ~ EDF, Therefore [ED and AB are corresponding side] 43

51 Or, Or ED2 324 Thus ED 18 cm Example15: In fig. 31, ABCD is a trapezium in which AB ratios of areas of triangles AOB and COD CD and AB 2CD. Find the Solution: In OAB and OCD, OAB OCD (Alternate angle) and OBA ODC (Alternate angle) So, OAB ~ OCD (AA similarity) Hence, Thus, the required ratio 4:1 Fig.31 Example 16: In fig 32, XY AC and XY divides triangular region into two parts equal in area Determine Solutions: Since the triangular region ABC has been divides by XY two parts of equal area, therefore ar ( or Fig.32 (1) Now, XY AC So, X A and Y C (corresponding angles) Hence, So, (2) Therefore from (1) and (2) 44

52 or or, 1-1- or i.e., Example 17: Prove that the ratio of the areas of two similar triangles is equal to ratio of the squares of their corresponding medians. Solution: Let the triangles be AM and DN their medians (Fig.33) We have: (Since ABC ~ DEF) or ( AM and DN are median) or (1) Also, B E (Corresponding angles of similar triangles ABC and DEF) (2) So, from (1) and (2) ABM ~ DEN (SAS Similarity) or (Corresponding sides must be proportional ) (3) Now, So, [From (3) Similarity, the result can be proved for other corresponding medians. Pythagoras Theorem Draw a right ABC, right angled at B. Measure sides AB, BC and AC. Find AB2, BC2 and AC2. You will find that AC2 AB2 + BC2. In fact this type of relation is true in all 45

53 right triangles and is known as Pythagoras theorem we may state the theorem as follows: Theorem: In a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. It can be proved as follows: Given: A right triangle ABC, right angled at B (Fig.34) To Prove: AC2 AB2 + BC2 Construction: Draw BD AC. Proof: In triangles ADB and ABC. Fig.34 A A (Common) ADB ABC (each 90 So, ADB ~ ABC (Why?) Thus, (1) Or, Similarly, DC (2) Adding (1) and (2), AC AC Hence, the required result. We now prove the Converse of the Pythagoras Theorem Theorem: In a triangle, if the square on one side is equal to the sum of the squares on the other two sides, then the angle opposite to the first side is a right angle. The theorem can be proved as shown below: 46

54 Given: A triangle ABC in which To prove: ABC 90 Construction: Let us construct a right triangle PQR right angled at Q such that PQAB and QRBC Proof: In Fig.35 or, But, (Given) So, or, PR AC Therefore, B Q So, But Q 90 Thus, B 90 Hence, Example 18: P and Q are the points on the sides CA and CB respectively of a ABC, right angled at C. Prove that: Solution: Join PQ In (By Pythagoras Theorem) (1) Similarly, in (2) + [As [As is a right angled triangle] 47 Fig.36

55 Example19: The sides of some triangles are given below. Determine which of them are right triangles. (i) (iii) 50 cm, 80 cm and 120 cm Solution: (i) 100, Since so, triangle will be a right triangle. (ii) 625 Since (iii), (80) Since Example 20: ABC is an isosceles triangle with AC BC. If Solution: prove that We are given AC BC. (1) Also or Fig.37 i.e., Thus, angle opposite the side AB must be a right angle. Hence. Example 21: In Fig. 38, AB BC CA 2a and AD BC. Show that (i) AD (ii) area ar ( ABC) Solution: (i) So, BDDCa. In Fig.38 48

56 Or So, AD a (ii) Area ( ABC) Example 22: A ladder reaches a window which is 12 metres above the ground on one side of the street. Keeping the foot at the same point, the ladder is turned to the other side of the street to reach a window 9 metres high. Find the width of the street if the length of ladder is 15 metres. Solution: In [Fig. 39] So, (why?) Thus, BC 9 metres In So, Thus, EB 12 Metres. Hence, EC EB + BC (12 + 9) Metres 21 Metres. 49

57 Solution: Fig. 39 Example 23: In Fig. 40, ABC is an obtuse triangle obtuse angled at B and side CB is in produced to D such that segment AD CD, Prove that Solution: So, 2 + Fig

58 Example24: In Fig. 41; AD BC. Prove that Solution: We have [ is right angled at D and by Pythagoras Theorem] Fig ) Example 25: In that BC. If AC > AB, then show + Solution: In right AB² AE² + BE² (AD² ² we have (Pythagoras Theorem) ED²) + (BD ED)² AD² ED² + BD² + ED² AD² 2BD 2BD +BD² Fig. 42 AD² (1) Similarly, In right AC² AE + EC² (AD² AD² (Pythagoras Theorem) ED²) + (ED +DC)² ED² + ED² + DC² + 2DC. ED AD² + 2 AD² + BC. ED + (2) Adding (1) & (2), AB²+AC² 2AB² + 2 2AD² +. 51

59 STUDENT S SUPPORT MATERIAL 52

60 STUDENT S WORKSHEET 1 WARM UP W1 CLASSIFICATION OF TRIANGLES Name of Student Date Classify the following triangles according to sides and angles: Figures Type of Triangle 53

61 SELF ASSESSMENT RUBRIC 1 WARM UP (W1) Parameter Can classify triangles on the basis of sides Can classify triangles on the basis of angles 54

62 STUDENT S WORKSHEET 2 WARM UP 2 REVISING CONGRUENCY Name of Student Date Which of the following figures are congruent? How will you get to know? 55

63 SELF ASSESSMENT RUBRIC 2 WARM UP (W2) Parameter Can recognise congruent figures 56

64 STUDENT S WORKSHEET 3 PRE CONTENT 1 RECOGNIZING SIMILAR FIGURES BASED ON ITS DICTIONARY MEANINGS Name of Student Date Dictionary Task: Look for the meaning of the word similar in the dictionary. Observe the following pictures and tell which of them are similar. 57

65 SELF ASSESSMENT RUBRIC 3 PRE CONTENT (P1) Parameter able to recognise similar objects. 58

66 STUDENT S WORKSHEET 4 PRE CONTENT 2 CORRESPONDING PARTS OF CONGRUENT TRIANGLES Name of Student Date Observe the following figures and fill the table: Pair of Congruent figures ABC DEF PQR LMN XYZ EFD Corresponding Sides 59 Corresponding Angles

67 Rectangle (DCBA) Rectangle (PQRS) SELF ASSESSMENT RUBRIC 4 PRE CONTENT (P2) Parameter Able to write corresponding sides of given congruent figures Able to write corresponding angles of given congruent figures 60

68 STUDENT S WORKSHEET 5 PRE CONTENT 3 RECAPITULATION OF RATIO AND PROPORTION Name of Student Date 1. Find x if 2:5 :: x:25 2. If x:7 :: 25: 35, find x 3. Find p if 12:p::120: SELF ASSESSMENT RUBRIC 5 PRE CONTENT (P3) Parameter Able to use the concept of proportion in finding unknown. 61

69 STUDENT S WORKSHEET 6 CONTENT WORKSHEET (CW1) SIMILAR FIGURS Name of Student Date I. Megha said, Two similar figures have the same shape but not necessarily the same size. Keeping in mind the concept of similar figures, examine the following pair of figures and find which of the following are similar. 62

70 63

71 II. Reflect on the following statements. i. All circles are similar. 64

72 ii. All squares are similar. iii. A circle can be similar to a square iv. A triangle can be similar to a square III. Activity: Draw any shape. Enlarge it. What do you say about the similarity of two shapes? IV. Exploration Work in a group of 4 students. Draw a line segment AB of any length. At A draw a ray at any angle of 30⁰, using a protractor. At B draw a ray at any angle of 55⁰, using a protractor. Let the two rays meet at C. Observe the triangles made by each member of your group. 65

73 What do you observe? Are all the triangles similar? Verify by measurement, whether the corresponding sides of any two triangles drawn in the group, are in the same ratio. Pair of Similar Triangles Ratio of Corresponding Sides 66

74 SELF ASSESSMENT RUBRIC 6 CONTENT WORKSHEET (CW1) Parameter Able to recognise similar figures Able to reflect on concept of similar figures Knows enlargement of a shape is done using proportion STUDENT S WORKSHEET 7 CONTENT WORKSHEET (CW2) BASIC PROPORTIONALITY THEOREM (BPT) Name of Student Date I. Hands on Activity Aim: To verify the Basic Proportionality theorem by paper cutting and pasting using a parallel line board. 67

75 Statement: If a line is drawn parallel to any side of a triangle, to intersect the other two sides at two distinct points, then the other two sides are divided in the same ratio. Material required: Coloured paper, parallel line board, pair of scissors, sketch pen, ruler, glue Procedure: Step 1. Draw a triangle on a coloured paper. Step 2. Label the triangle as ABC. 68

76 Step 3. Cut the triangle. Step 4. Take the parallel line board and place the triangle ABC on it such that the side BC coincides with any of lines on the board. Step 5. Draw a line parallel to BC using a ruler by the help of lines on the parallel line board. 69

77 Step 6. Let the parallel line drawn to BC intersect AB and AC at D and E respectively. 70

78 Step 7. Step 8. Step 9. Step 10. Source: Find AD/DB and AE/EC. What do you observe? Repeat the activity for two more triangles. Write the result. II. ICT Activity: Aim: To verify the Basic Proportionality theorem which states that, in a triangle, if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points then other two sides are divided in the same ratio. Previous Knowledge Assumed: Concept of lines, parallel lines and triangles Basic knowledge of working on GeoGebra Procedure: 1. Draw a line slider using the tool (Slider) min and 5 as max. 2. Draw a polygon ABC using the tool (polygon). Name it as f. Provide 1 as. 3. Draw a line parallel to line segment BC using tool (Parallel line) D is generated in the process.. Point 4. Now let us redefine the point D and make it as a function of our slider. Right click on D and choose redefine. Replace the Y coordinate of the point D with f. 5. Now let us define the points E & F as the intersection points of the parallel line and line segments AB and AC respectively using tool (Intersect two objects). 6. Let us draw line segment EF using tool (Segment between two points) 71.

79 7. Hide out parallel line and the point D by right clicking on the object in the algebra window and un-checking the show object attribute. 8. Let us determine the length of the line segments AE, EB, AF and FC using tool (Distance or length). Click on points A and then point E to create the length segment for AE. Do it for other line segments too. 9. In the input box at the bottom of the screen, create a ratio of the lengths of the line segments. E.g. ratio1distanceae/distanceeb. 10. Create text labels to show up the results on the screen using tool (insert text). 11. Now slide the slider and take down different observations. If the ratios remain same the theorem is verified. 12. Repeat the activity with three different sets. 13. Note down the observations. 14. Write the final result. Observations: Serial Number First time Second time Third time Ratio 1 Ratio 2 Observation Result: _ 72

80 SELF ASSESSMENT RUBRIC 7 CONTENT WORKSHEET (CW2) Parameter Able to verify BPT using hands on activity on a parallel line board. Able to explore BPT using GeoGebra STUDENT S WORKSHEET 8 CONTENT WORKSHEET (CW3) APPLICATION OF BPT Name of Student Date 1. Draw a diagram to explain the basic proportionality theorem 73

81 2. Solve the following questions. Make a list of results used in solving the problem. Figure Given To find AE DE BC, Solution AD 3 cm, DB 1.5 cm, EC 1 cm ST QR SQ 7.2 cm, PT 1.8 cm, TR 5.4 cm PS 3. Using the basic proportionality theorem, solve the following problems. In a triangle ABC, D and E are points on the sides AB and AC respectively such that DE BC. i) If AD4, AE8, DB x-4, and EC3x-19, find x. ii) If AD/BD 4/5 and EC2.5 cm, find AE. iii) If ADx, DBx-2, AEx+2 and ECx-1, find the value of x. iv) If AD2.5 cm, BD3.0 cm and AE3.75 cm, find the length of AC. 4. Using basic proportionality theorem, prove that any line parallel to the parallel sides of a trapezium divides the non parallel sides proportionally. 5. In the given figure, PA, QB and RC are each perpendicular to AC. Prove that 1/x + 1/z 1/y. 6. ABCD is a trapezium with AB CD. The diagonals AC and BD intersect each other at O. If AO 2x+4, OC2x-1, DO3 and OB 9x-21, Find x. 7. Prove that the line segments joining the mid points of adjacent sides of a quadrilateral form a parallelogram. 74

82 SELF ASSESSMENT RUBRIC 8 CONTENT WORKSHEET (CW3) Parameter Able to apply the basic proportionality theorem in solving problems 75

83 STUDENT S WORKSHEET 9 CONTENT WORKSHEET (CW4) CONVERSE OF BPT Name of Student Date 1. Observe the following figure: Based on your observation fill the following table: Triangles Ratio 1 Ratio2 ABC PQR XYZ Are the two ratios equal? Angles Measures ADE.. ABC.. PST PQR XLM.. XYZ.. 76 Are the two angles equal? Relation between the line drawn with the base line.

84 What can you conclude from the above table? 2. In a triangle ABC; D and E are points on the sides AB and AC respectively. For each of the following show that DE BC. i) AB 5.6 cm, AD 1.4 cm, AC 7.2 cm and AE 1.8 cm ii) AD 5.7 cm, BD 9.5 cm, AE 3.3 cm and EC 5.5 cm. 3. In a triangle ABC, P and Q are points on sides AB and AC respectively, such that PQ BC. If AP 2.4 cm, AQ 2 cm, QC 3 cm and BC 6 cm, Find AB and PQ. 4. In a triangle ABC, D and E are points on AB and AC respectively such that DE BC. If AD 2.4 cm, AE 3.2 cm, DE 2.0 cm and BC 5.0 cm, find BD and CE. SELF ASSESSMENT RUBRIC 9 CONTENT WORKSHEET (CW4) Parameter Able to apply the basic proportionality theorem Able to apply the converse of basic proportionality theorem in solving problems 77

85 STUDENT S WORKSHEET 10 CONTENT WORKSHEET (CW5) PYTHAGORAS THEOREM Name of Student Date Below are some jumbled up words which state the rule applicable on Pythagorean triplets. Rewrite it to form a meaningful sentence. _ Verify whether the following triplets satisfy the above rule. 78

86 (3,4,5) (5,12,13) (9,40,41) (7,24,25) (8,15,17) (11,60,61) (12,35,37) _ 79

87 Observe the following figures, each having a right angled triangle. Regular polygons or semicircles are made on the three sides of the triangles. Figure Area of I (in terms of side a ) Area of II (in terms of side b ) Area of III (in terms of side c ) Relation between the areas of I, II & III (if any) What can you conclude about the relationship between the sides of a right angled triangle? 80

88 Self Assessment Rubric 10 Content Worksheet (CW5) Parameter Able to write the Pythagoras theorem STUDENT S WORKSHEET 11 CONTENT WORKSHEET (CW6) HANDS ON ACTIVITY TO VERIFY PYTHAGORAS THEOREM Name of Student Date Hands on Activity: Activity 1: Aim-By paper cutting and pasting verify Pythagoras theorem. Pythagoras theorem states that in a right triangle square of hypotenuse is equal to the sum of squares of other two sides. Material Required - Coloured Paper, pair of scissors, Glue Procedure 1. Draw a right triangle of dimension a, b and c. c is the hypotenuse. 2. Make 3 more such triangles. 3. Cut a square of side c units. 81

89 4. Now arrange the 5 cut out pieces to make a square of side (a + b) 5. Compare the area of square of side (a + b) with the sum of all parts of the square. 6. Write your observations. source Activity 2: Objective: To verify the Pythagoras Theorem by the method of paper folding, cutting and pasting. Material Required: Card board, coloured pencils, pair of scissors, fevicol, geometry box. Previous Knowledge: 1. Area of a square. 2. Construction of parallel lines and perpendicular bisectors. Procedure: 1. Take a card board of size say 20 cm 20 cm. 2. Cut any right angled triangle and paste it on the cardboard. Suppose its sides are a, b and c. 82

90 3. Cut a square of side a cm and place it along the side of length a cm of the right angled triangle. 4. Similarly cut squares of sides b cm and c cm and place them along the respective sides of the right angled triangle. 5. Label the diagram as shown in Fig Join BH and AI. These are two diagonals of the square ABIH. The two diagonals intersect each other at the point O. 7. Through O, draw RS BC. 8. Draw PQ, the perpendicular bisector of RS, passing through O. 9. Now the square ABIH is divided in four quadrilaterals. Colour them as shown in Fig From the square ABIH cut the four quadrilaterals. Colour them and name them as shown in Fig 2. Fig 1 83

91 Fig 2 Observations _ Conclusion: 84

92 SELF ASSESSMENT RUBRIC 11 CONTENT WORKSHEET (CW6) Parameter Able to state Pythagoras Theorem. Able to verify Pythagoras Theorem using explorations or models. 85

93 STUDENT S WORKSHEET 12 CONTENT WORKSHEET (CW7) APPLICATION OF PYTHAGORAS THEOREM Name of Student Date 1. There is a staircase as shown in figure connecting points A and B. Measurements of steps are marked in the figure. Find the straight distance between A and B. 2. Show that the area of a rhombus on the hypotenuse of a right angled triangle with one of the angles as 60⁰ is equal to the sum of the area of rhombus with one of their angles as 60⁰, drawn on the other two sides. 86

94 3. An aeroplane leaves an airport and flies due north at a speed of 1800 km/h. At the same time, another plane leaves the same airport and flies due west at a speed of 1600 km/h. How far apart will the two planes be after 2.5 hours? 4. Find the lengths diagonals in the following figures. 5. Determine the perimeter and area of the given figure. 6. In the following figure, justify the relationship pc ab. 87

95 7. Prove that three times the sum of squares of the sides of a triangle is equal to four times the sum of the squares of the medians of the triangles. 8. A man drives south along a straight road for 17 miles. Then turns west at right angles and drives for 24 miles where from he turns north and continues driving for 10 miles before coming to a halt. What is the straight distance from his starting point to his terminal point? SELF ASSESSMENT RUBRIC 12 CONTENT WORKSHEET (CW7) Parameter Able to state Pythagoras Theorem. Able to apply Pythagoras Theorem in geometrical problems. 88

96 STUDENT S WORKSHEET 13 CONTENT WORKSHEET (CW8) ICT Activity on Pythagoras Theorem Name of Student Date ICT Activity- (Geogebra Activity) Pythagoras theorem Aim: To verify that in a right angled triangle the sum of squares of the perpendicular and base is equal to the square of the hypotenuse. Previous Knowledge Assumed: Concept of angles, circles and triangles Basic knowledge of working on GeoGebra Procedure: 1. Draw a line slider using the tool (Slider) and 3 as max Draw a point B using the tool (New point). Draw another point A using the same tool. Let us use the slider to make point A as dynamic. Let us right click on the point A and chose redefine. Now replace the Y coordinate with d Draw a line through points A and B using tool (Line through two points). Draw a perpendicular line to line passing through point A and point B using tool 7. (Perpendicular line). After selecting the tool click on line passing through point A and B and the click above it. Point C is also formed in this process. Let us now locate the intersection point of line AB and the line perpendicular to it 8. using tool (Intersect two points). Point D is formed now. Let us form a polygon out of the three points A, C and D using tool (polygon) 9.. Name it as d. Provide -3 as min. After selecting the tool, click on points A, C and D one by one. Let us hide lines a, b and c by right clicking on the these lines and un-checking show objects. 89

97 10. Let us quickly draw semicircles around these points using the tool (Semicircle through two points). After selecting the tool click on the two points A and C. 11. Similarly draw the semicircles around the line segments AD and DC. 12. Now draw point E on semicircle e using tool (New point). 13. Let us draw the points F and G on semicircles around CD and AD. 14. Lets us quickly draw circles using tool (Circle through three points) points A, E and C. After selecting the tool click on points A, E and C. 15. Similarly draw the circles using points C, F & D and D, G & A. 16. Let us calculate the area of the circles using the tool (Area) the circles and the area objects will appear. using. Just click inside 17. Now using tool (insert text), write area of semicircle with diameter AC + area h/2 18. Similarly write for other two semicircles. 19. Let us hide the circles by right clicking on the objects in the algebra window and un-checking show objects. 20. Now compare the sum of the areas of the semicircles. The sum of two is equal to the third one. This verifies our theorem. 21. Repeat the activity with three different sets. 22. Note down the observations. 23. Write the final result. Observations: Serial Number First time Second time Third time Area of semicircle1 Area of semicircle2 Area of semicircle3 Observation Result: 90

98 SELF ASSESSMENT RUBRIC 13 CONTENT WORKSHEET (CW8) Parameter Able to state Pythagoras Theorem. Able to verify Pythagoras Theorem using explorations on GeoGebra 91

99 STUDENT S WORKSHEET 14 CONTENT WORKSHEET (CW9) CONVERSE OF PYTHOGORAS THEOREM Name of Student Date Rohan eplored one result on triangles. Read carefully. Do you agree with Rohan? Comment _ Now, fill the following table: Numbers-(a, b, Verify, Construct triangle with sides Is triangle a Write c) a2+b2 c2 a, b, c right the side where c > a, b angled opposite triangle? to right angle. (3,4,5) Yes 92 5

100 (5,12, 13) (4,5,7) (6,8,10) (10,12,15) 93

101 Based on your exploration in the above table, rewrite the jumbled up words given below to form a meaningful sentence. _ 94

102 ICT Activity- Geogebra Activity Converse of Pythagoras theorem Aim: To verify, if in a triangle the sum of the squares of the two sides is equal to the square of the third side then it is a right angled triangle. Previous Knowledge Assumed: Concept of angles, circles and triangles Basic knowledge of working on GeoGebra Procedure: 1. Draw a line segment 3 cm long using tool (segment with given length from point). 2. Draw a circle with radius 5 cm with point A as center, using tool (circle with center and radius). 3. Draw a circle with radius 4 cm with point B as center and using the same tool. 4. Let us define the intersection point of the two circles using tool (Intersect two objects) 5.. Name it point C. Let us draw line segments AC and BC using tool (Segment between two points). 6. Now let us hide the two circles by going to the algebra window on the left hand side. Right click on the circles c and d one by one and uncheck show object. 7. Now let us measure the angle CBA by using tool (Angle) and clicking on point C, then point B and then point A. If it comes to 90 degrees the theorem is verified. 8. Repeat the activity with three different sets. 9. Note down the observations. 10. Write the final result. 95

103 Observations: Serial Number First time Second time Third time Degree measure of the Angle Observation Result: Do the following: 1. Which of the following triplets forms the sides of a right angled triangle? Support your answer with proper reasoning. (13, 12, 5) (50, 60, 70) 96

104 (4, 6, 8) (20, 60, 70) (7, 14, 19) 2. The sides of a triangle measure a, b, and c. Use the table to find the type of the triangle. 97

105 SELF ASSESSMENT RUBRIC 14 CONTENT WORKSHEET (CW9) Parameter Able to state converse of Pythagoras Theorem. Able to verify converse of Pythagoras Theorem. Able to apply Pythagoras Theorem in geometrical problems. 98

106 STUDENT S WORKSHEET 15 CONTENT WORKSHEET (CW10) APPLICATION OF SIMILAR TRIANGLES Name of Student I. Date Write the conditions for two triangles to be similar Creating Patterns using similar triangles: Recognizing and using congruent and similar shapes can make calculations and design work easier. For instance, in the design below, only two different shapes were actually drawn. The design was put together by copying and manipulating these shapes to produce versions of them of different sizes and in different positions. 99

107 Create your own design by using only similar triangles. Design I Design II Design III Design IV II. Hands on Activity III. Do the following: 1. What is the height of the telephone pole? 100

108 2. Triangles RST is similar to triangle XYZ. Find all the missing measures. 101

109 3. Determine whether each pair of triangles is similar. If they are similar, state the similarity criteria. 4. Find the value of x for each pair of similar triangles. 102

110 SELF ASSESSMENT RUBRIC 15 CONTENT WORKSHEET (CW10) Parameter Able to state and understand all the criterion of similaritysss, AA,SAS Able to verify all criterions of similarity. Able to apply the similarity criterion in problems. 103

111 STUDENT S WORKSHEET 16 POST CONTENT (PCW1) Name of Student Date Practice assignment-bpt and its Converse 1. In given Figure B C BC. Find the length AB. Justify your answer. 2. If AB CD in each of the following figure, find x. Write the statement of theorem used. 3. In triangle ABC, DE BC and 4. In the given figure, A. If AC 4.8 cm, find AE P and AD PM. Show that DM AP. 104

112 STUDENT S WORKSHEET 17 POST CONTENT (PCW2) Name of Student Date Practice assignment: Similarity of triangles 1. A girl of height 90 cm is walking away from the base of a lamp post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds. 2. A vertical pole of length 12m casts a shadow 8m long on the ground and at the same time a tower casts a shadow 54m long. Find the height of the tower. 3. AM and PN are medians of triangles ABC and PQR, respectively. Also similar to PQR. Show that ABC is. 4. Altitudes AD and CE of triangle ABC intersect at P. Which of the following hold true? a. Triangle AEP is similar to triangle CDP b. Triangle ABD is similar to triangle CBE c. Triangle AEP is similar to triangle ADB d. Triangle PDC is similar to triangle BEC 5. Given that AB CD, describe how do you know that triangle ABE is similar to triangle CDE. Also, find the value of x. 105

113 STUDENT S WORKSHEET 18 POST CONTENT (PCW3) Name of Student Date Practice assignment- Pythagoras theorem and its converse 1. L and M are the mid points of AB and BC respectively of triangle ABC, right angled at B. Prove that 4 LC2 AB2 + 4BC2. 2. In triangle ABC, right angled at C, Q is the mid- point of BC. Prove that AB2 4 AQ2 3 AC2. 3. In the given figure, AD is perpendicular to BC and BD is one-third of CD. Prove that 2CA2 2AB2+ BC2. 106

114 4. In the given figure, a triangle ABC is right angled at B. The side BC is trisected at points D and E. Prove that 8 AE2 3AC2 + 5AD2. 5. In the given figure, triangle ABC is right angled at C; and D is the mid- point of BC. Prove that AB2 4AD2 3 AC2. STUDENT S WORKSHEET 19 POST CONTENT (PCW4) Name of Student Date Do the following: 1. Fill in the blanks (a) All equilateral triangles are (Similar/Congruent) (b) If ABC ~ FED then AB/BC. (c) Circles with equal radii are (Similar/Congruent) 107

115 2. In given figure 3. Write the statement of Pythagoras Theorem. 4. An airplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another airplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far will be the two planes after one & half hours? 5. In a ABC right angled at C, AC BC. Then AB2 AC2 6. Find the value of x in each of the similar triangles. Justify your answer. 7. State whether the following pairs of polygons are similar or not: 8. In the figure, the line segment XY is parallel to side AC of ABC and it divides the triangle into two equal parts of equal areas. Find the ratio. 9. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ABE CFB. 10. In ABC, AD and AED ABC. Show that AB AC. BC. Prove that AB2 - BD2 BC2 - CD AD is a median of ABC. The bisector of F respectively. Prove that EF BC. 108 ADB and ADC meet AB and AC in E &

116 STUDENT S WORKSHEET 20 POST CONTENT (PCW5) Name of Student Date Solve the following crossword puzzle: Across 1. The ratio of areas of two similar triangles is equal to the ratio of the on their corresponding sides. 4. If the corresponding sides of two triangles are, their corresponding angles are equal and the two triangles are similar. 5. In a triangle, if the square on one side is equal to sum of the squares on the other two sides, the angle to the first side is a right angle. 8. If in two triangles, the corresponding angles are equal, their corresponding sides are proportional and the triangles are. 9. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the ratio. 109

117 Down 2. In triangle the square on the hypotenuse is equal to the sum of the squares on the other two sides. 3. Basic Proportionality theorem is also known as theorem. 6. If a line divides two sides of a triangle in the same ratio, the line is to the third side. 7. If one angle of triangle is equal to one angle of another triangle and the sides including these angles are proportional, the two triangles are similar by similarity. STUDENT S WORKSHEET 21 POST CONTENT (PCW6) Name of Student Date Multiple Choice questions: Choose the correct answer: In the following fig QA AB and PB AB, then AQ is (i) 15 units (ii) 8 units (iii) 5 units (iv) 9 units 110

118 3. The ratio of the areas of two similar triangles is equal to the (i) ratio of their corresponding sides (ii) ratio of their corresponding attitudes (iii) ratio of the squares of their corresponding sides (iv) ratio of the squares of their perimeter 4. The areas of two similar triangles are 144 cm2 and 81 cm2. If one median of the first triangle is 16 cm, length of corresponding median of the second triangle is (i) 9 cm (ii) 27 cm (iii) 12 cm (iv) 16 cm Given Quad. ABCD ~ Quad PQRS then the value of x is (i) 13 units (ii) 12 units (iii) 6 units (iv) 15 units 111

119 7. 8. If ABC ~ DEF, ar ( DEF) 100 cm2 and AB/DE 1/ 2 then ar ( ABC) is (i) 50 cm2 (ii) 25 cm2 (iii) 4 cm2 (iv) None of the above. If the three sides of a triangle are a, 3a, 2a, then the measure of the angle opposite the longest side is (i) 45⁰ (ii) 30⁰ (iii) 60⁰ (iv) 90⁰ 9. In a ABC, point D is on side AB and point E is on side AC such that DE BC. If AD 2x 3, DB x 1, AE 5x 7 and EC 2(x 1), then the value of x is (a) 1 (b) 1 or 1/2 (c) 1 (d) None of these 10 Bisector AD of A of ABC meets base BC at D. If AB 10cm, BC 8cm and AC 6cm, the length of BD is (a) 7 cm (b) 5cm (c) 4cm (d) 2cm 11. The areas of two similar triangles ABC and PQR are respectively 64 cm2 and 121 cm2. If QR 15.4 cm, then BC is (a) 11.2cm (b) 8cm (c) 11cm (d) 15.4cm 12. In two similar triangles ABC and PQR, if their corresponding altitudes AD and PS are in the ratio of 4:9, then, ar ( PQR)/ar( ABC) is (a) 16:81 (b) 81:16 (c) 9:4 (d) 4:9 13. In ABC, PQ is a line segment which cuts AB and AC at P and Q respectively such that PQ BC and it divides ABC into two equal parts. Then BP/AB is equal to (a) (b) (c) 112 (d)

120 STUDENT S WORKSHEET 22 POST CONTENT (PCW7) Name of Student Date Application of similar triangles in daily life Aim: To find height of any object using pocket height finder. Material Required: Cardboard, color sheet, thread, punching machine, marker, ruler, glue. Previous knowledge: criteria for similarity of triangles. Procedure: 1. Cut a card measuring 6 inches by 1 inches and mark off in inches. 2. Thread a piece of string through a hole as near to the bottom as possible, and make a knot so that it cannot slip through. 3. Cut the string so that it is 18 inches long after knotting. 4. The object to be measured must be covered exactly by the card when it is held out and the string pulled tight to the eye. 5. The ratio of the length of string to the height of the card will be the same as the distance to the height of the object. Observation: 1. Two imaginary similar triangles are formed. First is formed by height finder, string and line of sight and second is formed by object, distance between object and man and line of sight. 2. The ratio using these measurements is 1:3 and the height required is its distance multiplied by. Conclusion: We can find the height of any object using pocket height finder. 113

121 EXTRA READINGS: Pythagoras Theorem: gorean.html /pythagoras.html Videos Model of Pythagoras theorem- Water Proof: Moving model of Pythagoras theorem: Geometrical proof (Video) Pythagoras theorem 114