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MathematicsMathematics

CBSE-i

Similar Triangles Similar Triangles

UNIT-11UNIT-11

CLASS CLASS

XX

(Core)(Core)

Shiksha Kendra, 2, Community Centre, Preet Vihar,Delhi-110 092 India

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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.

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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.

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 11500 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

PREFACEPREFACE

ACKNOWLEDGEMENTSACKNOWLEDGEMENTS

Advisory Conceptual Framework

Ideators

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

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

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 Ms. Preeti Hans, Proof Reader

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

English :

Geography:

Ms. Sarita Manuja

Ms. Renu Anand

Ms. Gayatri Khanna

Ms. P. Rajeshwary

Ms. Neha Sharma

Ms. Sarabjit Kaur

Ms. Ruchika Sachdev

Ms. Deepa Kapoor

Ms. Bharti Dave Ms. Bhagirathi

Ms. Archana Sagar

Ms. Manjari Rattan

Mathematics :

Political Science:

Dr. K.P. Chinda

Mr. J.C. Nijhawan

Ms. Rashmi Kathuria

Ms. Reemu Verma

Dr. Ram Avtar

Mr. Mahendra Shankar

Ms. Sharmila Bakshi

Ms. Archana Soni

Ms. Srilekha

Science :

Economics:

Ms. Charu Maini

Ms. S. Anjum

Ms. Meenambika Menon

Ms. Novita Chopra

Ms. Neeta Rastogi

Ms. Pooja Sareen

Ms. Mridula Pant

Mr. Pankaj Bhanwani

Ms. Ambica Gulati

History :

Ms. Jayshree Srivastava

Ms. M. Bose

Ms. A. Venkatachalam

Ms. Smita Bhattacharya

Material Production Groups: Classes IX-X

English :

Ms. Rachna Pandit

Ms. Neha Sharma

Ms. Sonia Jain

Ms. Dipinder Kaur

Ms. Sarita Ahuja

Science :

Dr. Meena Dhami

Mr. Saroj Kumar

Ms. Rashmi Ramsinghaney

Ms. Seema kapoor

Ms. Priyanka Sen

Dr. Kavita Khanna

Ms. Keya Gupta

Mathematics :

Political Science:

Ms. Seema Rawat

Ms. N. Vidya

Ms. Mamta Goyal

Ms. Chhavi Raheja

Ms. Kanu Chopra

Ms. Shilpi Anand

Geography:

History :

Ms. Suparna Sharma

Ms. Leela Grewal

Ms. Leeza Dutta

Ms. Kalpana Pant

Material Production Groups: Classes VI-VIII

Content1. Syllabus 1

2. Scope document 2

3. Teacher's Support Material 4

Teacher Note 5

Activity Skill Matrix 9

Warm Up W1 11

Classification of Triangles

Warm Up W2 11

Revising Congruency

Pre -Content P1 12

Recognizing Similar Figures Based on its Dictionary Meaning

Pre -Content P2 12

Corresponding Parts of Congruent Triangles

Pre -Content P3 13

Recapitulation of Ratio and Proportion

Content Worksheet CW1 13

Similar Figures


Basic Proportionality Theorem


Application of BPT


Converse of BPT


Pythagoras Theorem


Hands on Activity to Verify Pythagoras Theorem


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Application of Pythagoras Theorem


ICT Activity on Pythagoras Theorem


Converse of Pythagoras Theorem


Application of Similar Triangles

Post Content Worksheet PCW1 20






Post Content Worksheet PCW7

Assessment Plan 21

Study Material 22

Student Support Material 52

SW1: Warm Up (W1) 53


SW2: Warm Up (W2) 55

Revising Congruency

SW3: Pre Content (P1) 57

Recognizing Similar Figures Based on its Dictionary Meaning





SW6: Content (CW1) 62

Similar Figures

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Basic Proportionality Theorem (BPT)


Application of BPT

SW9:Content (CW4) 76

Converse of BPT


Pythagoras Theorem











SW16: Post Content (PCW1) 104

SW17: Post Content (PCW2) 105

SW 18: Post Content (PCW3) 106





Suggested Videos & Extra Readings. 114

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1

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

2

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

3

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.

4

TEACHER’S

SUPPORT

MATERIAL

5

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:

6

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.

7

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. All congruent figures are similar but all similar figures are not congruent.

ii. 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

that-

i. 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.

8

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 32 + 4²= 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.

9

ACTIVITY SKILL MATRIX

Type of Activity Name of Activity Skill to be developed

Warm UP(W1) Classification of triangles

Recognition and segregation.

Warm UP(W2) Revising Congruency

Observation and Knowledge

Pre-Content (P1) Recognizing similar figures based on its dictionary meaning

Vocabulary building, Identifying relations, Application of knowledge.

Pre-Content (P2) Corresponding parts of congruent triangles

Observation and Analysis.

Pre-Content (P3) Recapitulation of ratio and proportion

Computational skills.

Content (CW 1) Similar figures Analytical thinking, Reflection, Concept Development

Content (CW 2) Basic Proportionality theorem

Observation, ICT skill, drawing inferences.

Content (CW 3) Application of BPT

Thinking skill, Analysis and synthesis of knowledge, Application.

Content (CW 4) Converse of BPT

Knowledge, Thinking skills, Problem solving.

Content (CW 5) Pythagoras theorem

Observation, Analytical skills, Reasoning, 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

Problem solving skills.

Content (CW 8) ICT Activity on Pythagoras theorem

ICT skills and observation

Content (CW 9) Converse of Pythagoras theorem

Observation, Analytical skills, Reasoning, Drawing Inferences.

10

Content (CW 10) Application of similar triangles

Problem solving skills

Post - Content (PCW 1)

Assignment (BPT) Problem solving skills.


Assignment (Similar triangles)

Problem solving skills.


Assignment based Pythagoras theorem and converse.

Conceptual knowledge.


Concept Check Knowledge and application.


Crossword Knowledge based


MCQ Knowledge application


Hands on Application

11

ACTIVITY 1 – WARM UP (W1)


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: http://www.basic-mathematics.com/types-of-triangles.html

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.


Has knowledge of concept of congruent figures

Able to find the congruent figures.

Extra reading:

http://www.basic-mathematics.com/types-of-triangles.html

12

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.


Has understood the meaning of the word similarity.

Able to find similar figures.

Extra reading:

ACTIVITY 4 – PRE CONTENT (P2)


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.


Can correctly mark pair of corresponding sides and corresponding angles in two

congruent figures.

Extra reading:

13

ACTIVITY 5 – PRE CONTENT WORKSHEET (P3)


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.


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).

14

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).


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:


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.

15

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.


Able to verify the basic proportionality theorem.

Extra reading:


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.


Able to apply the basic proportionality theorem.

Extra reading:

16

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.


Able to find that converse of BPT holds true

Able to apply the basic proportionality theorem and its converse in solving

problems

Extra reading:


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.


Can check whether a triplet is Pythagorean or not.

Can state Pythagoras Theorem.

Extra reading:

http://www.youtube.com/watch?v=Ng2EpkKooo4

http://www.cut-the-knot.com/pythagoras/

http://en.wikipedia.org/wiki/Pythagorean_theorem




17

http://jwilson.coe.uga.edu/emt668/emt668.student.folders/headangela/essay1/pytha

gorean.html

http://www.jimloy.com/geometry/pythag.htm

http://www.mathsisfun.com/pythagoras.html

http://www.sunsite.ubc.ca/LivingMathematics/V001N01/UBCExamples/Pythagoras

/pythagoras.html



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.


Able to verify Pythagoras theorem.

Extra reading:



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.


Able to apply Pythagoras theorem in solving problems

Extra reading:

http://jwilson.coe.uga.edu/emt668/emt668.student.folders/headangela/essay1/pythagorean.html




http://www.sunsite.ubc.ca/LivingMathematics/V001N01/UBCExamples/Pythagoras/pythagoras.html


18



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.


Able to verify Pythagoras theorem

Extra reading:

ACTIVITY 14 –CONTENT WORKHEET (CW9)


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.


Able to verify that converse of Pythagoras theorem is true

Able to apply the theorem in solving problems

Extra reading:

19



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.


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.

20

SUGGESTIVE RUBRIC FOR FORMATIVE ASSESSMENT

(EXEMPLARY)

Parameter Mastered Developing Needs motivation

Needs personal attention

Understanding of BPT (Thales Theorem).

Able to state Basic Proportionality Theorem

Able to state Basic Proportionality

Able to state BPT partially correct.

Not Able to state BPT

Able to prove the theorem.

Able to draw the figure correctly, able to write the proof partially correct.

Able to draw the figure according to the statement but can‟t prove.

Not Able to prove.

Able to apply BPT for the given figure and able to draw accurate figure for a given problem based on BPT.

Able to apply BPT for the given figure but not able to draw accurate figure for a given problem based on BPT.

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.

21

TEACHERS’ RUBRIC FOR SUMMATIVE ASSESSMENT

OF THE UNIT

Parameter 5 4 3 2 1

Understan-ding similarity.

Able to differentiate between similar and the congruent figures.

Able to identify similar triangles.

Able to identify the corresponding parts of similar triangles.

Able to define the conditions of similarity.

Able to state all similarity criterions – SSS, AA, SAS.

Able to apply all similarity criterions – SSS, AA, SAS in problems.

Not able to differentiate between similar and the congruent figures.

Not able to identify similar triangles.

Not able to identify the corresponding parts of similar triangles.

Not able to define the conditions of similarity.

Not able to state all similarity criterions – SSS, AA, SAS.

Not able to apply all similarity criterions – SSS, AA, SAS in problems.

Basic Proportion-ality theorem and Converse.

Can state BPT and its Converse.

Can apply BPT or its converse appropriately in problems of geometry.

Cannot state BPT and its Converse.

Cannot apply BPT or its converse appropriately in problems of geometry.

Theorem based on area of similar triangles.

Can state the theorem.

Can apply the theorem appropriately in problems of geometry.

Cannot state the theorem.

Cannot apply the theorem appropriately in problems of geometry.

Pythagoras theorem and its converse.

Can state Pythagoras Theorem and its Converse.

Can apply Pythagoras Theorem or its converse appropriately in problems of geometry.

Cannot state Pythagoras Theorem and its Converse.

Cannot apply Pythagoras Theorem or its converse appropriately in problems of geometry.

22

STUDY MATERIAL

23

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) (iii)

Fig.1

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

24

They are of same shape but are of different sizes. We call these squares as similar

figures.

• Now observe the following equilateral triangles

Fig.3

These are of same shape and same size and hence congruent equilateral triangles.

• Again observe the following three equilateral triangles

Fig. 4

Fig.4

These are of same shape but different sizes. We call then similar equilateral triangles

• Now see the following circles

Fig.5

(i) (ii) (iii)

Fig.3

(i) (ii) (iii)

25

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.

Look at the following photograph

(i)

(ii)

(iii)

Fig.7

From the above discussion, it can be said that all congruent figures

are similar but all similar figure need not be congruent

26

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

Example 1: State whether the following pairs of polygons are similar or not. Give

reasons.

(i)

(ii)

(ii)

Two polygons are similar if their corresponding

(i) angles are of same measure (equal)

(ii) sides are proportional

27

(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 Aand Dare corresponding angles, Band Eare 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.

28

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 (640-546 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.

29

Similarly: = = - (2)

But ar( BDE) = ar ( CDE) (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)]

+1 = + 1 (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. 11

30

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 intersecting AB and AC respectively

at D and E such that =

To prove: 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, + 1 = + 1

i.e., =

or, =

So, FC = EC

This is impossible. It is possible only when E and F coinside. i.e. DF and are the same

line.

Hence, BC or DE BC.

We now take some examples to illustrate the use of these results.

31

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,

= = 1.3

And = = = 1.5

Thus,

So, AB is not parallel to DF (By converse of BPT) Fig. 14

Example 4: In Fig.15, if AB RQ and AC RS, prove that = .

Solution: In PQR, we have:

AB RQ

= (Corollary of BPT) (1)

Again in PQR, We have:

AC RS

= (corollary of BPT) (2) Fig.15

So, From (1) and (2), we have:

=

32

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, APQ= ABC (Corresponding Angles)

and AQP= ACB (corresponding angles) (1) Fig. 16

Also it is given that AQP= 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 Fig.17

So, let us form AC and let line intersect AC at G.

Now, from ADC, we have:

EG DC (why?)

So, = (1)

Again, from CAB, we have:

GF AB (why?)

So, =

33

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 , and Fig. 18

You will observe that = =

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 A = P, B = Q and C = R.

To Prove: = = and hence

ABC PQR

34

Construction : Assuming that AB<PQ

and AC<PR, take points S and T on PQ

and PR respectively such that AB=PS and

AC=PT. Join ST (Fig 19).

Note: In case AB>PQ and AC>PR, we can

take AS=PQ and AT=PR and If AB=PQ,

then and hence similar.

Proof: (SAS congruence criterion) Fig. 19

So, B= S and C = T (CPCT)

But B = Q (Given)

So, S = Q

Hence, ST QR (Since corresponding angles are equal)

Therefore, = (corollary of BPT)

So, = (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,

i.e. ABC PQR.

SSS Similarity

Now, draw two triangles ABC and PQR Such that = = (Fig. 20)

35

Now, measure angle A, B , C , P , Q and R . You will see that A= P , B=

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: 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

To Prove: A= P , B= Q , and C = R

and hence ABC PQR.

Construction: Take Points S and T on

PQ and PR respectively such that AB=PS

and AC=PT. Join ST (Fig.21) Fig. 21

Proof: = (Given)

So, = (By construction)

Therefore, =

Hence, ST QR (converse of BPT)

So, S = Q and T = R (corresponding angles) (1)

Therefore, PST PQR (AAA Similarity, P is common)

So, = (Sides will be proportional)

So, = (Since AB = PS by construction) (2)

36

But it is since that =

So, from (2) , =

Therefore, ST = BC

Hence, ABC PST (SSS criterion for congruence)

So, therefore A = P , B= S and C = T

But S = Q and T = R [Already proved in (1) ]

So, we have 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 Fig. 22

= and A = P (Fig. 22)

Here, two pairs of sides are proportional and the angles included between these pairs of

sides are equal.

37

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 A = P and = .

To prove: ABC PQR

Construction: Take points S and T on

PQ and PR respectively such that AB

= PS and AC=PT. Join ST (Fig.23)

Proof: (SAS congruence criterion) Fig. 23

So, = S and C = T (CPCT) (1)

Also, Since = = , 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.

38

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:

39

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:

40

Example9: In Fig 26, if then show that AB CD.

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) Fig. 27

Hence, = (corresponding sides are proportional)

Example 11: In Fig. 28, AM and DN are

respectively the medians of triangle ABC

and DEF.

Such that = = , Show that

(i) ABM DEN

(ii) ABC 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)

41

Or,

Hence, or

(ii) (Proved in (i) above)

So, B = E (corresponding angles are equal) (1)

Now, is 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, AB=PC (CPCT) (1)

Also, congruence)

So, DE = QF (CPCT) (2)

Now, in

(Given)(Since AP=2AM and DQ=2DN) and (From (1) and (2)

So, = =

So, (SSS similarity of criterion)

Therefore, CAM = FDN (Corresponding angles are equal) (3)

42

And CPM = FQN (Corresponding angles are equal) (4)

Also, Since (Already proved)

So, BAM = CPM (CPCT) (5)

Further, Since (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 and BAC = EDF

So, (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. 30

43

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 (As ∆ABC~∆DEF) (3)

From (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 AB=12 cm

and PQ=8 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 AB=22cm, find the length of the side of ∆DEF corresponding to

side AB.

Solution: Since, ∆ABC ~ ∆EDF,

Therefore = [ED and AB are corresponding side]

44

Or, =

Or ED2 =

= 324 =

Thus ED = 18 cm

Example15: In fig. 31, ABCD is a trapezium in which AB CD and AB = 2CD. Find the

ratios of areas of triangles AOB and COD

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 ( Fig.32

or = (1)

Now, XY AC

So, X = A and Y = C (corresponding angles)

Hence,

So, = (2)

Therefore from (1) and (2)

45

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

46

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,

Or, (1)

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:

47

Given: A triangle ABC in which

To prove: ABC = 90

Construction: Let us

construct a right triangle

PQR right angled at Q such

that PQ=AB and QR=BC

Proof: In Fig.35

or,

But, (Given)

So,

or, PR = AC

Therefore,

So, B = Q

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) Fig.36

Similarly, in

(2)

= + [As

= [As is a right angled triangle]

48

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)2 + 502 =6400+2500=8900

Since

Example 20: ABC is an isosceles triangle with AC = BC.

If prove that

Solution: 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, BD=DC=a.

In Fig.38

49

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.

50

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. 40

51

Example24: In Fig. 41; AD⊥BC. Prove

that

Solution: We have

[ is right

angled at D and by Pythagoras

Theorem]

= 2 Fig. 41

= +

= )

Example 25: In BC. If AC > AB, then show

that + ²

Solution: In right we have

AB² = AE² + BE² (Pythagoras Theorem)

= (AD² ED²) + (BD ED)²

= AD² ED² + BD² + ED² 2BD

= AD² 2BD +BD² Fig. 42

=AD² (1)

Similarly, In right

AC² =AE + EC² (Pythagoras Theorem)

= (AD² ED²) + (ED +DC)²

=AD² ED² + ED² + DC² + 2DC. ED

=AD² + 2

= AD² + BC. ED + (2)

Adding (1) & (2), AB²+AC² = 2AB² + 2

= 2AD² + .

52

STUDENT’S SUPPORT

MATERIAL

53

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

54

SELF ASSESSMENT RUBRIC 1 – WARM UP (W1)

Parameter

Can classify triangles on

the basis of sides

Can classify triangles on

the basis of angles

55


WARM UP 2

REVISING CONGRUENCY


Which of the following figures are congruent? How will you get to know?

56

SELF ASSESSMENT RUBRIC 2 – WARM UP (W2)

Parameter

Can recognise congruent

figures

57


PRE CONTENT 1

RECOGNIZING SIMILAR FIGURES BASED ON ITS DICTIONARY

MEANINGS


Dictionary Task:

Look for the meaning of the word similar in the dictionary. Observe the following

pictures and tell which of them are similar.

58

SELF ASSESSMENT RUBRIC 3 – PRE CONTENT (P1)

Parameter

able to recognise similar

objects.

59


PRE CONTENT 2

CORRESPONDING PARTS OF CONGRUENT TRIANGLES


Observe the following figures and fill the table:

Pair of Congruent figures Corresponding Sides

Corresponding Angles

ABC DEF

PQR LMN

XYZ EFD

60

Rectangle (DCBA) Rectangle (PQRS)


Parameter

Able to write

corresponding sides of

given congruent figures

Able to write

corresponding angles of

given congruent figures

61


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::120:100

4.


Parameter

Able to use the concept of

proportion in finding

unknown.

62


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.

64

II. Reflect on the following statements.

i. All circles are similar.

________________________________________________________________________

________________________________________________________________________

65

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.

66

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

67

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



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.

68

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.

69

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.

70

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

71

Step 7. Find AD/DB and AE/EC.

Step 8. What do you observe?

Step 9. Repeat the activity for two more triangles.

Step 10. Write the result.

Source: http://mykhmsmathclass.blogspot.com

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) . Name it as “f”. Provide 1 as

min and 5 as max.

2. Draw a polygon ABC using the tool (polygon) .

3. Draw a line parallel to line segment BC using tool (Parallel line) . Point

D is generated in the process.

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) .

http://mykhmsmathclass.blogspot.com/

72

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. ratio1=distanceAE/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 Ratio 1 Ratio 2 Observation

First time

Second time

Third time

Result:

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

73

SELF ASSESSMENT RUBRIC 7 –


Parameter

Able to verify BPT using

hands on activity on a

parallel line board.

Able to explore BPT using

GeoGebra


CONTENT WORKSHEET (CW3) APPLICATION OF BPT

Name of Student________________ Date_____________

1. Draw a diagram to explain the basic proportionality theorem

74

2. Solve the following questions. Make a list of results used in solving the problem.

Figure Given To find

Solution

DE BC, AD = 3 cm, DB = 1.5 cm, EC = 1 cm

AE

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 AD=4, AE=8, DB = x-4, and EC=3x-19, find x.

ii) If AD/BD = 4/5 and EC=2.5 cm, find AE.

iii) If AD=x, DB=x-2, AE=x+2 and EC=x-1, find the value of x.

iv) If AD=2.5 cm, BD=3.0 cm and AE=3.75 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, OC=2x-1, DO=3 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.

http://3.bp.blogspot.com/_azMmeWfLWbo/SrTwoCTgQkI/AAAAAAAAAT0/9FXy4zLcp70/s1600-h/1.bmp

75

SELF ASSESSMENT RUBRIC 8


Parameter

Able to apply the basic

proportionality theorem

in solving problems

76



CONVERSE OF BPT


1. Observe the following figure:

Based on your observation fill the following table:

Triangles Ratio 1 Ratio2 Are the two ratios equal?

Angles Measures

Are the two angles equal?

Relation between the line drawn with the base line.

ABC = = ADE=…..

ABC = …..

PQR = = PST = ……

PQR = ……

XYZ = = XLM = …..

XYZ = …..

77

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.



Parameter

Able to apply the basic


Able to apply the

converse of basic


in solving problems

78



PYTHAGORAS THEOREM


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.

79

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

(3,4,5) (5,12,13) (9,40,41)

(7,24,25) (8,15,17) (11,60,61)

(12,35,37)

80

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?

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

81

Self Assessment Rubric 10 – Content Worksheet (CW5)

Parameter

Able to write the

Pythagoras theorem



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.

82

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 http://mykhmsmathclass.blogspot.com

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.

http://mykhmsmathclass.blogspot.com/

83

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 1.

6. 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 1.

10. From the square ABIH cut the four quadrilaterals. Colour them and name them as

shown in Fig 2.

Fig 1

84

Fig 2

Observations

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

Conclusion:

85



Parameter

Able to state Pythagoras Theorem.

Able to verify Pythagoras Theorem using explorations or models.

86



APPLICATION OF PYTHAGORAS THEOREM


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.

87

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.

88

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?



Parameter


Able to apply Pythagoras Theorem in geometrical problems.

89





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.


Concept of angles, circles and triangles


Procedure:

1. Draw a line slider using the tool (Slider) . Name it as “d”. Provide -3 as min

and 3 as max.

2. Draw a point B using the tool (New point) .

3. Draw another point A using the same tool.

4. 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.

5. Draw a line through points A and B using tool (Line through two points) .

6. Draw a perpendicular line to line passing through point A and point B using tool

(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.

7. Let us now locate the intersection point of line AB and the line perpendicular to it

using tool (Intersect two points) . Point D is formed now.

8. Let us form a polygon out of the three points A, C and D using tool (polygon)

. After selecting the tool, click on points A, C and D one by one.

9. Let us hide lines a, b and c by right clicking on the these lines and un-checking

show objects.

90

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) using

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) . Just click inside

the circles and the area objects will appear.

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.




Observations:

Serial Number

Area of semicircle1

Area of semicircle2

Area of semicircle3

Observation

First time

Second time

Third time

Result:

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

91



Parameter


Able to verify Pythagoras Theorem using explorations on GeoGebra

92



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, c) where c > a, b

Verify, a2+b2= c2

Construct triangle with sides a, b, c

Is triangle a right angled triangle?

Write the side opposite to right angle.

(3,4,5)

32+42

=9 + 16 =25 =52

Yes

5

93

(5,12, 13)

(4,5,7)

(6,8,10)

(10,12,15)

94

Based on your exploration in the above table, rewrite the jumbled up words given

below to form a meaningful sentence.

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

95

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.


Concept of angles, circles and triangles


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) . Name it point C.

5. 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.




96

Observations:

Serial Number Degree measure of the Angle Observation

First time

Second time

Third time

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)

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

97

(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.

98



Parameter

Able to state converse of

Pythagoras Theorem.

Able to verify converse of Pythagoras Theorem.

Able to apply Pythagoras Theorem in geometrical problems.

99



APPLICATION OF SIMILAR TRIANGLES


I. Write the conditions for two triangles to be similar.

1. ________________________________________________________________________

________________________________________________________________________

2. ________________________________________________________________________

________________________________________________________________________

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.

100

Create your own design by using only similar triangles.

II. Hands on Activity

III. Do the following:

1. What is the height of the telephone pole?

Design I

Design IV

Design III

Design II

101

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

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

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

102

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.

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

103



Parameter

Able to state and understand all the criterion of similarity- SSS, AA,SAS

Able to verify all criterions of similarity.

Able to apply the similarity criterion in problems.

104


POST CONTENT (PCW1)


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 . If AC = 4.8 cm, find AE

4. In the given figure, A = P and AD = PM. Show that DM AP.

105


POST CONTENT (PCW2)


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 ABC is

similar to PQR. Show that = .

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.

106


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.

107

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.


POST CONTENT (PCW4)


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)

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2. In given figure = and AED = ABC. Show that AB = AC.

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 BC. Prove that AB2 - BD2 = BC2 - CD 2.

11. AD is a median of ABC. The bisector of ADB and ADC meet AB and AC in E &

F respectively. Prove that EF BC.

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POST CONTENT (PCW5)


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.

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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.


POST CONTENT (PCW6)


Multiple Choice questions:

Choose the correct answer:

1.

2. In the following fig QA AB and PB AB, then AQ is

(i) 15 units

(ii) 8 units

(iii) 5 units

(iv) 9 units

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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

5.

6. Given Quad. ABCD ~ Quad PQRS then the value of x is

(i) 13 units

(ii) 12 units

(iii) 6 units

(iv) 15 units

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7. 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.

8. 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) (d)

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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.

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EXTRA READINGS:

Pythagoras Theorem:



http://jwilson.coe.uga.edu/emt668/emt668.student.folders/headangela/essay1/pytha

gorean.html



http://www.sunsite.ubc.ca/LivingMathematics/V001N01/UBCExamples/Pythagoras

/pythagoras.html

Videos

http://mathematicsvideos.blogspot.com/search/label/Videos-%20Similar%20triangles

Model of Pythagoras theorem- Water Proof:

http://www.youtube.com/watch?v=hbhh-9edn3c

Moving model of Pythagoras theorem:

http://www.youtube.com/watch?v=Z68uPQU2v3o&NR=1

http://www.youtube.com/watch?v=8R8b4NelWN4&feature=related

Geometrical proof (Video)

Pythagoras theorem










http://mathematicsvideos.blogspot.com/search/label/Videos-%20Similar%20triangles

http://www.youtube.com/watch?v=hbhh-9edn3c

http://www.youtube.com/watch?v=Z68uPQU2v3o&NR=1

http://www.youtube.com/watch?v=8R8b4NelWN4&feature=related


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Converse of Pythagoras Theorem:

http://www.youtube.com/watch?v=wxUnodhYGEQ&feature=related

http://www.winpossible.com/lessons/Geometry_Pythagorean_Theorem_Converse.ht

ml

Similar triangles

http://similartriangles3.pbworks.com/w/page/23053498/Applying-Similar-Triangles-

to-the-Real-World

http://www.mathopenref.com/similartriangles.html

http://www.youtube.com/watch?v=wxUnodhYGEQ&feature=related

http://www.winpossible.com/lessons/Geometry_Pythagorean_Theorem_Converse.html

http://www.winpossible.com/lessons/Geometry_Pythagorean_Theorem_Converse.html

http://similartriangles3.pbworks.com/w/page/23053498/Applying-Similar-Triangles-to-the-Real-World

http://similartriangles3.pbworks.com/w/page/23053498/Applying-Similar-Triangles-to-the-Real-World

http://www.mathopenref.com/similartriangles.html

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