Download Core - The New Indian Model School, Dubai

CLASS
X
CBSE-i
UNIT-11
Mathematics
Similar Triangles
(Core)
Shiksha Kendra, 2, Community Centre, Preet Vihar,Delhi-110 092 India
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PREFACE
The Curriculum initiated by Central Board of Secondary Education -International (CBSE-i) is a progressive step in
making the educational content and methodology more sensitive and responsive to the global needs. It signifies the
emergence of a fresh thought process in imparting a curriculum which would restore the independence of the learner
to pursue the learning process in harmony with the existing personal, social and cultural ethos.
The Central Board of Secondary Education has been providing support to the academic needs of the learners
worldwide. It has about 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
ACKNOWLEDGEMENTS
Advisory
Shri Vineet Joshi, Chairman, CBSE
Shri Shashi Bhushan, Director(Academic), CBSE
Ideators
Ms. Aditi Misra
Ms. Amita Mishra
Ms. Anita Sharma
Ms. Anita Makkar
Dr. Anju Srivastava
English :
Ms. Sarita Manuja
Ms. Renu Anand
Ms. Gayatri Khanna
Ms. P. Rajeshwary
Ms. Neha Sharma
Ms. Sarabjit Kaur
Ms. Ruchika Sachdev
Geography:
Ms. Deepa Kapoor
Ms. Bharti Dave
Ms. Bhagirathi
Ms. Archana Sagar
Ms. Manjari Rattan
Conceptual Framework
Shri G. Balasubramanian, Former Director (Acad), CBSE
Ms. Abha Adams, Consultant, Step-by-Step School, Noida
Dr. Sadhana Parashar, Head (I & R),CBSE
Ms. Anuradha Sen
Ms. Jaishree Srivastava
Ms. Archana Sagar
Dr. Kamla Menon
Ms. Geeta Varshney
Dr. Meena Dhami
Ms. Guneet Ohri
Ms. Neelima Sharma
Dr. Indu Khetrapal
Dr. N. K. Sehgal
Material Production Groups: Classes IX-X
Mathematics :
Science :
Dr. K.P. Chinda
Ms. Charu Maini
Mr. J.C. Nijhawan
Ms. S. Anjum
Ms. Rashmi Kathuria
Ms. Meenambika Menon
Ms. Reemu Verma
Ms. Novita Chopra
Dr. Ram Avtar
Ms. Neeta Rastogi
Mr. Mahendra Shankar
Ms. Pooja Sareen
Political Science:
Ms. Sharmila Bakshi
Ms. Archana Soni
Ms. Srilekha
Dr. Rajesh Hassija
Ms. Rupa Chakravarty
Ms. Sarita Manuja
Ms. Himani Asija
Dr. Uma Chaudhry
History :
Ms. Jayshree Srivastava
Ms. M. Bose
Ms. A. Venkatachalam
Ms. Smita Bhattacharya
Economics:
Ms. Mridula Pant
Mr. Pankaj Bhanwani
Ms. Ambica Gulati
Material Production Groups: Classes VI-VIII
English :
Ms. Rachna Pandit
Ms. Neha Sharma
Ms. Sonia Jain
Ms. Dipinder Kaur
Ms. Sarita Ahuja
Dr. Indu Khetarpal
Ms. Vandana Kumar
Ms. Anju Chauhan
Ms. Deepti Verma
Ms. Ritu Batra
Science :
Dr. Meena Dhami
Mr. Saroj Kumar
Ms. Rashmi Ramsinghaney
Ms. Seema kapoor
Ms. Priyanka Sen
Dr. Kavita Khanna
Ms. Keya Gupta
Mathematics :
Ms. Seema Rawat
Ms. N. Vidya
Ms. Mamta Goyal
Ms. Chhavi Raheja
Political Science:
Ms. Kanu Chopra
Ms. Shilpi Anand
Material Production Group: Classes I-V
Ms. Rupa Chakravarty
Ms. Anita Makkar
Ms. Anuradha Mathur
Ms. Kalpana Mattoo
Ms. Savinder Kaur Rooprai
Ms. Monika Thakur
Ms. Seema Choudhary
Mr. Bijo Thomas
Ms. Kalyani Voleti
Dr. Sadhana Parashar,
Head (I and R)
Shri R. P. Sharma, Consultant
Ms. Seema Lakra, S O
Geography:
Ms. Suparna Sharma
Ms. Leela Grewal
History :
Ms. Leeza Dutta
Ms. Kalpana Pant
Ms. Nandita Mathur
Ms. Seema Chowdhary
Ms. Ruba Chakarvarty
Ms. Mahua Bhattacharya
Coordinators:
Ms. Sugandh Sharma,
Dr. Srijata Das,
Dr. Rashmi Sethi,
E O (Com)
E O (Maths)
E O (Science)
Ms. Ritu Narang, RO (Innovation) Ms. Sindhu Saxena, R O (Tech) Shri Al Hilal Ahmed, AEO
Ms. Preeti Hans, Proof Reader
Content
1.
Syllabus
1
2.
Scope document
2
3.
Teacher's Support Material
4
v
Teacher Note
5
v
Activity Skill Matrix
9
v
Warm Up W1
11
Classification of Triangles
v
Warm Up W2
11
Revising Congruency
v
Pre -Content P1
12
Recognizing Similar Figures Based on its Dictionary Meaning
v
Pre -Content P2
12
Corresponding Parts of Congruent Triangles
v
Pre -Content P3
13
Recapitulation of Ratio and Proportion
v
Content Worksheet CW1
13
Similar Figures
v
Content Worksheet CW2
14
Basic Proportionality Theorem
v
Content Worksheet CW3
15
Application of BPT
v
Content Worksheet CW4
16
Converse of BPT
v
Content Worksheet CW5
16
Pythagoras Theorem
v
Content Worksheet CW6
17
Hands on Activity to Verify Pythagoras Theorem
v
Content Worksheet CW7
17
Application of Pythagoras Theorem
v
Content Worksheet CW8
18
ICT Activity on Pythagoras Theorem
v
Content Worksheet CW9
18
v
Converse of Pythagoras Theorem
v
Content Worksheet CW10
19
Application of Similar Triangles
v
Post Content Worksheet PCW1
20
v
Post Content Worksheet PCW2
20
v
Post Content Worksheet PCW3
20
v
Post Content Worksheet PCW4
20
v
Post Content Worksheet PCW5
20
v
Post Content Worksheet PCW6
20
v
Post Content Worksheet PCW7
v
Assessment Plan
21
v
Study Material
22
v
Student Support Material
52
v
SW1: Warm Up (W1)
53
Classification of Triangles
v
SW2: Warm Up (W2)
55
Revising Congruency
v
SW3: Pre Content (P1)
57
Recognizing Similar Figures Based on its Dictionary Meaning
v
SW4: Pre Content (P2)
59
Corresponding Parts of Congruent Triangles
v
SW5: Pre Content (P3)
61
Recapitulation of Ratio and Proportion
v
SW6: Content (CW1)
Similar Figures
62
v
SW7: Content (CW2)
73
Basic Proportionality Theorem (BPT)
v
SW8: Content (CW3)
75
Application of BPT
v
SW9:Content (CW4)
76
Converse of BPT
v
SW10: Content (CW5)
78
Pythagoras Theorem
v
SW11: Content (CW6)
85
Hands on Activity to Verify Pythagoras Theorem
v
SW12:Content (CW7)
86
Application of Pythagoras Theorem
v
SW13:Content (CW8)
89
ICT Activity on Pythagoras Theorem
v
SW14:Content (CW9)
92
Converse of Pythagoras Theorem
v
SW15:Content (CW10)
99
Application of Similar Triangles
v
SW16: Post Content (PCW1)
104
v
SW17: Post Content (PCW2)
105
v
SW 18: Post Content (PCW3)
106
v
SW 19: Post Content (PCW4)
107
v
SW 20: Post Content (PCW5)
109
v
SW 21: Post Content (PCW6)
110
v
SW 22: Post Content (PCW7)
113
v
Suggested Videos & Extra Readings.
114
SYLLABUS –UNIT 11
GEOMETRY – SIMILAR TRIANGLES (CORE)
Introduction to
similarity
Similar figures, similarity of two polygons, similarity of two
triangles.
Criteria of
similarity of two
polygons
Two polygons are similar when their a) corresponding angles are
same and.
b) corresponding sides are in proportion
Criteria of
similarity
AAA, SSS, SAS through exploration
Application problems
Basic
Proportionality
theorem
Proof and applications
Pythagoras
theorem
Proof and applications
Converse of
Pythagoras
theorem
Statement and simple applications
1
SCOPE DOCUMENT
Key Concepts
1.
Similarity of polygons
2.
Similarity of triangles
3.
Basic Proportionality Theorem
4.
Pythagoras Theorem
Learning Objective:
1.
To understand the difference between similar and congruent figures.
2.
To define similar triangles.
3.
To state Basic Proportionality Theorem (Thales Theorem)
4.
To verify BPT using explorations or models.
5.
To apply BPT in geometrical problems.
6.
To state converse of BPT.
7.
To verify converse of BPT using explorations or models.
8.
To apply converse of BPT in geometrical problems.
9.
To state Pythagoras Theorem.
10. To verify Pythagoras Theorem using explorations or models.
11. To apply Pythagoras Theorem in geometrical problems.
12. To state converse of Pythagoras Theorem.
13. To verify converse of Pythagoras Theorem using explorations or models.
14. To apply converse of Pythagoras Theorem in geometrical problems.
15. To state and understand all the criterion of similarity- SSS, AA,SAS.
16. To verify all criterions of similarity using exploration or models.
17. To apply the similarity criterion in problems
2
Extension Activities:
1.
Sierpinski triangle can be generated by continuously repeating equilateral triangles.
Create sierpinski triangle in series and find the ratio of the original equilateral
triangle to the tenth triangle in the series.
2.
Extension of Pythagoras theorem:
a)
Generalize the Pythagoras theorem to mean that if similar figures are drawn on
each side of the right triangle, the sum of the areas of two smaller figures
equals the area of larger figure.
Perspective:
Fractal is a new branch of geometry which is based on concept of self similarity. It
has been observed that the shape of a portion of the object, if magnified, look like
original object. For example clouds, mountains, tress etc. Students can find more on
fractals.
SEWA:
Similar triangles find their application in many real life situations. For example,
consider the following problems:
a)
Campers walking along the south side of a river want to fell a tree tall enough so
that they can walk on the tree to get across the river. How can they find the width
of the river at its narrowest point without swimming across the river?
b) What must be the distance between the camera and statue if statue of some height
is to be photographed?
Research:
The Greek mathematician Thales used the knowledge of similar triangles to
estimate the height of the Greek Pyramid? Research about his technique and
prepare a presentation.
b) Extend the result to similar polygons for four or more sides.
3
TEACHER’S
SUPPORT
MATERIAL
4
TEACHER’S NOTE
The teaching of Mathematics should enhance the child‟s resources to think and reason,
to visualize and handle abstractions, to formulate and solve problems. As per NCF
2005, the vision for school Mathematics includes:
1. Children learn to enjoy mathematics rather than fear it.
2. Children see mathematics as something to talk about, to communicate through,
to discuss among them, to work together on.
3. Children pose and solve meaningful problems.
4. Children use abstractions to perceive relationships, to see structures, to reason
out things, to argue the truth or falsity of statements.
5. Children understand the basic structure of Mathematics: Arithmetic, algebra,
geometry and trigonometry, the basic content areas of school Mathematics, all
offer a methodology for abstraction, structuration and generalisation.
6. Teachers engage every child in class with the conviction that everyone can learn
mathematics.
Students should be encouraged to solve problems through different methods like
abstraction, quantification, analogy, case analysis, reduction to simpler situations, even
guess-and-verify exercises during different stages of school. This will enrich the
students and help them to understand that a problem can be approached by a variety of
methods for solving it. School mathematics should also play an important role in
developing the useful skill of estimation of quantities and approximating solutions.
Development of visualisation and representations skills should be integral to
Mathematics teaching. There is also a need to make connections between Mathematics
and other subjects of study. When children learn to draw a graph, they should be
encouraged to perceive the importance of graph in the teaching of Science, Social
Science and other areas of study. Mathematics should help in developing the reasoning
skills of students.
Proof is a process which encourages systematic way of
argumentation. The aim should be to develop arguments, to evaluate arguments, to
make conjunctures and understand that there are various methods of reasoning.
Students should be made to understand that mathematical communication is precise,
employs unambiguous use of language and rigour in formulation. Children should be
encouraged to appreciate its significance.
At the secondary stage students begin to perceive the structure of Mathematics as a
discipline. By this stage they should become familiar with the characteristics of
Mathematical communications, various terms and concepts, the use of symbols,
precision of language and systematic arguments in proving the proposition. At this
stage a student should be able to integrate the many concepts and skills that he/she has
learnt in solving problems.
Unit on similar triangles focus on lots of exploration and geogebra activities in order to
attain the following learning objectives:
5
1.
To understand the difference between similar and congruent figures.
2.
To define similar triangles.
3.
To state Basic Proportionality Theorem (Thales Theorem)
4.
To verify BPT using explorations or models.
5.
To Prove BPT logically.
6.
To apply BPT in geometrical problems.
7.
To state converse of BPT.
8.
To verify converse of BPT using explorations or models.
9.
To apply converse of BPT in geometrical problems.
10. To state Pythagoras Theorem.
11. To verify Pythagoras Theorem using explorations or models.
12. To apply Pythagoras Theorem in geometrical problems.
13. To state converse of Pythagoras Theorem.
14. To verify converse of Pythagoras Theorem using explorations or models.
15. To apply converse of Pythagoras Theorem in geometrical problems.
16. To state and understand all the criterion of similarity- SSS, AA,SAS.
17. To verify all criterions of similarity using exploration or models.
18. To apply the similarity criterion in problems
Concept of similarly in geometry is analogous to algebraic concept of ratio and
proportion. It finds its application in making maps, scale drawings, enlargement of
photos and indirect measurements of distance e. g. height of a tall building etc.
To introduce the concept of similar figures teacher can use interesting computer
applications. He/She can perform the copy-paste operation on screen for various
figures and demonstrate that all figures identical in shape and size are congruent
figures.
Further teacher can increase of decrease the size of copied figure and show that every
time he/she performs this operation, size changes but shape is retained. When shape is
identical but size changes figures are known as similar figures.
6
Further the figures can be overlapped, inserted, rotated or flipped in order to refine the
understanding of relation between congruency and similarity and to thrust upon
following points:
i.
ii.
All congruent figures are similar but all similar figures are not congruent.
All similar figures hold reflexive, symmetric and transitive property.
Reflexive
Δ ABC ~Δ ABC
Transitive
Δ ABC ~ Δ DEF
and
Δ DEF ~ Δ GHI
=> Δ ABC ~ ΔGHI
Symmetric
Δ ABC ~ Δ DEF
Also Δ DEF ~ Δ ABC
Further, students can observe that when the figures are similar the corresponding sides
are in proportion. It can be again shown on the screen that when the size is not
increased or reduced proportionately the shape is changed.
They can further be allowed to explore why some polygons like square, circle, regular
hexagon etc. are always similar? Why all triangles are not similar? Why all equilateral
triangles are similar but right triangles are not similar? Under what conditions triangles
are similar? To find the answer to above questions students can be allowed to draw
various figures to measure them to cut, to overlap the sides or angles etc. Teacher must
facilitate the conditions which can help them to come up with the idea on their own
thati. All squares are similar, because the corresponding angles are always same i.e. 90˚.
ii. When all the corresponding angles are same triangles are similar.
iii. Equilateral triangles have all angles of 60˚ for any dimension of triangles, so all
equilateral triangles are similar.
iv. In right triangles all angles are not always same so all right triangles are not similar.
7
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.
8
ACTIVITY SKILL MATRIX
Type of Activity
Warm UP(W1)
Warm UP(W2)
Pre-Content (P1)
Pre-Content (P2)
Pre-Content (P3)
Content (CW 1)
Content (CW 2)
Content (CW 3)
Content (CW 4)
Content (CW 5)
Name of Activity
Classification of
triangles
Revising
Congruency
Skill to be developed
Recognition and segregation.
Recognizing
similar figures
based on its
dictionary
meaning
Corresponding
parts of congruent
triangles
Vocabulary building, Identifying
relations, Application of knowledge.
Recapitulation of
ratio and
proportion
Similar figures
Computational skills.
Observation and Knowledge
Observation and Analysis.
Analytical thinking, Reflection, Concept
Development
Observation, ICT skill, drawing
inferences.
Basic
Proportionality
theorem
Application of BPT Thinking skill, Analysis and synthesis of
knowledge, Application.
Converse of BPT
Knowledge, Thinking skills, Problem
solving.
Pythagoras
Observation, Analytical skills, Reasoning,
theorem
Drawing Inferences.
Content (CW 6)
Hands on activity
Pythagoras
theorem
Thinking skill, Analysis and synthesis of
knowledge, Application, ICT.
Content (CW 7)
Application of
Pythagoras
theorem
ICT Activity on
Pythagoras
theorem
Problem solving skills.
Converse of
Pythagoras
theorem
Observation, Analytical skills, Reasoning,
Drawing Inferences.
Content (CW 8)
Content (CW 9)
ICT skills and observation
9
Content (CW 10)
Application of
similar triangles
Problem solving skills
Post - Content
(PCW 1)
Post - Content
(PCW 2)
Post - Content
(PCW 3)
Assignment (BPT)
Problem solving skills.
Assignment
(Similar triangles)
Assignment based
Pythagoras
theorem and
converse.
Concept Check
Problem solving skills.
Crossword
Knowledge based
MCQ
Knowledge application
Hands on
Application
Post - Content
(PCW 4)
Post - Content
(PCW 5)
Post - Content
(PCW 6)
Post - Content
(PCW 7)
Conceptual knowledge.
Knowledge and application.
10
ACTIVITY 1 – WARM UP (W1)
Classification of Triangles
Specific objective: To motivate the learners to classify the triangles according to sides
and angles.
Description: Students has the knowledge of types of triangles according to sides and
angles. This is a warm up activity to gear up the students for further learning.
Execution: Distribute the worksheet (W1) in which various triangles are drawn.
Students may be asked to work in pairs. They will classify the given triangles into two
categories: (i) according to sides namely scalene, isosceles and equilateral and (ii)
according to angles namely acute angled, right angled and obtuse angled.
Parameters for assessment:
Able to classify the triangles according to sides
Able to classify the triangles according to angles
Extra reading: 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.
Parameters for assessment:
Has knowledge of concept of congruent figures
Able to find the congruent figures.
Extra reading:
11
ACTIVITY 3 – PRE CONTENT (P1)
Recognising Similar Figures Based on its Dictionary Meanings
Specific objective: To enable students to recognise similar figures based on dictionary
meaning.
Description: Students has the knowledge of concept of congruent figures.
Execution: Firstly ask the students to find meaning of the word similar from the
dictionary. Distribute the worksheet (P1) in which pair of geometrical figures are
drawn. Students may be asked to work in pairs. They will find from the given pairs, the
similar figures.
Parameters for assessment:
Has understood the meaning of the word similarity.
Able to find similar figures.
Extra reading:
ACTIVITY 4 – PRE CONTENT (P2)
Corresponding Parts of Congruent Triangles
Specific objective: To test the understanding of meaning of corresponding sides and
corresponding angles in two given congruent figures.
Description: Students has the knowledge of concept of congruent figures. They know
what the corresponding sides and corresponding angles are. Through this activity, their
knowledge and understanding about the same concept will be tested.
Execution: Draw two congruent figures on the board and ask a few students to label
pair of corresponding sides. Ask few students to mark pair of corresponding angles.
Distribute the worksheet (P2) in which pair of geometrical figures are drawn. Students
may be asked to work in pairs. They will write pairs of corresponding sides and
corresponding angles.
Parameters for assessment:
Can correctly mark pair of corresponding sides and corresponding angles in two
congruent figures.
Extra reading:
12
ACTIVITY 5 – PRE CONTENT WORKSHEET (P3)
Recapitulation of Ratio and Proportion
Specific objective: To test the understanding of concept of proportionality.
Description: Students has the knowledge of concept of proportion. In the applications
of basic proportionality theorem, we need to use the previous knowledge of this
concept. Through this activity, their knowledge and understanding about the same
concept will be tested.
Execution: In a specified time of (say 10 minutes) ask the students to solve the problems
based on the concept of proportion.
Parameters for assessment:
Can correctly find an unknown when a:b :: c:d and value of any 3 out of a, b, c
and d is given.
Extra reading:
ACTIVITY 6 –CONTENT WORKSHEET (CW1)
Similar Figures
Specific objective: To explore the concept of similar figures.
Description: This activity is based on exploring the concept of similarity. Students
will be encouraged to use strategies like paper cutting, overlapping, measurement
etc. to visualise the concept of similarity. They will learn, to enlarge a given shape
using the concept of proportion.
Execution: Firstly, ask the students to cut out pieces in task I given in (CW1).
13
Use pair and share strategy and have a discussion in the classroom. This will enhance
the communication skills as well.
In part II of (CW1), a reflection activity is given. Students are asked to brainstorm
and think on statements like all circles are similar, all squares are similar etc. You
may ask the students to draw the shapes and then reflect. In part III, students will be
asked to draw any shape on a squared paper or a dotted sheet, and then enlarge it.
It will enhance their drawing skills, thinking skills and application of Math
concepts. After this, the students will be asked to work in group and take up task IV
given in (CW1).
Parameters for assessment:
Able to find correct pairs of similar figures
Able to reflect correctly on similarity of two given geometrical shapes
Able to enlarge and make a similar figure
Extra reading:
ACTIVITY 7 –CONTENT WORKSHEET (CW2)
Basic Proportionality Theorem (BPT)
Specific objective: To verify, the basic proportionality theorem by using a parallel line
board activity
Description: This is a hands on activity based on exploring and learning the basic
proportionality theorem. Students will verify the fact that if a line is drawn parallel to
one side of a triangle to intersect the other two sides at distinct points, then the other
two sides are divided in the same ratio.
Execution: Ask the students to bring a parallel line board sheet. It is a sheet on which
parallel lines are drawn.
14
Distribute the instruction sheet given in (CW2). Students will verify the result following
the instructions given in the sheet.
In part II of (CW2), an instruction worksheet on verifying BPT using GeoGebra is given.
You may ask the students to work in a multimedia room and explore the result using
this software.
Parameters for assessment:
Able to verify the basic proportionality theorem.
Extra reading:
ACTIVITY 8 –CONTENT WORKSHEET (CW3)
Application of BPT
Specific objective: To learn to apply, the basic proportionality theorem.
Description: This is a worksheet based on exploring and learning the application of
basic proportionality theorem. Students will apply the fact that if a line is drawn
parallel to one side of a triangle to intersect the other two sides at distinct points,
then the other two sides are divided in the same ratio ordinary font sum one
goldness in solving the given problems.
Execution: Distribute (CW3) and ask the students to solve the given problems using
BPT. Use pair and share and discuss.
Parameters for assessment:
Able to apply the basic proportionality theorem.
Extra reading:
15
ACTIVITY 9 – CONTENT WORKSHEET (CW4)
Converse of BPT
Specific objective: To explore the converse of basic proportionality theorem.
Description: This is a worksheet based on exploring and learning the converse of basic
proportionality theorem.
Execution: Students will be given an activity sheet (CW4). In task I, they will explore
the fact that if a line divides any two sides of a triangle in the same ratio, the line
must be parallel to the third side.
Parameters for assessment:
Able to find that converse of BPT holds true
Able to apply the basic proportionality theorem and its converse in solving
problems
Extra reading:
ACTIVITY 10 –CONTENT WORKSHEET (CW5)
Pythagoras Theorem
Specific objective: To understand and apply Pythagoras theorem.
Description: In this activity, students will recall Pythagorean triplets. They are already
familiar with the statement of Pythagoras theorem. Now they will learn more about
Pythagoras theorem.
Execution: Teacher may write the triplets on black board. Students will work out to find
a rule satisfying all the triplets. A class discussion should be taken up as an extension
activity to verify whether multiplying a Pythagorean triplet with a number results in
another Pythagorean triplet.
Parameters for assessment:
Can check whether a triplet is Pythagorean or not.
Can state Pythagoras Theorem.
Extra reading:
http://www.youtube.com/watch?v=Ng2EpkKooo4
http://www.cut-the-knot.com/pythagoras/
http://en.wikipedia.org/wiki/Pythagorean_theorem
16
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
ACTIVITY 11 –CONTENT WORKSHEET (CW6)
Hands on Activity to Verify Pythagoras Theorem
Specific objective: To verify Pythagoras theorem by hands on activity.
Description: CW6 contains instructions to perform hands on activity for verifying
Pythagoras theorem using paper cutting and pasting.
Execution: Students will be asked to bring materials in advance and perform hands on,
in the class.
Parameters for assessment:
Able to verify Pythagoras theorem.
Extra reading:
ACTIVITY 12 –CONTENT WORKSHEET (CW7)
Application of Pythagoras Theorem
Specific objective: To learn to apply Pythagoras theorem in solving problems
Description: This is a worksheet based on problems on the application of Pythagoras
theorem.
Execution: Students will solve the given problems using Pythagoras theorem.
Parameters for assessment:
Able to apply Pythagoras theorem in solving problems
Extra reading:
17
ACTIVITY 13 –CONTENT WORKSHEET (CW8)
ICT Activity on Pythagoras Theorem
Specific objective: Verification of Pythagoras theorem using the software GeoGebra
Description: Instruction sheet for working on GeoGebra is given in CW8.
Execution: students will follow the steps given in the instruction sheet and verify the
Pythagoras theorem.
Parameters for assessment:
Able to verify Pythagoras theorem
Extra reading:
ACTIVITY 14 –CONTENT WORKHEET (CW9)
Converse of Pythagoras Theorem
Specific objective: To explore the converse of Pythagoras theorem and apply in solving
problems.
Description: In this worksheet (CW9), there is a brainstorming question, followed by
self exploration on the converse of Pythagoras theorem. The instructions for verifying
the converse of Pythagoras theorem using Geo Gebra are also given. Further problems
are given based on the application of converse of Pythagoras theorem.
Execution: Firstly, students will have a discussion on the given statement in CW9. They
will be asked to explore the converse of Pythagoras theorem. Using the instruction
sheet, further they will verify the result using GeoGebra software. After learning the
theorem, students will be asked to solve the problems based on the theorem.
Parameters for assessment:
Able to verify that converse of Pythagoras theorem is true
Able to apply the theorem in solving problems
Extra reading:
18
ACTIVITY 15 –CONTENT WORKSHEET (CW10)
Application of Similar Triangles
Specific objective: Exploring the criterions of similarity namely SSS, SAS and AA.
Description: This is a self exploratory worksheet. Students will explore the following
using triangular cut-outs:
1. If the corresponding sides of two triangles are proportional, then they are
similar. This is called SSS Similarity.
2. If in two triangles, corresponding angles are equal, then the triangles are similar.
This is called AA similarity.
3. If in two triangles one pair of corresponding sides are proportional and the
included angles are equal then the two triangles are similar. This is called SAS
Similarity.
Execution: To begin with ask the students to make creative designs using similarity.
One design is shown in the worksheet.
Ask them to take triangle cut-outs and verify the similarity criterions. Proceeding
further students will solve problems based on similarity of triangles.
Parameters for assessment:
Able to verify AAA, SSS and SAS Similarity in given triangles.
Able to apply similarity criterions in solving problems.
Extra reading:
ACTIVITY 16 – ACTIVITY 22- POST CONTENT (PCW1 TO PCW7)
Specific objective: To enhance the understanding of concepts learnt.
Description: PCW1 to PCW6 are designed for further practicing the concepts learnt in
the classroom.
Execution:
Assessment Plan
Assessment guidance plan for teachers
2.22 Assessment Plan
Assessment guidance plan for teachers
With each task in student support material a self –assessment rubric is attached for
students. Discuss with the students how each rubric can help them to keep in tune their
own progress. These rubrics are meant to develop the learner as the self motivated
learner.
19
SUGGESTIVE RUBRIC FOR FORMATIVE ASSESSMENT
(EXEMPLARY)
Parameter
Mastered
Understanding
of BPT (Thales
Theorem).
Able to state
Basic
Proportionality
Theorem
Able to prove the
theorem.
Developing
Needs
motivation
Able to state
Able to state
Basic
BPT partially
Proportionality correct.
Able to draw
the figure
correctly, able
to write the
proof partially
correct.
Able to apply BPT Able to apply
for the given
BPT for the
figure and able to given figure
draw accurate
but not able to
figure for a given draw accurate
problem based on figure for a
BPT.
given problem
based on BPT.
Needs personal
attention
Not Able to
state BPT
Able to draw the Not Able to
figure according prove.
to the statement
but can‟t prove.
Able to apply
BPT for some
problems with
given figures
but not able to
draw accurate
figure for a
given problem
based on BPT.
Not able to
apply BPT.
From above rubric it is very clear that
Learner requiring personal attention is poor in concepts and requires the
training of basic concepts before moving further.
Learner requiring motivation has basic concepts but face problem in calculations
or in making decision about suitable substitution etc. He can be provided with
remedial worksheets containing solutions, methods of given problems in the
form of fill-ups.
Learner who is developing is able to choose suitable method of solving the
problem and is able to get the required answer too. May have the habit of doing
things to the stage where the desired targets can be achieved, but avoid going
into finer details or to work further to improve the results. Learner at this stage
may not have any mathematical problem but may have attitudinal problem.
Mathematics teachers can avail the occasion to bring positive attitudinal changes
in students‟ personality.
Learner who has mastered has acquired all types of skills required to solve the
problems based on BPT.
20
TEACHERS’ RUBRIC FOR SUMMATIVE ASSESSMENT
OF THE UNIT
Parameter
Understanding
similarity.
5
4
3
2
1
Able to differentiate
between similar and the
congruent figures.
Not able to differentiate
between similar and the
congruent figures.
Able to identify similar
triangles.
Not able to identify
similar triangles.
Able to identify the
corresponding parts of
similar triangles.
Not able to identify the
corresponding parts of
similar triangles.
Able to define the
conditions of similarity.
Not able to define the
conditions of similarity.
Able to state all similarity
criterions – SSS, AA, SAS.
Not able to state all
similarity criterions –
SSS, AA, SAS.
Able to apply all similarity
criterions – SSS, AA, SAS
in problems.
Basic
Proportionality
theorem and
Converse.
Can state BPT and its
Converse.
Not able to apply all
similarity criterions –
SSS, AA, SAS in
problems.
Cannot state BPT and its
Converse.
Can apply BPT or its
converse appropriately in
problems of geometry.
Cannot apply BPT or its
converse appropriately
in problems of geometry.
Theorem
based on
area of
similar
triangles.
Pythagoras
theorem and
its converse.
Can state the theorem.
Cannot state the
theorem.
Cannot apply the
theorem appropriately in
problems of geometry.
Cannot state Pythagoras
Theorem and its
Converse.
Cannot apply
Pythagoras Theorem or
its converse
appropriately in
problems of geometry.
Can apply the theorem
appropriately in problems
of geometry.
Can state Pythagoras
Theorem and its Converse.
Can apply Pythagoras
Theorem or its converse
appropriately in problems
of geometry.
21
STUDY
MATERIAL
22
SIMILAR TRIANGLES
Introduction
You are already familiar with congruent figures. Recall that two figures are said to be
congruent if they are of same shape and same size, i.e. one figure can be considered as a
trace copy of the other. However, we come across many figures which are of the same
shape but not necessarily of the same size. Such figures are called similar figures.
In this chapter, we shall discuss such figure in general and triangles in particular. We
shall also discuss various criteria for similarity of triangles and use them to arrive at a
famous theorem related to a right angled triangle commonly known as Pythagoras
Theorem.
(1)
Congruent and similar Figures,
See the following figures:
(i)
(ii)
Fig.1
(iii)
They all are of same shape and same size. So, these squares are congruent
Now see the following figures.
(i)
(ii)
(iii)
(iv)
Fig. 2
23
They are of same shape but are of different sizes. We call these squares as similar
figures.
• Now observe the following equilateral triangles
(i)
(ii)
(iii)
Fig.3
Fig.3
These are of same shape and same size and hence congruent equilateral triangles.
• Again observe the following three equilateral triangles
(i)
Fig. 4(ii)
(iii)
Fig.4
These are of same shape but different sizes. We call then similar equilateral triangles
• Now see the following circles
Fig.5
24
These are of same shape and same size. Hence, congruent circles.
• Again observe the following circles.
Fig. 6
These are of the same shape but not of same size. We call then similar circles.
From the above discussion, it can be said that all congruent figures
are similar but all similar figure need not be congruent
Look at the following photograph
(i)
(ii)
(iii)
Fig.7
25
In (i) there are photographs of same monument Tajmahal. They are of the same shape.
They are similar, Note that angles in each photograph ore of same measure. Also the
length of pillars are proportional
In (ii), triangle are also similar. By measurement it can be seen that corresponding
angles of triangles are of the same measure and corresponding sides are proportional in
any two triangles taken at a time.
In (iii), quadrilaterals are also similar. Here again, corresponding angles are of same
measure and sides are proportional
In view of the above, we say that
Two polygons are similar if their corresponding
(i) angles are of same measure (equal)
(ii) sides are proportional
Example 1: State whether the following pairs of polygons are similar or not. Give
reasons.
(i)
(ii)
(ii)
26
(iv)
Solution: (i) Not similar. Corresponding angles are equal but sides are not proportional.
as
=
=1
(ii) Similar. Corresponding angles are equal and sides are proportional
=
=
=
=
(iii) Not similar. Corresponding sides are proportional but corresponding
angles are not equal.
(iv) Not Similar. Corresponding angles are equal but corresponding sides
are not proportional.
=
but
=
(2) Similarity of triangles
Recall that a triangle is also a polygon. Therefore, the conditions for similarity of
two polygons shall also be valid for the similarity of two triangles. Thus, we can say
Two Triangles are similar, if their
(i)
Corresponding angles are equal, i.e. of the same measure and (ii) Corresponding
sides are proportional, i.e., they are in the same ratio.
As in the case of congruence, the words corresponding vertices; corresponding angles
and corresponding sides are of great significance: For example, if A B C and DEF are
similar under a correspondence says A D , B
C and E F , then the two triangles
need not be similar under the correspondence A E , B D and C F.
It may also be noted that symbol „ ‟ is used to represent „ is similar to‟ Thus, if ABC
and DEF are similar under the correspondence A
D ,B
E and C
F, then we
write it as ABC
DEF (read as triangle ABC is similar to DEF). In this case, side
corresponding to AB is DE, side corresponding to BC is EF and side corresponding to
CA is FD. Similarly A and D are corresponding angles, B and E are corresponding
angles and C and F are corresponding angles. It will not be correct to write the
similarity of the above triangles as ABC
EDF.
27
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.
28
=
Similarly:
=
-
But ar( BDE) = ar ( CDE)
(2)
(3)
(Recall that triangles on the same base and between the same parallels are equal in area)
=
So,
[From (1) and (2)] Hence, Proved,
We can obtain the following corollary from the above theorem:
Corollary: If in ABC, a line parallel to BC intersects AB and AC at D and E
respectively, then
=
Proof: See fig.10. We have:
=
(1)
+1 =
+1
[From Basic proportionality theorem (BPT)]
(Adding on both sides
=
or
=
or
(2)
Dividing (1) by (2), We get
=
.
Hence, Proved
Fig. 10
The converse of the BPT is also true. It is as follows:
Converse of Basic Proportionality Theorem: If a line divides any two sides of a
triangle in the same ratio, then the line is parallel to the
third side.
It means that if in
ABC, D and E are points on sides AB and
AC respectively, such that
=
, then DE BC (Fig 11).
Fig. 11
29
This result can be verified by taking suitable points D and E on AB and AC respectively
and then.
Showing that
D=
B or
E=
C which will give
It can be proved as follows:
Given:
ABC and a line
=
at D and E such that
To prove:
intersecting AB and AC respectively
BC, i.e., DE BC.
Fig. 12
Proof: Let us assume that is not parallel to BC, i.e. DE is not parallel to BC. So, there
must be some line parallel to BC, through D. Let this line intersects AC at F (Fig. 12).
Thus, we have DF BC.
Since DF BC, therefore by BPT, we have
=
But it is given that
So,
=
Or,
+1=
+1
=
i.e.,
or,
=
=
So, FC = EC
This is impossible. It is possible only when E and F coinside. i.e. DF and
line.
Hence,
BC or DE
BC.
We now take some examples to illustrate the use of these results.
30
are the same
Example 2:
In Fig. 13, if PQ AB, find QC.
Solution: PQ AB (Given)
So,
=
(By BPT)
Or,
=
Or,
QC =
cm = 4.5 cm.
Fig. 13
Example 3: In Fig 14, find whether, AB is parallel to DF or not
Solution: From the figure,
And
=
= 1.3
=
= = 1.5
Thus,
So, AB is not parallel to DF (By converse of BPT)
Example 4: In Fig.15, if AB RQ and AC RS, prove that
Solution: In
Fig. 14
=
.
PQR, we have:
AB RQ
=
(Corollary of BPT)
(1)
Again in PQR, We have:
AC RS
=
(corollary of BPT)
(2)
So, From (1) and (2), we have:
=
31
Fig.15
Example 5: In Fig 16, if =
=
and
AQP =
ABC ,
prove that ABC is an isosceles triangle.
Solution:
=
(Given)
So, PQ BC (Converse of BPT)
Hence,
and
APQ =
AQP =
ABC (Corresponding Angles)
ACB (corresponding angles)
AQP =
Also it is given that
(1)
Fig. 16
ABC
So, from (1), we have
ACB =
ABC
Therefore AB = AC (Sides opposite to the equal angles of a triangle) i.e. ABC is an
isosceles triangle.
Example6: Prove that any line parallel to the parallel sides of a trapezium divides its
non-parallel sides in the same ratio.
Solution: Let ABCD is a trapezium in which line
AB CD intersects the sides AD and BC at E and F
respectively.
We are to prove that
=
To apply some knowledge of BPT, We must have a triangle
So, let us form AC and let line
Now, from
intersect AC at G.
ADC, we have:
EG DC (why?)
So,
=
Again, from
(1)
CAB, we have:
GF AB (why?)
So,
=
32
Fig.17
Or,
=
(2)
Therefore, from (1) and (2), we have
=
(3) Criteria for similarity of Two Triangles
AAA Similarity: Draw any
two triangles ABC and PQR,
such that A = P , B = Q
and C = R (Fig. 18) Now,
measure, AB, BC, CA, PQ, QR
and RP.
After this find
You will observe that
,
and
=
Fig. 18
=
In this way, you have verified that if angles of one triangle are equal to corresponding
angles of the other triangle, then the corresponding sides of the two triangles are in the
same ratio or proportional. In other words, if corresponding angles of two triangles
are equal then their corresponding sides are proportional.
Thus, we see that both the conditions of similarity of polygons are satisfied and hence
the two triangles are similar. So, we may state the following criterion for similarity of
two triangles.
If corresponding angles of two triangles are equal then their corresponding sides are
proportional and hence the two triangles are similar. This criterion is referred to as
the AAA criterion for similarity of two triangles
In fact, the above result can be proved as shown below:
Given: Two triangles ABC and PQR such that
To Prove:
ABC
=
=
and hence
PQR
33
A = P, B = Q and C = R.
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)
B = S and
So,
C = T (CPCT)
Q (Given)
B=
But
Fig. 19
So, S = Q
Hence, ST QR (Since corresponding angles are equal)
Therefore,
So,
=
=
(corollary of BPT)
(Since AB = PS and AC = PT)
(1)
Similarly, by taking points on other pair of sides PQ and QR, it can be seen that
=
(2)
Hence, from (1) and (2),
=
=
i.e, corresponding sides are proportional and hence, the triangles are similar,
PQR.
i.e. ABC
SSS Similarity
Now, draw two triangles ABC and PQR Such that
34
=
=
(Fig. 20)
A , B , C , P , Q and
Q and
C = R . That is if the
corresponding sides of the two triangles
are proportional then the corresponding
angles are equal. Thus, the two
conditions of similarity of two polygons
are satisfied and so the two triangles
ABC and PQR are similar. Thus, We can
say that:
Now, measure angle
R . You will see that
A= P,
B=
Fig. 20
If Corresponding sides of two triangles are proportional, then their corresponding
angles are equal and hence the two triangles are similar. This criterion is referred to
as the SSS criterion for similarity of two triangles.
In fact, the above result can be proved as shown below:
Given: Two triangles ABC and PQR such
that
A = P , B = Q , and
and hence ABC
PQR.
To Prove:
C= R
Construction: Take Points S and T on
PQ and PR respectively such that AB=PS
and AC=PT. Join ST (Fig.21)
Proof:
So,
=
=
Therefore,
Fig. 21
(Given)
(By construction)
=
Hence, ST QR (converse of BPT)
So,
S=
Q and T = R (corresponding angles)
Therefore, PST
PQR (AAA Similarity,
(1)
P is common)
So,
=
(Sides will be proportional)
So,
=
(Since AB = PS by construction)
35
(2)
But it is since that
So, from (2) ,
=
=
Therefore, ST = BC
Hence,
ABC
So, therefore
But
PST (SSS criterion for congruence)
A = P , B = S and
S = Q and
So, we have
C= T
T = R [Already proved in (1) ]
A= P,
B = Q and
C= R
i.e., corresponding angles are equal.
Hence,
PQR.
Note: In view of the above two results, it can be stated that
(i) If the corresponding angles of two triangles are equal, then the two triangles are
similar (AAA similarity criterion). Further, if two angles of one triangle are
respectively equal to the two angels of the other triangles, then their third angles
will automatically, be equal. Therefore, we may say that if two angles of a triangle
are equal to corresponding two angles of the other triangles then the triangles are
similar (AA similarity criterion).
(ii) If the corresponding sides of
two
triangles
are
proportional,
then
the
triangles are similar (SSS
criterion for similarity).
SAS similarity Criterion.
To understand this criterion,
draw two triangles ABC and PQR
such that
=
and
A=
Fig. 22
P (Fig. 22)
Here, two pairs of sides are proportional and the angles included between these pairs of
sides are equal.
36
Now measure the remaining angles and sides, of the triangle i.e, measure B , C , Q ,
R . Side BC and Side QR.
You will see that
B= Q ,
C=
R and
is the same as
or
Thus, two triangles are similar (By AAA as well as SSS criterion of similarity) . So, may
state the third criterion of similarity of two triangles as follows:
If one angle of a triangle is equal to one angle of the other triangle and the sides
including these angles are proportional, then the triangles are similar. This is referred
to as the SAS similarity criterion for similarity of triangles.
In fact, We can prove the above result as shown below:
Given:
ABC and PQR in which
To prove: ABC
A = P and
=
.
PQR
Construction: Take points S and T on
PQ and PR respectively such that AB
= PS and AC=PT. Join ST (Fig.23)
(SAS congruence criterion)
Proof:
So,
= S and C = T (CPCT)
Also, Since =
=
=
Fig. 23
(1)
, therefore
(By construction)
So, ST QR (Converse of BPT)
Hence S= Q and T= R (Corresponding angles)
(2)
So, B= Q and C= R (From (1) and (2)
i.e.,
(AAA similarity)
Let us now take some examples to illustrate the use of this criteria.
37
Example7: In Fig. 24, ABC is a right triangle, right angled at B and D is any point on
side AB. If DE AC,
prove that
Solution: In
A = A (Common angle)
B = E (Each = 90 )
So,
(AA similarity criterion)
Fig. 24
Example 8 : Examine each of the following pairs of triangles and state which of them
are similar. If similar, state the criterion used by you for it. Also, write there pairs of
similar triangles in symbolic notation:
38
Fig. 25
Solution: (i) Triangles are Similar.
AAA similarity Criterion.
Symbolic Notation:
(ii) Triangles are similar.
SSS similarity criterion.
Symbolic Notation:
(iii) Triangles are not similar, because
(iv) Triangles are not similar, because
(v)
Triangles are similar, because F = 30 and N = 80
AAA similarity criterion.
Symbolic Notation.
(vi) Triangles are similar.
SAS Similarity criterion
Symbolic Notation:
39
Example9: In Fig 26, if
AB CD.
then show that
Solution:
(Given)
So, A = C and B = D (Corresponding angles)
Since A = C (or B = D),
Therefore AB CD (Because A and C are alternate
angles)
Fig. 26
Example 10: ABCD is a trapezium in which AB CD and its diagonals meet each other
at O. Prove that
Solution: See Fig. 27
In
, We have:
A = C (alternate angles, AB CD) and B = D
(alternate angles, AB CD)
So,
(AA similarity)
Hence,
=
Fig. 27
(corresponding sides are proportional)
Example 11: In Fig. 28, AM and DN are
respectively the medians of triangle ABC
and DEF.
Such that
(i)
(ii)
=
=
ABM
ABC
, Show that
DEN
DEF
Solution: (i) BM =
BC and EN = EF
Fig. 28
Since AM and DN are medians)
It is given that
So,
=
=
(Since BM = BC and EN = EF)
40
Or,
Hence,
or
(ii)
(Proved in (i) above)
So, B = E (corresponding angles are equal)
Now, is
(1)
we have.
(Given)
And B = E (Proved is (1))
So,
(SAS similarity criterion)
Example 12: In Fig 29. AM and D N
are respectively medians of
Prove that
Solution: Produce AM to P and DN
to Q such that AM = MP and DN =
NQ
Now,
(SAS congruence)
Fig. 29
So, AB=PC (CPCT)
(1)
congruence)
Also,
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)
41
(3)
And CPM = FQN (Corresponding angles are equal)
Also, Since
(4)
(Already proved)
So, BAM = CPM (CPCT)
Further, Since
(5)
(Already proved)
So, EDN = FQN (CPCT)
(6)
Therefore, from (4), (5) end (6), we have:
BAM = EDN
(7)
Adding (3) and (7), We have
CAM + BAM = FDN + EDN
i.e., BAC = EDF
Now, we have
So,
and BAC = EDF
(SAS Criterion)
(4) Area of Similar Triangles
You have learnt that in two similar triangles, their corresponding sides are in the
same ratio. What can we say about the ratio of areas of two similar triangles? You
also know that area is measure in square units. Do we expect that the ratio of two
similar triangles has something to do with the ratio of sides of these triangles?
Answer to this question is provided by the following Theorem.
Theorem: The ratio of the areas of two similar triangles is equal to the ratio of
the squares of their corresponding sides.
Given: ∆ABC and ∆DEF
(Fig 30) Such that ∆ABC and
∆DEF are similar.
To Prove:
=
=
=
Fig. 30
42
Construction : Draw altitudes AM and DN (Fig.30)
Proof:
ar (∆ABC)
and ar
Thus
=
(1)
Also ∆ABM ~ ∆DEN (By AA similarity criterion)
Hence
(2)
Also
From
(As ∆ABC~∆DEF)
(3)
(1), (2) and (3)
=
=
=
Thus
=
=
Thus, we have proved the theorem.
Let us consider some examples illustrating the use of the theorem above.
Example13: Two triangles ABC and PQR are similar to each other, in which 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]
43
Or,
=
Or
ED2 =
= 324 =
Thus
ED = 18 cm
Example15: In fig. 31, ABCD is a trapezium in which AB
ratios of areas of triangles AOB and COD
CD and AB = 2CD. Find the
Solution: In ∆OAB and ∆OCD, OAB = OCD (Alternate angle) and OBA = ODC
(Alternate angle)
So, ∆OAB ~ ∆OCD (AA similarity)
Hence,
=
=
Thus, the required ratio = 4:1
Fig.31
Example 16: In fig 32, XY AC and XY divides
triangular region into two parts equal in area
Determine
Solutions: Since the triangular region ABC has been
divides by XY two parts of equal area, therefore
ar (
or
Fig.32
=
(1)
Now, XY AC
So, X = A and Y = C (corresponding angles)
Hence,
So,
=
(2)
Therefore from (1) and (2)
44
or
or, 1-
= 1-
or
i.e.,
Example 17: Prove that the ratio of the areas of two similar triangles is equal to ratio of
the squares of their corresponding medians.
Solution: Let the triangles be
AM and DN their
medians (Fig.33)
We have:
(Since ∆ABC ~ ∆DEF)
or
( AM and DN are median)
or
(1)
Also, B = E (Corresponding angles of similar triangles ABC and DEF)
(2)
So, from (1) and (2) ∆ABM ~ ∆DEN (SAS Similarity)
or
(Corresponding sides must be proportional )
(3)
Now,
So,
=
[From (3) =
Similarity, the result can be proved for other corresponding medians.
Pythagoras Theorem
Draw a right ABC, right angled at B. Measure sides AB, BC and AC. Find AB2, BC2
and AC2. You will find that AC2 = AB2 + BC2. In fact this type of relation is true in all
45
right triangles and is known as Pythagoras theorem we may state the theorem as
follows:
Theorem: In a right triangle, the square on the hypotenuse is equal to the sum of
the squares on the other two sides.
It can be proved as follows:
Given: A right triangle ABC, right angled
at B (Fig.34)
To Prove: AC2 = AB2 + BC2
Construction: Draw BD ⊥ AC.
Proof: In triangles ADB and ABC.
Fig.34
A = A (Common)
ADB = ABC (each 90
So,
∆ADB ~ ∆ABC (Why?)
Thus,
(1)
Or,
Similarly,
= DC
(2)
Adding (1) and (2),
= AC
= AC
=
Hence, the required result.
We now prove the Converse of the Pythagoras Theorem
Theorem: In a triangle, if the square on one side is equal to the sum of the
squares on the other two sides, then the angle opposite to the first side is a right
angle.
The theorem can be proved as shown below:
46
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,
B= Q
So,
But Q = 90
Thus,
B = 90
Hence,
Example 18: P and Q are the points on the sides CA
and CB respectively of a ∆ABC, right angled at C.
Prove that:
Solution: Join PQ
In
(By Pythagoras Theorem)
(1)
Similarly, in
(2)
=
=
+
[As
[As
is a right angled triangle]
47
Fig.36
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
Solution:
prove that
We are given
AC = BC.
(1)
Also
or
=
Fig.37
i.e.,
Thus, angle opposite the side AB must be a right angle. Hence
.
Example 21: In Fig. 38, AB = BC = CA = 2a and AD⊥BC.
Show that
(i) AD =
(ii) area ar ( ABC) =
Solution:
(i)
So, BD=DC=a.
In
Fig.38
48
Or
=
So, AD = a
(ii) Area ( ABC) =
=
=
Example 22:
A ladder reaches a window which is 12 metres above the ground on one side of the
street. Keeping the foot at the same point, the ladder is turned to the other side of the
street to reach a window 9 metres high. Find the width of the street if the length of
ladder is 15 metres.
Solution:
In
[Fig. 39]
So,
(why?)
Thus, BC = 9 metres
In
So,
Thus, EB =12 Metres.
Hence, EC = EB + BC = (12 + 9) Metres
= 21 Metres.
49
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
50
Example24: In Fig. 41; AD⊥BC. Prove
that
Solution:
We have
[
is right
angled at D and by Pythagoras
Theorem]
=
=
Fig. 41
2
+
=
)
Example 25: In
that
BC. If AC > AB, then show
+
Solution:
In right
AB² = AE² + BE²
= (AD²
²
we have
(Pythagoras Theorem)
ED²) + (BD
ED)²
= AD²
ED² + BD² + ED²
= AD²
2BD
2BD
+BD²
Fig. 42
=AD²
(1)
Similarly, In right
AC² =AE + EC²
= (AD²
=AD²
(Pythagoras Theorem)
ED²) + (ED +DC)²
ED² + ED² + DC² + 2DC. ED
=AD² + 2
= AD² + BC. ED +
(2)
Adding (1) & (2), AB²+AC² = 2AB² + 2
= 2AD² +
.
51
STUDENT’S
SUPPORT
MATERIAL
52
STUDENT’S WORKSHEET 1
WARM UP W1
CLASSIFICATION OF TRIANGLES
Name of Student________________
Date____________
Classify the following triangles according to sides and angles:
Figures
Type of Triangle
53
SELF ASSESSMENT RUBRIC 1 – WARM UP (W1)
Parameter
Can classify triangles on
the basis of sides
Can classify triangles on
the basis of angles
54
STUDENT’S WORKSHEET 2
WARM UP 2
REVISING CONGRUENCY
Name of Student________________
Date____________
Which of the following figures are congruent? How will you get to know?
55
SELF ASSESSMENT RUBRIC 2 – WARM UP (W2)
Parameter
Can recognise congruent
figures
56
STUDENT’S WORKSHEET 3
PRE CONTENT 1
RECOGNIZING SIMILAR FIGURES BASED ON ITS DICTIONARY
MEANINGS
Name of Student________________
Date____________
Dictionary Task:
Look for the meaning of the word similar in the dictionary. Observe the following
pictures and tell which of them are similar.
57
SELF ASSESSMENT RUBRIC 3 – PRE CONTENT (P1)
Parameter
able to recognise similar
objects.
58
STUDENT’S WORKSHEET 4
PRE CONTENT 2
CORRESPONDING PARTS OF CONGRUENT TRIANGLES
Name of Student________________
Date____________
Observe the following figures and fill the table:
Pair of Congruent figures
ABC
DEF
PQR
LMN
XYZ
EFD
Corresponding
Sides
59
Corresponding
Angles
Rectangle (DCBA)
Rectangle (PQRS)
SELF ASSESSMENT RUBRIC 4 – PRE CONTENT (P2)
Parameter
Able to write
corresponding sides of
given congruent figures
Able to write
corresponding angles of
given congruent figures
60
STUDENT’S WORKSHEET 5
PRE CONTENT 3
RECAPITULATION OF RATIO AND PROPORTION
Name of Student_______________
Date____________
1. Find x if 2:5 :: x:25
2. If x:7 :: 25: 35, find x
3. Find p if 12:p::120:100
4.
SELF ASSESSMENT RUBRIC 5 – PRE CONTENT (P3)
Parameter
Able to use the concept of
proportion in finding
unknown.
61
STUDENT’S WORKSHEET 6
CONTENT WORKSHEET (CW1)
SIMILAR FIGURS
Name of Student_______________
Date___________
I. Megha said, “Two similar figures have the same shape but not necessarily the same
size.”
Keeping in mind the concept of similar figures, examine the following pair of figures
and find which of the following are similar.
62
63
II. Reflect on the following statements.
i.
All circles are similar.
________________________________________________________________________
________________________________________________________________________
64
ii.
All squares are similar.
________________________________________________________________________
________________________________________________________________________
iii.
A circle can be similar to a square
________________________________________________________________________
________________________________________________________________________
iv.
A triangle can be similar to a square
________________________________________________________________________
________________________________________________________________________
III. Activity:
Draw any shape. Enlarge it. What do you say about the similarity of two shapes?
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
____________________________________________________________________________
IV. Exploration
Work in a group of 4 students.
Draw a line segment AB of any length.
At A draw a ray at any angle of 30⁰, using a protractor.
At B draw a ray at any angle of 55⁰, using a protractor.
Let the two rays meet at C.
Observe the triangles made by each member of your group.
65
What do you observe? Are all the triangles similar?
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
Verify by measurement, whether the corresponding sides of any two triangles drawn
in the group, are in the same ratio.
Pair of Similar Triangles
Ratio of Corresponding Sides
66
SELF ASSESSMENT RUBRIC 6 – CONTENT WORKSHEET
(CW1)
Parameter
Able to recognise similar
figures
Able to reflect on concept
of similar figures
Knows enlargement of a
shape is done using
proportion
STUDENT’S WORKSHEET 7
CONTENT WORKSHEET (CW2)
BASIC PROPORTIONALITY THEOREM (BPT)
Name of Student______________
Date____________
I. Hands on Activity
Aim: To verify the Basic Proportionality theorem by paper cutting and pasting using a
parallel line board.
67
Statement: If a line is drawn parallel to any side of a triangle, to intersect the other two
sides at two distinct points, then the other two sides are divided in the same
ratio.
Material required: Coloured paper, parallel line board, pair of scissors, sketch pen,
ruler, glue
Procedure:
Step 1. Draw a triangle on a coloured paper.
Step 2. Label the triangle as ABC.
68
Step 3. Cut the triangle.
Step 4. Take the parallel line board and place the triangle ABC on it such that the side
BC coincides with any of lines on the board.
Step 5. Draw a line parallel to BC using a ruler by the help of lines on the parallel line
board.
69
Step 6. Let the parallel line drawn to BC intersect AB and AC at D and E respectively.
70
Step 7.
Step 8.
Step 9.
Step 10.
Source:
Find AD/DB and AE/EC.
What do you observe?
Repeat the activity for two more triangles.
Write the result.
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)
min and 5 as max.
2. Draw a polygon ABC using the tool (polygon)
. Name it as “f”. Provide 1 as
.
3. Draw a line parallel to line segment BC using tool (Parallel line)
D is generated in the process.
. Point
4. Now let us redefine the point D and make it as a function of our slider. Right
click on D and choose redefine. Replace the Y coordinate of the point D with f.
5. Now let us define the points E & F as the intersection points of the parallel line
and line segments AB and AC respectively using tool (Intersect two objects)
.
6. Let us draw line segment EF using tool (Segment between two points)
71
.
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
First time
Second time
Third time
Ratio 1
Ratio 2
Observation
Result:
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
72
SELF ASSESSMENT RUBRIC 7 –
CONTENT WORKSHEET (CW2)
Parameter
Able to verify BPT using
hands on activity on a
parallel line board.
Able to explore BPT using
GeoGebra
STUDENT’S WORKSHEET 8
CONTENT WORKSHEET (CW3)
APPLICATION OF BPT
Name of Student________________
Date_____________
1. Draw a diagram to explain the basic proportionality theorem
73
2.
Solve the following questions. Make a list of results used in solving the problem.
Figure
Given
To
find
AE
DE BC,
Solution
AD = 3 cm,
DB = 1.5 cm,
EC = 1 cm
ST QR
SQ = 7.2 cm,
PT = 1.8 cm,
TR = 5.4 cm
PS
3. Using the basic proportionality theorem, solve the following problems.
In a triangle ABC, D and E are points on the sides AB and AC respectively such
that DE BC.
i)
If 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.
74
SELF ASSESSMENT RUBRIC 8
CONTENT WORKSHEET (CW3)
Parameter
Able to apply the basic
proportionality
theorem
in solving problems
75
STUDENT’S WORKSHEET 9
CONTENT WORKSHEET (CW4)
CONVERSE OF BPT
Name of Student________________
Date____________
1. Observe the following figure:
Based on your observation fill the following table:
Triangles
Ratio 1
Ratio2
ABC
=
=
PQR
=
=
XYZ
=
=
Are the
two
ratios
equal?
Angles
Measures
ADE=…..
ABC = …..
PST = ……
PQR =
……
XLM = …..
XYZ = …..
76
Are the
two angles
equal?
Relation
between the
line drawn
with the
base line.
What can you conclude from the above table?
_________________________________________________________________________
_________________________________________________________________________
2.
In a triangle ABC; D and E are points on the sides AB and AC respectively.
For each of the following show that DE BC.
i) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm
ii) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm.
3.
In a triangle ABC, P and Q are points on sides AB and AC respectively, such that
PQ BC. If AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm, Find AB and PQ.
4.
In a triangle ABC, D and E are points on AB and AC respectively such that DE BC.
If AD = 2.4 cm, AE = 3.2 cm, DE = 2.0 cm and BC = 5.0 cm, find BD and CE.
SELF ASSESSMENT RUBRIC 9
CONTENT WORKSHEET (CW4)
Parameter
Able to apply the basic
proportionality theorem
Able to apply the
converse of basic
proportionality theorem
in solving problems
77
STUDENT’S WORKSHEET 10
CONTENT WORKSHEET (CW5)
PYTHAGORAS THEOREM
Name of Student________________
Date_____________
Below are some jumbled up words which state the rule applicable on Pythagorean
triplets. Rewrite it to form a meaningful sentence.
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
Verify whether the following triplets satisfy the above rule.
78
(3,4,5)
(5,12,13)
(9,40,41)
(7,24,25)
(8,15,17)
(11,60,61)
(12,35,37)
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
79
Observe the following figures, each having a right angled triangle. Regular polygons or
semicircles are made on the three sides of the triangles.
Figure
Area of I
(in terms
of side
‘a’)
Area of II (in
terms of side
‘b’)
Area of III
(in terms of
side ‘c’)
Relation between
the areas of I, II &
III (if any)
What can you conclude about the relationship between the sides of a right angled
triangle?
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
80
Self Assessment Rubric 10 – Content Worksheet (CW5)
Parameter
Able
to
write
the
Pythagoras theorem
STUDENT’S WORKSHEET 11
CONTENT WORKSHEET (CW6)
HANDS ON ACTIVITY TO VERIFY PYTHAGORAS THEOREM
Name of Student_______________
Date_____________
Hands on Activity:
Activity 1: Aim-By paper cutting and pasting verify Pythagoras theorem.
Pythagoras theorem states that in a right triangle square of hypotenuse is equal to the
sum of squares of other two sides.
Material Required - Coloured Paper, pair of scissors, Glue
Procedure 1. Draw a right triangle of dimension a, b and c . c is the hypotenuse.
2. Make 3 more such triangles.
3. Cut a square of side c units.
81
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.
82
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
83
Fig 2
Observations
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
Conclusion:
84
SELF ASSESSMENT RUBRIC 11
CONTENT WORKSHEET (CW6)
Parameter
Able to state Pythagoras
Theorem.
Able to verify Pythagoras
Theorem
using
explorations or models.
85
STUDENT’S WORKSHEET 12
CONTENT WORKSHEET (CW7)
APPLICATION OF PYTHAGORAS THEOREM
Name of Student________________
Date_____________
1. There is a staircase as shown in figure connecting points A and B.
Measurements of steps are marked in the figure. Find the straight distance
between A and B.
2. Show that the area of a rhombus on the hypotenuse of a right angled triangle with
one of the angles as 60⁰ is equal to the sum of the area of rhombus with one of their
angles as 60⁰, drawn on the other two sides.
86
3. An aeroplane leaves an airport and flies due north at a speed of 1800 km/h. At
the same time, another plane leaves the same airport and flies due west at a
speed of 1600 km/h. How far apart will the two planes be after 2.5 hours?
4. Find the lengths diagonals in the following figures.
5.
Determine the perimeter and area of the given figure.
6. In the following figure, justify the relationship pc = ab.
87
7. Prove that three times the sum of squares of the sides of a triangle is equal to four
times the sum of the squares of the medians of the triangles.
8. A man drives south along a straight road for 17 miles. Then turns west at right
angles and drives for 24 miles where from he turns north and continues driving
for 10 miles before coming to a halt. What is the straight distance from his
starting point to his terminal point?
SELF ASSESSMENT RUBRIC 12
CONTENT WORKSHEET (CW7)
Parameter
Able to state Pythagoras
Theorem.
Able to apply Pythagoras
Theorem in geometrical
problems.
88
STUDENT’S WORKSHEET 13
CONTENT WORKSHEET (CW8)
ICT Activity on Pythagoras Theorem
Name of Student________________
Date_____________
ICT Activity- (Geogebra Activity) – Pythagoras theorem
Aim: To verify that in a right angled triangle the sum of squares of the perpendicular
and base is equal to the square of the hypotenuse.
Previous Knowledge Assumed:
Concept of angles, circles and triangles
Basic knowledge of working on GeoGebra
Procedure:
1.
Draw a line slider using the tool (Slider)
and 3 as max.
2.
3.
4.
Draw a point B using the tool (New point)
.
Draw another point A using the same tool.
Let us use the slider to make point A as dynamic. Let us right click on the point A
and chose redefine. Now replace the Y coordinate with d.
5.
6.
Draw a line through points A and B using tool (Line through two points)
.
Draw a perpendicular line to line passing through point A and point B using tool
7.
(Perpendicular line)
. After selecting the tool click on line passing through
point A and B and the click above it. Point C is also formed in this process.
Let us now locate the intersection point of line AB and the line perpendicular to it
8.
using tool (Intersect two points)
. Point D is formed now.
Let us form a polygon out of the three points A, C and D using tool (polygon)
9.
. Name it as “d”. Provide -3 as min
. After selecting the tool, click on points A, C and D one by one.
Let us hide lines a, b and c by right clicking on the these lines and un-checking
show objects.
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10. Let us quickly draw semicircles around these points using the tool (Semicircle
through two points)
. After selecting the tool click on the two points A and C.
11. Similarly draw the semicircles around the line segments AD and DC.
12. Now draw point E on semicircle e using tool (New point)
.
13. Let us draw the points F and G on semicircles around CD and AD.
14. Lets us quickly draw circles using tool (Circle through three points)
points A, E and C. After selecting the tool click on points A, E and C.
15. Similarly draw the circles using points C, F & D and D, G & A.
16. Let us calculate the area of the circles using the tool (Area)
the circles and the area objects will appear.
using
. Just click inside
17. Now using tool (insert text)
, write “area of semicircle with diameter
AC = “ + area h/2
18. Similarly write for other two semicircles.
19. Let us hide the circles by right clicking on the objects in the algebra window and
un-checking show objects.
20. Now compare the sum of the areas of the semicircles. The sum of two is equal to the
third one. This verifies our theorem.
21. Repeat the activity with three different sets.
22. Note down the observations.
23. Write the final result.
Observations:
Serial
Number
First time
Second time
Third time
Area of
semicircle1
Area of
semicircle2
Area of
semicircle3
Observation
Result:
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
90
SELF ASSESSMENT RUBRIC 13
CONTENT WORKSHEET (CW8)
Parameter
Able to state Pythagoras
Theorem.
Able to verify Pythagoras
Theorem
using
explorations on GeoGebra
91
STUDENT’S WORKSHEET 14
CONTENT WORKSHEET (CW9)
CONVERSE OF PYTHOGORAS THEOREM
Name of Student___________
Date________
Rohan eplored one result on triangles . Read carefully.
Do you agree with Rohan? Comment
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
Now, fill the following table:
Numbers-(a, b, Verify,
Construct triangle with sides Is triangle a Write
c)
a2+b2= c2 a, b, c
right
the side
where c > a, b
angled
opposite
triangle?
to right
angle.
(3,4,5)
32+42
=9 + 16
=25
=52
Yes
92
5
(5,12, 13)
(4,5,7)
(6,8,10)
(10,12,15)
93
Based on your exploration in the above table, rewrite the jumbled up words given
below to form a meaningful sentence.
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
94
ICT Activity- Geogebra Activity – Converse of Pythagoras theorem
Aim: To verify, if in a triangle the sum of the squares of the two sides is equal to the
square of the third side then it is a right angled triangle.
Previous Knowledge Assumed:
Concept of angles, circles and triangles
Basic knowledge of working on GeoGebra
Procedure:
1.
Draw a line segment 3 cm long using tool (segment with given length from point)
.
2.
Draw a circle with radius 5 cm with point A as center, using tool (circle with center
and radius)
.
3.
Draw a circle with radius 4 cm with point B as center and using the same tool.
4.
Let us define the intersection point of the two circles using tool (Intersect two
objects)
5.
. Name it point C.
Let us draw line segments AC and BC using tool (Segment between two points)
.
6.
Now let us hide the two circles by going to the algebra window on the left hand
side. Right click on the circles c and d one by one and uncheck show object.
7.
Now let us measure the angle CBA by using tool (Angle)
and clicking on
point C, then point B and then point A. If it comes to 90 degrees the theorem is
verified.
8.
Repeat the activity with three different sets.
9.
Note down the observations.
10. Write the final result.
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Observations:
Serial Number
First time
Second time
Third time
Degree measure of the Angle
Observation
Result:
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
Do the following:
1. Which of the following triplets forms the sides of a right angled triangle? Support
your answer with proper reasoning.
(13, 12, 5)
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
(50, 60, 70)
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
96
(4, 6, 8)
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
(20, 60, 70)
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
(7, 14, 19)
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
2.
The sides of a triangle measure a, b, and c. Use the table to find the type of the
triangle.
97
SELF ASSESSMENT RUBRIC 14
CONTENT WORKSHEET (CW9)
Parameter
Able to state converse of
Pythagoras Theorem.
Able to verify converse of
Pythagoras Theorem.
Able to apply Pythagoras
Theorem in geometrical
problems.
98
STUDENT’S WORKSHEET 15
CONTENT WORKSHEET (CW10)
APPLICATION OF SIMILAR TRIANGLES
Name of Student________________
I.
Date_____________
Write the conditions for two triangles to be similar.
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.
99
Create your own design by using only similar triangles.
Design I
Design II
Design III
Design IV
II. Hands on Activity
III. Do the following:
1. What is the height of the telephone pole?
100
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
2.
Triangles RST is similar to triangle XYZ. Find all the missing measures.
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
101
3.
Determine whether each pair of triangles is similar. If they are similar, state the
similarity criteria.
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
4.
Find the value of x for each pair of similar triangles.
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
102
SELF ASSESSMENT RUBRIC 15
CONTENT WORKSHEET (CW10)
Parameter
Able to state and
understand all the
criterion of similaritySSS, AA,SAS
Able to verify all
criterions of similarity.
Able to apply the
similarity criterion in
problems.
103
STUDENT’S WORKSHEET 16
POST CONTENT (PCW1)
Name of Student________________
Date_____________
Practice assignment-BPT and its Converse
1.
In given Figure B‟C‟ BC.
Find the length AB. Justify your answer.
2.
If AB CD in each of the following figure, find x. Write the statement of
theorem used.
3.
In triangle ABC, DE BC and
4.
In the given figure,
A =
. If AC = 4.8 cm, find AE
P and AD = PM. Show that DM AP.
104
STUDENT’S WORKSHEET 17
POST CONTENT (PCW2)
Name of Student________________
Date_____________
Practice assignment: Similarity of triangles
1. A girl of height 90 cm is walking away from the base of a lamp post at a speed of 1.2
m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4
seconds.
2. A vertical pole of length 12m casts a shadow 8m long on the ground and at the same
time a tower casts a shadow 54m long. Find the height of the tower.
3. AM and PN are medians of triangles ABC and PQR, respectively. Also
similar to PQR. Show that
=
ABC is
.
4. Altitudes AD and CE of triangle ABC intersect at P. Which of the following hold
true?
a. Triangle AEP is similar to triangle CDP
b. Triangle ABD is similar to triangle CBE
c. Triangle AEP is similar to triangle ADB
d. Triangle PDC is similar to triangle BEC
5. Given that AB
CD, describe how do you know that triangle ABE is similar to
triangle CDE. Also, find the value of x.
105
STUDENT’S WORKSHEET 18
POST CONTENT (PCW3)
Name of Student_______________
Date_____________
Practice assignment- Pythagoras theorem and its converse
1. L and M are the mid points of AB and BC respectively of triangle ABC, right angled
at B. Prove that 4 LC2 = AB2 + 4BC2.
2. In triangle ABC, right angled at C, Q is the mid- point of BC. Prove that AB2 = 4 AQ2
– 3 AC2.
3. In the given figure, AD is perpendicular to BC and BD is one-third of CD. Prove that
2CA2 = 2AB2+ BC2.
106
4. In the given figure, a triangle ABC is right angled at B. The side BC is trisected at
points D and E. Prove that 8 AE2 = 3AC2 + 5AD2.
5. In the given figure, triangle ABC is right angled at C; and D is the mid- point of BC.
Prove that AB2 = 4AD2 – 3 AC2.
STUDENT’S WORKSHEET 19
POST CONTENT (PCW4)
Name of Student________________
Date____________
Do the following:
1.
Fill in the blanks
(a) All equilateral triangles are ________________ (Similar/Congruent)
(b) If ∆ABC ~ ∆FED then AB/BC = _________ .
(c) Circles with equal radii are ______________ (Similar/Congruent)
107
2.
In given figure
3.
Write the statement of Pythagoras Theorem.
4.
An airplane leaves an airport and flies due north at a speed of 1000 km per hour. At
the same time, another airplane leaves the same airport and flies due west at a
speed of 1200 km per hour. How far will be the two planes after one & half hours?
5.
In a ∆ABC right angled at C, AC = BC. Then AB2 = _____AC2
6.
Find the value of x in each of the similar triangles. Justify your answer.
7.
State whether the following pairs of polygons are similar or not:
8.
In the figure, the line segment XY is parallel to side AC of ∆ABC and it divides the
triangle into two equal parts of equal areas. Find the ratio .
9.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects
CD at F. Show that ∆ABE ≈ ∆CFB.
10. In ABC, AD
=
and AED = ABC. Show that AB = AC.
BC. Prove that AB2 - BD2 = BC2 - CD 2.
11. AD is a median of ABC. The bisector of
F respectively. Prove that EF BC.
108
ADB and
ADC meet AB and AC in E &
STUDENT’S WORKSHEET 20
POST CONTENT (PCW5)
Name of Student________________
Date____________
Solve the following crossword puzzle:
Across
1.
The ratio of areas of two similar triangles is equal to the ratio of the __________ on
their corresponding sides.
4.
If the corresponding sides of two triangles are ___________________, their
corresponding angles are equal and the two triangles are similar.
5.
In a triangle, if the square on one side is equal to sum of the squares on the other
two sides, the angle ___________ to the first side is a right angle.
8.
If in two triangles, the corresponding angles are equal, their corresponding sides
are proportional and the triangles are ___________ .
9.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in
distinct points, the other two sides are divided in the _____________ ratio.
109
Down
2.
In _________ triangle the square on the hypotenuse is equal to the sum of the
squares on the other two sides.
3.
Basic Proportionality theorem is also known as ____________ theorem.
6.
If a line divides two sides of a triangle in the same ratio, the line is ______________
to the third side.
7.
If one angle of triangle is equal to one angle of another triangle and the sides
including these angles are proportional, the two triangles are similar by
____________ similarity.
STUDENT’S WORKSHEET 21
POST CONTENT (PCW6)
Name of Student________________
Date_____________
Multiple Choice questions:
Choose the correct answer:
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
110
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
111
7.
8.
If ∆ABC ~ ∆DEF, ar (∆DEF) = 100 cm2 and AB/DE= 1/ 2 then ar (∆ABC) is
(i) 50 cm2
(ii) 25 cm2
(iii) 4 cm2
(iv) None of the above.
If the three sides of a triangle are a, √3a, √2a, then the measure of the angle opposite
the longest side is
(i) 45⁰
(ii) 30⁰
(iii) 60⁰
(iv) 90⁰
9.
In a ∆ABC, point D is on side AB and point E is on side AC such that DE BC.
If AD = 2x – 3, DB = x – 1, AE = 5x – 7 and EC = 2(x –1), then the value of x is
(a) –1
(b) 1 or –1/2
(c) 1
(d) None of these
10 Bisector AD of A of ∆ABC meets base BC at D. If AB = 10cm, BC = 8cm and
AC = 6cm, the length of BD is
(a) 7 cm
(b) 5cm
(c) 4cm
(d) 2cm
11. The areas of two similar triangles ABC and PQR are respectively 64 cm2 and 121
cm2. If QR = 15.4 cm, then BC is
(a) 11.2cm
(b) 8cm
(c) 11cm
(d) 15.4cm
12. In two similar triangles ABC and PQR, if their corresponding altitudes AD and PS
are in the ratio of 4:9, then, ar (∆PQR)/ar(∆ABC) is
(a) 16:81
(b) 81:16
(c) 9:4
(d) 4:9
13. In ∆ABC, PQ is a line segment which cuts AB and AC at P and Q respectively such
that PQ BC and it divides ∆ABC into two equal parts. Then BP/AB is equal to
(a)
(b)
(c)
112
(d)
STUDENT’S WORKSHEET 22
POST CONTENT (PCW7)
Name of Student_______________
Date____________
Application of similar triangles in daily life
Aim: To find height of any object using pocket height finder.
Material Required: Cardboard, color sheet, thread, punching machine, marker, ruler,
glue.
Previous knowledge: criteria for similarity of triangles.
Procedure:
1.
Cut a card measuring 6 inches by 1 inches and mark off in inches.
2. Thread a piece of string through a hole as near to the bottom as possible, and make
a knot so that it cannot slip through.
3.
Cut the string so that it is 18 inches long after knotting.
4.
The object to be measured must be covered exactly by the card when it is held out
and the string pulled tight to the eye.
5.
The ratio of the length of string to the height of the card will be the same as the
distance to the height of the object.
Observation:
1.
Two imaginary similar triangles are formed. First is formed by height finder, string
and line of sight and second is formed by object, distance between object and man
and line of sight.
2.
The ratio using these measurements is 1:3 and the height required is its distance
multiplied by
.
Conclusion:
We can find the height of any object using pocket height finder.
113
EXTRA READINGS:
Pythagoras Theorem:
http://www.cut-the-knot.com/pythagoras/
http://en.wikipedia.org/wiki/Pythagorean_theorem
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
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://www.youtube.com/watch?v=Ng2EpkKooo4
114
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-Trianglesto-the-Real-World
http://www.mathopenref.com/similartriangles.html
115