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6 9 Class VII CBSE-i Unit-7 mathematics UNDERSTANDING 1 shAPES 5 4 Student's Section 2 5 1 4 3 0 9 Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 092 India CBSE-i mathematics UNDERSTANDING shAPES Preet Vihar,Delhi-110 092 India The CBSE-International is grateful for permission to reproduce and/or translate copyright material used in this publication. The acknowledgements have been included wherever appropriate and sources from where the material may be taken are duly mentioned. In case any thing has been missed out, the Board will be pleased to rectify the error at the earliest possible opportunity. All Rights of these documents are reserved. No part of this publication may be reproduced, printed or transmitted in any form without the prior permission of the CBSE-i. This material is meant for the use of schools who are a part of the CBSE-International only. 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 outside India. 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 nonevaluative 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 Advisory Conceptual Framework Shri Vineet Joshi, Chairman, CBSE Dr. Sadhana Parashar, Director (Training), Shri G. Balasubramanian, Former Director (Acad), CBSE Ms. Abha Adams, Consultant, Step Dr. Sadhana Parashar, Director (Training), Ideators VI-VIII Ms Aditi Mishra Ms Guneet Ohri Ms. Sudha Ravi Ms. Himani Asija Ms. Neerada Suresh Ms Preeti Hans Ms Neelima Sharma Ms. Gayatri Khanna Ms. Urmila Guliani Ms. Anuradha Joshi Ms. Charu Maini Dr. Usha Sharma Prof. Chand Kiran Saluja Dr. Meena Dhani Ms. Vijay Laxmi Raman Material Production Groups: Classes VI-VIII English : Physics : Mathematics : Ms Neha Sharma Ms. Vidhu Narayanan Ms. Deepa Gupta Ms Dipinder Kaur Ms. Meenambika Menon Ms. Gayatri Chowhan Ms Sarita Ahuja Ms. Patarlekha Sarkar Ms. N Vidya Ms Gayatri Khanna Ms. Neelam Malik Ms. Mamta Goyal Ms Preeti Hans Ms. Chhavi Raheja Biology: Ms Rachna Pandit Mr. Saroj Kumar Ms Renu Anand Hindi: Ms. Rashmi Ramsinghaney Ms Sheena Chhabra Mr. Akshay Kumar Dixit Ms. Prerna Kapoor Ms Veena Bhasin Ms. Veena Sharma Ms Trishya Mukherjee Ms. Seema Kapoor Ms. Nishi Dhanjal Mr. Manish Panwar Ms Neerada Suresh Ms. Kiran Soni Ms. Vikram Yadav Ms Sudha Ravi Ms. Monika Chopra Ms Ratna Lal Ms Ritu Badia Vashisth Ms. Jaspreet Kaur CORE-SEWA Ms Vijay Laxmi Raman Ms. Preeti Mittal Ms. Vandna Ms. Shipra Sarcar Ms.Nishtha Bharati Chemistry Ms. Leela Raghavan Ms.Seema Bhandari, Ms. Poonam Kumar Ms. Seema Chopra Mendiratta Ms. Madhuchhanda Ms. Rashmi Sharma MsReema Arora Ms. Kavita Kapoor Ms Neha Sharma Ms. Divya Arora Ms. Sugandh Sharma, E O Mr. Navin Maini, R O (Tech) Shri Al Hilal Ahmed, AEO Ms. Anjali, AEO Shri R. P. Sharma, Consultant (Science) Mr. Sanjay Sachdeva, S O Coordinators: Dr. Srijata Das, E O Dr Rashmi Sethi, E O (Co-ordinator, CBSE-i) Ms. Madhu Chanda, R O (Inn) Mr. R P Singh, AEO Ms. Neelima Sharma, Consultant (English) Ms. Malini Sridhar Ms. Leela Raghavan Dr. Rashmi Sethi Ms. Seema Rawat Ms. Suman Nath Bhalla Geography: Ms Suparna Sharma Ms Aditi Babbar History : Ms Leeza Dutta Ms Kalpana Pant Ms Ruchi Mahajan Political Science: Ms Kanu Chopra Ms Shilpi Anand Economics : Ms. Leela Garewal Ms Anita Yadav CORE-Perspectives Ms. Madhuchhanda, RO(Innovation) Ms. Varsha Seth, Consultant Ms Neha Sharma Ms.S. Radha Mahalakshmi, EO CONTENT 1. Study Material 1 2. Student's support material (Student's worksheets) 41 C SW 1: Warm Up Activity W1 42 Recall parts of an angle C SW 2: Warm Up Activity W2 45 Different types of angles C SW 3: Pre Content Worksheet P1 48 Different types' of pairs of angles C SW4: Pre Content Worksheet P2 50 Angles Puzzle C SW5: Content Worksheet C1 51 Angles and angles C SW 6: Content Worksheet C2 54 Angles in nature C SW 7: Content Worksheet C3 59 Angle sum property of a triangle C SW 8: Content Worksheet C4 63 Applying Angle sum property of a triangle C SW 9: Content Worksheet C5 67 Exterior angle property of a triangle C SW 10: Content Worksheet C6 Exterior angle skill drill 69 C SW 11: Content Worksheet C7 71 Triangle inequality property C SW 12: Content Worksheet C8 74 Pythagoras property for a right angled triangle C SW 13: Content Worksheet C9 76 Application of Pythagoras Theorem C SW 14: Post Content Worksheet Pc1 79 Independent Practice C SW 15: Post Content Worksheet Pc2 81 Test your progress Acknowledgments C 88 Suggested videos/ links/ PPT's C 89 STUDY MATERIAL 1 Understanding shapes Introduction In the Class VI, you have learnt some basic concepts of geometry such as a point, line, ray, line segment, angles and their classifications, triangles and their classifications, etc. In this unit, we shall first briefly review the angles, and their types and then extend the study to certain pairs of angles. After this, you will be introduced to different angles formed by a transversal with two lines. Relationship between these angles will also be discussed in the case when a transversal intersects two parallel lines. In this unit, we shall also discuss some simple properties of triangles including Pythagoras theorem. 1. Angles : A Review Angles : Angle is a figure formed by two rays with a common initial point. The common initial point is called the vertex and the two rays forming the angle are called its arms (See Fig 1) Fig. 1 POQ is an angle whose vertex is O and arms are rays OP and OQ. An angle is measured in degrees. Types of angles : (i) Acute angle : An angle whose measure is more that O0 and less than 900 is called an acute angle. In Fig. 2, POQ is an acute angle. 2 Fig. 2 (ii) Right angle : An angle whose measure is 900, is called a right angle (See Fig. 3). POQ is a right angle. Fig. 3 (iii) Obtuse angle: An angle whose measure is more than 900 and less than 1800, is called an obtuse angle. (See Fig.4) Fig. 4 (iv) Straight angle: An angle whose measure is 1800, is called a straight angle. (See Fig. 5) Fig. 5 (v) Reflex angle : An angle whose measure is more than 1800 and less than 3600 is called a reflex angle (See Fig.6) 3 Fig. 6 (vi) Complete angle : An angle whose measure is 3600, is called a complete angle (See Fig. 7) Fig. 7 (vii) Zero angle : An angle whose measure is 00, is called a zero angle. Fig. 8(i) 2. Pairs of Angles : Two angles are related to each other in many ways. We discuss some of these relations below: (i) Complementary angles : Two angles are said to be complementary if the sum of their measures is 900. For example, the angles of measures 270 and 630 are complementary as 270 + 630 = 900. Similarly, angles of measures 350 and 550 are also complementary angles as 350 + 550 = 900, and so on. But angles of measures 370 and 630 are not complementary angles as 370 + 630 = 1000 900. When two angles are complementary each angle is said to be complement of the other. Thus the angle of measure 270 is the complement of the angle of measure 63o and vice-versa. Similarly, the angle of measure 350 is complement of the angle of measure 550 and vice-versa. 4 Fig. 8(ii) (ii) Supplementary angles Two angles are said to be supplementary if the sum of their measures is 1800. For example, the angle of measure 700 and the angle of measure 1100 are supplementary angles as 700 1100 = 1800. Similarly, the angle of measure 530 and 1270 are also supplementary angles as 530 + 1270 = 1800 and so on. But angles of measure 1250 and 650 are not supplementary angles as 1250 + 650 = 1900 1800. Fig. 9 5 If two angles are supplementary, then each angle is said to be supplement of the other. Thus, the angle of measure 700 is the supplement of the angle of measure 1100 and vice-versa. Similarly, the angle of measure 530 is supplement of the angle of measure 1270 and vice-versa. (iii) Adjacent angles Two angles are said to be adjacent if (i) they have a common vertex, (ii) they have a common arm, and (iii) their non common arms are on the opposite side of the common arm (See Fig. 10). Fig. 10 For example in Fig.10, POQ and ROP are adjacent angles as they have a common vertex O, common arm OP and their non common arms OQ and OR are on opposite sides of the common arm OP. However, ROP and ROQ are not adjacent as they have a common vertex O, common arm OR but their non common arms OP and OQ are on the same side of common arm OR. (iv) Linear Pair Two angles are said to form a linear pair if (i) They are adjacent angles (ii) their non common arms are in the same line, i.e., they form opposite rays. 6 Fig. 11(i) Angles 1 and 2 in Fig.11 (i) form a linear pair as (i) they are adjacent angles (ii) their non common arms OR and OQ are opposite rays. Angles 1 and 2 in Fig. 11(ii) do not form a linear pair as Fig. 11(ii) (i) they are adjacent angles, but (ii) their non-common arms are not opposite rays. You can observe that angles of a linear pair are always supplementary. But the converse is not true. For example, angles in Fig. 12, are supplementary but they do not form a linear pair. 7 Fig. 12 (v) Vertically opposite angles. In Fig. 13, two lines intersect each other at the point O forming four angles 1, 2, 3 and 4. 1 and 2 are adjacent angles. What about adjacent. 1 and 3? They are not These angles are called vertically opposite angles. Similarly, 2 and 4 are also vertically opposite angles. Fig. 13 1 and 3, vertically opposite angles 4 and 2, vertically opposite angle Activity 1 : In Fig. 13, Measure 1 and 3. Also measure 2 and 4. 8 Repeat this activity by drawing two more pairs of intersecting lines and name them as before. Fig. 14 Fig. 15 Complete the following table: Figure 1 3 2 4 13 14 15 What do you observe? Each time, you will find that 1= 3 and 2= 4 In fact, it is true in general also. 9 Is 1= 3 Is 2= 4? Thus, we can say that If two lines intersect each other, then vertically opposite angles are equal We now consider some examples to explain these concepts. Example 1: State whether the following statements are True or False: (i) Angles of measures 180 and 1620 are complementary (ii) Angles of measures 290 and 1510 are supplementary (iii) Two right angles are always supplementary (iv) Two acute angles are always supplementary (v) Angels of measure 450 and 450 are complementary angles. (vi) Two adjacent angles always form a linear pair. (vii) Vertically opposite angles are equal. (viii) Vertically opposite angles are adjacent. (ix) Two supplementary angles always form a linear pair. Solution: (i) False as 180+1620 = 1800 (ii) True (iii) True (iv) False (v) True (vi) False (vii) True 900 (viii) False (ix) False Example 2: Find: (i) Complement of the angle whose measure is 470. (ii) Supplement of the angle where measure is 770. 10 Solution: (i) Complement of the angle of 470 = (900 470) = 430 (ii) Supplement of the angle of 770 = (1800 770) = 1030 Example 3: Which of the following pairs of angles are adjacent? Fig. 16 Solution: (i) 1 and 2 are not adjacent as they do not have a common vertex. (ii) 1 and 2 are adjacent. (iii) 1 and 2 are adjacent. (iv) 1 and 2 are not adjacent as they do not have common vertex. Example 4: Find the values of a, b, c, 11 Fig. 17 Solution: (i) a = 410 (vertically opposite angles equal) a + c = 1800 (Linear pair) So, c = 1800 410 = 1390 b = 1390 (vertically opposite angles) (ii) b = 660 (vertically opposite angles) (a + 350)+b = 1800 (Angles at a point on a straight line) or a + 350+660 = 1800 or a + 1010 = 1800 or a = 1800 1010 = 790 c + 660 = 1800 (Linear pair) So, c = 1800 660 = 1140 12 3. Intersection of Lines by a Transversal Transversal A line which intersects two or more lines in distinct points is called a transversal. Fig. 17 In Fig 17 (i), line p intersects two lines and m in two points A and B (distinct points) and therefore, p is a transversal to and m. Also, in (ii), line p intersects three lines , m and nth three points A, B and C (distinct points) and, therefore, p is a transversal to lines , m and n. In (iii), line p intersects three lines, , m and n in two points A and B. not distinct points and therefore, p is not a transversal to , m and n. Similarly, in (iv), p is not a transversal to and m because it intersects two lines in one point A only. Angles made by a Transversal with two lines Look at Fig.18, in which a transversal p intersects two lines A and B. and m at points In all, eight angles ( 1, 2, 3, 4, 5, 6, 7, and 8) have been formed. These angles are related to each other in some way. Let us see how! 13 Fig. 18 (i) Exterior Angles and Interior Angles 1, 2, angles. (ii) 7, and 8 are exterior angles, while 3, 4, 3, and 6, are interior Corresponding Angles 1, and 5, are two angles formed at different vertices A and B and they lie on the same side of the transversal p. In these two angles, one angle is interior angle and the other is exterior angle. Such a pair of angles is called a pair of corresponding angles. 2, and 6, are also corresponding angles. Similarly, other pairs of corresponding angles are: 3, and 7, 4, and 8. (iii) Alternate Interior Angles and Alternate Exterior Angles In Fig 18, 3, and 5, are interior angles lying on the opposite of the transversal p. They are known as alternate interior angles. Another pair sides alternate interior angles is 4, and 6. 2, and 8, are exterior angles lying on the opposite sides of the transversal p. They are known as alternate exterior angles. Another pair of alternate exterior angles is 1, and 7. Sometimes, the phrase „alternate angles‟ is used to mean alternate interior angles. (iv) Interior Angles on the same side of the Transversal 4, and 5, are two interior angles that lie on the same (right) side of the transversal p. They are known as interior angels on the same side of the transversal. Another pair of interior angles on the same side of the transversal is 3 and 6. 14 Sometimes, the phrase „co-interior angles‟, consecutive angles or allied angles‟ is also used to mean‟ “interior angles on the same side of the transversal”. 4. Intersection of Parallel lines by a Transversal Recall that two lines in a plane are said to be parallel, if they do not intersect each other. When two parallel lines are intersected by a transversal, then the pairs of different angles exhibit some intersecting properties. Let us explore. Activity 1: Draw two parallel lines and m using the opposite edges of a ruler or the opposite edges of a geometry box (Fig.19) Fig. 19 Now, draw a transversal p intersecting these lines at A and B. Measure 1, and 5, 2, and 6, 4, and 8, and 3, and 7. What do you observe? Repeat this activity two more times, naming the lines and angles in the same manner as before. Record your observations in the form of a table as given below: S. 1 5 2 6 4 8 3 7 Is No 1 = 5? 1. 2. 3. From the above observations, you may say that: 15 Is 2 = 6? Is 4 = 8? Is 3 = 7? If two parallel lines, are intersected by a transversal, then angles in each pair of corresponding angles are equal. Now, measure 3 and 5, angle 4 and 6 of Fig. 19. What do you observe? Repeat the activity two more times as before and record your observations in the form of a table as given below: S. No 3 5 4 6 Is 3 = 5? Is 4 = 6? 1. 2. 3. From the above observations, you may say that If two parallel lines are intersected by a transversal, then angles in each pairs of alternate interior angles are equal. Now, measure 4 and 5 and 3 and 6. What do you observe? Repeat the activity two more times as before and record your observations in the form of a table as given below: S. No 4 5 3 6 Is 4 + 5 = 1800 Is 3 + 6 = 1800 1. 2. 3. From the above observations, you may say that: If two parallel lines are intersected by a transversal, then sum of the two interior angles on the same side of the transversal is 1800, i.e., the two angles are supplementary. We may summarise the above discussion in the following form. 16 If two parallel lines are intersected by a transversal, then (i) Each pair of corresponding angles is equal; (ii) Each pair of alternate interior angles is equal. (iii) Sum of the interior angles on the same side of the transversal is 1800. Activity 2: Now draw any two lines and m such that they are not parallel to each other. Then, draw a transversal p (Fig 20). Fig. 20 Now, measure each pair of corresponding angles ( 1 and and 3 and 7). 5, 2 and 6, 4 and 8 Are they equal? Repeat the activity two more times. What do you observe? You may say that If two non-parallel lines are intersected by a transversal, then no pair of corresponding angles is equal. After this, measure each pair of alternate interior angles ( 3 and 5 and 4 and 6). Are they equal? Repeat the activity two more time. What do you observe? You may say that: If two non-parallel line are intersected by a transversal, then no pair of alternate interior angles is equal. 17 Now, measure each pair interior angles on the same side of the transversal ( 4 and 5 and 3 and 6). Is 4 + 5 = 180 ? Is 3+ 6 = 1800 ? Repeat the activity two more times. What do you observe? You may say that: If two non-parallel lines are intersected by a transversal, then in none of the pair of interior angles on the same side of the transversal, their sum is 1800. We can summarise the above results as follows: If two non-parallel lines are intersected by a transversal, then (i) angles in pairs of corresponding angles, are not equal, (ii) angles in pairs of alternate interior angles are not equal and (iii) sum of the angels in pairs of interior angles on the same side of the transversal is not 1800. In view of the above discussion, we say now say that: If two lines are intersected by a transversal and if any of the following three conditions is satisfied: (i) angles in any pair of corresponding angles are equal. (ii) angles in any pair of alternate interior angles are equal. (iii) the sum of the interior angles on the same side of the transversal is 1800, then the two lines are parallel. We now take some examples to illustrate the use of these results/properties. 18 Example 5: In Fig. 21, if find the values of x and y. Fig. 21 Solution: x+130o = 180o (Linear pair) So, x = 180o – 130o = 50o Now, y = x (Alternate interior angle) So, y = 50o Thus, values of x and y are 50o each. Example 6: In Fig.22, if p q, find all the angles, 1, 2, 3, 4, 5, 6 and 7. Fig. 22 Solution: 1 + 700 = 1800 (Linear pair) So, 1 = 1800 700 = 1100 4 = 700 (Corresponding angles) 19 5 = 1 (Corresponding angles) So, 5 = 1100 2 = 4 (Alternate interior angles) So, 2 = 700 3 + 700 = 1800 (Linear pair) So, 3 = 1800 700 = 1100 7 = 3 (Corresponding angles) So, 7 = 1100 Also, 6 = 2 (Corresponding angles) So, 6 = 700 Thus, 1 = 1100, 2 = 700, 3 = 1100, 4 = 700, 5 = 1100, 6 = 700, and 7 = 1100 Note : Question can be solved in different ways. Example 7: Are lines p and q in Fig.23 parallel? Fig. 23 Solution: Lines p and q are intersected by a transversal t. Further, sum of interior angles on the same side of the transversal = 1050 + 720 = 1770 20 It is not equal to 1800. So, lines p and q are not parallel. Example 8: In Fig.24, if find the values of x and y in each case. Fig. 24 Solution: In (i) ; y = 1300 (Corresponding angles) and x = 1100 (Alternate interior angles) In (ii) ; x = 500 (Alternate interior angles) and y = 800 (Corresponding angles) Example 9: In Fig. 25, if BA ED and BC EF, find DEF. Fig. 25 21 Solution: BA ED (Given) So, DPC = ABC = 800 (Corresponding angles) Also, BC EF (Given) So, DEF = DPC = 800 (Corresponding angles) Example 10: In Fig.26, if ED QP and EF QR, find DEF. Fig. 26 Solution: EF QR (Given) So, PQR + QSF = 1800 (Interior angles on the same side of the transversal) or or Also, 1100 + QSF = 1800 QSF = 1800 1100 = 700 (1) ED QP (Given) So, DEF = QSF (Corresponding angles) or DEF = 700 [From (1)] Example 11: When a transversal intersects two lines and m, then the interior angles on the same side of the transversal are equal. Is it true that the two lines will always be parallel? If no, under what condition the lines will be parallel? Solution: No, because for lines to be parallel, the sum of the two interior angles on the same side of the transversal is 1800. Thus, lines will be parallel only when each equal interior angles is 900. 22 5. Triangle and its Properties You have already learnt about triangles and their classifications in Class VI. Recall that triangle is a polygon made up of three line segments. In Fig. 27, ABC is a triangle with vertices A, B and C. It has three sides AB, BC and CA and three angles A, B and C. Fig. 27 You may also recall that triangles are classified in two ways as follows: (i) According to sides : (a) Equilateral Triangle: A triangle with all three sides equal is known as an equilateral triangle. (b) Isosceles Triangle: A triangle having two sides equal is known as an isosceles triangle. (c) Scalene Triangle : A triangle having no two sides equal is known as a scalene triangle. Fig. 28 23 (ii) According to Angles : (a) Acute angled Triangle: A triangle having all angles acute is known as an acute angled triangle or simply acute triangle. (b) Right angled triangle: A triangle with one angle a right angle is known as a right angled triangle or simply a right triangle. (c) Obtuse angles triangle: A triangle with one angle an obtuse angle is known as an obtuse angled triangle or simply an obtuse triangle. Fig. 29 Angle Sum Property of a Triangle Activity : Draw a triangle on a thick sheet of paper and mark it angles as 1, 2 and 3 [Fig.30 (i)]. Cut out the triangle and then cut it into three parts such that each part represents one of its angles [Fig. 30]. Now, rearrange these three parts at a point O such that any two angles are adjacent to each other and there is no overlapping as shown in Fig.30 (iii). Fig. 30 24 What do you observe? You will observe that these three angles together make a straight angle. Repeat the activity by drawing two more triangles. Each time, you will find that the three cut out angles form a straight angle. So, you may say that 1+ 2+ 3 = 1800 In other words, the sum of the three angles of a triangle is 1800. You can also arrive at the above property by drawing different triangles ABC, measuring angles, of each triangle and then finding their sum. Each time, you will see that A+ B+ C = 1800. The above property can be better understood in the following way; Let ABC a triangle and let be a line, drawn through A and parallel to BC. Angles has been marked as shown in Fig.31. Fig. 31 Now, you may say that 2 = 4 (Alternate interior angle) (1) 3 = 5 (Alternate interior angles) (2) Adding the above two, we set 2 + 3= 4+ 5 So, 1 + 2 + 3 = 1 + 4 + 5 (3) 25 But 1 + 4 + 5 = 1800, because they form a straight angle. So, from (3), we get 1 + 2 + 3 = 1800 Note that in the above method, a reason has been provided for each statement. Obtaining a property through reasoning is known as a proof. This property is usually known as the Angle Sum Property of a Triangle. Let us explain the use of this property through some examples. Example12: Two angles of a triangle ABC are of measures 750 and 350 (Fig.32). Find the third angle of the triangle. Fig. 32 Solution: By angle sum property, A+ B+ C = 1800 or , 350 + 750 + C = 1800 or, 1100 + or C = 1800 C = 1800 1100 = 700 Thus, measure of the third angle is 700. Example 13: The ratio of the angles P, Q and R of a PQR is 2 : 3 : 4. Find the angles. Solution: Let P = 2x, So, Q = 3x and R = 4x. Therefore, by angle sum property of a triangle, 2x+3x+4x = 1800 or, 9x = 1800 26 or, x= = 200 So, P = 2x = 2 x 200 = 400, Q = 3x = 3 x 200 = 600, R = 4x = 4 x 200 = 800. Example 14: Is it possible to have triangle in which (i) two angles are acute? (ii) two angles are right? (iii) two angles are obtuse? (iv) all angles are greater than 600? (v) each angle is equal to 600? Solution: (i) Yes. for example, 400, 600, 800. (ii) No. sum cannot be more than 1800. (iii) No. sum cannot be more than 1800. (iv) No. sum cannot be more than 1800. (v) Yes. sum is equal to 1800. Exterior Angle Property Consider a triangle ABC. Produce its side BC to form ray BD as shown in Fig.33. Fig. 33 Observe that an angle ACD has been formed in the exterior of the triangle. This angle is adjacent to 3 of ABC. 27 An angle formed in the exterior of the triangle and adjacent to one of its angles is called an exterior angle of the triangle. Thus, ACD is an exterior angle of the ABC. (Fig.33). Note that by producing side AC, another exterior angle 5 is formed at C, because it is adjacent to 3 of ABC. Fig. 34 However, 6 is not an exterior angle of ABC as angles. 3 and 6 are vertically opposite 1 and 2 (angles other than 3 which is adjacent to exterior angle 4) are known as interior opposite angles (or remote interior angles) corresponding to exterior angle ACD. Measure 1, 2 and 4. Then find the sum 1+ 2. Repeat this activity by taking two more triangles ABC and producing side BC. Now, complete the following table. S. No 1 2 1+ 2 4 Is 4= 1 + 2? 1. 2. 3. What do you observe? You will find that in each case, 28 Is 4= 6? 4= 1+ 2 Thus, you may say that: An exterior angle of a triangle is equal to the sum of its two interior opposite angles. The above property is known as the exterior angle property of a triangle. We can also obtain the exterior angle property of the triangle as follows. 1+ 2+ 3 = 1800 (Angle sum property) (1) Also, 3+ 4 = 1800 (Linear pair) (2) So, from (1) and (2), 3+ 4= 1+ 2+ 3 or, 4= 1+ 2+ or, 4= 1+ 2 3 3 Here, also, we have provided a proof for the above property. It is also obviously clear from the exterior angle property of triangle that an exterior angle of a triangle is greater than either of its interior opposite angles. Thus, 4> 1 and 4> 2. Let us explain the use of this property through some examples. Example 15: In Fig.35, find the value of x. Fig. 35 29 Solution: x + 700 = 1500 (Exterior angle property) or x = 1500 700 = 800 Thuse, value of x is 800 Example 16: Find the value of x and y in Fig.36. Fig. 36 Solution: x + 1300 = 1800 (Linear pair) or, x = 1800 1300 = 500 Now, y = 600 + x (Exterior angle property) or, y = 600 + 500 = 1100 Thus, values of x and y are 500 and 1100 respectively. Example 17: One of the exterior angles of a triangle is 700. If its interior opposite angles are equal, find the angles of the triangle. Solution: Let each of the interior opposite angles be x. So, x + x = 700 (Exterior angle property) or 2x = 700 or x = = 350 So, two angles of the triangle are 350 and 350. Therefore, the third angle of the triangle = 1800 350 350 = 1100. Sum of Any two sides of a Triangle 30 Activity : Draw any triangle ABC and measure its three sides AB, BC and CA. Now find AB + BC, BC + CA and CA + AB. Compare AB+BC with CA, BC+CA with AB and CA+AB with BC. Fig. 37 Repeat the activity by drawing two more triangles and labeling them as ABC. Write your observations in the form of a table as given below: S. AB No. BC CA AB+BC Is BC+CA AB + BC > CA? Is BC + CA >AB? CA+AB 1. 2. 3. What do you observe? You will find that in each case, AB+BC > CA, BC+CA >AB and CA+AB>BC. So, you may say that: Sum of any two sides of a triangle is greater than the third side. This property is known as Triangle Inequality Property. You have been above that in ABC, AB+BC > CA So, BC > CA AB, i.e., CA AB < BC Also, BC+CA > AB So, CA > AB BC, i.e., AB BC < CA Similarly, CA – BC < AB. 31 Is CA+AB> BC? In view of the above, we may say that: Difference of any sides of a triangle is less than the third side. We now take some examples to illustrate the use of the above properties. Example 18: State which of the following can be the possible lengths (in cm) of a triangle. (i) 10, 12, 6 (ii) 7.5, 2.5, 5 (iii) 3, 8, 4 (iv) 5, 9, 7 Solution: (i) We have: 10+12 = 22 and 22 > 6, 12 + 6 = 18 and 18 > 10 Also, 6+10 = 16 and 16 > 12 Thus, sum of two sides is greater than the third side in each case. Hence, it is possible to have a triangle. (ii) 7.5 + 2.5 = 10 and 10 > 5, 7.5 + 5 = 12.5 and 12.5 > 2.5. But 2.5 + 5 = 7.5 and 7.5 = 7.5 That is, sum of the sides 2.5 and 5 is equal to the third side 7.5. Hence, it is not possible to have a triangle. (iii) 3+8 = 11 and 11 > 4, 8+4 = 12 and 12 > 3 But 3 + 4 = 7 and 7 < 8 Hence, it is not possible to have a triangle (iv) 5+9 = 14 and 14 > 7, 9+7 = 16 and 16 > 14 Also 7+5 = 12 and 12 > 9. 32 So, it is possible to have a triangle Example 19: Lengths of two sides of a triangle are 9cm and 12cm. Between which two lengths the third sides of triangle can be? Solution: Sum of the two sides = (9+12)cm = 21cm So, 21cm > Third side Therefore, third side must be less than 21cm Also, difference of two sides = 12cm 9cm = 3 cm So, 3 cm < Third side Therefore third side must be greater than 3cm. Thus, length of the third side lies between 3cm and 21cm In other words, 3cm < third side < 21cm Example 20: In Fig. 38, ABCD is a quadrilateral and AC and BD are its diagonals. Is AB+BC+CD+AD > AC+ BD? Give reasons. Fig. 38 Solution: Yes. From ABC, AB+BC > AC (Sum of two sides greater than third side) (1) From BCD, BC+CD > BD (Why?) (2) From ADC, AD+CD > AC (Why?) (3) From (4) ABD, AB+AD > BD (Why?) Adding (1), (2), (3) and (4), we get AB+BC+BC+CD+AD+CD+AB+AD > AC+BD+AC+BD or 2(AB+BC+CD+AD) > 2(AC+BD) i.e., AB+BC+CD+AD > AC+BD 33 6. Pythagoras Theorem Recall that a triangle is called a right angled triangle or right triangle, if one of its angles is a right angle. Side opposite the right angle is called its hypotenuse. In Fig.39, ABC is right triangle right angled at B. Clearly, its hypotenuse is AC. Fig. 39 Activity: Draw 3 different right triangle. Name each triangle as ABC, right angled at B. Measure, AB, BC, CA of each triangle. Now find AB2, BC2 and CA2, and complete the following table: Triangle AB BC CA AB2 BC2 CA2 AB2+BC2 Is AB2+BC2 = AC2? 1. 2. 3. What do you observe? You will find that in each case, AB2 + BC2 = AC2 i.e, the square on the hypotenuse = Sum of squares on other two sides In fact, this is true for all the right triangles. This property may be stated as follows: 34 In a right triangle, The square on the hypotenuse = Sum of the squares on the other two sides (legs) This property is known as Pythagoras theorem. This property is named after Pythagoras, a Greek mathematician of 6th century BC. In fact, this property was also known to ancient Indians. The Indian mathematician Baudhayan (800 BC) has stated this theorem in the following form: “The square described on the diagonal of a rectangle has an area equal to the sum of the areas of the squares described on its two sides” The above property can also be verified through the following activity: Activity: 1. Make eight identical copies of a right triangle with hypotenuse ‘c’ and sides a and b units. 2. Arrange these triangle in two different ways in a square of side (a+b) as shown below (Fig 40) Fig. 40 The Converse of Pythagoras Theorem is also true, i.e., If in a triangle, square of a side is equal to the sum of squares of other two sides, then angle opposite the first side is a right angle, that is, it is a right triangle. You have seen in a right ABC, right angled at B, AC2 = AB2 + BC2 So, AC2 > AB2 and AC2 > BC2 35 which means AC > AB and also AC > BC i.e., hypotenuse is the longest side of a right triangle Let us explain the use of Pythagoras theorem through some example: Example 21: The length of the sides (legs) of a right triangle are 5 cm and 12 cm. Find length of its hypotenuse. Solution: Let PQR be a right triangle in which PQ = 5 cm, QR = 12 cm and Fig. 41 Using Pythagoras theorem, PR2 = PQ2+QR2 = 52 + 122 = 169 = 132 So, PR = 13 cm Hence, the length of hypotenuse of PQR = 13 cm. 36 Q = 900. (Fig. 41) Example 22: In ABC, B is a right angle and AB = 6 cm, AC = 10 cm. Find BC. Solution: As ABC is a right triangle with B = 900, So, hypotenuse is AC. Using Pythagoras theorem AC2 = AB2 + BC2 or, 102 = 62 + BC2 or, 102 62 = BC2 or, 100 36 = BC2 or, 64 = BC2 or, 82 = BC2 or, BC = 8 Thus, side BC = 8 cm Example 23: Determine whether the triangles with lengths of side given below are right triangles: (i) 7 cm, 24 cm, 25 cm (ii) 4.5 cm, 6 cm, 7.5 cm (iii) 2.5 cm, 6 cm, 7.2 cm Solution: (i) Let a = 7, b = 24, c = 25 a2 = 49, b2 = 242 = 576, c2 = 252 = 625 Since 625 = 49 + 576 i.e., c2 = a2+b2 So, the given triangle is a right triangle. (ii) Let a = 4.5, b = 6, c = 7.5 We have a2 = 4.52 = 20.25 b2 = 62 = 36 37 c2 = 7.52 = 56.25 Since, c2 = a2 + b2, So, the given triangle is a right triangle. (iii) Let a = 2.5 b = 6, c = 7.2 a2 = 2.52 = 6.25 b2 = 62 = 36 c2 = 7.22 = 51.84 Since c2 = 51.84 (6.25 + 36) = a2 +b2 So, the given triangle is not a right triangle. Example 24: A tree is broken at a height 9 m from the ground and its top touches the ground at a distance of 12 m from its base. Find the original height of the tree. Solution: Let ABC be the tree which is broken at B such that BD is its broken part = AB Also BC=9 m and DC = 12 m (Fig 42) Fig. 42 Using Pythagoras Theorem in right triangle BD2 = BC2 + CD2 BD2 = 92 + 122 38 BCD, = 81 + 144 = 225 = 152 So, BD = 15 m. Hence, original height of the tree = 15 m + 9 m = 24 m. Example 25: A ladder of length 13 m placed on the ground window of a building. The distance of the foot of leader from the wall is 5m. Find the height of the window from the ground. Solution: Let W be the window, and WY be the leader (See Fig.43) Fig. 43 Clearly xy = 5 m ad WY = 13 m . Using Pythagoras Theorem in WXY (Fig 43), WY2 = XY2 + WX2 132 = 52 + WX2 or, WX2 = 132 - 52 = 169 – 25 = 144 = 122 So, WX = 12 Hence, the height of the window = 12 m. Example 26: A pole of height 15 m is erected vertically on the ground with the support of a wire which is tied from the top of the pole to a peg on the ground. If the distance of the peg from the foot of the pole is 8 m, find the length of the wire, assuming there is no slag in the wire. 39 Solution: Let AB be the pole, C be the peg on the ground (See Fig 44). Fig. 44 Clearly, AB = 15 m, BC = 8 m. Using Pythagoras theorem in right triangle ABC, AC2 = AB2 + BC2 = 152 + 82 = 225 + 64 = 289 = 172 So, AC = 17 Hence, the length of wire = 17m. 40 STUDENT’S SUPPORT MATERIAL 41 STUDENT’S WORKSHEET – 1 Recall parts of an angle Warm up w1 Name of the student ______________________ Activity 1 – Recall and name each figure (line, line segment, ray): 42 Date ______ Activity 2 – Scratch your head Name the two arms of the angle, the vertex and the name of the angle in three different ways for each of the following figures: 43 44 The two arms of the angles are the two …………………… with the vertex as the ……………………… point. Activity 2: Organise the letters in the given words and form meaningful words related to angle. i) HIRTG …………………………………………….. ii) UTCAE ……………………………………………. iii) TESOUB ……………………………………………. iv) ATTGSIR v) LRXEEF vi) LTMCEEPO ……………………………………………. ……………………………………………. ……………………………………………. STUDENT’S WORKSHEET – 2 Different types of angles Warm up w2 Name of the student ______________________ Activity 1– Identify the angles: i) …………………………………….. ii) …………………………………….. 45 Date ______ iii) …………………………………….. iv) …………………………………….. …………………………………….. v) …………………………………….. Activity 2– Three friends are playing in the field. Identify, label the different types of angles in the picture and name them and write in the box. You may put the vertices for the angles yourself to name them: 46 47 STUDENT’S WORKSHEET – 3 Different types’ of pairs of angles PRECONTENT WORKSHEET P1 Name of the student ______________________ Date ______ Activity 1– Get ready for fun!! Take an origami sheet and follow the steps- Now unfold and mark the center as Also name the lines formed by folds. Observe and find different types of pairs of angles formed at the centre, say O (you may put the names of the points for naming the angles yourself): Linear Pairs are ……………………………………………………………………. …………………………………………………………………… …………………………………………………………………… …………………………………………………………………… 48 Complementary angles are ……………………………………………………………………. …………………………………………………………………… …………………………………………………………………… …………………………………………………………………… Supplementary angles are ……………………………………………………………………. …………………………………………………………………… …………………………………………………………………… …………………………………………………………………… Vertically Opposite angles ……………………………………………………………………… ……………………………………………………………………… ……………………………………………………………………… ……………………………………………………………………… 49 STUDENT’S WORKSHEET – 4 Angles Puzzle PRECONTENT WORKSHEET P2 Name of the student ______________________ Date ______ Activity 1– Brush your thinking skills to solve the puzzle and circle the solutions of the given clues: 50 Measure of angle is more than 180 degrees and less than 360 degrees. Measure of angle is less than 90 degrees. Measure of angle is 180 degrees. Sum of two angles is 180 degrees. Measure of angle is more than 90 degrees and less than 180 degrees. Sum of two adjacent angles is 180 degrees. Sum of two angles is 90 degrees. Measure of angle is 90 degrees. Measure of angle is 360 degrees. STUDENT’S WORKSHEET – 5 Angles and angles CONTENT WORKSHEET c1 Name of the student ______________________ Activity 1– Angle Relationships Name the relationship between angle ’a’ and angle ‘b’: i) ii) iii) 51 Date ______ iv) v) Activity 2 – Evaluate the measure of angle ‘b’ and give reasons: i) b = ……………. ii) b = ……………. 52 iii) b = ……………. iv) b = ……………. Activity 3 – Write the pair of adjacent angles, complementary angles, supplementary angles and vertically opposite angles: i) ii) 53 iii) iv) iv STUDENT’S WORKSHEET – 6 Angles in nature CONTENT Worksheet C2 Name of the student ______________________ Date ______ Activity 1 – Label and name different types of angles and pair of angles made in the pictures by a transversal with parallel lines: 54 Join the crows to make transversal ! 55 56 Activity 2 - Calculate and write the angles next to each of the question mark on the shapes below. Justify your answer. i) ii) iii) 57 iv) v) 58 STUDENT’S WORKSHEET – 7 Angle Sum Property of a Triangle CONTENT Worksheet C3 Name of the student ______________________ Date ______ Activity - Club 180 A) Ants are at work. They are eager to join ‘CLUB 180’ and decided to collect 180 sugar crystals. Three ants decided to store their sugar crustals in corners of a triangle. Will they succeed? Draw three triangles here and measure their angles(angles represent their collection of sugar) and check wether they qualify for ‘CLUB 180’ or not? Total collection of sugar Total collection of sugar Total collection of sugar = Sum of the three angles = Sum of the three angles = Sum of the three angles = = = ________ join CLUB180. ( Can / Cannot) ________ join CLUB180. ________ join CLUB180. ( Can / Cannot) ( Can / Cannot) 59 B) Ants are happy and celebrating with their choice of shape, as they joined CLUB 180. They want to investigate further. Go to the following link and help them arrive at a conclusion. http://www.mathopenref.com/triangleinternalangles.html Conclusion: C) Geogebra Activity Start on a new geogebra worksheet. Draw a triangle using the Polygon tool. Find the measure of angles using Angle tool(in clockwise direction) 60 Adjust the rounding of decimal in angle, using option and selecting 0 decimal. Use the command shown below in the input bow of the Geogebra worksheet to get the sum of angles of the triangle. (Remember! if you choose the vertices in the wrong order (clockwise direction), you would get the wrong answer.) Move the vertices of triangle one by one to change the angles of the triangle. Conclusion: Hence, The sum of angles of triangle is always 1800. Write three different angle selections that you have got. 1. 2. 3. 61 Extension Activity -I Cut a triangle and shade the three angles of triangle with different color. Cut the triangle into three pieces so that each one has an angle of the triangle. Arrange the three angles of triangle, adjacent to each other about point O, on the given line. Paste here O What do you observe? ______________________________________________________________________________ ______________________________________________________________________________ Are these three angles together forming a straight angle? ______________ (Yes/No) So, __________________________________________________________________ Extension Activity-II Trace ∠ 2 and ∠3 on a trace paper and paste their copy on ∠ 5 and ∠ 4. Observe ∠ 2, ∠ 1 and ∠ 3. Is ∠ 2 = ∠ 5? Give reason. Is ∠ 3 = ∠ 4? Give reason. Hence what do you conclude? (Use Linear Pair) ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ 62 STUDENT’S WORKSHEET – 8 Applying Angle Sum Property of a Triangle CONTENT Worksheet C4 Name of the student ______________________ Date ______ Activity - Appreciate Your Knowledge 1. Answer the following. Give reason. a) Can a triangle have two right angles? ________________________________________________________________________ _______________________________________________________________________ b) Can a triangle have two obtuse angles? ________________________________________________________________________ _______________________________________________________________________ c) Can an isosceles triangle have an obtuse angles? ________________________________________________________________________ _______________________________________________________________________ d) Can a right angled triangle be an isosceles triangle? ________________________________________________________________________ _______________________________________________________________________ 63 e) Can an equilateral triangle have an angle of 650? ________________________________________________________________ ________________________________________________________________ f) Can a triangle have all angle < 600? ________________________________________________________________ ________________________________________________________________ 2. Can you construct a triangle whose angles are given here. If triangle is possible write their type on the basis of angles. i. 450, 650 and 730 ii. 700, 480 and 620 iii. 420, 600 and 780 iv. 710, 830 and 450 v. 1080, 210 and 710 vi. 420, 480 and 900 vii. 300, 800 and 900 viii. 320, 420 and 960 3. Find the unknown angle. (i) (ii) (iii) 64 4. Find ∠CBE, ∠CEB and ∠ABE. 5. The angles of triangle are given in ratios. Find the measure of these angles. i. 2 : 3 : 5 ii. 1: 2 : 2 Find the measure of the angles of a triangle: a. The three angles are equal to each other. b. One is twice the smallest and another is three times the smallest. c. the three angles are multiples of 3. d. If one of the angle is 500 and the other two angles are in ratio 6 : 7. e. Two angles are equal and the third angle is greater than each of these angles by 500. 65 6. The vertex angle of an isosceles triangle is 580. Find the other two angles. 7. One angle of an isosceles triangle is 1080. Find the other two angles. Draw a figure. 8. In a right angled triangle, one acute angle is of 560 , find the third angle. 9. ΔPQR is an isosceles triangle with PQ=QR. If ∠ Q is 500. Find the other two angles. 10. Find the value of x, y, and z.( hint: x and y form a linear pair) A 50° 50° C 30° y x B z D 11. Two angles of a triangle are 700 and 200 , find the third angle and name the triangle. 66 STUDENT’S WORKSHEET -9 Exterior Angle Property of a Triangle CONTENT WORKSHEET C5 Name of the student ______________________ Date ______ Activity- Solution Search A) In the TRIANGLE JUNGLE CAMP, Pussy is too excited to see her tent. Is she inside her tent? _______________ Name the angle to give her location. _______________ Write name of: The exterior angle : The interior adjacent angle : The interior opposite angles: B) Go to the following link and help Pussy find the property of Exterior angle of a triangle. http://www.mathopenref.com/triangleextangle.html I conclude,______________________________________________________ _______________________________________________________________ 67 Write three different sets of exterior angle and related angles to suggest the above property and draw diagrams with proper label. 1. 2. 3. C) Extension Activity Trace ∠ 1 and ∠2 on a trace paper and paste their copy in the space of ∠ 4. Observe relation between ∠ 1, ∠ 2 and ∠ 4. Observe relation between ∠ 3 and ∠ 4. Is there any Linear Pair in the figure? Explain. ______________________________________________________________________ Hence what do you conclude? ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ 68 STUDENT’S WORKSHEET -10 Exterior angle Skill Drill CONTENT WORKSHEET C6 Name of the student ______________________ Date ______ Activity- Test your knowledge 1. In the given figures find the exterior angle. (i) (ii) 2. Find the other two angles in the given figure. 3. Find x and y. 69 4. Find ACD and CED 5. One of the exterior angles of a triangle is 1000 and its interior opposite angles is equal to each other. What is the measure of each of these two angles? 6. One of the exterior angles of a triangle is 1100 and the interior opposite angles are in ratio2:3. Find the angles of the triangle. 7. Find BCE and 8. Find PRQ, BDE. RDE and DEF. 70 STUDENT’S WORKSHEET -11 Triangle Inequality Property CONTENT WORKSHEET C7 Name of the student ______________________ Date ______ Activity- Triangle Trouble A) Ants are back from CLUB180 bash and are really delighted to find coloured candies. They are so fond of tringles so they are trying to make a Candy triangle , but there seems to be some trouble. Here are some number combinations. Take matchsticks in place of candy and try to form triangles of sides having as many matchsticks as the number in each combination. Go ahead. 3,4,5 6,4,3 2 , 4 ,1 4,3,4 71 What do you observe? ______________________________________________________________________________ ______________________________________________________________________________ B) Go to the following link and help Ants find the Triangle Inequality Property. http://www.mathopenref.com/triangleinequality.html Write three different sets of lengths of sides to suggest the above property and draw diagrams with proper label. 1. 2. 3. C) Solve the following questions: 1. In each of the following, there are three numbers. State if they could possibly be the lengths of the sides of a triangle. i) 2.5, 2.5, 6 ii) 4, 5, 8 2. The lengths of two sides of a triangle are 12cm and 15cm. Between what two measures should the length of the third side fall? 3. Draw a triangle as ABC, put < , > , or = to complete the statement: i) AB ______ BC + CA ii) AC ______ AB + CB 72 4. In ABC, O is a point in its interior. Put < , > , or = to complete the statement: a) AB ______ AO + OB b) AC ______ AO + OC c) BC ______ BO + OC d) Is AO + AB > BO? 5. If D is a point in the exterior of ABC, a) Is AB + BD > AD? b) Is AC + CD > AD? 6. Join the interior point to the vertices of given polygon to form triangles and complete the following statements. i) In ______, JS + SK ____JS ii) In ______, KS + SH ____ KH iii) In ______, SH + SI _____ IH iv) In ______, JS + SI _____ IJ 7. Join the interior point Q, to the vertices of given polygon to form triangles and show that : AQ + BQ + CQ + DQ + EQ > (AB + BC + CD + DE + EA) 73 STUDENT’S WORKSHEET – 12 Pythagoras Property for a Right angled Triangle CONTENT Worksheet C8 Name of the student ______________________ Date ______ Activity -Farm Area A) Mr. Trio has two small square plots. He got an offer from his friend, whose farm is also square in shape and is attached to his farms about its corners. His friend offered him to exchange his plot with Mr. Trio’s two plots as his friend’s plot has the same area as the combined area of Mr. Trio’s two plots. A B C He needs your help. Is Area A = Area B + Area C? Explain. ______________________________________________________________________________ ______________________________________________________________________________ Observe that these three plots enclose a right angled triangle. Here is another such diagram. A Is Area A = Area B + Area C? Explain. B _________________________________________________ C 74 _________________________________________________ _________________________________________________ B) Many plots in that area have the right angled triangle enclosed between them. Count their area and complete the table. Count two half squares as one square to calculate their area. (i) (ii) Is Area A = Area B + Area C? Explain. (i) ________________________________________________________________________ ________________________________________________________________________ (ii) ________________________________________________________________________ ________________________________________________________________________ What do you conclude from this activity? _____________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________ a2 + b2 = c2 75 EXTENSION ACTIVITY Go to the following link to understand Pythagoras Property for a Right angled Triangle. http://www.mathopenref.com/pythagorastheorem.html STUDENT’S WORKSHEET – 13 Application of Pythagoras Theorem CONTENT Worksheet C9 Name of the student ______________________ Date ______ Activity - Right Angle Magic 1. Decide whether the given three numbers form a Pythagorean Triplet. a) 6, 8 , 10 b) 10, 15, 32 c) 9 , 40 , 41 2. Given a right angled triangle. a Calculate to complete the table. c b a b 6 8 c 16 20 16 34 Working 76 3. The longest side of a right triangle is 13cm. If one of the other two sides is 5 cm , find the length of the third side. 4. Calculate the length of the wire tied from a tower of church to the ground, if the length of base is 21m and the height of the tower is 20m. 5. A ladder 13m is placed against the wall at a distance 5m from its base. How high up on the wall would it reach? 6. Can 6,8,11 be the lengths of the sides of a right triangle? 7. What will be the length of the diagonal of the given rectangles? i) ii) 8 X 15 12 X 16 8. A tree is broken at a height of 8m from the ground but did not separate. If its top touches the ground at a distance of 15m from the base of the tree then find the original height of the tree. 77 8. A ladder 2.5 m long is placed against a wall. If its foot is 0.7 m away from the wall, how high up the wall does it reach? Draw a diagram. 9. A man goes 6m due east and then 8m due earth. Find the distance between starting point and the terminal point. 10. Find the perimeter of a rectangle having one side measuring 15m and the diagonal of 17m. 12. A farmer bought a triangular plot. The sides of a triangular plot are 15cm, 36cm and 39 cm, show that it is a right Δ plot. ? ? ? ? 13. A yatch leaves a port and travels 15km due east. Then it turns and travels 8km due north. How far is the yatch from the port? 14. A frog is chasing a fly if its movement is denoted in the given figure then how many meters he is from his pond? 78 STUDENT’S WORKSHEET – 14 Independent Practice Post CONTENT Worksheet PC1 Name of the student ______________________ Date ______ 1. The angles are supplementary and the smaller angle is one half of the larger. Find the angles. 2. If two angles forming a linear pair are in the ratio 6:4, then find the angles. 3. Find value of x, y and z. y (3x-20)° 70° Z 4. Solve and find x, y and z. 4x° 2x° 3x° Z 5. y l Name the following angles F A a) Interior and exterior angles B m Q b) Alternate interior angles n P D C c) Pairs of corresponding angles E 6. If two angles of a triangle are 450, 650 then find the third angle. 7. If AB // CD and , then find CDP 550 , ABP 450 . D C 55° Then find DPB. P 45° 79 A B 8. If the two interior angles on the same side of a transversal which cuts two parallel lines are (3x – 4)0 and (5x – 8)0, find the value of x. 9. Is it possible to construct a triangle having angles 490, 810, 500? 10. Can the numbers given below be the length of sides of a triangle? Give reasons. (a) 5,7,9 (b) 3,4,5 (c) 5,2,7 (d) 5,8,20 11. If two angles of a triangle are of 400 each, then find the third angle. 12. Is 6,8,10 a Pythagorean triplet? 13. Is it possible to construct a triangle having sides 6cm, 2.5cm, and 9cm? 14. In each of the following the measures of three angles are given. State in which cases, the angles can possibly be those of a Δ. Give reasons (a) 15. 630,370,800 (b) 590,720,610 (c) 450,610,840 (d) 300, 1200, 200 If line l||m, then find the value of x and y. l 40° 60° X y m A B 42° 16. In given figure AB//CD and PRD 1320 if APQ 42 0 then find x and y. X Y C Q R D 17. The lengths of two sides of a triangle are 9cm and 5cm. Between what two measures should the length of the third side fall? 18. A ship leaves a port and travels 12km due east. Then it turns and travels 9km due north. How far is the ship from the port? 80 19. A ladder 2.5 m long is placed against a wall. If its foot is 0.7 m away from the wall. How high up the wall does it reach? STUDENT’S WORKSHEET – 15 Test Your Progress Post CONTENT Worksheet PC2 Name of the student ______________________ Date ______ Activity - Check your progress: 1. Find the measure of the angle which is the supplement of angle 123º. 2. Find the complements of the following angles. a. 45º b. 8º c. 20º d. 85º e. 1 º f. 25º g. 78º h. 43º i. 18º j. 4º 3. Find the supplements of the following angles. a) 50º b) 130º c) 120º 81 d) 95º e) 104 º f) 125º g) 48º h) 143º i) 118º j) 6º 4. Identify the pairs of adjacent angles in the following figures a) b) c) 5. Find all angles in the following figures. a) b) c) 6. Find the value of x in the following figures. a) b) c) 82 d) 7. Given lines e) and m which are parallel in the following figures, find the measure of x a) d) f) b) c) e) f) 8. Are line or rays l and m parallel to each other in the following figures? Give reasons for your answer. a) b) c) 83 9. Peter has written the measure of two angles on to her drawing. Is she correct, say why? 10. Find the value of x a) 11. Choose the appropriate answer for the question asked a) a triangle has i) six elements ii) three elements iii) both of them iv)none of these b) If the exterior angle of an isosceles triangle is 110º and its exterior opposite angles are equal then each angle is equal to i) 65º ii) 50º iii)60º iv) 55º c) in a triangle PQR, m is the mid point of PR, the median is i) PM ii) RM iii)QM 84 iv) none of these d) if angleA=30 º angleB=70 º, then the exterior angle formed by producing BC is equal to i) 70 º ii) 30 º iii) 100 º iv) 180 º e) can the exterior angle of triangle be a straight angle i) Yes ii) No iii)sometimes iv) none of these f) Sum three angles of a triangle is equal to i) one right angle ii) two right angles iii) three right angles iv) none g) If two sides of a triangle are equal, then angles opposite to these sides are i) supplementary ii) not equal iii) equal iv) nothing can be said h) Can a triangle with 3 cm, 6 cm, and 9 cm be constructed i) yes ii) no iii) none i) If two angles of a triangle are 30 º and 70 º, then the third angle if i) 60 º ii) 50 º iii)80 º iv) 100 º iii) scalene iv) isosceles j) A triangle having none of its sides equal is i) equilateral triangle ii) right triangle 12. One of the acute angle of a right triangle is 49º. Find the other acute angle 13. Find x, y, z in the following figures: a) 85 b) 14. When the three sides of a triangle are produced as shown: show that x+y+z = 360 º 15. State true or false for the following statements: 86 16. a) AD>AC+CD b) BC<AB+AC c) AB>AC+BC d) AC<AD+CD e) DE<AD+AE 17. A man goes 20 m due west and then 15 m due north. How far is it from the starting point? 18. A ladder 25 m long reaches a window of a building 20 m above the ground. Determine the distance of the foot of the ladder from the building 19. In a right angle triangle ABC, the lengths of legs one is given. Find the length of hypotenuse: a) a= 6cm, b= 8cm b)a= 8 cm, b= 15 cm 20. How much shorter is to walk diagonally across a rectangular field 80 m long and 60 m wide than along two of its adjacent sides? 87 Acknowledgements http://www.youtube.com/watch?v=vw-rOqDBAvs Angle sum prop. Video http://www.mathopenref.com in aknwldgmnt http://www.scribd.com/examville/d/11579155-Properties-of-a-Triangle 88 Suggested Video links Name Title/Link Video Clip 1 Types of angles http://www.youtube.com/watch?v=NAGsqs7AQRw&feature=related Video Clip 2 Types of angles www.youtube.com/watch?v=RSqrwN77L9o Video Clip 3 Types and Pairs http://www.youtube.com/watch?v=QBAfSo_9XBo&feature=related Video Clip 4 Special Angles http://www.youtube.com/watch?v=qhV1SjAE3mM&feature=related Video Clip 5 Classifying Angles http://www.youtube.com/watch?v=xfdDU0cvYss&feature=related Video Clip 6 Corresponding and Alternate angles http://www.youtube.com/watch?v=3S9Js06dSjM&feature=related Video Clip 7 Pairs of angles http://www.youtube.com/watch?v=Vvrac2aG6Y4&feature=related Video Clip 8 Properties of Triangles www.youtube.com/watch?v=Goa6zsWpG6w Weblink1 westernreservepublicmedia.org/phi/images/AnglesAndTriangles.ppt Weblink 2 www.youtube.com/watch?v=NAGsqs7AQRw Weblink 3 http://www.authorstream.com/Presentation/paramjeet-96480properties-triangles-propperties-education-ppt-powerpoint/ 89 CENTRAL BOARD OF SECONDARY EDUCATION Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 092 India