Download student`s worksheet – 4 - CBSE

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
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may be reproduced, printed or transmitted in any form without the
prior permission of the CBSE-i. This material is meant for the use of
schools who are a part of the CBSE-International only.
The Curriculum initiated by Central Board of Secondary Education -International (CBSE-i) is a progressive step in making the
educational content and methodology more sensitive and responsive to the global needs. It signifies the emergence of a fresh
thought process in imparting a curriculum which would restore the independence of the learner to pursue the learning
process in harmony with the existing personal, social and cultural ethos.
The Central Board of Secondary Education has been providing support to the academic needs of the learners worldwide. It has
about 11500 schools affiliated to it and over 158 schools situated in more than 23 countries 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
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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
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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
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_________________________________________________
_________________________________________________
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.
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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