Maths magazine

E-MATHAZINE
CLASS – X A
SESSION-2013-2014
Teacher Incharge – Ms.Sushma Singh
Group Leaders – Mansi, Chetna
Front page and Logo- Chetna

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Page 2
We, the students of class 10-A
would like to thank our maths
teacher Mrs. Sushma Singh for
her precious guidance. We would
also like to thank each and every
member of our group for their
co-operation which helped us to
complete this project successfully.

 Advisor-Mrs.Sushma Singh
 Chief Editor- Mayank Saini
 Editor-
*Gaurav Dabas
* Mansi Manwal
*Chetna
*Shivam Arora
*Diksha
*Himanshi
*Sandeep
*Nishant
*Kamaljeet
*Yasmin
*Priyanka

WORK BREAKDOWN
ROLL NO. NAME WORK GIVEN
1 Akram Story and News
2 Alisha Raptiles and saree designs
3 Anuradha Painting and Geometrical designs
4 Ashish Cartoons and paintings
5 Chetna Work breakdown and limitations
6 Deepika Logical reasoning and series
7 Deeksha Essential questions
8 Gaurav Mental maths Q/A and activity
9 Gaurav Dabas Conclusion and product
10 Himanshi Introduction and lessons
11 Kamaljeet Analysis and data interpretation

12 Kaustubh Geometrical designs and quiz
13 Mansi Manwal Sugesstions and bibilography
14 Mayank Saini Historical background
15 Md. Shahid Rza Story and puzzle
16 Naveen Activity and small project
17 Naveen Singh Cartoons and game
18 Nishant Methodology
19 Nitin Bisht Mathematician and article
20 Nitish Mathematician and graphs
21 Payal Arora Aims and objective
22 Pooja Graphs and article
23 Prabhat Article and news
24 Pragati Story and puzzle

25 Preeti Sudha Series and mathematician
26 Priyanka Abstract of the project
27 Rajat Makkar Geometrical designs and topics
28 Rupam Activity and article
29 Sagar Rana Puzzles and news
30 Sandeep Reference
31 Sahil Dash Graphs and painting
32 Shivam Arora Tools used
33 Shivam Gulati Mental maths Q/A and
reasoning
34 Sumit Quiz and viva
35 Yasmin Topics and presentation

Our project is in a form of magazine. We all give our 100% to complete this
project. We faced many problems but Sushma Singh mam help us a lot to
make this project successfully. This magazine describe topics like:- story,
cartoon, mathematics, games, article, newspaper, group activity,
geometrical design, hots, reptiles, mathematical designs and many other
topics. You can look up all topics and get knowledge about them by
this magazine. Our whole class did a tremendous job . The work was
distributed by the chief editor, editor, and our teacher
Mrs.Sushma Singh Mam. We think that
our magazine is easily understandable because we used easy language
and some interesting facts so you don't loses your interest. Everyone gives
his and her all effort to the project to make it awesome,
easily understandable and as well as interesting also. Such that everyone
want to read this magazine either he/she be a teacher or student or even
pass out from college. We hope that all of you will like this magazine and all
the teachers and student who read get a vast knowledge from
this magazine.

Page 9
Mathematics first arose from the practical
need to measure time and to count. The
earliest evidence of primitive forms of
counting occurs in notched bones and
scored pieces of wood and stone. Early
uses of geometry are revealed in patterns
found on ancient cave walls and pottery.
As civilisations arose in Asia and the Near
East, sophisticated number systems and
basic knowledge of arithmetic, geometry,
and algebra began to develop.

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For more than two thousand years,mathematics has been a part of
the human search for understanding. Mathematical discoveries
have come both fromthe attempt to describe the natural worldand
fromthe desire to arrive at a formof inescapable truth fromcareful
reasoning.These remainfruitful and importantmotivations for
mathematical thinking, but in the last century mathematics has
been successfully applied to many other aspects of the human
world: voting trends in politics,the dating of ancient artifacts,the
analysis of automobile traffic patterns,and long-termstrategies for
the sustainable harvestof deciduous forests,to mention a few.
Today, mathematics as a mode of thought and expressionis more
valuable than ever before.Learning to think in mathematical terms
is an essential part of becoming a liberally educated person.

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REFRENCES

Indian mathematicians have made a
number of contributions to mathematics
that have significantly influenced
scientists and mathematicians in the
modern era. These include place-value
arithmetical notation, the ruler, the
concept of zero, and most importantly,
the Arabic-Hindu numerals
predominantly used today.

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ARYABHATTA

Aryabhata was born in Taregna, which is a small
town in Bihar, India, about 30 km from Patna
(then known as Pataliputra), the capital city of
Bihar State. Evidences justify his birth there. In
Taregna Aryabhata set up an Astronomical
Observatory in the Sun Temple 6th century.
There is no evidence that he was born outside
Patliputra and traveled to Magadha, the centre of
instruction, culture and knowledge for his studies
where he even set up a coaching institute. However,
early Buddhist texts describe Ashmakas as being
further south, in dakshinapath or the Deccan, while
other texts describe the Ashmakas as having
fought Alexander.

It is fairly certain that, at some point, he went
to Kusumapura for advanced studies and that he
lived there for some time. A verse mentions
that Aryabhatta was the head of an institution
at Kusumapura, and, because the university
of Nalanda was in Patliputra at the time and
had an astronomical observatory, it is
speculated that Aryabhata might have been the
head of the Nalanda university as well.
Aryabhata is also reputed to have set up an
observatory at the Sun temple in Taregana,
Bihar.

ARYABHATTA’S
Contribution
in math’s
Place value
system
Trigonometry
Algebra
Approximation
of pie
Numeration
Indeterminate
equation

The Aryabhatta numeration is a system of
numerals based on Sanskrit phonemes. It
was introduced in the early 6th century by
Āryabhaṭa, in the first chapter titled Gītika
Padam of his Aryabhatiya. It attributes a
numerical value to each syllable of the
form consonant vowel possible in Sanskrit
phonology, from ka = 1 up to hau = 10

In Ganitapada 6, Aryabhata gives the area of a
triangle as
tribhujasya phalashariram samadalakoti
bhujardhasamvargah
that translates to: for a triangle, the result of a
perpendicular with the half-side is the area.
Aryabhata discussed the concept of sine in his
work by the name of ardha-jya. Literally, it
means "half-chord". For simplicity, people started
calling it jya. When Arabic writers translated his
works from Sanskrit into Arabic, they referred it
as jiba

The place-value system, first seen in the 3rd
century Bakhshali Manuscript, was clearly in place in his work.
While he did not use a symbol for zero, the French
mathematician Georges Ifrah explains that knowledge of zero
was implicit in Aryabhata's place-value system as a place holder
for the powers of ten with null coefficients
However, Aryabhata did not use the Brahmi numerals. Continuing
the Sanskrit tradition from Vedic times, he used letters of the
alphabet to denote numbers, expressing quantities, such as the
table of sine's in a mnemonic form.

Page 24

Srinavasa ramanujan is one of the
celebrated Indian mathematician . His
important contribution to the field
includes hardy-ramanujan little wood
circle, method in number theory, rode
v-ramanujan’s identities in the partition
in numbers, work on algebraic
inequalities, elliptic function, continued
faction, partial sums and product of
helper geometric series.etc. and many
more.
Srinivasa ramanujan born on 22nd
December 1887 in madras, India. He
born in middle very little poor family.
Like sophism gasman he received no
formal education in mathematics but
made important contribution to the
advancement of mathematics. He gave
us many formulas and to all over the
world by which we are able today to

Page 27
Archimedes

Archimedes was born in Syracuse,
Greece in 287 BC and died 212
BC . He was the son of an
astronomer : Phidias. Archimedes
received his formal education in
Alexandria, Egypt which at the
time was considered to be the
'intellectual center' of the world.
When he completed his formal
studies in Alexandria, he returned
and stayed in Syracuse for the
rest of his life

Discovered
how to find
the volume
of a sphere
and
determined
the exact
value of Pi.

It is believed that
he was actually
the first to have
invented integral
calculus, 2000
years before
Newton and
Leibniz

Powers of Ten,
a way of
counting that
refers to the
number of 0's
in a number
which
eliminated the
use of the
Greek alphabet
in the
counting
system

A formula to
find the area
under a curve,
the amount of
space that is
enclosed by a
curve.

Page 33
Pythagoras

The Pythagoreans' believed “All is Number,”
meaning that everything in the universe depended
on numbers. They were also the first to teach that
the Earth is a Sphere revolving around the sun.
Pythagoras is often
considered the first
true
mathematician.

Pythagoras was born on Samos a Greek
island off the coast of Asia Minor. He
was born to Pythais (mom) and
Mnesarchus (dad).
As a young man, he left his native city for
Southern Italy, to escape the tyrannical
government. Pythagoras then headed to
Memphisin Egypt to study with the priests
there who were renowned for their wisdom. It
may have been in Egypt where he learned
some geometric principleswhich eventually
inspired his formulation of the theorem that
is now called by his name.
Towards the end of his life he fled to Metapontum because
of a plot against him and his followers by a noble of Croton
named Cylon. He died in Metapontum around 90 years old
from unknown causes.
Life

Many of Pythagoras’
beliefs reflect those of
the Egyptians. The
Egyptian priests were
very secretive. The
refusal to eat beans or
wear animal skins and
striving for purity were
also characteristics of
the Egyptians.

Page 38
a 2 + b 2 = c 2

Proof of Pythagorean theorem by rearrangement of 4
identical right triangles. Since the total area and the areas of
the triangles are all constant, the total black area is constant.
But this can be divided into squares delineated by the
triangle sides a, b, c, demonstrating that a2 + b2 = c2 .

Page 40
The sum of the angles of a triangle is equal to two right
angles or 180 degrees
The five regular solids
Venus as an evening star was the same planet
as Venus as a morning star.
The abstract quantity of
numbers. There is a big step
from 2 ships + 2 ships = 4
ships, to the abstract result 2
+ 2 = 4

Mathematics is one kind of science. We cannot do a single moment
without mathematics. Mathematics has made our everyday life easy
and comfortable. In official and personal life
become paralyzed without mathematics.
Different kinds of functions are performed by
mathematics. Mathematics works with numbers, counting, and
numerical operations. It is used to calculate something. It can be
done both technically and manually. Large andcomplicated
mathematical problems is solved by the computer software. On the
other hand, easy mathematical operations are performed without
the help of any machine or computer software. In official works like
banking, policy, school, college and universities mathematical
calculations are done by technically.

• The merits of mathematics in our everyday life cannot be
described in words. It has opened a new dimension to us. We
cannot do a single day without mathematics. Mathematics helps us
to solve difficult mathematical problems. It has enriched our
life. Mathematics helps us to decide if something is a good, risky or
not. Mathematics helps us to create everything as without the
application of mathematics. We cannot create any building,
picture, furniture, good art, wallpaper, your room, bridge etc. It
shows us to become beneficial in life.
.
• In the end, it can be said that mathematics helps us to take any
kinds of decision. It works just like a mentor. Without
mathematics, we never take any decision. Our everyday life
depends much on mathematics. We cannot go even an inch without
mathematics. Our everyday life becomes paralyzed without
mathematics. Therefore, it can be said that mathematics is a part
and parcel in our everyday life.

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TYPES OF GRAPH
Bar graph
Double bar graph
Histogram
Frequency polygon
Equation graph
More than Ogive
Less than ogive

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Regular Solids
 Tetrahedron
 Cube
 Octahedron
 Dodecahedron
 Icosahedron

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Circles

Page 75

INTRODUCTION
 Mark three non-collinear
point P, Q and R on a paper.
Join these points in all
possible ways. The
segments are PQ, QR and RP.
A simple close curve formed
by these three segments is
called a triangle. It is named
in one of the following ways.
 Triangle PQR, Triangle PRQ,
Triangle QRP, Triangle RPQ
or Triangle RQP

Triangle
 A triangle is a polygon of three sides. In fact, it is the polygon with the least
number of sides. A triangle PQR consists of all the points on the line
segment PQ, QR and RP.
 Sides: The three line segments, PQ, QR and RP that form the triangle PQ, are
calledthe sides of the triangle PQR.
 Angles: A triangle has three angles. In figure 4-1, the three angles are ∠PQR,
∠QRP and ∠RPQ.
 Parts of triangle: A triangle has six parts, namely, three sides, PQ, QR
and RP, and three angles ∠PQR, ∠QRP and ∠RPQ. These are also known as
the elements of a triangle.
Vertices of a Triangle
 The point of intersection of the sides of a triangle is known as its vertex. In
figure4-1, the three vertices are P, Q and R. In a triangle, an angle is formed
at the vertex. Since it has three vertices, so three angles are formed. The
word triangle =tri + angle ‘tri’ means three. So, triangle means closed figure
of straight lines having three angles.

Classification of Triangles :
Triangles can be classified in two groups
A. Triangles differentiated
on the basis of their
sides.
1. Equilateral Triangle: A
triangle with all sides equal to
one another is called Equilateral
Triangle. Here, PQ = QR = RP.
Therefore Triangle PQR is an
Equilateral Triangle.

2. Isosceles Triangle : A triangle
with a pair of equal sides is called
an Isosceles Triangle.
Here, PQ = QR.
Isosceles Triangle.

3. Scalene Triangle: A triangle in
which all the sides are of different
lengths and no two sides are equal,
the triangle is called a scalene
triangle.
Here PQ ≠ QR ≠ PR.
Scalene Triangle.

B. Triangles differentiated
on the basis of their angles.
1. Acuteangled triangle: A
triangle whose all angles are acute
is called an acute-angled triangle or
simply an acute triangle.
Or
If all the three angles of a triangle
are less than 90 degree then it is an
AcuteAngled Triangle.

2. Right-angled triangle: A
triangle whose one of the
angles is a right angle is called a
right-angled triangle, or simply
a right triangle.
The side opposite to the right
angle is called Hypotenuse and
the other two sides are called
the legs of the triangle. In a
right triangle, hypotenuse is the
greatest side.
Or
If one angle of a triangle is
equal to 90 degree then it is a
Right Angled Triangle.

3. Obtuse-angled triangle: A
Triangle one of whoseangle is
Obtuseis called an Obtuse-
angled triangle or simply an
ObtuseTriangle.
Or
If one angle of a triangle is
greater then 90 degree then it is a
Obtuse Angled Triangle.

Properties of a Triangle
1. Angle Sum Property: The sum of the
measures of the three angles of a triangle is 180.
Let us prove this property
Proof: Let PQR be any triangle. Draw a line AB
parallel to QR, passing through P. Mark the
angles as shown in figure 4-13.
Now, ∠4 = ∠2 (Alternate angles)
∠5 = ∠3 (Alternate angles)
Therefore
∠1 + ∠2 + ∠3 = ∠1 + ∠4 + ∠5 = 180
(Linearity property)
Therefore, Sum of the angles of a triangle is 180
degree (Two right angles). Hence proved

2. Exterior Angle Property: In a triangle, the
measure of an exterior angle equals the sum of
the measures of the remote interior angles.
In figure 4-14, Triangle PQR is shown. The
side QR is extended to S. Now, ∠PRS is the
exterior angle. With reference to the exterior
∠PRS, ∠PQR and ∠ RPQ are remote interior
angles (or opposite interior angles). Let us
prove this property
Proof: Refer to figure 4-14
∠PRS or ∠4 is an exterior angle. We have to
prove that ∠4 = ∠1 + ∠2
We know that the sum of the angles of a
triangle equals 180 degree
In Triangle PQR, ∠1 + ∠2 + ∠3 = 180 … (1)
Since ∠3 and ∠4 form a linear pair ∠3 + ∠4 =
180 … (2)
From (1) and (2), we have
∠3 + ∠4 = ∠1 + ∠2 + ∠3 or
∠4 = ∠1 + ∠2 Hence Proved
Since ∠4 = ∠1 + ∠2, this implies that ∠4 > ∠1
and ∠4 > ∠2, Therefore, the exterior angle of a
triangle is greater than each remote interior

3. Triangles Inequality: The sum of the lengths of any two sides of
a triangle is greater than the length of the third side.
Let us prove this property Here, we have to prove that PQ + PR >
QR, PQ + QR > PR and QR + PR > PQ.
Extend side QP to S such that PS=PR. Join SR.
Proof: In Triangle PQR, we have PS=PR (By construction)
Thus Triangle PSR is an isosceles triangle. We know that in an
isosceles triangle, angles opposite equal sides are equal
Therefore ∠ PSR = ∠ PRS… (1)
Or ∠ PRS = ∠ PSR (Identity congruence)
∠QRP + ∠PRS > ∠PRS (The sum of two non-zero numbers is
greater than each individual number) ∠ QRP + ∠ PRS > ∠ PSR
(From 1)
Or ∠ QRS > ∠ PSR (Since ∠QRP and ∠PRS are adjacent angles)
Or ∠QRS > ∠QSR (Since ∠PRS and ∠QSR is the same angle)
or QS > QR (Since, side opposite to greater angle is greater)
or QP+PS > QR (Since QS=QP+PS)
or QP+PS > QR (Since PS=PR by construction)
or PQ+PR > QR (Since QP and PQ the same side of the triangle
taken in different order)

Congruence of Triangles
XYZABC 
Definition :Two triangles are congruent if threesides and three angles of
one triangle are equal to the correspondingsides and angles of other triangle.
The congruence of two triangles follows immediately from the congruence
of three lines segments and three angles.
Given two triangles ΔABC and ΔXYZ.
If AB is congruent to XY, Ais congruent to X ,
BC is congruent to YZ, B is congruent to Y ,
CA is congruent to ZX, C is congruent to Z
then we say that ΔABC is congruent to ΔXYZ, and we write
X
Y
Z
A
B
C

1. Side-Angle-Side Principle
If AB is congruent to XY
B is congruent to Y
BC is congruent to YZ
then ΔABC is congruent to ΔXYZ
A
B
C
X
Y
Z

2. Angle-Side-Angle Principle
If Ais congruent to  X
AC is congruent to XZ
C is congruent to Z
A
B
C
X
Y
Z
((

3. Side-Side-Side Principle
If AB is congruent to XY
CA is congruent to ZX
A
B
C
X
Y
Z

Theorem
If ΔABC is congruent to ΔXYZ ,
then
AB is congruent to XY
CA is congruent to ZX
and
A is congruent to  X
B is congruent to  Y
C is congruent to  Z
In short, corresponding parts of
congruent triangles are congruent.

Example 14.5
Show that the diagonals in a kite is perpendicular to each other.
Recall that a kite is a quadrilateral with 2 pairs of
congruent adjacent sides. In particular for the following
figure, AB = AD and CB = CD.
A
BD
C
E

A
BD
C
We first need to show that ΔADC and ΔABC are congruent.
This is true because
AD = AB
DC = BC
AC = AC
and we have the SSS
congruence principle.
Therefore, (click)
1 2
1 is congruent to  2
(b/c it is a kite)
(b/c it is a kite)
(b/c they are the same side)

A
BD
C
E
1 2
Now we only consider ΔADE and ΔABE.
They should be congruent because
AD = AB
 1 =  2
AE = AE
Hence SAS principle applies.
 AED is congruent to  AEB, and they both add up
to 180, hence each one is 90.

Similarity of Triangles
Definition
Given ΔABC and ΔXYZ.
 B is congruent to  Y
C is congruent to Z
and AB : XY = BC : YZ = CA : ZX
then we say that ΔABC is similar to ΔXYZ, and the
notationis
ΔABC ~ ΔXYZ
A
B
C
X
Y
Z

1. SSS similarity principle
If AB : XY = BC : YZ = CA : ZX
then ΔABC is similar to ΔXYZ.
A
B
C

2. AAA similarity principle
C is congruent to  Z
then ΔABC is similar to ΔXYZ
C
X
Y
ZA
B

3. AA similarity principle
(because the angle sum of a triangle is always
180o)
A
B
C
X
Y
Z

4. SAS similarity principle
If AB : XY = BC : YZ and
B is congruent to Y
A
B
C
X
Y
Z

Some More Properties of Triangle
1. The angles opposite to equal
sides are always equal.
Example: In figure
Given: ▲ABC is an isosceles triangle in which
AB = AC
TO PROVE: ∠B = ∠C
CONSTRUCTION : Draw AD bisector of ∠BAC
which meets BC at D
PROOF: IN ▲ABD & ▲ACD
AB = AC (GIVEN)
∠BAD = ∠CAD (GIVEN)
AD = AD (COMMON)
▲ABD and ▲ACD are similar triangles (BY
SAS RULE)
Therefore, ∠B = ∠C
A
B C
D

2. The sides opposite to equal angles
of a triangle are always equal.
Example : In Figure
Given: ▲ ABC is an Isosceles triangle in which
∠B = ∠C
TO PROVE: AB = AC
CONSTRUCTION : Draw AD the bisector of
BAC which meets BC at D
Proof : IN ▲ ABD & ▲ ACD
∠B = ∠C (GIVEN)
AD = AD (GIVEN)
Therefore, ▲ ABD &▲ ACD are similar
triangles (BY ASA RULE)
Therefore, AB = AC
A
B CD

Inequality
When two quantities are unequal then on comparing these
quantities we obtain a relation between their measures called
Inequality relation.
THEOREM 1 . If two sides of a triangle are unequal the larger side
has the greater angle opposite to it.
Given : IN ▲ABC , AB >AC
TO PROVE : ∠C = ∠B
Draw a line segment CD from vertex such that AC = AD
Proof : IN ▲ACD , AC = AD
∠ACD = ∠ADC --- (1)
But ADC is an exterior angle of ▲BDC
∠ADC > ∠B --- (2)
From (1) & (2)
∠ACD > ∠B --- (3)
∠ACB > ∠ACD ---4
From (3) & (4)
∠ACB > ∠ACD > ∠B , ∠ACB > ∠B ,
Therefore, ∠C > ∠B
A
B C
D

THEOREM 2. In a triangle the greater angle has a large
side opposite to it
Given: IN ▲ ABC ∠B > ∠C
TO PROVE : AC > AB
PROOF : We have the three possibility for sides AB and AC
of ▲ABC
(i) AC = AB
If AC = AB then opposite angles of the equal sides are
equal than
∠B = ∠C
But we know AC ≠ AB
(ii) If AC < AB
We know that larger side has greater angles opposite to
it.
AC < AB , ∠C > ∠B, we know that AC is not greater then
AB
(iii) If AC > AB
We have left only this possibility
AC > AB
A
CB

THEOREM 3. The sum of any two angles is
greater than its third side
TO PROVE : AB + BC > AC
BC + AC > AB
AC + AB > BC
CONSTRUCTION: Produce BA to D such that AD
+ AC .
Proof: AD = AC (GIVEN)
∠ACD = ∠ADC (Angles opposite to equal sides
are equal ) --- (1)
∠BCD > ∠ACD ----(2)
From (1) & (2)
∠BCD > ∠ADC = BDC
BD > AC (Greater angles have larger opposite
sides )
BA + AD > BC (BD = BA + AD)
BA + AC > BC (By construction)
Therefore, AB + BC > AC and BC + AC >AB
CB
D
A

THEOREM 4. Of all the line segments
that can be drawn to a given line from
an external point , the perpendicular
line segment is the shortest.
Given : A line AB and an external
point.
Join CD and draw CE perpendicular to
AB
TO PROVE CE < CD
PROOF : IN ▲CED, ∠CED = 900
THEN ∠CDE < ∠CED
CD < CE ( Greater angles have larger
side opposite to them. )
BA
C
ED

1. If the altitude from one vertex of a triangle
bisects the opposite side, then the triangle
is isosceles triangle.
Given : A ▲ABC such that the altitude AD from
A on the opposite side BC bisects BC I. e. BD =
DC
To prove : AB = AC
SOLUTION : IN ▲ ADB & ▲ADC
BD = DC
∠ADB = ∠ADC = 90
AD = AD (COMMON )
Therefore, ▲ADB & ▲ ADC are similar triangles
(BY SAS RULE )
Hence, AB = AC
A
CDB

2. In a isosceles triangle altitude from the vertex
bisects the base .
EXAMPLE: (in fig. 2.6)
GIVEN: An isosceles triangle AB = AC
To prove : D bisects BC i.e. BD = DC
Proof: IN ▲ ADB & ▲ADC
∠ADB = ∠ADC
AD = AD
∠B = ∠C ( Given: AB = AC)
Therefore, ▲ADB & ▲ ADC are similar triangles (By ASA)
Hence, BD = DC (BY CPCT)
A
CDB

3. If the bisector of the vertical angle of a triangle bisects the base of the
triangle, then the triangle is isosceles.
GIVEN: A ▲ABC in which AD bisects ∠A meeting BC in D such that BD = DC,
AD = DE
To prove : ▲ABC is isosceles triangle .
Proof: In ▲ ADB & ▲ EDC
BD = DC
AD = DE
∠ADB = ∠EDC
Therefore, ▲ADB & ▲EDC are similar triangles (By SSA)
Therefore, AB = EC
∠BAD = ∠CED (BY CPCT)
Hence, ∠CAD = ∠CED
And AC = EC (SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL)
AC = AB , HENCE ▲ABC IS AN ISOSCELES TRIANGLE.
E
D
C
B
A

MATHEMATICAL GAMES
Times hitori……
1 2 3 6 6 8 2 9
1 4 3 5 6 9 2 8
8 8 2 8 9 1 2 3
5 5 4 8 3 5 7 9
2 3 2 7 2 4 9 6
1 9 1 4 2 8 3 5
5 1 3 2 6 6 5 7
9 6 8 3 1 7 5 4
How to play
(1) A number may repeat just once in each row or
column. Eliminate repeating number by darkening
cells.

(2) Darkened cells must never be adjacent in a row or
column.
(3) Unmarked cells must create a single continuous
area undivided by darkened cells.
(4)
Every time you darken a cell you can
automatically circle its vertical and
horizontal neighbours which
may cannot be eliminated.

Any cell ‘SEND WICHED’
between neighbours of the
some value can be
circled.
(5)

Example:
8 5 5 7 3 5 4 6
4 5 1 6 7 6 5 2
2 5 7 1 5 4 6 8
5 6 2 4 3 7 6 3
2 7 5 5 1 7 2 7
1 6 6 6 4 3 5 7
7 8 4 6 2 3 3 7
8 4 5 2 3 5 7 1
(6) A ‘triple’ is a special case of sand which circle the
centre cell and darken the ends. e.g. 666.

cartoons
Hello Mickey. How
are you?
I am fine Ben.
Mickey do you know
about real numbers?
Yes I know in class
Xth. I had read about
real numbers.

Can you
describe me
about this?
Yes why not?
Thanks.
First you should know
what are two very
important propertiesof
real numbers.

What are these
two properties ?
(1) Euclid’s division lemma
(2)Fundamental theorem
of arithmetic.
What are the main
role of these
properties?
Let 2 positive integers a
and b. There exist unique
integers q and r satisfying
a=bq+r,0<r<b

Step (1) We find the whole no. q and r such that c=dq+r,
0<r<d.
Step (2) If r=0,d is the HCF of c and d . If r is not= 0, apply
division lemma to d and r.
Step (3) Continue the process till the remainder is 0.
For example: HCF of 4052 and 12576
Here, 12576=4052.8+420
4052=420 .9+272
420=272.1+148
272=148.1+124
148=24.5+4
24=4.6+0
So, 4 is the HCF of 12576 and 4052.

According to this theorem , every
composite number can be expressed as a
product of prime and this factorization is
unique a par from the order in which
prime factors occur.
For example, 32760=2.2.2.3.3.5.13
It is very easy. Now tell me
about Fundamental theorem
of Arithmetic.
Yo! I understand very
formly about real numbers.
Thanks.

Page 119
Mathematical series is a sequence or correct
ordering of mathematical figures or number
according to there pattern.
π+=
-

Page 120
Basic concepts
Here we work at mathematical series.
As we know mathematical series is a
sequence or correct ordering of
mathematics figures or numbers by
according to there pattern. Here we
use two types of mythical series
• Mathematical drawing series
• Number series.

Page 121
Mathematical Drawing Series
?
?
-
-

Page 122
-
-
?
?
- ?

Page 123
Number series
4 , 9 , 16 , 25 ,- ?
1 , 3 , 6 , 10 , - ?
8 ,27,64,125, - ?

• Objective
• To verify using the method of paper cutting, pasting and folding
that the lengths of tangents drawn from an external point are
equal.
• Pre-requisite knowledge
• Meaning of tangent to a circle.
• Materials required
• coloured papers,
• pair of scissors,
• ruler,
• sketch pens,
• compass,
• pencil.
To verify that length of tangents drawn from an
external point to a circle are equal

• 1. Draw a circle of any radius on a coloured paper and cut it. Let
O be its centre.
• 2. Paste the cutout on a rectangular sheet of paper.[Fig 10(a)]
• 3. Take any point P outside the circle.
• 4. From P fold the paper in such a way that it just touches the
circle to get a tangent
• PA (A is the point of contact). [Fig 10(b)]. Join PA.
• 5. Repeat step 4 to get another tangent PB to the circle (B is the
point of contact).
• [Fig 10(c)]. Join PB.
• 6. Join the centre of the circle O to P, A and B. [Fig 10(d & e)]
• 7. Fold the paper along OP. [Fig 10(f)] What do you observe?
• Observations
• Students will observe that
• 1. Δ OPA and Δ OPB completely cover each other.
• 2. Length of tangent PA = Length of tangent PB.

Page 131

Mentalmaths questions
1) Express 1000 as a product of prime factor ?
➢ 23 x 53 = 1000.
2) Which prime numbers will be repeatedly
multiplied in prime factorizatation of 3200 ?
➢ 2.
3) Find the digit at units place of 8n if n is a
multiple of five ?
➢ 8.
4) What are the prime factors of denominators
of fraction 7/80 ?
➢ 7/24 x 5

5) If HCF of two no is 68 & 85. what is the
LCM of two numbers ?
➢ 340.
6) What is the HCF of 95 & 152 ?
➢ 14.
7) Find the no which when divided by 18 gives
the quotient and reminder as 7 & 4?
➢ 130.
8) 176 when divided by a no gives the
reminder 5 & quotient 9 ? What is the
no?
➢ 19.

9) By which smallest irrational number±27 be
multiplied so as to get a rational number ?
➢ should be multiply by ±3 to get a rational no.
10) What is the product of (±7 + ±5) and (±7-±5)
?
➢ 2.

QUIZ
Ques1.A number consists of two digits whose sum is
5,When the digits are reversed, the number becomes
greater by 9. Find the number?
Answer: 23
Ques2.The diameter of a circle whose area is equal to
the sum of the areas of the two circles of radii 40cm
and 9cm is:
Answer: 82cm

Ques3.The number of cubes of side 2cm which can
be cut from a cube of side 6cm is:
Answer:27
Ques4.The value of K for which the pair of
equations K x- y=2 and 6x-2y=3 has a unique
solution.
Answer; k is not equal to 3

Ques5.In a triangle right angle at Q .In which
PR=13cm and PQ=12cm,find tan P –cot
R.
Answer; tan P – cot R=0
Ques6.If sin A=3/5,calculate Cos A.
Answer; Cos A=4/5.

Ques7.The value of tan1,tan2,tan3….tan89 is;
Answer;1
Ques8.Draw a figure in which triangle ACB is
similar to triangle APQ .If BC=8cm,
PQ=4cm,BA=6.5cm,AP=2.8cm,find CA and
AQ.
Answer; AC=5.6cm,AQ=3.25cm.

Ques9.Draw a triangle in which d is the mid point
of side AB and e is the mid point of side AC
AD=3cm,BD=4cm and DE=2cm ,if DE is
parallel to BC then x is equal to:
Answer;4.7cm
Ques10.(sec A +tan A)(1-sinA)
Answer: cos A

Ques11.All circles are ;
Answer; Similar
Ques12.Half the perimeter of a rectangular garden
,whose length is 4cm more than its width , is
36m.Find the dimensions of the garden.
Answer; Length=11m and width=7m

VIVA Questions
 Ques1.If a+1,2a+1,4a-1 are in A.P then value of a is:
(A)1 (B)2
(C)3 (D)4
Answer;(B)2
Ques2.If the 3rd and 9th terms of an AP are 4 and -8 respectively ,which term
of this AP is 0?
(A)4th (B)5th
(C)6th (D)7th
Answer;(B)5th

 Ques3.The lengths of the diagonals of a rhombus are 24cm,and 32cm The
perimeter of the rhombus is O?
(A)9cm (B)128cm
(C)80cm (D56cm
Answer; (C)80cm
Ques4.What is the value of sin A at the thirty degree angle;
(A)0 (B)1/2
(C)1 (D)2
Answer: (B)1/2

 Ques5.If the diameter of a protractor is 7cm then its perimeter is:
(A)18cm (B)20cm
(C)22cm (D)26cm
Answer: (A)18cm
Ques6.If the surface area of a sphere is 144 pie, then its radius is:
(A)6cm (B)8cm
(C)12cm (D)10cm
Answer: (A)6cm

 Ques7.The circumference of a circle is 44cm. Then the area of the circle is:
(A)276cm square (B)44cm square
(C)176cm square (D)154cm square
(D)154cm square
Ques8.If sin 3A= Cos (A-26), where 3A is an acute angle find the value of
A.
(A)26 (B)27
(C)28 (D)29
Answer: (D)29

 Ques9.Triangle ABC and triangle PQR are similar triangle such that angle
A=32 degree and angle R=65 degree then angle B is;
(A)83degree (B)32 degree
(C)65 degree (D)97 degree
Ques10.Which of the following are not the sides of a right triangle?
(A)9cm,15cm,12cm (B)2cm,1cm,10cm
(C)400cm,300cm,500cm (D)9cm,5cm,7cm
Answer;(D)9cm,5cm,7cm

Ques11.Solve for x and y : x / a + y / b =2 ax-by=a*a-b*b
(A) x=a ,y=b (B) x=b, y=a
(C) x=a-b ,y= a+ b (D) x=a + b, y=a-b
Answer; (A )x=a ,y=b
Ques12.Triangle ABC is similar to triangle PQR, in which
QP=3cm,QR=6cm,BC=8cm,AC=4 under root 3
(A)2+under root 3 (B)4+3under root 3
(C)4+under root 3 (D)3+4 under root 3
Answer; (B)4+3 under root 3.

Q-1 Find the area of square whose perimeter is 84.
ANS. Side=perimeter/4=84/4=21
Area=21 21=441
Q-2 Product of two numbers is 8192.If one number is
twice the other , find smallest number.
ANS. Let one number=x
Then, x 2x=8192
x x=4096
x=64
Q-3 What will be next: 123,234,345,…..,……
ANS. 456,567,678
Q-4 Average of four numbers is 30. If sum of 1st three
numbers is 85. Find fourth number.

ANS. 85+X/4=30
X=120-85
4TH no.=35
Q-5 The next line will be:
25 50 53
26 52 55
27 54 57
ANS. 28 56 59
Q-6 In a year the 1st April was Monday . What will
be the day on 18th April in the same year?
ANS. The day on 18th April will be Thursday.
Q-7 In figure 36490, digits 6 and 9 are replaced.
The difference between the new formed and
original number……..
ANS. 39460-36490=2970

Q-8 What will be the next:
4 16 64
6 36 216
8 64 512
ANS. 10 100 1000
Q-9 Number of prime numbers between 10 and 20 is:
ANS. 4 prime numbers:
11, 13, 17, 19
Q-10 The length of a rectangle is 4m. The breadth is
half of it . What will be its perimeter?
ANS. Perimeter=2(L+B)
=2(4+2)
=2 6=12m

 Wikipedia
 Learnnext
 Mypbworks.com
 britannica.com
 mathsisfun.com
 math.com
 libraryquest.org

REFRENCES
 Applied Mathematics
 Basic Mathematics
 Calculus
 Complex Analysis
 Constants & Numerical
Sequences
 Functional Analysis
 R.D. Sharma
 R.S. Aggarwal

Page 153
Thank You

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