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LEARNER’S CHOICE
( In this section, you can get the solutions of our mathematical questions )
Question by:- Alok Gupta
Rasra (ballai), Uttar Pradesh.
Q1.
Four times the first of three consecutive odd integers is six more than the product of two and
the third integer. Find the integers.
Solution by:- Shyam Thapa, Kohima.
Soln:- Let the three consecutive odd integers be x, x + 2 and x + 4.
As per the question:
4Xx =[2X(x+4)]+6
4x = 2x + 8 + 6
4x = 2x + 14
4x – 2x = 14
X
‗ 14
2
7
x=7
The required three consecutive odd integers are:7, 9 and 11.
Originality in Algebra
(I)
The original form of the present word “algebra” was “al-jebrw” almugabala: that was
introduced by Mohammad ibu Musa Al Khowarizmi, an Arab astronomer in ninth century.
Some of the original exponential forms:Year
Mathematician
Nationalism
Exponential form.
1550
Rafael Bombelli
Italian
1 (for 2x)
2
2 (for 2x2)
2
3 (for 2x3)
2
1600
Nicolas Chuquet
France
20 (for 2), 21(for 2x)
22 (for 2x2)
Ø (for x0), 2 (x2), 3 (for x3)
1610
Pietro Cataldi
1593
Adrianus Romanus
Dutch
1 ( 15 ) (for x15)
1619
Jobst Burgi
Swiss
7iii – 2v + 9iv
(for 7x3 – 2x5 + 9x4)
Question by:- RuopfÜvinuo Khezhie
Class:- 10
Little Flowe Higher Secondary School, Kohima.
Q:-
Prove that.
(I)
Sin A – Cos A + 1 ‗ Cos A
Sin A + Cos A – 1
1 – Sin A
(II)
( 1 – 2Sin2A)2 ‗ 2Cos A – 1
Cos4A – Sin4A
2
Solution by:- H. Rameswar Khuman
Christ Joti School
Mantri Pukhri, Imphal East.
Soln:(I)
Sin A – Cos A + 1 ‗ Cos A
Sin A + Cos A – 1
1 – Sin A
LHS ‗ Sin A – Cos A + 1
Sin A + Cos A – 1
=
Sin A – Cos A + 1
Cos A
Sin A + Cos A - 1
Cos A
Sin A – Cos A + 1
= Cos A Cos A
Cos A
Sin A + Cos A 1
Cos A Cos A
Cos A
‗ Tan A – 1 + Sec A
Tan A + 1 – Sec A
‗ (Sec A + Tan A) – 1
Tan A + 1 – Sec A
Dividing num. & deno. By Cos A
‗ (Sec A + Tan A) – (Sec2A – Tan2A)
Tan A + 1 – Sec A
‗ (Sec A + Tan A) – (Sec A – Tan A) (Sec A + Tan A)
Tan A + 1 – Sec A
‗ (Sec A + Tan A) (1 – Sec A + Tan A)
(1 – Sec A + Tan )
= Sec A + Tan A
1 + Sin A
Cos A
Cos A
=
= 1 + Sin A
Cos A
= (1 + Sin A) (1 – Sin A)
Cos A (1 – Sin A)
=
1 – Sin2A
Cos A (1 – Sin A)
=
Cos2A
Cos A (1 – Sin A)
=
Cos A = RHS
1 – Sin A
Multiplying num & done by (1 – Sin A)
(a + b) (a – b) = a2 – b2
Proved.
( 1 – 2Sin2A)2 ‗ 2Cos2A – 1
Cos4A – Sin4A
(II)
2 2
LHS ‗ ( 1 – 2Sin A)
(Cos4A – Sin4A)
‗ (1 – Sin2A – Sin2A)2
(Cos2A)2 – (Sin2A)2
=
(Cos2A – Sin2A)2
(Cos A – Sin2A) (Cos2A + Sin2A)
2
2
2
2
A – Sin2A)
= (Cos A – 2Sin A) (Cos
2
(Cos A – Sin A) X 1
= Cos2A – Sin2A
= Cos2A – (1 – Cos2A)
= Cos2A – 1 + Cos2A
= 2cos2A – 1 = RHS
Proved
Sin2Ɵ + Cos2Ɵ = 1
a2 – b2 = (a – b) (a + b)
Sin2Ɵ + Cos2Ɵ = 1
Continuation from the previous issue:Question by:-
Q:-
Thoubal District
How to do a geometrical proof easily?
Solution:-
(a)
M. Imtiaz Ahmad
Thoubal Moijng,
Okram Sunder Singh.
St. Joseph’s School, Kohima.
Theorem:-
It is a statement of a geometrical truth that has to be proved deductively (by
deductive method)
Example:-
The sum of the three angles of a triangle is 180o.
The statement need a deductive proof:X
A
Y
Now, let us consider a ABC.
XY BC is drawn through A.
Then, it is found that
B
C
< XAB + <BAC + <YAC = 180o
[ Forming a straight angle ]
<B + <BAC + <C = 180o
[ <XAB = <B, <YAC = <C; all < , XY BC ]
s
in ABC, <A + <B + <C = 180o
So, the statement is a theorem, because it needs a proof and the fact is supported by a
deductive proof.
(b)
Corollary/Criterion:It is a truth statement derived from a theorem (s)
EX:“A triangle cannot have more than one right angle”.
From the above theorem it can be extended that <A + <B = 180o - <C
i.e, <A + <B
180o
[ ( <A + <B ) is less than 180o ] [ <C ≠ 0 ]
It shows that angles can neither be both right angles, an obtuse angle and a right angle nor
both obtuse angles. So the statement is a corollary.
Parts of a theorem:(a)
General enunciation:- It is the general statement without having a reference to the parts
(angles/points/segment) of a particular figure.
(b)
Particular enunciation:- It is a statement referring to a particular figure. It consists of the
hypothesis / what is given and the conclusion / what is to be proved.
EX:(i) The sum of all the angles of a triangle is 180o. This statement is an example of the general
enunciation.
(ii) In ABC, <A + <B + <C = 180o is an example of particular enunciation.
 Steps in performing a geometrical proof:I.
II.
III.
Figure:- Transformation of the statement by a diagram using letters.
Given:- It should be a particular enunciation referring to the figure drawn above already.
To prove:- [ Originally a phrase was used as “It is required to prove that ….” ]
What is required to be proved for the particular figure has to mention.
IV.
V.
Construction:- If needed, the figure can be transformed into an equivalent figure or can be
extra parts of the figure. Proper statements should be mentioned for every part of
construction.
Proof:- It is an analytical part leading to the required conclusion.
Proof:It is an analytical part done by using suitable axiom (s) Theorem (s)/any other universal facts (s)
of deductive reasoning. This section consists of two parts:(a)
Argument:It is a stepwise arrangement of logical facts that connects between the given and the
conclusion.
(b)
Reasons:Each reason is a short statement expressed by an acceptable abbreviation (s) for every
application of an assumption to be written just to the right side of the particular step in the
Argument.
This section is the end part of a geometrical proof.
Originally, an abbreviation, Q.E.D (Quod erat demonstrandum) was written at the end showing
that the proof had been completed. But, in modern geometry it is replaced by hence
proved/proved.
A Format for Geometrical Proof.
Figure
Given:-
________
________ ________ ________
To prove:-
________
________ ________ ________
Construction:-
________
________
________ ________
Proof:-
Argument
Reasons
Hence proved
To be contd..
This section contains messages from resource persons
 Articles/Reports that can help students in learning of general
mathematics are cordially invited.
Continuation from the previous issue:-
Vedic mathematics
Part – II
Okram Sunder Singh
Now, let us have a generalized algorithm by observing the examples of the multiplication of whole
number with base 100 from previous issue.
Algorithm ( For base 100 )
(I)
Base:- 100
(II)
Subtraction of the base (100) from the whole numbers W1 and W2.
W1 – 100 = S1
W2 – 100 = S2
(III)
Addition of ( W1 and S2 ) / ( W2 and S1 )
( W1 + S2 ) = ( W2 + S1 ) = A
(IV)
Multiplication of A and 100.
A X 100 = M
(V)
To find the product of S1 and S2.
S1 X S2 = P
(VI)
Sum of M and P.
( M + P ) = Required product.
Multiplication of whole numbers with base 1000/10000/………
It has similar algorithm to that of base 100. Multiplication of A by 100 in the algorithm of base -100
should be replaced by 1000/10000/…….
Now let us see an example:Calculation of 957 X 1024
975 , 1024
(+)
24 X -43 = -1032
981000
- 1032
979968
So, 957 X 1024 = 979968
Base = 1000
(a) Subtraction of base 957 – 1000 = - 43
1024 – 1000 = 24
(b) cross addition
957 + 24 = 981
© Multiplication by 1000
( as base is 1000 )
981 X 1000 = 981000
(d) Addition of ( -1032 )
981000 + ( -1032 )
= 979968
Multiplication of whole numbers with base 50:Algorithm for base 50 has some modifications from that of base 100 as given below.
Algorithm ( for base 50 ):(I)
Original base:- 50
(II)
Subtraction of the bases from the whole numbers W1 and W2.
W1 – 50 = S1
W2 – 50 = S2
(III)
Addition of ( W1 and S2 ) / ( W2 and S1 ):( W1 + S2 ) = ( W2 + S1 ) = A
(IV)
To adopt a reference base ( RB ):RB is a multiple of 10, i.e. 10/100/100…..
According to our convenience, we can choose any of these as RB.
(V)
To find adjusting factor ( AF ):AF =
Original base ( OB )
Reference base ( RB )
For example:-
(VI)
RB
AF ‗ OB
RB
AF
10
50
10
5
100
50
100
1
2
Multiplication of A and AF:A X AF = M1
(VII)
Multiplication of M1 and 10/100/100…..
Multiplication is done according to RB.
RB
10
100
M1 X ( 10/100/1000 ) = M2
Multiplier
10
100
(VIII) To find the product of S1 and S2:S1 X S2 = P
(IX)
Sum of M2 and P
(M2 + P ) = Required Product.
Illustrative explanation
Calculation of 42 X 47
(i)
Choosing 10 as RB:-
(a)
Subtract the base 50.
42 – 50 = -8
47 – 50 = -3
(b)
Add the difference ( -8 ) to 47 or ( -3 ) to 42. (cross addition)
42 , 47
(+) - 3 , - 8
39
©
Find AF as,
AF ‗ OB ‗ 50 ‗ 5
RB 10
(d)
Multiply 39 from step (b) by 5 (AF)
42 , 47
-3 , -8
39
X 5 (AF)
195
(e)
Multiply 195 (from above) by 10 (as RB is 10)
42 , 47
-3 , -8
39
X 5
195
X 10
1950
(f)
Find the product of ( -3 ) and ( -8 ).
42 , 47
- 3 X - 8 = 24
39
X 5
195
X 10
1950
(g)
Add the product 24 [ from (d) ] to 1950.
42 , 47
- 3 X - 8 = 24
39
X 5
195
X 10
1950
(+) 24
1974
So, 42 X 47 = 1974.
(ii)
Choosing 100 as RB.
a) Same as in (i)
b) Same as in (i)
c) Find
AF ‗ 50 ‗ 1
100 2
(d) Multiply 39 from (b) by ½ (AF)
42 , 47
-3 , -8
39
(X) ½ (AF)
19.5
(e)
Multiple 19.5 by 100 (as RB is 100)
42 , 47
-3 , -8
39
(X) ½
19.5
(X) 100
1950
(f)
Find the product of ( -3 ) and ( -8 )
42 , 47
- 3 X - 8 = 24
39
(X) ½
19.5
(X) 100
1950
(g)
Add 24 from (f) to 1950
42 , 47
- 3 X - 8 = 24
39
(X) ½
19.5
(X) 100
1950
(+)
24
1974
So, 42 X 47 = 1974
Multiplication of whole numbers greater than 50.
Calculation of 53 X 57, using RB = 10
53 , 57
(+) 7 X 3 = 21
60
X 5
300
X 10
3000
+ 21
3021
Subtraction of base (50)
53 – 50 = 3
57 – 50 = 7
RB = 10 , AF = 50 = 5
10
Multiplication by RB (10)
So, 53 X 57 = 3021
Multiplication of whole numbers, one is less than 50 and the other is greater than 50.(RB = 10)
Calculation of 44 X 56
Subtraction of base (50)
44 , 56
6 X - 6 = -36
50
X 5 (AF)
250
X 10
2500
(+)
- 36
2464
44 – 50 = -6
56 – 50 = 6
So, 44 X 56 = 2464
AF = 50 = 5
10
multiplication by 10 (as RB is 10)
To be contd..
Learning from our mistakes
Parts of the contents in this section have errors either in questions or answers or in both.
Your work is to find out the errors and then to do corrections.
Polygon
 A closed figure formed by joining line segments is known as a polygon.
 A polygon has at least three sides.
 Depending upon the number of sides/amount of the angle measures, specific names are given as:-
No. of sides
3
4
5
6
Name
Triangle
Quadrilateral
Pentagon
Hexagon
No. of sides
7
8
9
10
Name
Heptagon
Octagon
Nonagon
Decagon
 A segment joining between two non consecutive vertices is known as a diagonal.
In fig (a), in quadrilateral ABCD
AC and BD are diagonals.
D
C
A
B
Fig (a)




Every polygon has at least a diagonal.
Every two diagonals of a polygon intersect each other.
Number of diagonals of a polygon cannot be greater than the number of sides.
In a heptagon, every vertices has four diagonals, so, the total number of diagonals of a heptagon is 4 X 7
or 28
 The polygon that has 35 diagonals is a decagon.
 A polygon is a convex if all the diagonals are inside it
D
C
but it is a concave if any of its diagonals is outside it
 In the fig (b), ABCDA, BOCB are convex polygons
O
but ABCDOA, AODCOBA are concave polygons.
A
B
Fig (b)
 A polygon is a regular polygon if all sides are equal in length.
 Each angle inside a polygon is known as interior angle but any angle outside polygon formed by any two
sides of the polygon is known as an exterior angle.
A
1
In fig ©,
<1, <2, <3 are interior angles
But a and b are exterior angles.
a
2
3
B
b
C
Fig ©
 Exterior angle theorem:In a triangle an exterior angle is equal to the sum of interior opposite angles.
So,
B
In fig (d), ABC
<a = <b + <c.
b
a
c
D
C
B
Fig (d)
Similarly, in fig (e),
In quadrilateral PQRS
<5 = <1 + <2 + <3.
P
Q
1
5
4
S
2
3
Fig (e)
R
 Angle sum property:The sum of all the interior angles of a polygon of n sides is (n – 2) 180o.
 In case of a triangle, n = 3, so (3 – 2) X 180o or 180o is the sum of all the three angle of a triangle.
 The sum of interior angles of a polygon has a minimum value of 180o.
 100o cannot be the sum of all the interior angles of a polygon as it is less than 180 o but 1275o can be the
sum of all the interior angles of a polygon.
 If each interior angle of a regular polygon is 144o, then the number of sides is 8.
 Each exterior angle of a regular polygon of sides is given by 360o
N
Please do correction as it may misguide.
Research and Developments.
This section consists of logical and creative works done by young learners.
Submitted by:Name:-
Reshikala Jamir
Class:-
7 ‘A’
Roll. No:-
24
(Year – 2014)
G. Rio Higher Secondary School, Kohima.
Dear Sir, I have presented one of the activities from mathematics that I have done as my school
work.
Aim:- To determine the area of a circle.
Requirements:-
Woolen thread, glue, glass sheet, pencil, scale, protector, scissors etc.
Procedures:1st step:Making a circular plate of coil with a piece of woolen thread that is moistened with glue by
circulating it closely on a glass plate. The woolen circular plate is allowed to dry completely.
2nd step:-
p
P1
P2
O
o
A
P
1
A
P
2
The woolen circular plate is cut along PO from outer to the center (o) of the circular plate. Then
the cut circular plate is slowly and carefully rollled out (without separating the threads) so that
the pieces of the threads become straight.
Observation:The whole circular woolen plate is transformed in a triangle, i,e
Here, OA P1P2
OP 1P2 .
Height of the triangle = OA = r (radius of the woolen circular plate)
Base = P1P2 = 2π r (circumference)
Calculation:Area of the circular plate
= Area of P1OP2
= ½ X P1P2 X OA
[ Area of a triangle = ½ X base X height ]
=½X2πrxr
= π r2
Observation:Area of a circular plate = π r2
Thank you Reshikala Jamir for your sharing. You have done it with clear and proper
statements. We are very proud of you and your teachers efforts. We hope that you will present
some more in future I order to share mathematical intelligence.
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