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RASTHRIYA MADHYAMIKA SHIKSHA ABHIYAN (RMSA)
Orientation Programme to the District Resource Persons &
Teachers handling classes IX and X On
Content & Methodology
MATHEMATICS
Department of School Education Government of Andhra Pradesh
Hyderabad.
RASHTRIYA MADHYAMIKA SHIKSHA ABHIYAN COMMISSIONER & DIRECTOR OF SCHOOL EDUCATION
ANDHRA PRADESH - HYDERABAD.
Chief Adviser
Smt. Poonam Malakondaiah IAS.,
Commissioner & Director Of School Education And
Ex-Officio Project Director RMSA, Andhra Pradesh, Hyderabad.
Editorial Board
Dr. R. Satyanarayana Sri. D.Seshagiri Rao Additional Director, Consultant, RMSA, Hyderabad. RMSA, Hyderabad. Ms. G. N .Vidya, Sri N. Srirama Murthy Additional Director (Retd.) Joint Director (Retd), Consultant, RMSA. School Education.
Material Prepared by
Sri. Padala Suresh Kumar
SA., G.H.S, Nallakunta, Hyd.
Sri. K.L.N. Shastry
SA, G.H.S, Karimnagar.
PREFACE
“ The secondary stage ( lX and X class) marks beginning for the transition from
functional mathematics studied till the upper primary stage to the study of mathematics
as a discipline. The logical proofs of proposition, Theorems etc is introduced at this
stage. A part from being a specific subject it is to be treated as a Concommitant to any
subject involve in analysis as Reasoning.
With their recent introduction of Computers in schools educational computing and
the emergence of learning through the understanding of cause-effect relationships and
enter play of the variables, the teaching of mathematics will have to be suitably
redesigned to bring to in line with modern technological devises? –NPE 1986, POE
1992.
As envisaged in National Policy of Education 1986 and through its programme of
action 1992, mathematics has to be tangle as a discipline by giving opportunity to the
children to arrive at logical proofs of any proposition, Theorems etc.
In this training Programme care has been taken to see that hard topics are
identified with the help of class room teachers, later involving class room teachers and
experts in mathematics, enrichment material has been developed for use in this training
programme and further training programmes later with the main objective to see that
children get confidence in learning mathematics proficiently.
Hope you would all go through the material carefully and offer your valuable
specific suggestions to benefit the children.
CONTENT
SL No TOPIC Page Number
1 Trigonometry. 5-19
2 Limits. 20-23
3 Progressions. 24-29
4 Surface area of cones [mensuration]. 30-35
5 Binomial Theorem. 36-39
6 Mathematical induction. 40-44
7 Coordinate Geometry. 45-57
8 Logarithms. 58-62
9 Linear Programming. 63-87
10 Projects. 88-107
i. Path of pursuits.
ii. Building Trigonometrical Tables.
iii. Solids of Revolution
iv. Explore the magic.
v. Surface area of the cone
vi. Model Ratio.
vii. Quadrilateral and Parallelogram.
viii. Three Circles
ix. A Transformation group.
x. Path of the moving chord inside a circle.
xi. Square root by Guess average method.
xii. Conic sections.
1
1.TRIGONOMETRY
Definition:
Angle : an angle is the amount of a revolving line with reference to a fixed
line (called initial line).
If the rotation is in anti-clockwise direction, the angle measured is taken as
positive and if the rotation is in clock-wise direction, the angle measured is taken as
negative.
Measurement of angles:
1. Sexagesimal system of angles 1 rt angle = 90 degrees = 90o
01 = 60 minutes = 60l 1l = 60 seconds = 60ll
2. Centesimal systems of angles
1 rt angle = 100 grades = 100g 1 grade = 100 minutes = 100l 1l = 100 seconds = 60ll
2
3 Circular system
2rt angles 1 radian = _________ i.e, π radians = 2rt angles π
4. 1
rθ = [θ = angle subtended at the centre of the circle by an arc of the length l and
radius of the circle =r] Trigonometric ratios of angles: Definition: If θ is an acute angle of a right angled triangle, then the six t-ratios are
sin ,cos , tan ,cot ,secθ θ θ θ θ and cosecθ .These are called circular functions or
Trigonometric functions of an angle.
• some relation between T-ratios
a) 2 2sin cos 1θ θ+ =
b) 2 21 cot secθ θ+ = , 2 21 tan secθ θ+ = .
c) 2 21 cot cos ecθ θ+ =
i.e, 2 2sin 1 cosθ θ= − , 2 2tan sec 1θ θ= − , 2 2cot cos 1ecθ θ= −
d) 1
cos .sin
ecθθ
= , 1
sec .cos
θθ
= , 1
cot .tan
θθ
=
e) sin
tan .cos
θθθ
= , cos
cot .sin
θθθ
=
f) 1
sec tansec tan
θ θθ θ
+ =−
( )2
nπθ π≠ +
g) 1
cos cot .cos cot
ecec
θ θθ θ
+ =−
3
• Signs of T-ratios
In 1st quadrant, all t-ratios are +ve , in 2nd quadrant, only sin ,cscθ θ
are +ve; in 3rd quadrant, only tan ,cotθ θ +ve, in 4th quadrant, only cos ,secθ θ are +ve.
• a) Domain and Range of Trigonometrical functions: Trigonometric function Domain Range
sinθ [ 1,1]−
cosθ � [ 1,1]−
tanθ {(2 1) , }
2n n z
π− + ∈� ( , )α α−
cscθ { : }n n Zπ− ∈� ( ,1] [1, )−∞ ∞U
secθ {(2 1) , }
2n n z
π− + ∈� ( , 1] [1, )−∞ − ∞U
cotθ { : }n n Zπ− ∈� ( , )α α−
Hence -1 ≤ sinθ ≤ 1 i.e | sinθ | ≤ 1 -1 ≤ cosθ ≤ 1 i.e | cosθ | ≤ 1
sec 1θ ≥ or sec 1θ ≤ −
cos 1ecθ ≥ or cos 1ecθ ≤ −
4
Graph of sign curve: sinθ
b) Maximum and minimum values of T-ratios;
For all values of θ
1 sin 1θ− ≤ ≤ , and 1 cos 1θ− ≤ ≤ 1) Max( sinθ ) = 1, Min( sinθ ) = -1.
2) Max( sin cosθ θ )= Max (sin 2
2
θ) =
1
2
Min (sin cos )θ θ = Min (sin 2
2
θ) =
1
2
3) If sin cosy a x b x c= + + then , ,a b c R∀ ∈
We can write 2 2y c a b= + + sin( )x α+ Where a= r cosθ
} ⇒2 2r a b= + +
b = r sinθ
tanb
aα =
Since 1 sin( ) 1x α− ≤ + ≤
∴2 2 2 2c a b y c a b− + ≤ ≤ + +
5
Hence Max 2 2( cos sin )a x b x c c a b+ + = + +
and Min 2 2( cos sin )a x b x c c a b+ + = − +
θ 0o 15o 18o 30o
sinθ 00
0=
6 2
4
−
5 1
4
−
1 1
24=
cosθ 41
4=
6 2
4
+ 10 2 5
4
+
3
4=
3
2
θ 36o 45o 60o 90o
sinθ 10 2 5
4
−
2 1
4 2=
3 3
4 2=
1
cosθ 5 1
4
+
2 1
4 2=
1 1
4 2=
0
sin
tan .cos
θθθ
= , cos
cot .sin
θθθ
=
1
cos .sin
ecθθ
= , 1
sec .cos
θθ
= ,
6
• Allied angles θ θ− 90 θ−
90 θ+
180 θ−
180 θ+ 270 θ−
270 θ+
360 θ−
360 θ+
sinθ - sinθ cosθ
cosθ sinθ - sinθ -cosθ
- cosθ - sinθ sinθ
cosθ cosθ sinθ
- sinθ - cosθ - cosθ - sinθ sinθ cosθ cosθ
tanθ - tanθ cotθ - cotθ - tanθ tanθ cotθ -cotθ - tanθ tanθ
cosecθ
- cosecθ secθ secθ cosecθ - cosecθ -secθ -secθ -cscθ cscθ
secθ secθ cscθ -cscθ - secθ -secθ -cscθ cscθ secθ secθ
cotθ - cotθ tanθ - tanθ -cotθ cotθ tanθ - tanθ -cotθ cotθ
• Addition formulae
i) sin( ) sin cos cos sinA B A B A B+ = + ii) cos( ) cos cos sin sinA B A B A B− = −
iii) tan tan
tan( )1 tan tan
A BA B
A B
++ =−
• Subtraction formulae
i) sin( ) sin cos cos sinA B A B A B− = − ii) cos( ) cos cos sin sinA B A B A B− = +
iii) tan tan
tan( )1 tan tan
A BA B
A B
−− =+
* 2 2sin( )sin( ) sin sinA B A B A B+ − = −
7
* 2 2 2 2cos( )cos( ) cos sin cos sinA B A B A B A+ − = − = −
* 3 1
sin 75 cos152 5
+= =o o
* 3 1
cos75 sin152 5
−= =o o
* tan 75 2 3 cot15= + =o o
* cot 75 2 3 tan15= − =o o
• Sum and difference formulae (i) 2cos sin sin( ) sin( )A B A B A B= + + − (ii) 2cos sin sin( ) sin( )A B A B A B= + − − (iii) 2cos cos cos( ) cos( )A B A B A B= + + − (iv) 2sin sin cos( ) cos( )A B A B A B= − − + • Product Formulae
(i) ( ) ( )
sin sin 2sin cos2 2
C D C DC D
+ −+ =
(ii) ( ) ( )
sin sin 2cos sin2 2
C D C DC D
+ −− =
(iii) ( ) ( )
cos cos 2cos cos2 2
C D C DC D
+ −+ =
(iv) ( ) ( )
cos cos 2sin sin2 2
C D D CC D
+ −− =
8
• Multiple angles formulae
(i) sin 2 2sin cosA A A=
(ii) 2 2cos 2 cos sinA A A= −
= 22cos 1A −
= 21 2sin A−
= 4 4cos sinA A−
=
2
2
1 tan
1 tan
A
A
−+
=
2
2
cot 1
cot 1
A
A
−+
(iii) 2
2 tantan 2
1 tan
AA
A=
−
(iv) 3sin 3 3sin 4sinA A A= −
(v) 3cos3 4cos 3cosA A A= −
(vi)
3
2
3 tan tantan 3
1 3tan
A AA
A
−=−
* 21 cos 2sin2
AA− =
* 21 cos 2cos2
AA+ =
* 21 costan
1 cos 2
A A
A
− =+
* 01 2 2
sin 222 2
−=
9
* 01 2 2
cos 222 2
+=
* 01 2 2
tan 22 2 12 2 2
−= = −+
* 0 3 5 5 5sin 9 cos81
4
+ − −= = o
* 0 3 5 5 5cos9 sin81
4
+ + −= = o
* 0 0 3 5cos9 sin 9 sin 81 cos81
2
++ = = +o o
* 0 0 5 5cos9 sin 9 sin81 cos81
2
−− = = −o o
* 1
sin .sin(60 ).sin(60 ) sin 34
θ θ θ θ− + =
* 1
cos .cos(60 ).cos(60 ) cos34
θ θ θ θ− + =
* tan .tan(60 ). tan(60 ) tan 3θ θ θ θ− + =
* 0 0 1cos36 cos 72
2− =
* 0 0 1cos36 .cos 72
4=
10
• Hints and Tricks
1. Greatest and Least values of sin cosa bθ θ+ Put a= cosr θ , b= sinr θ
2 2 2a b r∴ + =
∴ sin cosa bθ θ+ = sin( )r θ α+ = 2 2 sin( )a b θ α+ +
Its greatest value = 2 2a b+
Lowest value = 2 2a b− + [∴ 1 sin( ) 1θ α− ≤ + ≤ ]
Similarly greatest value of sin cosa bθ θ− is also 2 2a b+ .
2. 1 1cos cos 2 cos 4 cos8 ......cos 2 sin(2 )
2 sinn n
nA A A A A A
A− =
3. sin .
2cos cos( ) cos( 2 ) .... cos[ ( 1) ]2sin
2
nnterms n
ββα α β α β αβ+ + + + + = + −
4. If A,B,C are angles of a triangle
Then sin( ) sinB C A+ =
cos( ) cosB C A+ = −
Also ( )
sin cos2 2
B C A+ =
( )
cos sin2 2
B C A+ =
5. If A+B+C =180o then
sin 2 sin 2 sin 2 4sin sin sinA B C A B C+ + = cos 2 cos 2 cos 2 1 4cos cos cosA B C A B C+ + = − −
11
PROPERTIES OF TRIANGLES AND CIRCLES
1. Sine formula In any ∆ABC
2sin sin sin
a b cR
A B C= = = is called sine formula.
2. Cosine formula
2 2 2
cos2
b c aA
bc
+ −=
2 2 2
cos2
c a bB
ca
+ −=
2 2 2
cos2
a b cC
ab
+ −=
3. Projection formula
cos cosa b C c B= +
cos cosb c A a C= +
cos cosc a B b A= +
4. Napier analogies
In ∆ABC
tan cot2 2
A B a b C
a b
− −=+
tan cot2 2
B C b c A
b c
− −=+
tan cot2 2
C A c a B
c a
− −=+
13
7. Formulae for circumscribed radius R
(i) 2sin 2sin 2sin
a b cR
A B C= = =
(ii) R = 4
abc
�
(iii) rs
= �
(iv) ( ) tan ( ) tan ( ) tan2 2 2
A B cr s a s b s c= − = − = −
(v) 4 sin sin sin2 2 2
A B Cr R=
Where r= In radius
(vi) cos cos cos 1r
A B CR
+ + = +
14
8. The radii of the described circles are given by
1 2 3, ,r r rs a s b s c
= = =− − −
�
1 tan2
Ar =
2 tan2
Br s=
3 tan2
Cr s=
1 2 3 4r r r r R+ + − =
1 2 3
1 1 1 1
r r r r+ + =
15
2.LIMITS
Consider the function 2 1
1
xy
x
−=−
. The value of the function at x=1 is of the form
0/0 which is meaningless. In this case we cannot divide the numerator by denominator
since x-1 is zero.
Now suppose x is not actually equal to 1 but very nearly equal to 1
then x-1 is not equal to zero. Hence in this case we can divide the numerator by
denominator.
2 11
1
xx
x
− = +−
Now if x is little greater than 1; then value of y will be greater than 2 and as x gets
nearer to 1, y comes nearer to 2. Now the difference between y and 2 is
2 2 21 2 1 ( 1)2 1
1 1 1
x x x xx
x x x
− − + −− = = = −− − −
The difference (x-1) can be made as small as we please by letting x tend to 1.
Thus we see that when x has fixed value 1, the value of y is meaningless but
when x tends to 1, y tends to 2 we say that the limit of y is 2 when x tends to 1.
Thus we write as
2
1
1l i m 2
1x
x
x→
− =−
16
Definition of limit : The number A is said to be the limit of f(x) at x=a if for any arbitrary
chosen positive number e, however small but not zero, there exists a corresponding
number and greater than zero such that
| ( ) |f x A− <∈
For all values of x for which
0 | |x a δ< − <
Where | |x a− means the absolute value of x-a without any regard to sign.
Various Methods for evaluation of limits:
1. Factorization a substitution.
2 2
l i m l i m ( ) 2x a x a
x ax a a
x a→ →
− = + =−
We have cancelled the factor ( x a− ) as ( x a− ) ≠0 because x≠a but x→a.
2. Expansion rule:
20
1 1lim log(1 )x
xx x→
− +
= 20
log(1 )limx
x x
x→
− +
=
2 3
20
{ .......}2 3lim
x
x xx x
x→
− − +
17
=
2
20
1 1[ . . . . . . ]
2 3l i mx
x x
x→
− −
= 0
1 1lim ......
2 3xx
→− −
=
1
2
3. Rationalization
3
3l i m
2 4x
x
x x→
−− − −
Multiply numerator as denominator by rationalization factor.
= 3
3 2 4l i m .
2 4 2 4x
x x x
x x x x→
− − + −− − − − + −
= 3 3
3lim .lim( 2 4 )
( 2) (4 )x x
xx x
x x→ →
− − + −− − −
= 3
3lim .(1 1) 1
2( 3)x
x
x→
− + =−
4. Limit of algebraic expression
3 2
3 20
2 5 7 3 3lim
55 3 2 5x
x x x
x x x→
+ + + =−− + −
4 3 2
3 20
2 5 7 3 0lim 0
55 3 2 5x
x x x x
x x x→
+ + + = =−− + −
18
5. Limits when x → ∞ are calculated by replacing x by 1/y and where x → ∞ ,
0y → . Hence lim ( ) lim ( )x x
f x yφ→∞ →∞
=
Where f(x) becomes ( )yφ , where x is put 1/y.
x = ∞ has no meaning as ∞ is not a fixed number on the real line. Hence we
always say x → ∞ . Also we cannot apply ordinary laws of algebraic on ∞ . i.e., 5-
5=0 or 5/5=1 but ∞ - ∞≠0 and ∞ / ∞ ≠ 1..since these are indertminate.
Some important questions:-
•
1lim .n n
n
x a
x an a
x a−
→
− =−
• lim .
m mm n
n nx a
x a ma
x a n−
→
− =−
• 0
sinlim 1x
x
x→=
• 0
tanlim 1x
x
x→=
• 0
1lim log
x
x
aa
x→
− =
• 1
lim[1 ] logx
xe
x→∞+ =
19
3.PROGRESSIONS
A sequence is an arrangement of numbers in a definite order according to some
rule.
Arithmetic Progression(A.P) :
A sequence 1 2 3, , .......... na a a a is called an arithmetic progression. if
there exists a constant number such that
a2=a1+d
a3=a2+d
……
An=an-1+d and so on
The constant‘d’ is called the common difference of the A.P
Ex: 2,5,8,11,14......
1t = 2 = 2+ 0= 2+(3x0) = 2+3(1-1)
2t =5 = 2+3 = 2+(3x1) = 2+3(2-1)
3t = 8= 2+6 = 2+(3x2) = 2+3(3-1)
4t = 11= 2+9 = 2+(3x3) =2+3(4-1)
5t =14= 2+12 = 2+(3x4) = 2+3(5-1)
.............................................................
nth term = 2+ 3(n-1)
20
if we represent the first term as ‘a’ and common difference as ‘d’ then the nth term
( 1)nt a d n= + −
2+5+8+11+14 = 40 = 5x8
5s = 5 x (2+14)/2
( )
2n
n a ls
+= l is the term.
Generally ( 1)l a n d tn= + − =
( ( 1) )[2 ( 1) ]
2 2n
n a a n d ns a n d
+ + −= = + −
2,5,8 are in A.P
5= (2+8)/2
a,b,c are in A.P
2
a cb
+=
2b a c⇒ = +
b is called A.M if a&c.
• if 1 2 3, , .......... na a a a are in A.P then
21
1 2 3, , ........... na a a a
A Mn
==
ia
n∑
• 1 2 3, , .......... na a a a are in A.P
• 1 2 1 ........n na a a a −+ = + = The sum of n terms of an A.P is the same as the
sum of n-terms. Then the sum of its (m+n) terms is zero.
Geometric Progression:
A sequence is said to be a Geometric Progression if the first
term is a non zero real number and each term is generated by multiply it by a non
zero number.
Ex:
1) 2,4,8,16,32,.........
2) 5,10,20,40,80,......
3) 1 1 1 1
, , , ,.......3 9 27 81
2,6,18,54,..... is a Geometric Progression.
0 1 11 2 2 1 2 3 2 3 −= = × = × = ×
1 2 12 6 2 3 2 3 2 3 −= = × = × = ×
2 2 3 13 18 2 3 2 3 2 3 −= = × = × = ×
3 3 4 15 54 2 3 2 3 2 3 −= = × = × = ×
---------------------------------------------
12 3n
nt−= ×
22
If the first term is 2 and the common ratio is 3, then
1n
nt a r −= ×
2 6 18 54 80+ + + =
= 2 40×
=43 1
23 1
−×−
1
1
n
n
rs a
r
−= ×−
2,6,18 are in G.P
26 2 18⇒ = ×
i.e., if a,b,c are in G.P 2b ac⇒ = or b ac=
Her b is called Geometric mean of a and c.
If 1 2, 3, ,....... na a a a are in G.P
⇒ G.M = 1 2 3....nna a a a
= n
iaπ
23
Harmonic Progression
In a sequence the reciprocals of all terms are in arithmetic progression then the sequence is said to be a Harmonic Progression.
Ex: 1 1 1 1
, , , ,.......2 5 8 11
2,5,8,11,......... are in A.P
⇒ 1 1 1 1
, , , ,.......2 5 8 11
are in H.P
The nth term of a H.P
1
( 1)nt a n d=
+ −
The relation between A.M, G.M and H.M is
2[ . ] [ . ][ . ]G M A M H M=
And [ . ] [ . ] [ . ]A M G M H M≥ ≥
24
Special Sequences
( 1)1 2 3 4 5 ..........
2
n nnterms n
++ + + + + + = =∑
22 2 2 2 2 ( 1)(2 1)1 2 3 4 5 ..........
6
n n nnterms n
+ ++ + + + + + = =∑
2 233 3 3 3 3 ( 1)
1 2 3 4 5 ..........4
n nnterms n
++ + + + + + = =∑
1 1 1 1 2...........
1 1 2 1 2 3 1 2 3.... 1
n
n n+ + + + =
+ + + + + + +
2 2 2 2 2 2 ( 1)1 2 3 4 5 6 ..........
2
n nnterms
− +− + − + − + = if n is even.
= ( 1)
2
n n + if n is odd
1 2 3 4 5 6 ..........2
nnterms
−− + − + − + = if n is even.
= 1
2
n + if n is odd.
21 3 5.......... nterms n+ + + =
2 4 6.......... ( 1)nterms n n+ + + = +
1 1 1 1 1(1 )(1 )(1 )............(1 )
2 3 4 n n− − − − =
1 1 1 1 1
(1 )(1 )(1 )............(1 )2 3 4 2
n
n
++ + + + =
2 2 2 2
1 1 1 1 1(1 )(1 )(1 )............(1 )
2 3 4 2
n
n n
+− − − − =
25
4.SURFACE AREAS OF CONES
A cone can be thought of as a ‘circular pyramid’ laving a flat circular base and a
curved surface.
However, while the two solids have similarities, a cone is not a pyramid
because all pyramids have flat faces.
Working Mathematically
In this activity you will make 4 cones from a circle pieces of the same radius.
1. Use compasses or a circular object to draw 4 identical circles of radius 8cm to
10cm.(Tracing around a coffee mug gives a circle of radius about 8cm) you can
make larger circles, if you wish.
2. Cut out these circles and locate their centers.
3. Cut out and discard these sectors.
26
4. For each remaining sector, calculate the : a) area b) length
5. Join together the cut radii to form a cone from each sector.
6. What is the relationship between the arc length of each sector and the
circumference of the base of the cone formed from it?
7. Measure the diameter of the base of each cone and then find the radius of each.
27
8. Using the original arc length of each sector and c= 2 rπ calculate the radius of
each cone.
9. How do these calculated and measured values compare?
10. Measure the slant height of each cone. What do you notice?
11. What is the area of each sector informs of the cone?
12. Copy and complete this table.
Original flat sector Arc length(cm)
Area of sector
(cm2)
Radius of cone(measured) (cm)
Radius of cone(calculated) (cm)
Slant height of cone
(cm)
13. Write an appropriate conclusion to summarise this activity.
Include a formula.
In this activity you discovered that the curved surface area of
the cone equals the area of the sector cone equals the area of the sector used to
construct the cone also the radius of the sector, l is the slant height of the once and
the arc length of the sector is equal to the circumference of the cone’s base.
28
Area of sector Area length of sector
__________________ = __________________________
Area of original circle circumference of the original circle
Area of sector
---------------------------- =2
2
r
l
ππ
2lπ
Area of sector = 2lπ 2
2
r
l
ππ
× rlπ=
So the curved surface area of a cone is rlπ . The surface area of a cone is equal
to the area of the circular base plus the curved surface area.
29
Surface area = 2r rlπ π+
2rπ rlπ
(Where r is the radius of the cone and l is the slant height of the cone).
Find the total surface area of each solid to the nearest square centimeter.
30
31
5.BIONOMIAL THEOREM 0[ ] 1x y+ =
1[ ]x y x y+ = +
2 2 2[ ] 2x y x xy y+ = + +
3 3 2 2 3[ ] 3 3x y x x y xy y+ = + + +
4 4 3 2 2 3 3[ ] 4 6 4x y x x y x y xy y+ = + + + +
5 5 4 3 2 2 3 4 5[ ] 5 10 10 5x y x x y x y x y xy y+ = + + + + +
.....................................................
1 1 2 20 1 2[ ] ....... ......n n n n n r r n
r nx y nc x nc x y nc x y nc x y nc y− − −+ = + + + + + (1)→
A formula by which any power of Binomial expression can be expressed as a
series [finite or infinite] is called the “Binomial Theorem” , which was first discovered
by Sir Isaac Newton.
∗ The coefficient 0nc , 1nc ,....... nnc are called binomial coefficients.
Properties of the Binomial expansion
1. There are (n+1) terms in the expansion ‘1’.
2. In any term of ‘1’ this sum of the exponents of x and y is always n.
3. The general term 1 r
n r rr cT n x y−+ = .
4. 0cn , 1cn ,....... ncn where 0cn =1, ncn =1 are called Binomial coefficients
which are shortly denoted by 1 2, ,......... ,... .r nc c c c respectively
32
5. 0 1 2 3......... 2nnc c c c c+ + + + = .
6. r n rc cn n−
=
7. The middle terms in the expansion of ( )nx y+ is
2
2 2
12
n
n n
n cT n x y+
= if n is even.
1
2
1 1
2 21
2n
n n
n cT n x y−
− +
+ = if n is odd.....
1
2
1 1
2 21
12
n
n n
n cT n x y+
+ −
+ +=
8. The greatest coefficient in the expansion of [ ]nx y+
= 2
ncn if n is even..
= 1
2
ncn− and 1
2
ncn+ if n is odd.
9. 0 1 2 3......... ( 1) 0nnc c c c c− + − + − =
10. for n>1 1
1 2 32 3 ......... ( 1) 0n
ncc c c n−− + + − =
33
11. Multinomial Theorem: The number of terms in the expansion
1 2 3[ ...... ]nrx x x x+ + + + is 1
( 1)rcn r−
+ −
Ex: The number of terms in the expansion 10( )x y z+ +
Sol: r=3, n=10
The number of terms = 3 1(10 3 1)c −+ − = 12 11
1 2
××
= 66
12. Binomial Theorem (for any real index)
If n is a real number and |x| <1 then
2 3 41 1 1(1 ) 1 ( 1) ( 1)( 2) ( 1)( 2)( 3) ......
2! 3! 4!nx nx n n x n n n x n n n n x+ = + + − + − − + − − − +
13. Some particular expansion:
For |x| < 1
1. 1 2 3 4(1 ) 1 ......x x x x x−+ = − + − +
2. 1 2 3 4(1 ) 1 ......x x x x x−− = + + + +
3. 2 2 3 4(1 ) 1 2 3 4 5 ......x x x x x−+ = − + − +
4. 2 2 3 4(1 ) 1 2 3 4 5 ......x x x x x−− = + + + +
5. 1 2 3
2 3(1 ) 1 ( 1) ( 2) ......nc c cx n x n x n x−− = + + + + +
34
14. Pascal Triangle: In ( )nx a+ , when expanded the various coefficients which
occur are 0nc , 1nc ,...... . The Pascal triangle gives the values of these coefficients for
n= 0,1,2,3,4,5.....
n=0 1
n=1 1 ∇ 1
n=2 1 ∇ 2 ∇ 1
n=3 1 ∇ 3 ∇ 3 ∇ 1
n=4 1 ∇ 4 ∇ 6 ∇ 4 ∇ 1
n=5 1 ∇ 5 ∇ 10 ∇ 10 ∇ 5 ∇ 1
35
6.MATHEMATICAL INDUCTION
Some times it is difficult to prove a theorem (or a formula) by direct method. In
this case, we use an indirect method known as the Principle of Mathematical Induction
or simply principle of Induction. Mathematical Induction is a special technique of proving
a statement or a theorem (or a formula) for all positive integers in two steps.
Step 1: Verify that the statement (or formula) is true for Particular values of n say
for n=1,2....
Step 2: Assuming that the statement (or formula) is true for n=m, where m is any
positive integer, show that it is true for n=m+1.
Conclusion: Since the statement (or formula) is true for n=2, it is true for n=2+1 =3 and
since it is true for n=3, it is true for n=3+1 i.e. 4 and so on. Hence the
statement (or formula ) is true for all positive integral values of n.
We shall denote a statement concerning the natural numbers
n=1,2,3,... by the symbol p(n). Putting n=1,2,3,..... in p(n) we get particular statements.
Principle of Mathematical Induction:
Principle of mathematical induction states that If p(n) be a
Statement such that
(i) P(1) is true.
(ii) P(m+1) is true whenever P(m) is true, m being any positive integer, then P(n)
is true for all positive integers n”.
Remark: We emphasize that proof by mathematical induction requires the
fulfillment of both the conditions (1) and (2) as stated above. Even if we prove a
36
certain statement for larger number of values of n say n=1,2,3...100 . we cannot
say that the statement is true for all n unless we establish condition (2). For example
consider trinomial f(n) = 2 41n n+ + . Substituting 1,2,3,4,5,6,7,8,9,10, in turn we
obtain 43,47,53,61,,71,83,97,113,131,151 which are all prime numbers. On the
basis of these results, we assert that the substitution of nay positive integer for n in
f(n) will always yield a prime number. But this reasoning is fallacious . In fact f(n)
yields a prime number for n=1,2.....39 but for n=40 , we have
f(40) = 240 40 41+ +
= 240 2.40 1+ +
= 2(40 1)+ = 41x41
Which is a composite number.
This example shows that we cannot make general assertion which respect to any n
unless we prove condition(2).
We now show that the condition (1) cannot be omitted either.
For example, we make the following assertion
“Every natural number is equal to the next natural number”.
To prove this, we assume n = m+1
Where m is a natural number. On the basis of this, we prove m+1 = m+2.
In fact adding 1 to each side of (1) we obtain the equation (2).
This shows that if our assertion is valid for n=m, then it is valid for n=m+1. Hence
we conclude that the assertion hold’s for all natural numbers n. But this arrangement
is again fallacious since we have drawn the conclusion proving the condition (2) only
and omitted condition (1) . But for n=1 , the statement is clearly false since 1 ≠ 2.
The above remarks shows that inorder to prove a certain statement for all natural
numbers n. It is essential to establish both the conditions (1) and (2) .
37
Illustration:
1. (66666.....6)2+(8888....8)2= ?
------------ ------------
n times n times
26 8 36 8 44+ = + =
266 88 4356 88 4444+ = + =
2666 888 443556 888 444444+ = + =
......................................................................
2 2(66666.......6) (888888.....8) (44444444........4)+ =
-----------------
2n-times
2. ( 1)
1 2 3 4 ....2
n nn
++ + + + + =
For n=1, LHS=1, RHS= 1(1 1)
2
+
LHS = RHS
Hence for n=1 the statement is true.
Let for n=m the above statement is true.
i.e., ( 1)
1 2 3 4 ....2
m mm
++ + + + + =
38
for n= m+1
( 1)
1 2 3 4 .... 1 ( 1)2
m mm m m
++ + + + + + + = + +
= 2 2( 1)
2
m m m+ + +
= 2 2 2
2
m m m+ + +
= 2 3 2
2
m m+ +
= ( 1)( 2)
2
m m+ +
= ( 1)( 1 1)
2
m m+ + +
For n=m+1 also the statement is true.
n N⇒ ∀ ∈ ( 1)
1 2 3 4 ....2
n nn
++ + + + + =
4. Prove by method of induction that
1
2 5 8 .... (3 1) (3 1)2
n n n+ + + + − = +
Sol: let p(n) = 2 5 8 .... (3 1)n+ + + + −
For n=1 LHS p(1) = 2
RHS = 1
(1)(3 1 1) 22
× + =
LHS = RHS
The statement is true.
39
Let p(n) be true for n=m that is, we suppose that
P(m) = 2 5 8 .... (3 1)n+ + + + −
= 1
(3 1)2
m m +
Now p(m+1) = 1( ) mp m T ++
= 1
(3 1) [3( 1) 1]2
m m m+ + + −
= 21[3 6 6 2]
2m m m+ + + −
= 21[3 7 4]
2m m+ +
= 1
( 1)(3 4)2
m m+ +
= 1
( 1)(3( 1) 1)2
m m+ + +
Above relation show that p(n) is true for n=m+1 . Hence p(n) is universally true by mathematical induction.
40
7.COORDINATE GEOMETRY
,l lX oX Y oY are two mutually ⊥ st.lines (known as axis) divide the XoY plane
into four parts. Each part is known as Quadrant.
∗ Distance Formulae
(i) Distance between two points (x1,y1) and (x2,y2) is 2 22 1 2 1( ) ( )x x y y− + − .
(ii) Distance of a point (x1,y1) from the origin = 2 2x y+ .
(iii) Distance of a point (x1,y1) from a straight line ax+by+c=0 is 1 1
2 2
| |ax by c
a b
+ ++
(iv) Distance of a straight line ax+by+c=0 from the origin = 2 2
| |c
a b+
(v) Distance between the parallel lines ax+by+c=0 and ax+by+d=0 is 2 2
| |c d
a b
−+
.
41
∗ Section formulae:
(i) The coordinates of a point which divides the line joining the points (x1,y1) and
(x2,y2) is the ratio m1: m2 is
Internally is 1 2 2 1 1 2 2 1
1 2 1 2
[ , ]m x m x m y m y
m m m m
+ ++ +
Externally is 1 2 2 1 1 2 2 1
1 2 1 2
[ , ]m x m x m y m y
m m m m
− −− −
(ii) The point p(x,y) divides the line joining the points (x1,y1) and (x2,y2) is =
1 2( ) : ( )x x x x− −
1 2( ) : ( )y y y y− −
(iii) The x-axis divides the line joining the points (x1,y1) and (x2,y2) is 1 2:y y−
(iv) The y-axis divides the line joining the points (x1,y1) and (x2,y2) is 1 2:x x−
(v) The mid point of the line joining the points (x1,y1) and (x2,y2) is
1 1 2 2[ , ]2 2
x y x y+ +
(vi) Centroid of ∆ABC with vertices (x1,y1) (x2,y2) and (x3,y3) is
1 2 3 2 3[ , ]3 3
x x x y y yG
+ + + +=
(vii) InCenre of ∆ABC with vertices A(x1,y1), B(x2,y2) and c(x3,y3) and whose sides
are a,b,c is 1 2 3 1 2 3[ , ]ax bx cx ay by cy
Ia b c a b c
+ + + +=+ + + +
42
Area formulae
* Area of ABC� with vertices A(x1,y1) B(x2,y2) c(x3,y3) is
∆ = 1
2
1 1
2 2
3 3
1 1
x y
x y
x y
x y
= 1 2 3 1
1 2 3 1
1
2
x x x x
y y y y
= 1
2
1 1
2 2
3 3
1
1
1
x y
x y
x y
= 1 2 3 2 3 1 1 2
1( ) ( ) ( )
2x y y x y y x y y− + − + −
= 1 2 2 1 2 3 3 2 3 1 1 3
1)
2x y x y x y x y x y x y− + − + −
* The area of a Quadrilateral whose vertices are (x1,y1) (x2,y2) (x3,y3) and (x4,y4) is
1 1
2 2
3 3
4 4
1 1
1
2
x y
x y
x y
x y
x y
=
= 1 2 2 1 2 3 3 2 3 4 4 3 4 1 1 4
1)
2x y x y x y x y x y x y x y x y− + − + − + −
43
* The area of a polygon whose vertices are (x1,y1) (x2,y2) (x3,y3)....(xn,yn) is
=
1 1
2 2
3 3
4 4
1 1
1
2 .......
......
n n
x y
x y
x y
x y
x y
x y
1 2 2 1 2 3 3 2 3 4 4 3 1 1
1...... )
2 n nx y x y x y x y x y x y x y x y= − + − + − + + −
* The area of a triangle enclosed by three lines 1 1 1 0a x b y c+ + = , 2 2 2 0a x b y c+ + =
and 3 3 3 0a x b y c+ + = is
1 1 1
2 2 1
3 3 3
a b c
a b c
a b c
=
= 1 2 3 3 2 1 2 3 3 2 1 2 3 2 1[ ] [ ] [ ]a b c b c b a c a c c a b a b− − − − −
44
* Area of a triangle enclosed by a straight line ax+by+c=0 with the coordinate axes is
1
.2
c c
a b=
2
2 | |
c
ab=
* Area of a triangle enclosed by a straight line 1x y
a b+ = with the coordinate axis is =
1
2ab .
* Area enclosed by a curve | |ax by c+ = is 22
| |
c
ab .
45
* Area of ABC� = 4 Area of ∆DEF where D,E,F are the midpoints of the sides of ∆ABC.
* Area of ∆ABC = 3. Area of ∆AGB
= 3. Area of ∆BGC
= 3. Area of ∆AGC
Where G is the Centroid of ∆ABC.
(x2-x1)
Area of ABC� 2 1
1| ( ) |
2k x x= −
46
Area of ∆ABC 2 1
1| ( ) |
2h y y= −
Note:
• If Area of ∆ABC =0 ⇔ Three points A,B,C are collinear.
• If area enclosed by three straight lines 1 1 1 0a x b y c+ + = , 2 2 2 0a x b y c+ + = ,
3 3 3 0a x b y c+ + = is zero. ⇔ The Three straight lines are concurrent.
Equations of Various kinds of straight lines:
• The equation of X-axis is Y=0.
• The equation of Y-axis is X=0.
• The equation of a straight line parallel to X-axis is y=K.
• The equation of a straight line parallel to Y-axis is x=K.
47
• The equation of a straight line passing through the origin (slope form or
Gradient form) is y mx= , where m is the slope.
• The equation of a straight line making an intercept on y-axis at the point
(0,c) is y mx c= + , where m=slope , c=y-intercept.[ slope – intercept
form].
48
• The equation of a straight line making intercepts on both the coordinate
axis is 1x y
a b+ = [intercept form]
Where a= X-intercept
b= Y-intercept
• The equation of straight line making intercept on X-axis at (a,o) is
( )y m x a= − m=slope, a=X-intercept.
49
• The equation of straight line passing through the point p(x1,y1) having
slope m is 1 1( )y y m x x− = − . [point-slope form]
• The equation of a straight line passing through the points (x1,y1) and
(x2,y2) is 1 2 1
1 2 1
y y y y
x x x x
− −=− − (The Two point form)
• The equation of a straight line passing through the point (x1,y1) and
parallel to the line ax+by+c=0 is 1 1( ) ( ) 0a x x b y y− + − =
• The equation of a straight line passing through the point (x1,y1) and
perpendiction to the line ax+by+c=0 is 1 1( ) ( ) 0b x x a y y− − − =
• The equation of a straight line passing through the point of intersection of
two straight line 1 1 1 0a x b y c+ + = and 2 2 2 0a x b y c+ + = is
1 1 1 2 2 2( ) ( ) 0a x b y c a x b y c+ + + + + =
50
• The equation of a straight line which is angle bisector of an angle
between the straight lines 1 1 1( ) 0a x b y c+ + = and 2 2 2( ) 0a x b y c+ + = is
1 1 1 2 2 2
2 2 2 21 1 2 2
( ) ( )a x b y c a x b y c
a b a b
+ + + += ±+ +
(+ for acute, _ve for obtuse angle)
• An equation of a line passing through a fixed point A(x1,y1) and making
an angle θ , 0 θ π≤ ≤ , 2
πθ ≠ , with the positive direction of the x-axis is
1 1
cos sin
x x y yr
θ θ− −= =
1 cosx x r θ= + 1 siny y r θ= + (parameter form)
• An equation of a line such that the length of the perpendicular from
the origin on it is p and the angle which this perpendicular makes
with the positive direction of the x-axis is α is
cos sinx y pα α+ = (normal form).
51
• ax+by+c=0 where a,b,c are real numbers and a and b cannot both
the zero simultaneously, is (general from) of a straight line.
Conditions and other formulae:
* Two line 1 1 1a x b y c+ + =0, 2 2 2a x b y c+ + =0
(1) are parallel if 1 1 1
2 2 2
a b c
a b c= ≠ , or 1 2 2 1 0a b a b− =
(2) are perpendicular if 1
1 2
2a b
b a= − or 1 2 1 2 0a a b b+ =
(3) are coincident if 1 1 1
2 2 2
a b c
a b c= = .
(4) are intersecting if 1 2 2 1 0a b a b− ≠ , i.e., 1 1
2 2
a b
a b≠
* Three points A(x1,y1), B(x2,y2) , C(x3,y3) are collinear if
(i) AB+BC = AC (ii) Slope of AB = slope of BC = slope of AC. (iii) Area of ∆ABC =0
* If m1,m2 are the slopes of two straight lines then
(i) lines are parallel if m1 = m2.
(ii) Lines are perpendicular if m1.m2=-1.
* The angle between two straight lines 1 1 1a x b y c+ + =0 and 2 2 2a x b y c+ + =0 is θ
then
52
1 2 2 1
2 2 2 21 1 2 2
sina b a b
a b a bθ −=
+ +
1 2 2 1
2 2 2 21 1 2 2
cosa a b b
a b a bθ +=
+ +
1 2 2 1
1 2 1 2
tana b a b
a a b bθ −=
+ or 1 2
1 2
| |1
m m
m m
−+
Where m1,m2 are the shapes of the straight lines.
* Three lines 1 1 1a x b y c+ + =0 , 2 2 2a x b y c+ + =0, 3 3 3 0a x b y c+ + = are concurrent
then
1 1 1
2 2 2
3 3 3
0
a b c
a b c
a b c
=
53
8.LOGARTITHMS
Definition: Let there be a number a>0 and a ≠ 1. A number x is called the logarithm
of another number y > 0 to the base a if xa y= .
We first observe that the logarithm of a number satisfying the conditions of the
definition is unique. For if andα β are two distinct logarithms of the number y to base a,
then by the above definition we have,
a yα = and a yβ =
Hence a aα β= (1)→
But, by the properties of powers with positive base different from 1, we conclude
from (1) that α β= . Thus , if the number y has a logarithm to base a then this logarithm
is unique. We denote it by the symbol loga y .
Thus by definition
logax y= if xa y=
54
Logarithm can be expressed as follows
_______________________________________________________________
Range Location of 10log x The value of 10log x
_______________________________________________________________
5 410 10x− −< < 105 log 4x− < < − 1 2 35 0. .....a a a− +
4 310 10x− −< < 104 log 3x− < < − 1 24 0. .....b b− +
3 210 10x− −< < 103 log 2x− < < − 1 23 0. .....c c− +
2 110 10x− −< < 102 log 1x− < < − 1 22 0. .....d d− +
1 010 10x− < < 101 log 0x− < < 1 21 0. .....e e− +
0 110 10x< < 100 log 1x< < 1 20 0. .....f f+
1 210 10x< < 101 log 2x< < 1 21 0. .....g g+
2 310 10x< < 102 log 3x< < 1 22 0. .....h h+
3 410 10x< < 103 log 4x< < 1 23 0. .....i i+
Logarithm = (integer) + 1 2 3 4(0. .....)r r r r
55
Characteristics and mantissa:
The integral part of a logarithm is called the characteristic and the decimal part is called mantissa. Logarithms to the base 10 are called common logarithms. The characteristic of common logarithms can be written down by inspection by the following rule.
Rule: The characteristic of the logarithm (base 10) of a number greater than 1 is
less by one than the number of digits in the integral part, and is positive. The
characteristic of the logarithm of a positive decimal of consecutive zero immediately
after the decimal point, and is negative.
Properties of Logarithms:
The students should commit to memory the following we
assume a>0, a ≠ 1 m>0, n>0
1. xa y= then logax y=
LHS is called exponential form where as RHS is called corresponding logarithmic
form.
2. 1a a= , 1b b= etc... log log 1a ba b= =
3. 0 01, 1a b= = etc. log 1 0, log 1 0a b= = .
4. 1
log .log 1, loglogb a b
a
a b ab
= =
Let logb a x= . ∴ xa b= loga b y= ∴ yb a=
Putting the value of b, we get ( )y x xya a a= =
∴ xy=1 or x=1/y.
56
5. Base changing formula
log log logb c ba a c= =
In general from log log .log .log ......logb c d e ba a e d k=
6. log log loga a amn m n= +
7. log log loga a a
mm n
n= −
8. log logna am n m=
9. log logq
paa
pn n
q=
10. loga na n=
11. log logp pa b>
⇒ a ≥ b if base p is +ve and > 1
or a ≤ b if base p is -ve and <1
i.e. 0< p < 1
12. if a>1 then 0 α β< < ⇔ log loga aα β<
13. if 0<a<1 then 0 α β< < ⇔ log loga aα β>
14. If a>1, α > 1 then log 0a α >
15. if 0<a<1, 0<α <1 then log 0a α >
16. if 0<a<1, 1α > then log 0a α <
57
17. if a>1. 1α > and aα > , then log 1a α >
18. if a>1. 1α > and aα < then 0 log 1a α< < .
19. if 0<a<1, 0<α <1 and aα > then 0 log 1a α< < .
20. if 0<a<1, 0<α <1 and aα < then log 1a α < .
58
9.LINEAR PROGRAMMING
The technique followed by mathematicians to solve such problems is called LINEAR
PROGRAMMING.
Historical Background: The technique of linear programming is of recent origin. It
started during the Second World War, when the war operations had to be planned to
economise the expenditure, minimize the losses and maximize the damage to the
enemy.
The technique of linear programming was used in order to minimize the shipping
expenditure for the Indian Cotton to be transported to different mills sin England, during
the second world war, when the oil prices were high.
An American Economist Dr. F.L.Hitchcock published a paper ‘Distribution of a
product from several sources to numerous locations’ in the journal of Mathematical
Physics in 1941. At the same time a Russian Mathematician L.V.Kantorovich published
a paper ‘ On the Translocation of Masses’ in the Doklady Academy Nauk U.S.S.R. It is
also believed that he solved a linear programming problem dealing with the organization
and planning of production in 1939. The above two are supposed to be the first papers
that appeared on the transportation problem which forms a branch of linear
programming.
In 1945, an English economist G.J.Stigler solved a rather long and complicated
det problem by the simplex method of linear programming. Mainly this problem was to
determine the qualities of 77 foods that are to be bought not only at the minimum cost,
but also to satisfy minimum requirement of nine nutritive elements like Vit A, thiamine
etc. This particular linear programming problem had 9 equations and 86 variables. This
paper ‘The cost of a subsistence’ appeared in the journal of Farm Economics in 1945.
However, the credit of making this branch popular, goes to an American
Economist G.B.Dantzig who has published more than 37 papers in this field.
59
For the work on these problems, the pioneer mathematician L.V.Kantorovich and
Economist T.C.Koopmann were awarded noble prize in 1974.
At present the technique of linear programming is widely used in production,
trade,commerce, industry and government operations. The specific examples of its uses
will appear in the subsequent lessons of this unit.
Objectives of the Course :
(1) To familiarize you with the process of maximization and minimization.
(2) To enable you to obtain algebraic models for the given problem.
(3) To enable you to use the graphical methods to obtain a solution.
In this lesson, you are introduced to the graphical method of solving a linear
programming problem. In the second lesson, the linear programming method is used to
solve transportation problems. We have used programmed instruction method for this
purpose. In the third lesson, the question of existence of a solution to a linear
programming problem is raised and considered with the help of examples.
Before we formally introduce the topic of linear programming it will be of interest
to discuss the linear equations and inequations in one or two variable. The equations
and inequations play and important role in obtaining the graphical solution of the
problems of linear programming.
1.1 FIRST DEGREE EQUATIONS IN ONE OR TWO VARIABLES
The graph of the equation x=4 on the number line is a point.
60
In the Cartesian plane it represents a straight line parallel to y-axis.
The equation x=4 is of first degree in one variable. The equation y=3 is also of
first degree in one variable and represents a line parallel to x-axis at a distance 3 above
the axis of x.
Let us consider the equation 2x+3y=6 . This equation is of first degree in two
variables x and y. We also call it a linear equation in two variables.
The equations x+4xy+3=0 and 2 22 4x xy y− − = are not linear as the
terms xy,2x ,
2y are of second degree. The graph of the equation 2x+3y=6 is a
straight line. Since two points determine a line uniquely, we assign at least two sets of
values for x and y which satisfy the above equation. Giving values to x, from the given
equation we get the corresponding values of y.
Usually we choose x=0 and calculate corresponding value of y and choose y=0
and calculate the corresponding value of x, to obtain two sets of values. (This method
fails, if the line is parallel to either of the axes or passes through the origin. In that case,
we can choose any arbitrary value for x and choose y so as to satisfy the equation).
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Plotting the points (0,2) and (3,0) and joining them by a straight line, we obtain
the graph of the line as given below.
Let us now consider the equation 2x+y=7. This equation may be considered a
mathematical model for the following real life situation.
A pen costs Rs.2.00 and a note book costs Re.1/-. How many pens and
notebooks can be purchases using all 7 rupees?
Let us suppose, x pens and y note books can be purchases out of Rs.7/-. Then
the mathematical statement for the above situation will be:
2x+3y=7
This equation has several solutions i.e.,
x=0, y=7
x=1, y=5
x=2, y=3
x=3, y=1
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These are the non-negative integral solutions of the above equation. They are
non-integral values of x and y which also satisfy the equation. For example, for x=3/4,
y= 11/2, and this is non-integral solution. In fact, for each value of x, we get a values of
y, and vice-versa. Similarly we may choose negative values for x or y to get the solution
of the above equation.
It is obvious that the equation 2x+y=7 has infinitely many solution. Some of them
are integral solutions, while others are not integral. For the given problem the only
admissible solutions are (0,7), (1,5),(2,3) and (3,1) because pens and notebooks cannot
be bought in fractions. The other solutions (non-integral positive/or negative/or integral
negative) of the mathematical equation 2x+y=7 are extraneous or spurious solutions.
They are solutions of the mathematical equation, but not the solutions of the real life
problem.
Ex : Two whole number are such that five times the first number minus three
times second number is equal to 72. Formulate mathematical relation and draw its
graph.
Solution: Let x and y be the whole numbers, then x and y are related by 5x-3y=72.
The graph is shown below. Observe that it consists of isolated points (a few of
which are shown) with integral coordinates, spread over the line 5x-3y=72.
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The mathematical equation 5x-3y=72 will have infinitely many solutions. Some of
them will be admissible solutions which will pertain to the given problem and others will
be spurious solutions. To determine all integral solutions of the above problem,
following algebraic technique is used.
From the equation 5x-3y=72 we see that
72 3 70 2 3 2 314
5 5 5
y y yx
+ + + += = = +
In order that x and y may be both integers, (2+3y) must be divisible by 5. The
possible value of y for which 2+3y is divisible by 5 are y=1,6,11,16…. and thus the
corresponding values of x are 15,18,21,24,…..
The coordinates of any point on the line 5x-3y=72 is a solution of the equation .
The coordinates of points, other than integral ones, will be solution of the mathematical
equation but not of the problem form which the mathematical equation is derived.
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1.2 Inequalities and their graphs
We have seen earlier that the statement x=4 represents an equation of first
degree in one variable and its graph is a point on the number line. Now let us
consider the statement 4x ≥ .
The statement 4x ≥ is a composite statement consisting of two simple
statements.
4x > or 4x =
connected with the word ‘or’ . Thus 4x ≥ means 4x > or x=4 or both, we call
4x ≥ an inequation. If x is an integer, the graph on the number line consists of isolated points a few of which are shown encircled below.
If x is a real number, it represents a ray as shown below:
We can also plot the graph of the inequation 4x ≥ in the
Cartesian plane. The graph of 4x ≥ will be the set of ordered pairs
{( , ) : 4}x y x ≥ . Here again we may visualize two distinct situations.
i) where x, y are integers. and
ii) x ,y are any real numbers.
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Case I: Let x, y be integers.
The graph will consist of all those points of the plane for which 4x ≥ , where x,y are integers. Some of the points which belong to the graph are marked with the circle O. The arrows indicate the graph consists of many other points – not shown in the figure.
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Case II: Let x, y be real numbers.
Now the graph will consist of the region for which 4x ≥ . The graph is shown by shaded portion in the diagram given below.
Now we consider the following inequation.
4 5 20x y+ ≥
To draw the graph , first of all we consider the equation 4 5 20x y+ = . We see that
for x=0 ,y=4 and for y=0 , x=5.
Thus the line passes through points (0,4) and (5,0). The graph of the line is as
follows:
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The graph of the line is as follows:
The line 4 5 20x y+ = divides the whole plane into three sets : R1, line AB and 2R .
We have to determine which set corresponds to the inequation 4 5 20x y+ > .
Let us consider a point not lying on the line, say the origin (0,0) . For x=0, y=0 we
find 4x+5y=0, 0>20 is absurd. Thus the point(0,0) does not belong to the region given
by 4 5 20x y+ > . Obviously the region 2R in which (0,0) is located does not correspond
to 4 5 20x y+ > , and thus, we conclude that R1 corresponds to the inequation
4 5 20x y+ > .
Tick the correct answer:
(1) Point (4,2) belongs to 1 2/ /R R lineAB .
(2) (4,2) satisfies 4 5 20x y+ >
4 5 20x y+ <
4 5 20x y+ =
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Note A:
In the above procedure, it is assumed that any line ax+by+c=0 divides the plane
into two regions, with the line as the common boundary, where the linear expression
ax+by+c, has the same sign for all points within the region.
Can you prove this result?
Conclusion :
I. The line AB corresponds to the equation 4 5 20x y+ = .
II. The Region 1R corresponds to the inequation 4 5 20x y+ > .
III. The line AB along with the region 1R corresponds to the inequation 4 5 20x y+ ≥ .
IV. The Region 2R corresponds to the inequation 4 5 20x y+ < .
V. The line AB along with the region 2R corresponds to the inequation 4 5 20x y+ ≤ .
Assignments
Exercise -1 : Draw the graph of the inequation 3x-y ≤ 6.
Exercise-2: Draw the graph of the inequation 3x y≤ .
1.3 Systems of equations and inequations :
We have seen that 0x ≥ represents a region lying towards the right of y-axis
including the y-axis. Similarly the region represented by 0y ≥ , lies above the x-
axis including the x-axis . The question arises what region will be represented by
0x ≥ and 0y ≥ simultaneously.
Obviously , the region given by 0x ≥ , 0y ≥ will consists of those points which
are common to both 0x ≥ and 0y ≥ . It is the first quadrant of the plane.
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Next we consider the region bounded by 0x ≥ , 0y ≥ and 2 8x y+ ≤ .
We have already seen that 0x ≥ and 0y ≥ represents the 1st quadrant. The
graph given by 2 8x y+ ≤ lies towards that side of the line x+2y=8 in which the origin is
situated. Hence the shaded region below in figure 1.31 represents 0x ≥ , 0y ≥ and
2 8x y+ ≤ simultaneously.
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If we have to consider the region bounded by 0x ≥ , 0y ≥ and 2 8x y+ ≤ , then it
will lie in the 1st quadrant and on that side of the line x+2y=8 in which the origin is not
located. The graph is shown by the shaded region, in figure 1.32.
1.4 Linear Progamming
A problem (i): A dealer has Rs.1500/- only for a purchase of rice and wheat. A
bag of rice costs Rs.180/- and a bag of wheat costs Rs.120/-.
(i)a. How many bags of rice/wheat can the dealer purchase if he does not wish
to spend all the money at his disposal?
(i)b. How many bags of rice/wheat can the dealer buy if he wants to spend the
full amount of Rs.1500/-.?
It is evident that if the dealer purchases rice only, he can buy at most 8 bags of
rice. Similarly he can purchase at most 12 bags of wheat. If the dealer buys both rice
and wheat, the total cost should not exceed Rs.1500/-. All possible combinations are
represented in the graph below by points marked by a dot or by a dot in a circle.
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Interpret each marked point in terms of the number rice and wheat bags that can
be purchases. For example : The point G(3,5) shows that 3 bags of rice and 8 bags of
wheat can be bought.
Coordinates of each point marked below the line AB represents an answer to the
question (i)a.
Coordinates of each point marked below the line AB gives an answer to the
question (i)b.
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Study the following table
Table 1
The table shows the possible combinations corresponding to the ping marked by
a dot in a circle. These points are either on the line AB or near to it. Observe also that
the dealer cannot by an extra bag of rice or wheat with the unutilized money (if any) in
the above cases.
Let us look at the problem algebraically. You have seen that there are various
possibilities of purchasing rice and wheat bags. Let the number of rice bags bought be x
and the number of wheat bags purchases be y.
The cost of rice bags purchases =180 x (in rupees)
The cost of wheat bags purchases = 120 y(in rupees)
Total cost of purchase = 180 x+ 120 y (in rupees)
The dealer has only Rs.1500/- to buy these grains.
Therefore 180x + 120 y ≥ 1500
Or 180 x+ 120 y ≤ 1500 ………………….(1)
x and y being the number of bags, they are non-negative integers.
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i.e. x ≥ 0 and y ≥ 0, x,y integers ………………..(2)
Thus our problem is to find x and y satisfying the conditions (1) and (2).
You know that 180x+120y=1500 represents a straight line. Removing the
common factor, this line is also represented by 3x+2y=25. You can recognize this line
as the line AB in the graph 1 of figure-1.40.
All the marked points on the line AB and below it in the first quadrant correspond
to the solution of problem (i). of all the solutions, the possibilities corresponding to
(1,11),(3,8),(5,5) and (7,2) are the ones when the dealer spends the entire amount he
has. All these points lie on the line AB. For this reason, the line AB can be called the
line Maximum Investment. All the other possibilities correspond to the points in the first
quadrant lying below this line and having integral coordinates.
Please refer back to problem (i)
Problem (ii) : The dealer in problem (i) has a storage capacity of ten bags only. How
many bags of rice and wheat can he buy utilizing his money?
It is obvious that some of the solutions found earlier have to be deleted due to
this additional restriction. Since he cannot afford to buy more than ten bags,
symbolically this means:
10x y+ ≤ …………………………(3)
Thus the possibilities corresponding to (1,11), (2,9) and (3,8) do no satisfy the
condition (3).
Are there any other possibilities that do no satisfy this additional restriction and
hence must be deleted? (yes)
With the additional restriction (3), the solution to the problem (ii) are represented
by points marked in the region OAED in graph 1.
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The dealer is in business to earn his living and expects to earn some profit.
Problem(iii): Let us now suppose that the dealer gets a profit of Rs.11/- and Rs.8/- per
bag of rice and wheat respectively. How many bags of rice and wheat should he buy so
as to expect maximum profit subject to the conditions (1),(2),(3) already imposed?
Let us list the possibilities that are likely to maximize the profit. (please refer to
Table 1 also).
Why can’t the dealer buy 6 bags of rice and 4 bags of wheat or 7 bags of rice and
3 bags of wheat?
(In either case the amount required to make the purchase exceeds Rs.1500/-).
It is seen from the table 2 that the dealer can expect maximum profit if he buys 5
bags each of rice and wheat.
Algebraically the profit is given by 11 x + 8 y. Hence the problem (iii) can be
expressed algebraically as:
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To determine integers x and y as to maximize the profit P = 11 x + 8 y, subject to
the conditions
180 120 1500x y+ ≤ …………(1)
0x ≥ , 0y ≥ ……………..(2)
And 1 0x y+ ≤ ……(3)
The conditions (1),(2) and (3) are known as constraints. Please observe that the
constraints are linear inequations.
The function P presents the profit and is to be maximized. It is called the
objective function for the given problem.
In the problem discussed above, P = 11 x + 8 y. For a given value of P, this
equation represents a straight line. As P varies, the equation P = 11 x + 8 y represents
a system of a parallel lines having slope – 11
8. For this reason, P is known as a linear
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objective function.
In graph 2, below, the region defined by the constraints is shown.
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The region OAED in which given constraints are all satisfied is called the
feasibility region .
Check the following:
Draw the line joining the points (8,0) and (0,11). It is one of the line in the system
corresponding to the objective function, because its slope is equal to 11/8. You should
verify that the objective function has the same value for any two points on this line.
Due to this property, any line belonging to the system of parallel lines given by
the objective function , is called an iso-profit line. (iso=equal).
Verify that the position of the iso-profit line through the vertex(5,5) corresponds
to the maximum profit subject to the constraints.
Note : The method of maximizing or minimizing a linear function of several variable
(called objective function), subject to the conditions that the variables are non-negative
and satisfy a set of linear equation or inequations is given the name LINEAR
PROGAMMING.
Consider now the following linear programming problem:
Problem (iv) : In a small-scale industry a manufacturer produces two types of book
cases. The first type of book case requires 3 hours on machine A and 2 hours on
Machine B for completion, where as the second type of book case requires 3 hours on
machine A and 3 hours on machine B. The machine A can run at the most for 18 hours
while the machine B for at the most 14 hours per day. He earns a profit of Rs.30/- in
each book case of the 1st type and Rs.40/- for each 2nd type book case. How many book
cases of each type should he make each day so as to have a maximum profit?
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Solution :
Let x be the number of 1st type book cases and y be the number of the 2nd type
book cases that the manufacturer will produce each day.
It is obvious that x and y must be whole numbers. The other conditions are
1) 0x ≥
2) 0y ≥
3) Since the 1st type book case requires 3 hours on machine A
and 2nd type book case also requires 3 hours on the same
machine A and the working capacity of this machine 1st at
the best 18 hours per day, we have,
3x+3y ≤ 18
i.e x + y ≤ 6.
4) On the machine B, since the 1st type book case takes 2
hours and the 2nd type book case takes 3 hours for completion and
the machine has the working capacity of 14 hours per day, we
have,
2 3 14x y+ ≤ .
5) His profit P, per day is given by
P = 30x + 40y
Now we will have to determine x and y subject to the conditions.
0x ≥
0y ≥
6x y+ ≤
And 2 3 14x y+ ≤
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So that the profit P is maximum. We also require that x and y be integers. Here P
is the objective function.
We use the graphical method to find the solution of the problem. First of all we
draw the graph of these inequations. The graph is as follows.
The shaded region OABC in the graph satisfies all the conditions and is the FEASIBILITY REGION. Every point in this region satisfies all the mathematical inequations and hence is known as feasible solution. Of course, for the given linear programming problem, only feasible points with integral coordinates represent a solution.
Take any point in this region, say (4,0) and calculate the profit at this point. We have:
P = 30 40x y+
= 30 X 4 + 40 X 0 for the point (4,0) = 120.
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Thus the equation : 30 40x y+ = 120
Or 3 4 12x y+ = is satisfied at (4,0).
The above equation represents the straight line FG. Every point of this line yields a profit of Rs.120/-. For example, the point (0,3) lies on this plane. For the point (0,3), which corresponds to no 1st type book case and three second type book cases, the profit is Rs.120/-. This line is an iso-profit line.
If we calculate the profit for a point (2,1) which corresponds to 2 first type book cases and 1 second type case, the profit comes out to be
P = 30 X 2 + 40 X 1
= Rs.100/-
The line 30 40 100x y+ =
Or 3 4 10x y+ = ………….(B)
is drawn as the line HM.
The line 3 4 10x y+ = is parallel to the line 3 4 12x y+ = . In fact all the iso-profit
lines are parallel, and on each point of the iso-profit line the profit is the same.
The profit line HM lies towards the origin side of the line FG and Yields less profit. Will be profit line lying towards the other of FG yield more profit?
Let us calculate the profit at appoint (3,2) which corresponds to 3 first type book cases and 2 second type book cases. The profit
P = 30 40x y+
= 30 X 3 + 40 X 2
= 90+80
= 170 rupees.
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Thus the profit is more as compared to the points which lie on the line FG. Hence it is obvious that as the iso-profit line moves away from the origin the profit goes on
increasing. In order to get the maximum profit we calculate the profit at the vertices of the feasibility region, namely O, A, B and C.
Profit at 0 is zero.
Profit at A = 30 X 6 + 40 X 0 = 180.
Profit at B, which is the point of intersection of
The line x + y =6 and 2x + 3y = 14 is equal to 30 * 4+ 40 * 2 =200
Since the coordinates of the pint C are not integers , we calculate the profit at the
point (1,4) which is a neighboring feasible point of C, and lies on the boundary of the
feasibility region. The profit is 30 * 1 + 40 * 4 = 190.
Verify that for any feasible point in the region OABC, the profit obtained lies
between the minimum value 0 and maximum values 200 of the profit calculated at the
vertices o, A, B and C.
Thus the small scale manufacturer gains maximum profit of Rs.200/- if he
prepares 4 first type book cases and 2 second type book cases.
A profit of Rs.200/- per day seems to be large for an industrialist. Don’t you think,
he/she should reduce the price of the book cases ?
ASSIGNMENTS
1. A mother wishes her children to obtain certain amount of nutrients from their
Breakfast cerials. The children has the choice of eating Idly or Upma or a mixture of
two. From their breakfast they must get 1 gram of thiamine, 5 gms of niacin and 400 cal
, 10 gms of Idly contains 0.10 mg of thiamine, 1mg of niacin and 110 cal, 10 gms of
Upma contains 0.25 mg of thiamine, 50 gms of Upma cost 60 paisa.
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Write down the objective function and constraints to calculate the minimum
expenditure on breakfast.
2. Maximize 45 80x y+
Subject to
5 20 400x y+ <
10 15 450x y+ <
0x ≥
0y ≥
Fit a problem to illustrate the mathematical model.
3. A factory manufactures two types of soaps each with the help of two machine A
and B. A is operated for two minutes and B for 3 minutes to manufacture the first type,
while the second type is manufactured by operating A for 3 minutes and B for 5
minutes. Each machine can be used for at most 8 hours on any day. The two types of
soaps are sold at a profit of 25p and 50p each respectively. Assuming that the
manufacturer can sell all the soaps he can manufacture how many soaps of each type
should the factory produce in a day so as to maximize the profit.
4. A business man has Rs.6000/- at his disposal and wants to purchase cows and
buffaloes to takes up a business. The cost prices of a cow is Rs.300/- and that of a
buffalo is Rs.1200/-. The man can store fodder for the live stock to the extent of 40
quintals per week. A cow gives 10 liters of milk and a buffalo gives 20 liters of milk per
day. Profit per liter of milk cow is 50 paisa and per liter of milk of a buffalo is 90 p. If
consumption of fodder per cow is 1 quintal and per buffalo is 2 quintals a week. Find the
number of live stock of each king the man has to purchase so as to get maximum profit
assuming that he can sell all the quantity of milk, he gets from the livestock.
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10.Projects
1.Path of Pursuits
Objective: To find the paths of four ants placed at the corners of the square , each one
moving in the direction of the ant in front of it. (This path is called the path pursuits).
Take a piece of stiff card board and mark a square ABCD of side 10 cm. Mark
the point A1 on AB at ½ on distance from A. Similarly mark B1,C1 and D1 at 1/2 cm
B,C,D respectively. Now mark A2 at a distance of ½ cm from A1, on the line A1 B1, B2
at ½ from B1 on the line B1C1 and so on. Continue in this way until the centre of the
square is reached. These envelopes are known as curves of pursuit. Since they are the
paths which four ants originally placed at the corners of the square, would follow if they
were always to walk in the direction of the ant infront of them.
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1. Can you stitch the path of pursuits on a black colored cloth using white thread?
2. Where is the point at which all four ants meet each other in the end?
3. Read the chapter on envelopes and evolutes (geometry ) to understand the
significance of this path.
2.Building Trigonometrical Tables Objective : A simple device can be constructed by the students that will enable them to
make their own table of trigonometric ratios for the sine and cosine.
Procedure
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1. On a graph paper, draw a circle with a radius of 10 cm.
2. Out thin strip of cardboard atleast 12cm long.
3. Draw a line down the center of the strip.
4. Attach one end of the strip to the center of the circle.
5. At the other end of the strip, 10 cm from the point where it is attached to the
circle, make a small hole and attach a piece of thread.
6. At the opposite end of the string, Attach a weight to serve as a plumb line.
The strip OB can be rotated around the point O so that OB makes different
angles θ with x-axis. The hanging plummet BD cuts the x-axis at the point C. Count the
number of spaces of length of the cord BC. Since hypotenuse is fixed at 10 cm we can
easily determine since ratio. Sin θ = BC/10. As we change the angle by moving the
cardboard strip we can observe the change in the value of sin θ . Similarly the value of
cos θ also can be read by counting the number of spaces of horizontal axis OA,
cos θ = OC/10.
From this we can get the value of tan θ , cot θ , sec θ and cosec θ . There may
be come error in counting the lengths of BC and OC. Therefore students are asked to
compare these values of sin θ , cos θ , etc. With the standard values given in the
trigonometric tables.
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The students can make different shapes and attach them to the motor to find the shape of the solids of revolution . For example,
1. A circle is attached as follows and rotated about x-axis.
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2. A rectangle is attached as shown and rotated about z-axis.
3.Solids of Revolution Objective: To show that various geometrical figures when revolved around a particular axis give various solids. How to use this aid The teaching aid consists of a motor and various objects of following shapes.
a) Circular.
b) Parabolic.
c) Triangular or angular.
d) Square or rectangular.
The objects are fixed to a pen refill, that should be attached to the motor Rotates about its axis. We get the following solids of revolution.
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4.Explore the Magic Objectives:
1. To enable the students to find out the constant sum after adding any rows and columns or any possible squares.
2. To enable the student to predict the numbers for the empty boxes provided.
Procedure
1. Tow magic squares are drawn on a thermocol and explanation is provided on a chart.
2. First look at the first magic square and you will observe four empty spaces
provided. And you have to fill the expected number inside it.
96 89
88 69 91 16
86 99
19 98 66 81
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3. Since the magic square shows that every rows, columns or diagonals and every
possible squares add upto the same constant, you will be able to find out the
expected constant.
4. Taking the constant sum into consideration you will be able to fill all the boxes
respectively.
5. If you didn’t get the numbers in the empty box then rotate the box and you will
see the correct answer. If case you found it, you turn it and check your answer.
6. In this way the square will become complete magic squares.
18 99 86 61
66 81 98 19
91 16 69 88
89 68 11 96
7. Now turn the numbers of first magic square upside down and you will be really
surprised to observe the name phenomena in the second magic square as well.
8. Thus you will get the constant as 264 !
5.Surface area of the cone Objective: To find the surface area of the cone. Description Take a circle of radius 1. Cut a sector from the circle as shown in the
figure. Then fold the sector to make a cone. Here, the radius 1 of the sector becomes
the slant height of the cone. The arc length of the sector becomes the circumstance of
the circular base of the cone.
Area of the curved surface = Area of the sector = A of the cone.
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Area of the Sector Arc length of the sector
------------------------- = 21
A
π = -----------------------------------
Area of the circle Circumference of the circle
= 2
2 1
rππ
A = 1rπ where r is the radius of the base of the cone and 1 is the slant height. The students can cut the sectors of different sizes from the circle and watch the
relationship between slant height of the cone and the radius of the circle.
1. When you cut a sector from a circle and fold it, the sector takes the shape of the
cone. What about the other sector of the circle?
2. If 1rπ is the area of the curved surface of the cone, then what is the total surface
area for the cone ?
3. If θ is the central angle of the sector of area A as shown aside, what is the
relation between θ and A?
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6.Model Ratio Objective : To find the ratio of the areas of two rectangles.
Problem ABC is an equilateral triangle and PQR is the equilateral triangle inscribed
in the incircle of the triangle ABC. What is the ratio of the areas of the tow triangles ?
Procedure
Rotate the inner triangle PQR by 180o and get the figure as shown above.
Take the other triangle provided in the pocket and keep it on the inner triangle PQR and
other spaces marked 1,2,3. All the four triangles are equal in area to each other. So
area of ∆ ABC = 4 * area of ∆ PQR. Area of ABC 4 ------------------ = --- Area of PQR 1 This ratio is called the model ratio.
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Proof Let the base of triangle ABC = b
Height of ∆ ABC = h = 2
2
4
bb − =
3
2 b
Area of ∆ ABC = 21 3 3
2 2 4b b b× × =
Base of ∆ PQR = 2
b
Height of ∆ PQR = 3
2 2b
×
Area of ∆ PQR = 21 3 3
2 2 4 16
bb b× × =
Area of ∆ ABC 23
4b
------------------------- = ---------
Area of ∆ PQR 23 16
b
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7.Quadrilateral and Parallelogram Objective: To show that the line joining the mid points of a quadrilateral is a
parallelogram.
Given
ABCD is a quadrilateral and E,F,G, H are the mid points of AB,BC, CD and DA respectively.
How to use Make a quadrilateral ABCD by fixing the proper screws to the holes in the
enclosed teaching aid. Find the mid points of the sides in the teaching aid and connect
midpoints by a thread. You will get a parallelogram . Similarly construct another
quadrilateral and join the corresponding midpoints and verify the result.
To prove : EFGH is a parallelogram. Construction : Join BD
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Proof: In ∆ ABD , E and H are the mid points of AB and DA.
EH is parallel to BD and EH = 1
2BD …………………………..(1)
In ∆ BCD, F and G are mid points of BC and CD.
FG is parallel to BD and FG = 1
2BD ……………………………..(2)
EH || BD and FH || BD …… EH||FG. From 1 and 2, EH||FG and EH|=| FG; EFGH is a parallelogram. 8.Three circles Objective: To show that if three circles of equal radii touch each other externally then
the triangle formed by joining the centers of these circles is an equilateral triangle.
In the teaching aid; A,B and C are the centers of the three circles which
Touch the each other externally at the points P,Q and R. The students verify whether
AB, BC and CA are equal by the actual measurements.
Repeat the experiments by changing the circles. In all the cases, the students find that AB = BC= CA.
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Analysis. AB = AP + PB = r + r = 2r BC = BQ + QC =r + r = 2r CA = CR + RA = r + r = 2r ABC is an equilateral triangle.
1. Can you justify that the points A, p and B lie on a single straight line ? (if not, the length of AB will not be equal to AP + PB).
2. If the radii of the three circles are different, then can you guess the result? 9.A Transformation Group Objective: To show that certain transformations of an index card (reflection about x-
axis, reflection about y-axis, rotation about the centre by 180o leads to the formation of
an abelian group.
Procedure Take a blank index card. Place it on an empty sheet of paper and trace around it.
Make a little square in upper right hand corner of both index cord and shade the
rectangle for front and leave it as it is in back side
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Step 1 : Flip the card over about a horizontal axis. Call this move x.
Step 2 : Flip the card over about a vertical axis. Call this move y.
Step 3 : Rotate the card about the centre by 180o . Call this move R.
Step 4: Leave it alone. Call this move I.
XoY : Start
YoR : Start
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RoX : Start
RoR : Start
O I X Y R
I I X Y R
X X I R Y
Y Y R I X
R R Y X I
From the above table, it is evident that (i) system is closed, (ii) it is associates,
(iii) it has an identity element I, (iv) each element has a unique inverse, i.e., itself, and
the system is commutative.
Hence this system of transformation (Reflection about x-axis , reflection about y-
axis, rotation about the centre by 180o ) from an abelian group.
Can you extend this idea to the transformation of an equilateral triangle ?
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10.Path of the Moving Chord Inside a Circle Objective: To illustrate that, the path of the moving chord of constant length inside a
circle is a circle and to find out the radius of this inner circle.
PQ is a chord of constant length which moves inside the circle of a radius R,
centered at the point A. What is the path of PQ ? The students can move the stick PQ
inside the circle and convince themselves tht the path of the moving chord PQ of
constant length inside a circle in a circle. They can repeat the experiment and verify the
above fact. It is also clear that the center of the new circle is also A. What is the radius
of this inner circle?.
To find the radius of the inner circle see Fig. (2) . In which , BD= a (length of the chord) AB = R ( radius of the outer circle) AC = r (radius of the inner circle)
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By Pythagoras theorem,
2 2 2AB AC BC= +
2
2 2
2
aR r
⎛ ⎞= + ⎜ ⎟⎝ ⎠
r= 2
2
4
aR −
1. What happens if the length of the chord PQ is equal to the diameter of the bigger
circle ?
2. What happens if the length of the chord PQ is equal to the radius of the bigger
chord?
11.Square Root by Guess Average Method This is a process in which the square root of a number can be found out correct
to certain number of decimal places by repetition of the process.
Procedure.
1. Take a number and guess a square root of it.
2. Divide the number by the guessed square root.
3. In order to refine the guess, take the average of the guessed square root and the quotient.
4. Repeat the process as often as necessary to gain accuracy to any desired place.
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For eg : Find 12 to connect to the nearest tenth.
1. Guess : 3.3 is the square root of 12.
2. Divide 12 by 3.3.
3.3)120(3.636 (round off to 3.64) 99 ……….. 210 198 ……….. 120 99 …………. 210
3. Compare guessed square root is 3.3, quotient is 3.64.
4. Find the average of guessed square root and the quotient.
3.3 + 3.64 6.94 Average = ---------------------------- = ------- = 3.47 2 2
5. Repeat the entire process using 3.47 as the new guess. Divide 12 by 3.47 3.47) 1200 (3.458 1041 -------- 1590 1388 --------- 2020 1735 --------- 2850
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6. Compare 2 lies between 3.47 and 3.458
3.47 + 3.458 6.928
Average = ---------------------------- = ------- = 3.465 2 2 When we compare we can see that to the nearest tenth 12 = 3.5.
7. To require greater accuracy the process may again the repeated with 3.46 as our
new guess.
This method brings one close to the desired square root rather quickly. If a student failed to see that 12 lies between 3 and 4, choose any number as a
guess and repeat the process from 1 to 7.
12.Conic Sections Objective: To show that when a right circular cone is cut in four specific ways we get
Conic sections namely (1) circle, (2) parabola, (3) ellipse and (4) hyperbola.
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How to use
• Hold the model and chart side by side, disjoint the right circular cone at the place
marked ‘1’ and see that the edge of surface is a circle, i.e. we get a circle by
cutting the right circular cone perpendicular to tis axis by a plane.
• Similarly disjoint the cone at the place marked ‘2’ and see that edge of the
surface is a parabola, i.e. when we cut the cone parallel to one of its side we
obtain parabola.
• Disjoint the cone at the place marked ‘3’ and see that the edge of the surface is
an ellipse, i.e, when we cut the cone at an inclined angle we get ellipse.
• Disjoint the cone at the place marked ‘4’ and see that the edge of the surface is a
‘hyperbola’, i.e, when we cut the cone parallel to the axis we get a hyperbola.
Result
The conic sections are shown by cutting a right circular cone in four specific ways.
1. Is it possible to cut the right circular cone in any other way to get a section other
than the four sections mentioned above?
2. Can you get all the four conic sections by cutting any other object ?
3. Is there any relation between the four conic sections?
