S.A.2 MATHS PROJECT: 2011

Monday, February 14, 2011

FUN IN MATHS

THEREFORE PUT THE ABOVE THEORUM AND GET CHILLED OUT..LOLZ NOT ALWAYS

SEQUENTIAL INPUTS

NO. 2519

MAKE "EM FOOL

SHORTEN YOUR BURDEN

GET SOME FUN

Trick 1: 2's trick

Step1: Think of a number .

Step2: Multiply it by 3.

Step3: Add 6 with the getting result.

Step4: divide it by 3.

Step5: Subtract it from the first number used.
ANSWER WILL ALWAYS BE 2

Trick 2: Any Number

Step1: Think of any number.

Step2: Double the number.

Step3: Add 9 with result.

Step4: sub 3 with the result.

Step5: Divide the result by 2.

Step6: Subtract the number with the number with first number started with.
ANSWER WILL ALWAYS BE 3

MISSION RELEASE PRESSURE.

Tips 2: Any 2 Digit Number

Ex: 47

Step1: Look for the nearest 10 boundary.

3 from 47 to 50

Step2: Since we went up 3 to 50, now go down 3 from 47 to 44.

Step3: Now mentally multiply 44 x 50 = 2200 - 1st answer.

Step4: 47 is 3 away from the 10 boundary 50, Square 3 as 9 - 2nd answer.

Step5: Add the first and second answer, 2200 + 9

TRICK 2
2 Digit Number which ends in 5

Step1: Multiply the first digit of number, with the next to its number. Ex: 35 is the number you want to square. 3 x 4 = 12

Step2: Finally add 25 at the end of the result

MAKE "EM FOOL

Day of the Week:

January has 31 days. It means that every date in February will be 3 days later than the same date in January(28 is 4 weeks exactly). The below table is calculated in such a way. Remember this table which will help you to calculate. January 0

February 3

March 3

April 6

May 1

June 4

July 6

August 2

September 5

October 0

November 3

December 5

Step1: Ask for the Date. Ex: 23rd June 1986

Step2: Number of the month on the list, June is 4.

Step3: Take the date of the month, that is 23

Step4: Take the last 2 digits of the year, that is 86.

Step5: Find out the number of leap years. Divide the last 2 digits of the year by 4, 86 divide by 4 is 21.

Step6: Now add all the 4 numbers: 4 + 23 + 86 + 21 = 134.

Step7: Divide 134 by 7 = 19 remainder 1.

The reminder tells you the day. Sunday 0

Monday 1

Tuesday 2

Wednesday 3

Thursday 4

Friday 5

Saturday 6

Answer: Monday

NUMBER 2519

2519 Mod n

2519 Mod 2 = 1

2519 Mod 3 = 2

2519 Mod 4 = 3

2519 Mod 5 = 4

2519 Mod 6 = 5

2519 Mod 7 = 6

2519 Mod 8 = 7

2519 Mod 9 = 8

2519 Mod 10 = 9

Sequential Numbers with 2519

1259 x 2 + 1 = 2519

839 x 3 + 2 = 2519

629 x 4 + 3 = 2519

503 x 5 + 4 = 2519

419 x 6 + 5 = 2519

359 x 7 + 6 = 2519

314 x 8 + 7 = 2519

279 x 9 + 8 = 2519

251 x 10 + 9 = 2519

SEQUENTIAL INPUTS (BEAUTY OF MATHEMATICS)

Sequential Inputs of numbers with 8

1 x 8 + 1 = 9

12 x 8 + 2 = 98

123 x 8 + 3 = 987

1234 x 8 + 4 = 9876

12345 x 8 + 5 = 98765

123456 x 8 + 6 = 987654

1234567 x 8 + 7 = 9876543

12345678 x 8 + 8 = 98765432

123456789 x 8 + 9 = 987654321

Sequential 1's with 9

1 x 9 + 2 = 11

12 x 9 + 3 = 111

123 x 9 + 4 = 1111

1234 x 9 + 5 = 11111

12345 x 9 + 6 = 111111

123456 x 9 + 7 = 1111111

1234567 x 9 + 8 = 11111111

12345678 x 9 + 9 = 111111111

123456789 x 9 + 10 = 1111111111

Sequential 8's with 9

9 x 9 + 7 = 88

98 x 9 + 6 = 888

987 x 9 + 5 = 8888

9876 x 9 + 4 = 88888

98765 x 9 + 3 = 888888

987654 x 9 + 2 = 8888888

9876543 x 9 + 1 = 88888888

98765432 x 9 + 0 = 888888888

Numeric Palindrome with 1's

1 x 1 = 1

11 x 11 = 121

111 x 111 = 12321

1111 x 1111 = 1234321

11111 x 11111 = 123454321

111111 x 111111 = 12345654321

1111111 x 1111111 = 1234567654321

11111111 x 11111111 = 123456787654321

111111111 x 111111111 = 12345678987654321

Without 8

12345679 x 9 = 111111111

12345679 x 18 = 222222222

12345679 x 27 = 333333333

12345679 x 36 = 444444444

12345679 x 45 = 555555555

12345679 x 54 = 666666666

12345679 x 63 = 777777777

12345679 x 72 = 888888888

12345679 x 81 = 999999999

Sequential Inputs of 9

9 x 9 = 81

99 x 99 = 9801

999 x 999 = 998001

9999 x 9999 = 99980001

99999 x 99999 = 9999800001

999999 x 999999 = 999998000001

9999999 x 9999999 = 99999980000001

99999999 x 99999999 = 9999999800000001

999999999 x 999999999 = 999999998000000001

......................................

Sequential Inputs of 6

6 x 7 = 42

66 x 67 = 4422

666 x 667 = 444222

6666 x 6667 = 44442222

66666 x 66667 = 4444422222

666666 x 666667 = 444444222222

6666666 x 6666667 = 44444442222222

66666666 x 66666667 = 4444444422222222

666666666 x 666666667 = 444444444222222222

LINEAR EQUATIONS IN TWO VARIABLES

In this chapter, you have studied the following points:

1.An equation of the forma x +b y +c = 0, wherea,b andc are real numbers, such thata and

bare not both zero, is called a linear equation in two variables.

2.A linear equation in two variables has infinitely many solutions.
3.The graph of every linear equation in two variables is a straight line.
4.x = 0 is the equation of the y-axis and y = 0 is the equation of the x-axis.
5.The graph ofx =a is a straight line parallel to they-axis.
6.The graph ofy =a is a straight line parallel to thex-axis.
7.An equation of the typey =m x represents a line passing through the origin.
8.Every point on the graph of a linear equation in two variables is a solution of the linear

equation. Moreover, every solution of the linear equation is a point on the graph of the

linear equation.

Friday, February 11, 2011

AREA OF PARALLELOGRAMS AND TRIANGLES

Theorem 9.1 : Parallelograms on the same base and between the same parallels

are equal in area.

Proof : Two parallelograms ABCD and EFCD, on

the same base DC and between the same parallels

AF and DC are given (see Fig.9.12).

We need to prove that ar (ABCD) = ar (EFCD).

In Δ ADE and Δ BCF,

∠ DAE = ∠ CBF (Corresponding angles from AD

BC and transversal AF) (1)

∠ AED = ∠ BFC (Corresponding angles from ED

FC and transversal AF) (2)

Therefore, ∠ ADE = ∠ BCF (Angle sum property of a triangle) (3) Also, AD = BC (Opposite sides of the parallelogram ABCD) (4)

So, Δ ADE ≅ Δ BCF [By ASA rule, using (1), (3), and (4)]

Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas) (5)

Now, ar (ABCD) = ar (ADE) + ar (EDCB)

= ar (BCF) + ar (EDCB) [From(5)]

= ar (EFCD)

So, parallelograms ABcD and EFCD are equal in area.

Example 2 : If a triangle and a parallelogram are on the same base and between the
same parallels, then prove that the area of the triangle is equal to half the area of the
parallelogram.
Solution : Let Δ ABP and parallelogram ABCD be
on the same base AB and between the same parallels
AB and PC (see Fig. 9.14).
You wish to prove that ar (PAB) =
1 ar (ABCD)
2
Draw BQ || AP to obtain another parallelogram ABQP. Now parallelograms ABQP
and ABCD are on the same base AB and between the same parallels AB and PC.
Therefore, ar (ABQP) = ar (ABCD) (By Theorem 9.1) (1)
But Δ PAB ≅ Δ BQP (Diagonal PB divides parallelogram ABQP into two congruent
triangles.)
So, ar (PAB) = ar (BQP) (2)
Therefore, ar (PAB) =
1
2
ar (ABQP) [From (2)] (3)
This gives ar (PAB) =
1 ar (ABCD)
2
[From (1) and (3)]

Theorem 9.2 : Two triangles on the same base (or equal bases) and between the
same parallels are equal in area.Now, suppose ABCD is a parallelogram whose one of the diagonals is AC
(see Fig. 9.20). Let AN⊥ DC. Note that
Δ ADC ≅ Δ CBA (Why?)
So, ar (ADC) = ar (CBA) (Why?)
Therefore, ar (ADC) =
1 ar (ABCD)
2
=
1 (DC AN)
2
× (Why?)
So, area of Δ ADC =
1
2 × base DC × corresponding altitude AN
In other words, area of a triangle is half the product of its base (or any side) and
the corresponding altitude (or height). Do you remember that you have learnt this
formula for area of a triangle in Class VII ? From this formula, you can see that two
triangles with same base (or equal bases) and equal areas will have equal
corresponding altitudes.
For having equal corresponding altitudes, the triangles must lie between the same
parallels. From this, you arrive at the following converse of Theorem 9.2 .
Theorem 9.3 : Two triangles having the same base (or equal bases) and equal
areas lie between the same parallels.
Let us now take some examples to illustrate the use of the above results.

FOR REFERENCE OF DIAGRAMS PLEASE REFER THE NCERT BOOK OF CLASS 9TH.....WE HAVE STATED ABOUT THE FIGURE.

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PROBABILITY

It is remarkable that a science, which began with the consideration of

games of chance, should be elevated to the rank of the most important

subject of human knowledge.

—Pierre Simon Laplace

An event for an experiment is the collection of some outcomes of the experiment.

P(E) =

Total number of trials

Example 9 :
Chapter 14), which gives the weights of 38 students of a class.
(i) Find the probability that the weight of a student in the class lies in the interval 46-50 kg.
(ii) Give two events in this context, one having probability 0 and the other having probability 1.
Consider the frequency distribution table (Table 14.3, Example 4,
Solution :
weight in the interval 46 - 50 kg is 3.
So, P(weight of a interval 46 - 50 kg) =
(i) The total number of students is 38, and the number of students with3   = 0.079
                                                  38
38 = 1. Example 10 :
were kept under standardised conditions favourable to germination. After 20 days, the
number of seeds which had germinated in each collection were counted and recorded
as follows:
Fifty seeds were selected at random from each of 5 bags of seeds, and
Table 15.11
<><><><>

Bag 1 2 3 4 5

Number of seeds germinated 40 48 42 39 41

What is the probability of germination of
(i) more than 40 seeds in a bag?
(ii) 49 seeds in a bag?
(iii) more that 35 seeds in a bag?
Solution :
(i) Number of bags in which more than 40 seeds germinated out of 50 seeds is 3.
P(germination of seeds in a bag) = 3= 0.6
                                               5
Total number of bags is 5.
(ii) Number of bags in which 49 seeds germinated = 0.0 = 0.5 = 1.                                      5 In all the examples above, you would have noted that the probability of an

Remark :
event can be any fraction from 0 to 1.
P(germination of 49 seeds ) =
                                        5
(iii) Number of bags in which more than 35 seeds germinated = 5.
So, the required probability =

(ii) For instance, consider the event that a student weighs 30 kg. Since no student has
this weight, the probability of occurrence of this event is 0. Similarly, the probability
of a student weighing more than 30 kg is
                                                                  38

The Probability of an event lies between 0 and 1 (0 and 1 inclusive).The empirical (or experimental) probability P(E) of an event E is given byNumber of trials in which E has happened

Wednesday, February 9, 2011

CONSTRUCTIONS.

Construction 11.1 : To construct the bisector of a given angle.

Given an angle ABC, we want to construct its bisector.

Steps of Construction :

1. Taking B as centre and any radius, draw an arc to intersect the rays BA and BC,

say at E and D respectively [see Fig.11.1(i)].

2. Next, taking D and E as centres and with the radius more than

1 DE, draw arcs to

intersect each other, say at F.

3. Draw the ray BF [see Fig.11.1(ii)]. This ray BF is the required bisector of the angle

ABC.Let us see how this method gives us the required angle bisector.

Join DF and EF.

In triangles BEF and BDF,

BE = BD (Radii of the same arc)

EF = DF (Arcs of equal radii)

BF = BF (Common)

Therefore, ∆BEF ≅ ∆BDF (SSS rule)

This gives ∠EBF = ∠ DBF (CPCT)

Construction 11.2 : To construct the perpendicular bisector of a given line

segment.

Given a line segment AB, we want to construct its perpendicular bisector.

Steps of Construction :

1. Taking A and B as centres and radius more than

AB, draw arcs on both sides of the line segment

AB (to intersect each other).

2. Let these arcs intersect each other at P and Q.

Join PQ (see Fig.11.2).

3. Let PQ intersect AB at the point M. Then line

PMQ is the required perpendicular bisector of AB.

L e t u s s e e h ow t h i s me t h o d g i v e s u s t h e

perpendicular bisector of AB.

Join A and B to both P and Q to form AP, AQ, BP

and BQ.

In triangles PAQ and PBQ,

AP = BP (Arcs of equal radii)

AQ = BQ (Arcs of equal radii)

PQ = PQ (Common)

Therefore, ∆ PAQ ≅ ∆ PBQ (SSS rule)

So, ∠ APM = ∠ BPM (CPCT)

Now in triangles PMA and PMB,

AP = BP (As before)

PM = PM (Common)

∠ APM = ∠ BPM (Proved above)

Therefore, ∆ PMA ≅ ∆ PMB (SAS rule)

So, AM = BM and ∠ PMA = ∠ PMB (CPCT)

As ∠ PMA + ∠ PMB = 180° (Linear pair axiom),

we get

∠ PMA = ∠ PMB = 90°.

Therefore, PM, that is, PMQ is the perpendicular bisector of AB.

Construction 11.3 : To construct an angle of 60

at the initial point of a given ray.

Let us take a ray AB with initial point A [see Fig. 11.3(i)]. We want to construct a ray

AC such that ∠ CAB = 60°. One way of doing so is given below.

Steps of Construction :

1. Taking A as centre and some radius, draw an arc

of a circle, which intersects AB, say at a point D.

2. Taking D as centre and with the same radius as

before, draw an arc intersecting the previously

drawn arc, say at a point E.

3. Draw the ray AC passing through E [see Fig 11.3 (ii)].

Then ∠ CAB is the required angle of 60°. Now,

let us see how this method gives us the required

angle of 60°.

Join DE.

Then, AE = AD = DE (By construction)

Therefore, ∆ EAD is an equilateral triangle and the ∠ EAD, which is the same as

∠ CAB is equal to 60

Construction 11.4 : To construct a triangle, given its base, a base angle and sum

of other two sides.

Given the base BC, a base angle, say ∠B and the sum AB + AC of the other two sides

of a triangle ABC, you are required to construct it.

Steps of Construction :

1. Draw the base BC and at the point B make an

angle, say XBC equal to the given angle.

2. Cut a line segment BD equal to AB + AC from

the ray BX.

3. Join DC and make an angle DCY equal to ∠BDC.

4. Let CY intersect BX at A (see Fig. 11.4).

Then, ABC is the required triangle.

Let us see how you get the required triangle.

Base BC and ∠B are drawn as given. Next in triangle

ACD,

∠ACD = ∠ ADC (By construction)

Therefore, AC = AD and then

AB = BD – AD = BD – AC

AB + AC = BD

Alternative method :

Follow the first two steps as above. Then draw

perpendicular bisector PQ of CD to intersect BD at

a point A (see Fig 11.5). Join AC. Then ABC is the

required triangle. Note that A lies on the perpendicular

bisector of CD, therefore AD = AC.

Remark : The construction of the triangle is not

possible if the sum AB + AC ≤ BC

Construction 11.5 : To construct a triangle given its base, a base angle and the

difference of the other two sides.

Given the base BC, a base angle, say ∠B and the difference of other two sides

AB – AC or AC – AB, you have to construct the triangle ABC. Clearly there are

following two cases:

Case (i) : Let AB > AC that is AB – AC is given.

Steps of Construction :

1. Draw the base BC and at point B make an angle

say XBC equal to the given angle.

2. Cut the line segment BD equal to AB – AC from ray BX.

3. Join DC and draw the perpendicular bisector, say PQ of DC.

4. L e t i t i n t e r s e c t BX a t a p o i n t A. J o i n A C

(see Fig. 11.6). then ABC is the required triangle.

Let us now see how you have obtained the required triangle ABC.

Base BC and ∠B are drawn as given. The point A lies on the perpendicular bisector of

DC. Therefore, AD = AC

So, BD = AB – AD = AB – AC.

Case (ii) : Let AB < AC that is AC – AB is given.

Steps of Construction : 1. Same as in case (i).

2. Cut line segment BD equal to AC – AB from the

line BX extended on opposite side of line segment BC.

3. Join DC and draw the perpendicular bisector, say PQ of DC.

4. Let PQ intersect BX at A. Join AC (see Fig. 11.7).

Then, ABC is the required triangle.

You can justify the construction as in case (i).

To construct a triangle, given its perimeter and its two base
Given the base angles, say
the triangle ABC.
∠ B and ∠ C and BC + CA + AB, you have to construct
Steps of Construction :
1. Draw a line segment, say XY equal to BC + CA + AB.
2. Make angles LXY equal to
3. Bisect
[see Fig. 11.8(i)].
∠B and MYX equal to ∠C.∠ LXY and ∠ MYX. Let these bisectors intersect at a point A
Fig. 11.8 (i)
4. Draw perpendicular bisectors PQ of AX and RS of AY.
5. Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC
[see Fig 11.8(ii)].
Fig. 11.8 (ii)
Then ABC is the required triangle. For the justification of the construction, you
observe that, B lies on the perpendicular bisector PQ of AX.
Therefore, XB = AB and similarly, CY = AC.
This gives BC + CA + AB = BC + XB + CY = XY.
Again
∠BAX = ∠AXB (As in Δ AXB, AB = XB) and
∠
Similarly,
ABC = ∠BAX + ∠AXB = 2 ∠AXB = ∠LXY∠ACB = ∠MYX as required.

Construction 11.6 :

angles.

CIRCLES

Theorem 10.1 : Equal chords of a circle subtend equal angles at the centre.

Proof : You are given two equal chords AB and CD

of a circle with centre O . You want

to prove that ∠ AOB = ∠ COD.

In triangles AOB and COD,

OA = OC (Radii of a circle)

OB = OD (Radii of a circle)

AB = CD (Given)

Therefore, ∆ AOB ≅ ∆ COD (SSS rule)

This gives ∠ AOB = ∠ COD

(Corresponding parts of congruent triangles) „

Theorem 10.2 : If the angles subtended by the chords of a circle at the centre

are equal, then the chords are equal.

The above theorem is the converse of the Theorem 10.1.

if you take ∠ AOB = ∠ COD, then

∆ AOB ≅ ∆ COD (Why?)

Can you now see that AB = CD?

Theorem 10.2 : If the angles subtended by the chords of a circle at the centre

are equal, then the chords are equal.

The above theorem is the converse of the Theorem 10.1.

if you take ∠ AOB = ∠ COD, then

∆ AOB ≅ ∆ COD (Why?)

Can you now see that AB = CD?

Theorem 10.3 : The perpendicular from the centre of a circle to a chord bisects

the chord.

What is the converse of this theorem? To write this, first let us be clear what is

assumed in Theorem 10.3 and what is proved. Given that the perpendicular from the

centre of a circle to a chord is drawn and to prove that it bisects the chord. Thus in the

converse, what the hypothesis is ‘if a line from the centre bisects a chord of a

circle’ and what is to be proved is ‘the line is perpendicular to the chord’. So the

converse is:

Theorem 10.4 : The line drawn through the centre of a circle to bisect a chord is

perpendicular to the chord.

Is this true? Try it for few cases and see. You will

see that it is true for these cases. See if it is true, in

general, by doing the following exercise. We will write

the stages and you give the reasons.

Let AB be a chord of a circle with centre O and

O is joined to the mid-point M of AB. You have to

p r o v e t h a t O M ⊥ A B . J o i n O A a n d O B

(see Fig. 10.16). In triangles OAM and OBM,

OA = OB (Why ?)

AM = BM (Why ?)

OM = OM (Common)

Therefore, ∆OAM ≅ ∆OBM (How ?)

This gives ∠OMA = ∠OMB = 90° (Why ?)

Theorem 10.5 : There is one and only one circle passing through three given

non-collinear points

Theorem 10.6 : Equal chords of a circle (or of congruent circles) are equidistant

from the centre (or centres).

Next, it will be seen whether the converse of this theorem is true or not. For this,

draw a circle with centre O. From the centre O, draw two line segments OL and OM

of equal length and lying inside the circle Then draw chords PQ

and RS of the circle perpendicular to OL and OM respectively

Measure the lengths of PQ and RS. Are these different? No, both are equal. Repeat

the activity for more equal line segments and drawing the chords perpendicular to them. This verifies the converse of the Theorem 10.6 which is stated as follows:

Theorem 10.7 : Chords equidistant from the centre of a circle are equal in

length.

We now take an example to illustrate the use of the above results:

Theorem 10.8 : The angle subtended by an arc at the centre is double the angle

subtended by it at any point on the remaining part of the circle.

Proof : Given an arc PQ of a circle subtending angles POQ at the centre O and

PAQ at a point A on the remaining part of the circle. We need to prove that

∠ POQ = 2 ∠ PAQ

Consider the three different cases as In (i), arc PQ is minor; in (ii),

arc PQ is a semicircle and in (iii), arc PQ is major.

Let us begin by joining AO and extending it to a point B.

In all the cases,

∠ BOQ = ∠ OAQ + ∠ AQO

because an exterior angle of a triangle is equal to the sum of the two interior opposite

angles.

Theorem 10.9 : Angles in the same segment of a circle are equal.

Again let us discuss the case (ii) of Theorem 10.8 separately. Here ∠PAQ is an angle

in the segment, which is a semicircle. Also, ∠ PAQ = 1/2∠ POQ = 1/2× 180° = 90°.

If you take any other point C on the semicircle, again you get that

∠ PCQ = 90°

Therefore, you find another property of the circle as:

Angle in a semicircle is a right angle.

The converse of Theorem 10.9 is also true. It can be stated as:

Theorem 10.10 : If a line segment joining two points subtends equal angles at

two other points lying on the same side of the line containing the line segment,

the four points lie on a circle (i.e. they are concyclic).You can see the truth of this result as follows:

In Fig. 10.30, AB is a line segment, which subtends equal angles at two points C and

D. That is∠ ACB = ∠ ADB

To show that the points A, B, C and D lie on a circle

let us draw a circle through the points A, C and B.

Suppose it does not pass through the point D. Then it

will intersect AD (or extended AD) at a point, say E

(or E′).

If points A, C, E and B lie on a circle,

∠ ACB = ∠ AEB (Why?)

But it is given that ∠ ACB = ∠ ADB.

Therefore, ∠ AEB = ∠ ADB.

This is not possible unless E coincides with D. (Why?)

Similarly, E′ should also coincide with D.

10.8 Cyclic Quadrilaterals

A quadrilateral ABCD is called cyclic if all the four vertices

of it lie on a circle .You will find a peculiar

property in such quadrilaterals. Draw several cyclic

quadrilaterals of different sides and name each of these

as ABCD.

T h e o r e m 1 0 . 11 : T h e s u m o f e i t h e r p a i r o f o p p o s i t e a n g l e s o f a c y c l i c

quadrilateral is 180º.

In fact, the converse of this theorem, which is stated below is also true.

Theorem 10.12 : If the sum of a pair of opposite angles of a quadrilateral is

180º, the quadrilateral is cyclic.

You can see the truth of this theorem by following a method similar to the method

adopted for Theorem 10.10.

FOR REFERENCE OF DIAGRAMS PLEASE REFER THE NCERT BOOK OF CLASS 9TH.....WE HAVE STATED ABOUT THE FIGURE.