   ### Video Learning 4 Kids - 5 ( SIMILAR TRIANGLES )

SIMILARITY OF TRIANGLES

What actually is similarity ?
It just means that the angles are same. So two triangles are similar if all three angles of first triangle are equal to the three angles of the second triangle.
Suppose I say that the two angles A & B of first triangle are equal to the two angles say P & Q of the second triangle, then it is obvious that the third angle of both will be same because the sum of angles in triangle is 180.
That is why it is enough to say Angle-Angle theorem (AA) and not AAA.
We here must also note that the ratio of sides of triangles is same corresponding to the equal angles.
We must also be aware that capital letters are used for angles and small letters are used for the sides.
So if A=P, B=Q & C=R, then we will have a/p = b/q = c/r
Suppose this ratio turns out to be one. Then these become congruent triangles.
Another way to show similarity is SAS, in this the two sides of the two triangles are in same ratio and the angle included between them is equal. This becomes the SAS theorem (Side Angle Side).

### VIDEO LEARNING 4 KIDS - 4

ANGLES AND TRIANGLES
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FINDING ANGLES

video (On mathematics )- This video shall explain for how to calculate the angles of the trianlges.

CONGRUENCY OF TRIANGLES
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﻿

VIDEO ON MATHEMATICS
This video shall explain how you can prove traingles are congruent or not !

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### VIDEO LEARNING 4 KIDS - 3 ( ALGEBRA & RATIO )

VIDEO LEARNING FOR KIDS

R A T I O- Triangles & Angles

ALGEBRA

Mathematics Plus:

Video Source:
yourteacher.com

### Video Learning 4 Kids - 2

VIDEO LEARNING FOR KIDS - SQUARE ROOT

SQUARE ROOT FOR 40

SIMILARLY SQUARE ROOT FOR 90

Mathematics Plus.

Source : www.yourteacher.com

### VIDEO LEARNING 4 KIDS - 1 (PYTHAGORAS)

HELLO.
THIS VIDEO IS MAINLY FOR KIDS WHO ARE LEARNING MATHS. IF YOU HAVE KIDS IN YOUR FAMILY THEN YOU CAN SHOW THEM >>>> THEY MIGHT FIND IT INTERESTING TO WATCH IT >> AND AT THE SAME TIME THEY WILL LEARN.....

NOT FOR THE ONES LOOKING FOR HIHGER MATHS

THE BELOW VIDEO SHOWS THE PYTHAGORAS THEOREM

THE BELOW VIDEO SHOWS HOW TO GET AREA OF RECTANGLE - PYTHAGORAS THEROEM USED

### Ratio And Proportions - Selected Problems in Ratio and Proportion

Ratio And Proportions

Well if you are planning to go for any interview, sitting for any placement, or any competitive exam for any job application, pin pointing to especially Bank PO exams / Bank Clerical Exams or any exam for recruitment to govt posts, you cannot miss out this topic of Mathematics. Then better have a look at it (Ratio and Proportion). Atelast 3 to 4 Questions in mathematics under algebra will surely be framed from this section because this is neither tough nor lengthy but good and concise questions can be framed from here to test your calculations and understanding. SO DONT MISS OUT THE EASY MARKS IN MATHEMATICS....Just take pain of 10 minutes to digest this capsule on Ratio & Proportion and make yourself little aware / refresh.

Problem in Ratio and Proportion (Mathematics) :
Three positive numbers are in the ratio 2:3:4 and the sum of their square is 1421. Find the numbers?

Solution:
Let the numbers in the ratio 2:3:4 be 2x,3x and 4x
Sum of their squares = (2x)2 + (3x)2 + (4x)2 =4x2 + 9x2 + 16x2 = 29x2 = 1421
x2 = 1421/29 = 49
x = 7

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Continued Proportion –
a, b and c are said to be in continued proportion if a/b =b/c or means b2=ac
Similarly we can also have a/b=b/c=c/d=d/e.......... for more numbers

Problem in Ratio and Proportion (Mathematics):
If a, b, c, d are in continued proportion, the prove that a. b. d. e = c4

Solution :
Since they are in continued proportion, we can have a/b =b/c =c/d =d/e =say k
so we have
a=bk
b=ck
c=dk or d=c/k
d=ek or e=d/k

We have to prove a. b. c. e = (bk).(ck). (c/k) . (d/k) = (ckk).(ck).(c/k).(c/kk) = c4 k3 / k3 = c4

Problem in Ratio and Proportion (Mathematics):
If a/b=c/d=e/f
prove that, (a+3c-5e)/(b+3d-5f) = a/b = c/d = e/f

Solution :
Let a/b=c/d=e/f = k(say)
(bk+3dk – 5fk) / (b+3d – 5f) = k (b+3d – 5f)/(b+3d – 5f) = k
Hence Proved

Problem in Ratio and Proportion (Mathematics) :
If p/(b-c) = m/(c-a) = n/(a-b) then prove that ap+bm+cn = 0

Solution:
Take each ratio = k
p=(b-c)k
m=(c-a)k
n=(a-b)k

Then ap+bm+cn = a(b-c)k + b(c-a)k + c(a-b)k = abk -ack + bck – abk +ack -cbk = 0

Problem in Ratio and Proportion (Mathematics) :
Ram, Sham and Suresh start business investing in the ratio 1/2 : 1/3: 1/4. The time for which each of them invested their money was in the ratio 8:6:12 respectively. If they get profit of Rs 18000 from the business. Then how much share of profit will each one get ?

Solution:
ratio of Investment = 1/2:1/3:1/4=6:4:3 and time for which investment is done = 8:6:12
So in actual their amount was in business in the ratio 6*8 : 4*6 : 3*12 =48:24:36 =4:2:3
So they should also share the profit in this ratio 4:2:3
So Ram gets = 4/9 *18000 =8000
Sham gets = 2/9*18000 = 4000
Suresh gets = 3/9*18000 = 6000

{This is actually based on the concept that a small sum put in for large time and large sum put in for small time, so actually we see for how much period we can have money on same scale. You can compare by money * time }

Problem in Ratio and Proportion (Mathematics) :
p,q and r are three positive numbers and Q=(p+q+r)/2
If (Q-p):(Q-q):(Q-r) = 2:5:7, then find the ratio of p,q and r ?

Solution
:
Substitute the value of Q =(p+q+r)/2 everywhere
you will get
(p+q+r-2p)/2 : (p+q+r-2q)/2 : (p+q+r -2r)/2 = 2:5:7
(p+q+r-2p) : (p+q+r-2q) : (p+q+r -2r) = 2:5:7
(q+r-p) : (p-q+r) : (p+q -r) = 2:5:7
or (q+r-p) : (p-q+r) : (p+q -r) = 2k:5k:7k
To solve this you can assume the following equations
q+r-p =2k --------( 1 )
p-q+r =5k --------( 2 )
p+q-r =7k --------( 3 )

q+r-p + p-q+r = 7k
2r = 7k
r = 7k/2

2p = 12k
p=6k

2q = 9k
q=9k/2

So p:q:r = 6k: 9k/2 : 7k/2 = 6 : 9/2 : 7/2 =12:9:7

### LOGARITHMS

Logarithm and logarithmic based functions and questions can be framed in every section of mathematics from algebra to calculus and even geometry. But the most confusing and difficulty faced is in the calculus and differentiation and continuity when the question is combined with multiple concepts in the scene. It becomes difficult to handle the question if you do not know in detail about how to move around with logarithms.

Most of you must have got confused with logarithms raised to powers and powers raised to log functions. Yes, it will become more confusing if such things appear in calculus, at that point of time you will have to also keep in mind the continuity, domain, range and base. Yes, another important thing is 'Base'.
Many of you might not even know how the base is changed. Remember if you are not used to such simple simple things, you will get stuck in the middle of any question and to your surprise even you might get stuck at the last stage of your solution because you are helpless to change base, as you dont know how to do it !!

So why not just spend a few minutes, to understand how actually you can mover around playing with indices, properties of log functions and changing bases and raising powers to numbers and even bases of log functions.

Important Points
The logarithmic function is the inverse of exponential function. N = ax → logaN = X
Two of the most important logarithmic base used are 10 and e.
Log to the base e is called Natural Log.
Log of negative number is not defined. You will never find log(-5). Where you find log(-x), it is clear that here x < 0, so that inside parameter of log function becomes positive.

Click on the question to enlarge  __________________________________

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IIT JEE EXAM 2011- EXAM DATE 10 April 2011, Last date to Apply = 15 December 2010

### SIMPLE TEST FOR 10+2 Students

Here is a simple test on mathematics for class 10+2. These 5 questions will test how much hold you have on the concepts & how well prepared you are with mathematics !!!!! Don't think just jump in ! After all Mathematics never killed anyone !!!!

### Simple Excercise

For those who are weak in mathematics it is not recommended that you jump to higher concepts question very quickly. If your base is not strong, we cannot expect you to yield results by studying 10 or 20 hours and produce miracle results. It actually is more important first to focus on your foundation and work on it. This starts from some simple simple questions of mathematics which will increase your confidence level, then moving towards the second level which is a little tricky type and more interesting problems in mathematics but not too much difficult to require core mathematics.When finally you build a more stronger base in the mathematics, then at this level it is advisable to go for higher concepts and problems in mathematics.So lets start by checking ourselves at the first two level, that do we indeed to work more before moving up or not.
Question: Try to Simplify
(5√150 + 6√24 - 2√96) : √3
Ans :
(5√150 + 6√24 - 2√96) : √3
=(5√25√2√3 + 6√8√3 - 2√16√2√3) : √3
=(25√2√3 + 12√2√3 - 8√2√3) : √3
=(25√2 + 12√2 - 8√2)√3 : √3
=(25√2 + 12√2 - 8√2) : 1
=29√2 : 1

Question :  Prove that 0/0 ≠ 1
Ans:

Find the area of the unshaded region ?:
Are of the Square =
102
Area of the semi circle = ½ R2 = ½ 52
Area of unshaded region = 100 - 25/2

Solution

IIT JEE EXAM 2011- EXAM DATE 10 April 2011, Last date to Apply = 15 December 2010

### Equations and Inequations - I

Problem : Find the smallest value of k for which both the roots of the equation  x2 - 8 k x+ 16 (k2 – k + 1) = 0 will be real roots ?
Solution : For both roots to be real first of all the Discriminant must be greater than zero
Hence,   ∆ ≥ 0
64k2-4 (16(k2-k+1) ) 0
64k2 -64k2 + 4k-4 0
k1
So the minimum value is k = 1

Problem : Find the maximum and minimum value of x such that x2 - 8 x + 6 < 0
Solution : We can first find where the expression x2 - 8 x + 6 = 0 holds true, thereon we can better decide which direction to go.
x2 - 8 x + 6 = 0 gives us,
x = -(-8) ± √64-4*6
x = 8 ± √64-16
x = 8 ± √48
x = 8 ± 4√3

So the minimum value of x is 8 – 4√3 and the maximum value is 8 + 4√3.
In between these two points the expression will be less than zero.

+ve ‹‹‹‹ +ve ‹‹‹ (8- 4√3) ›››››› -ve ‹‹‹‹ (8 + 4√3) ›››› +ve ›››› +ve

Problem: Find the values of x satisfying the following,
x2- 5x + 6 > 0, x2- 15x + 16 < 0 and x2 – 6x + 16 < 0
Solution: We shall solve each of the inequations seperately and then take the common values in each.
First equation, x2- 5x + 6 > 0
On solving you will get, x >2 & x < 3

Second equation, x2- 15x + 16 < 0
x = [15 ± √225-64 ] /2
x = [15 ± √161] /2
Thus we get,
x > [15 - √161] /2 & x <[15 + √161] /2

Third equation, x2 -6x - 16 < 0
On solving we get, x > - 2 & x < 8

The confusing but not difficult part now, taking the common values. We will better do it on a number line to avoid any confusion.

Hence from above it is very clear that the common region is 2 < x < 3

IIT JEE EXAM 2011- EXAM DATE 10 April 2011, Last date to Apply = 15 December 2010

### Satellite View of India on Diwali

SATELLITE VIEW OF INDIA ON THE OCCASSION OF DIWALI﻿.......

BEFORE DIWALI
﻿ India just 2 days Before Diwali

ON DIWALI NIGHT Viewing India from sky on Deepawali
.......... SIMPLY AWSOME...... WONDERFUL............

### Indian Bank PO Question

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The following questions appeared in Indian Bank PO exam. I shall solve these here to show you how easy it is. So if you are planning to get into the Bank Sector at good position it is important that you stop fearing about mathematics and approach in a more tactful & logical way. The data has been manipulated to create different questions.
_________________________________________________

Problem : Study the following pie chart & table carefully and answer the questions that follow:

The below figure shows the % breakup of total employees working in various departments of an organisation and the ratio of men to women working in them.
Total number of Employees working = 1800

Ration of Men to Women
 Department Men Women Production 11 1 HR 1 3 IT 5 4 Marketing 7 5 Accounts 2 7

Question : What is the number of Men working in the Marketing Department ?
1. 132
2. 174
3. 126
4. 189
5. None of these
Solution : Marketing dept holds 18% of the total 1800 employees. So total number of people in Marketing are = 0.18* 1800 = 324

In the Marketing department the ratio of men is to women is 7 : 5. This means that total number of men in the department = ( 7 / 12 ) x 324 = 189

Number of men = ( men / total men  + women  )  X   Total people in department
This can be weel understood by the fact that out of total 12 people 7 are men and 5 are women. So we use 7/12 of 324 as all the men in the department.

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Question : What is the ratio of number of men working in Accounts to the number of Women working in IT department ?

Solution :  No of men in the Accounts department = 2 / 9  X   (17% of 1800) = 2/9 x 0.17 x 1800 = 68
No of women in the IT department = 4 / 9  X (23% of 1800) = 4/9 x 0.23 x 1800 = 184

Ratio of Men in Accounts to the Ratio of Women in IT = 68 / 184 = 17 / 46

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Question : The number of women working in IT department as a percentage of number of women in HR department ?
Solution : Women in IT = 4 /9   X  ( 23% of 1800 ) = 4 / 9 X 0.23  X  1800 = 184

Women in HR deptt = 3/4   X  14% of 1800 = 3/4  X  0.14  X  1800 = 189

Women working in IT as percentage of Women in HR = Women in IT / Women in HR   X  100 %
= 184 / 189 X 100%  =  94.77 %

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Question : What is the ratio of  number of women exceeding men in accounts department to the number of men exceeding women in production ?
Solution  : Women exceeding men in accounts = Women - Men

Women in Account =  7 / 9 X 0.17  X 1800 = 238
Men in Account = 2/9 X 0.17 X 1800 = 68
Number of Women Exceeding Men = 238 - 68 =  170

Number of Men in Production = 11/12  X  28%  X  1800 = 11/12 x 0.28 x 1800 = 462
Number o Women in Production = 1 /12  X 28% X 1800 = 1/12  x  0.28  x 1800 = 42
Number of Men Exceeding Women = 462 - 42 =  420

Ratio of Women exceeding men in Acc to the number of Men exceeding Women in Production = 170 /420
= 17 : 42

Alternatively, this can also be done as :
Number of Women exceeding in Acc = ( 7 - 2 ) / 9   X   Number of people in Accounts
= 5/9   X  0.17 X 1800 = 170

Similarly number of Men Exceeding in Production = ( 11 - 1 ) / 12   X  Number of people in Production
=10/12 X  0.28  X 1800 = 420
Required ratio is thus  = 170 / 420 = 17 : 42

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IIT JEE EXAM 2011- EXAM DATE 10 April 2011, Last date to Apply = 15 December 2010

### Salary Day at Various Places -

Hello everybody......This post doesn't relate to mathematics or any question, but just to share the lighter side of salary day. I recieved this mail and thought why not share this too.....Not many of you might realize the importance of salary day today but sooner or later you will... when you start earning.........

SALARY DAY IN RUSSIA

﻿﻿ SALARY DAY IN RUSSIA
﻿﻿

﻿ SALARY DAY IN USA
﻿
﻿ UK
﻿ SALARY DAY IN AUSTRALIA SALARY DAY IN HONG KONG SALARY DAY IN TEXAS

﻿ SALARY DAY IN INDIA
﻿﻿ SALARY DAY IN KERELA
﻿ SALARY DAY IN PAKISTAN SALARY DAY IN AFGHANISTAN

### Geometry - Straight Lines & Triangle

Co-Ordinate Geometry

• Centroid : Medians are concurrent (G)
• Incentre: Int angle bisector are concurrent (I)
• Orthocentre : Altitudes are concurrent (H)
• Circumcentre : Perpendicular bisector are concurrent (O)
• H,G,O are collinearG divides HO in the ratio 2:1
• In an isoceles triangle all centres are collinear.

Problems:
1. Find the straight line passing through the point of intersection of straight lines x+y-5 = 0 and 4x-7y+10 = 0 and which is farthest from the point (2,2).
2. Prove that a Δ having all vertices with integer co-ordinates cannot be equilateral.??

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IIT JEE EXAM 2011- EXAM DATE 10 April 2011, Last date to Apply = 15 December 2010
Profit Loss & Percentage - 1

A person sold a table at a gain of 15 %. Had he bought for 25% less and sold it for Rs. 80 less, then he would have made a profit of 32%. Find the Cost Price of Table ?

Let the CP be x.
So his SP is at 15% profit means , S.P. = 1.15x
If CP was 25% less, it says if CP was 0.75x, and the SP was Rs 80 less, means if the SP was 1.15x-80 , then the Profit % would have been 32%
Profit% = ( SP - CP ) /  CP = (1.15x-80 - 0.75x) / 0.75x = 32 /100
=>  0.4x - 80 = 0.32 * 0.75x
=>  0.4x - 80 =  0.24x
or 0.4x - 0.24x = 80
0.16x = 80
Thus we get x= 80 / 0.16 =  500

Hence the Cost Price of the table is Rs 500