3.If ab = xy , then (a +b) ≠ (x+y) ,which smaller ? what will be its difference ?

     If the two numbers in the pair are widely separated, its sum will be larger  .  e.g., 8x 3 = 6x4 = 24

8 and 3 are widely separated than 6 and 4, hence the sum  8 + 3 =11 is larger than the sum 6 + 4 = 10.

Let ab = xy

(a+b)²  =  a² + b² + 2ab  or  2ab = (a+b)²  -  a² - b²  and

(x+y)²  =  x² + y² + 2xy  or  2xy = (x+y)²  - x² - y²

Invariably It satisfies the relation (a+b)²  -  a² - b²   = (x+y)²  - x² - y² or (a+b)² -  (x+y)² = a² + b² - x² - y²

If  ab = xy ,then  (a+b)²  + (x-y)² =  (a-b)²  + (x+y)²

 2. If a + b = x + y  then ab ≠ xy. If ab  ≠ xy which is smaller ?  what will be its difference ?

 If the two numbers in the pair are widely separated, its product will be smaller.  e.g., in 7+11 = 5 +13 , 5 and 13 are widely separated than 7 and 11,  hence  the product 5 x 13 = 65 is smaller than the product 7 x 11= 77. 

Let a + b = x+y = 2k

a = (k-n) and b = (k+n) , ab = k² – n²

x = (k+m) and y = (k-m), xy = k² -  m² 

which shows that n= {(b-a)/2] and m = [(x-y)/2]

(a+b)²  =  (x+y)²

 

a² + b² + 2ab =  x²  + y²  + 2xy

(k-n)²  + (k+n)²  + 2ab = (k +m)²  + (k – m)²  + 2xy

 

2(k² + n² ) + 2ab =  2(k² + m² ) + 2xy

Or ab – xy = m²  -  n²  = [(x-y)/2]² – [(b-a)/2]²

 1.  A----------------------------------------------------------------------B

                                                  <-L->

    A rope AB with length L is given. With which one can construct many different rectangles ,but all with same perimeter P. These different rectangles have different area A.

                                  

                                                                           

                                                                           

 

Which rectangle will have maximum area ?

These rectangles have same perimeter P = 2(l+b)

Area of a rectangle is the product of its length and breadth A = l x b

This geometrical figures predict an arithmetical fact – If the  sum of  any two numbers equals with sum of two different numbers  then their product will not be equal and vice-versa.

If a + b = x +y , then ab ≠ xy

If ab = xy then a+b ≠ x + y

          a + b = x+y = 24     ab              ab =  xy =24    a+b                                                                                     .        ………………………………………………

             1 +23                  23                   1x24               25

             2+22                   44                    2x12               14

             3+21                   63                    3 x 8               11

             4 +20                  80                    4 x 6               10

              5 + 19                 95   

             6 + 18               108 

             7 + 17               119

            8 +  16               128

            9 + 15                135

          10 + 14               140

          11 + 13               143

          12 + 12               144  

  ………………………………………………………………

From the table we find that when the difference between the pair of two numbers giving fixed sum is smaller greater will be its product and when the difference between the pair of two numbers giving fixed product is smaller will be its sum. 

Any increase in the length will have equal decrease in the breadth of the rectangle to keep up its perimeter same.To have maximum area, the length of the rectangle must be equal to its breadth, l =  b, where the rectangle becomes a square.  If P = L, then l = b = L/4 for a square and l =(L/4- x) , b = (L/4+x) for any rectangle .The area of the rectangle  A = (L/4 – x)( L/4+x) =( L² /16)  - x²   . To be maximum x = 0 .For a given perimeter, the area of a rectangle will be maximum when it becomes a square, 

Logical answer: The area of the rectangle depends upon two parameters  length and breadth. If length is greater ,breadth will be smaller ,if length is shorter, the breadth will be greater. Which make the area smaller than the maximum. To have maximum are, both the parameter must have optimum value that is they must be equal.

  Thavakai Academy of Creative Thinking and Skill Development

 In the mechanized world, everyone has to spend more time to improve their quality of life, even if they earn and provide the expenses of the family, personal care and individual attention towards the family is decreasing. The time spent with children is reducing . Children are not being taught morals and life education by parents, the first teacher to all children . Most of the children spend more time on their cell phones than their parents. Children are not only wasting their time but also they get some irrevocable  losses which affect the later part of their life.  They are prone to eye sight defect even in the childhood stage. They fail to do the other important works and neglect to develop skills that are required to protect life in the future. What best a parent can do in this issue? .

Stopping the habit all of a sudden is a reactive action and usually  our children don’t accept and follow our words.  Our approach must make our children to get some benefits without banning the unwanted habit once allowed freely. The reactive mode must become proactive for the continuous practice of children.  Children should be trained to use the cell phone and computer for blended learning. When they are introduced to improve problem solving skill through mathematical puzzles they use digital devices not only just for entertainment but also   to stimulate creative thinking with self-interest. With this idea in mind I would like to start a you-tube channel with a self organization called 'Thavakai Academy of Creative thinking and Skill Development' for the welfare of children worldover. After knowing the solution of a problem  its variants are discussed  in successive issue.       

 பக்காத் திருடன்

 செட்டிநாட்டில் உள்ள ஒரு பழங்கால பெண்மணியின் உருவப்படம் அருங்காட்சி யகத்தில் வைக்கப்பட்டுள்ளது. அதன் சட்டகம் 32 விலையுயர்ந்த கற்களால் அலங்கரிக்கப்பட்டுள்ளது, இதனால் ஒவ்வொரு பக்கத்திலும் 12 கற்கள் உள்ளன. ஒரு இருண்ட இரவில் ஒரு புத்திசாலியான திருடன் அருங்காட்சியகத்திற்குள் நுழைந்து நான்கு கற்களை எடுத்துச் சென்றான். மறுநாள் காலை காவலர் வந்து ஒவ்வொரு பக்கத்திலும் உள்ள கற்களை எண்ணினார். அதே 12 கற்கள். 4 கற்கள்  திருடப்பட்டதை  தாமதமாகஅருங்காட்சியக அதிகாரிகள் கண்டறிந்தனர். அது எப்படி சாத்தியமாக இருக்கும்?

சட்டத்தில் பாதிக்கப்பபெற்ற ரத்தினக் கற்கள்

                                                                4            4             4  = 12

                                                                4                            4

                                                                4            4              4   = 12

                                                               12                          12

ஒவ்வொரு பக்கத்திலும் மையத்தில் பாதிக்கப்பட்டுள்ள இரண்டு ரத்தினக் கற்கள் திருடப்பட்டு , காண்பவர்களால் உடனடியாக கண்டுபிடிக்கமுடியாதவாறு மீதியுள்ள கற்களைச்  சற்று மாற்றி திருடனால் பதிக்கப்பட்டன

                                                                                                                                                  .                                                                5                         2                           5             = 12

            2                                                      2

           5                          2                          5               = 12

          12                                                   12