Q. 2 B5.0( 2 Votes )

Take nine numbers forming a square in a calendar and mark the four numbers at the corners.



Add the squares of diagonal pairs and find the difference of the sums.

32 + 192 = 370

172 + 52 = 314

370 – 314 = 56

Explain using algebra, why the difference is always 56. (it is convenient to take the number at the centre of the squares as x – see the section, Another trick of the lesson, Unchanging Relation of the Class 7 textbook)

Answer :

Let’s use algebra to see this.

Taking the first number in the square as x, the others can be filled as below



The squares of diagonal are


x2 and (x+16)2


(x+14)2 and (x+2)2


Add the squares of the diagonal pair


= x2 + (x+16)2


= x2 +(x2+256+ 2 [Using identity = + + 2]


=2 x2 +256+ 32x


Add the squares of the other diagonal pair


= (x+14)2+(x+2)2


=(x2 +142 +2 (x2+22+2


= (x2 +196 +28 (x2 +4 +4)


= 2x2 +32+ 200


find the difference of these sums:


(2 x2 +256+ 32x) - ( 2x2 +32+ 200)


= 2 x2 +256+ 32x - 2x2 - 32- 200


= 256-200


= 56


Hence the difference is 56, we can take any number as x; which means this hold in any part of the calendar.


Yes we can take x in the center but this will complicate the calculations.


since then the numbers will be



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