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)
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
Couldn't generate an explanation.
Generated by AI. May contain inaccuracies — always verify with your textbook.
