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 = 370172 + 52 = 314370 – 314 = 56Explain 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 Rate this question :

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