Mark four numbers forming a square in a calendar:
Add the squares of the diagonal pair and find the difference of these sums:
42 + 122 = 160
112 + 52 = 146
160 – 146 = 14
Explain using algebra, why the difference is 14 always.
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+8)2
(x+7)2 and (x+1)2
Add the squares of the diagonal pair
x2 + (x+8)2
= x2 + (x2+82+2
[Using identity 
= 
+ 
+ 2
]
= 2x2 + 64+16
Add the squares of the other diagonal pair
= (x+1)2+(x+7)2
= (x2 +1 +2
(x2+72+2
= (x2 +1 +2
(x2 +49 +14
)
= 2x2 +16
+ 50
find the difference of these sums:
= (2x2 + 64+16
- (2x2 +16
+ 50)
= 2x2 + 64+16
- 2x2 -16
- 50
= 64-50
= 14
Hence the difference is 14, we can take any number as x; which means this holds in any part of the calendar.
Couldn't generate an explanation.
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