Math, asked by basvarajramchandra, 1 month ago

Construct the following magic squares. a) 3x3 magic square, with its central number 10. b) 4x4 magic square, with a magic sum as 34.​

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Answered by riyanshiyadav2009
0

Answer:

a) 3x3 magic square, with its central number 10. b) 4x4 magic square, with a magic sum as 34

Answered by Ristar
7

Answer:

❥Yøur-ᎪղՏ

solution 7: (i)

Sum of the rows

Row 1 = 5+\left(-1\right)+\left(-4\right)5+(−1)+(−4)

5-1-4\ =\ 05−1−4 = 0

Row 2 = \left(-5\right)+\left(-2\right)\ +7(−5)+(−2) +7

-5-2+7\ \ =0−5−2+7 =0

Row 3 = 0+3+\left(-3\right)0+3+(−3)

0+3-3\ \ =0\0+3−3 =0

Sum of the columns

Column 1 = 5+\left(-5\right)+05+(−5)+0

5-5+0\ =\ 05−5+0 = 0

Column 2 = \left(-1\right)+\left(-2\right)+\ 3\(−1)+(−2)+ 3

-1-2+3\ \ =\ 0−1−2+3 = 0

Column 3 = \left(-4\right)+7\ +\left(-3\right)(−4)+7 +(−3)

-4\ +7\ -3\ \ =\ 0−4 +7 −3 = 0

Sum of the diagonals

Diagonal 1

\left(5\right)+\left(-2\right)+\left(-3\right)(5)+(−2)+(−3)

5-2-3\ \ =\ 05−2−3 = 0

Diagonal 2

\left(-4\right)+\left(-2\right)\ +0\(−4)+(−2) +0

-4-2+0\ =\ -6−4−2+0 = −6

Box (i) is not a square because all the sums not equal.

(ii)

Sum of the rows

Row 1 = 1+\left(-10\right)\ +01+(−10) +0

1-10+0\ =-91−10+0 =−9

Row 2 = \left(-4\right)+\left(-3\right)+\left(-2\right)(−4)+(−3)+(−2)

-4-3-2\ \ \ =\ -9−4−3−2 = −9

Row 3 = \left(-6\right)+4+\left(-7\right)(−6)+4+(−7)

-6+4-7\ \ =\ -9−6+4−7 = −9

Sum of the columns

Column 1 = 1+\left(-4\right)+\left(-6\right)1+(−4)+(−6)

1-10\ =-91−10 =−9

Column 2 = \left(-10\right)+\left(-3\right)+4(−10)+(−3)+4

-10\ -3+4\ \ =\ -9−10 −3+4 = −9

Column 3 = 0+\left(-2\right)+\left(-7\right)0+(−2)+(−7)

0-2-7\ \ =\ -90−2−7 = −9

Sum of the diagonals

Diagonal 1

1+\left(-3\right)+\left(-7\right)1+(−3)+(−7)

1-10\ =\ -9\1−10 = −9

Diagonal 2

0+\left(-3\right)+\left(-6\right)0+(−3)+(−6)

-3-6\ =\ -9−3−6 = −9

Box (ii) is a magic square because all the sums are equal.

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