Answer :

Let us consider a square ABCD with diagonals intersecting at ‘O’.

OP is perpendicular to any of its sides such that

OP = 2√2 cm

Let the side of square be ‘L’

Then diagonal of square, BD = L√2


Diagonals of square bisect each other, therefore

Also, By symmetry

Now, In ΔOPD, By Pythagoras theorem

Hypotenuse2 = Perpendicular2 + Base2

OD2 = OP2 + DP2

L2 = 32

L2 = 4√2 cm

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