Really tricky question,

Initially marked answer E, as got all sides of quadrilateral formed inside equal but have no detail about diagonals.

But after read that correct option is A, came up with below solution. Seems it was an easy one.

Easy and simple steps, but to explain, i have written everything in detail.

Attached the hand drawn image.

Solution:Given Final image is a square. SO all sides are equal

Taking point 1:BQ=CR=DS=AP = a (suppose)

let remaining portion of side = b

So total square side =a+b

Draw perpendicular from Q on AD => QE . So QE = side of square= a+b

same way Draw perpendicular from P on CD => PF . So PF = side of square= a+b

Now, as its square, all vertices are 90 degree. As BE perpendicular on AD => BQ=AE =a and BA=QE= a+b

As AD= a+b and now AE=a and given SD=a there fore ES = a+b -a -a = b-a

So right triangle triangle QES has 1 side QE=a+b and 2nd side ES=b-a. So we can find hypotenuse QS .................(1)

Same way PFR makes a right triangle with 1 side PF= a+b and 2nd side RF= b-a. So we can find hypotenuse PR ...................(2)

By above 2 equations (1) and (2),

both triangles have same sides so hypotenuse length will be same

There for PR = QS

So option 1 is sufficient.Taking point 2:Given side of square =16. Doesn't give us any information

So option 2 not sufficientAnswer = A Hope above explanation makes sense.

Attachments

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