Square ABCD has an area of 9 square inches. Sides AD and BC

[VJesus12]

Square ABCD has an area of 9 square inches. Sides AD and BC are lengthened to x inches each. By how many inches were the AD and BC sides lengthened?

(1) The diagonal of the resulting rectangle measures 5 inches.
(2) The resulting rectangle can be cut into three rectangles of equal size.

[spoiler]OA=A[/spoiler]

Source: Manhattan GMAT

[deloitte247]

Statement 1
The diagonal of the resulting rectangle measures 5 inches.
Given that the area = 9 then each side of the square = 3
The Lengthened side will have Lenght = 3+x each and the diagonal of the obtained rectangle being 5
Using pythagoras $$\left(x+3\right)^2+3^2=5^2$$ $$\left(x+3\right)^2+9=25$$ $$\left(x+3\right)^2=25-9$$ $$\left(x+3\right)^2=16$$ $$\left(x+3\right)=\sqrt{16}$$ $$\left(x+3\right)=4$$ $$\left(x\right)=4-3$$ $$\left(x\right)=1$$
Statement 1 is SUFFICIENT. Sides are lengthened by +1

Statement 2
The resulting rectangle can be cut into the three rectangles of equal sizes.
The information given here is not specific enough as any rectangle can be divided into three rectangles of equal sizes hence statement 2 is INSUFFICIENT.
Statement 1 above is SUFFICIENT.

$$answer\ is\ Option\ A$$