Area of square = \(a^2 = 4.5 = \frac{9}{2}\)
i.e. \(a = \frac{3}{\sqrt{2}}\)

We know, diagonal of a square is \(\sqrt{2}a\) i.e. \(Diagonal = 3\)

We are also given that \(\frac{DQ}{QB} = \frac{1}{2}\)
DQ + QB = DB (diagonal)
x + 2x = 3
x = 1
i.e. DQ = 1 and QB = 2

Now, drop a perpendicular CE on DB (refer to the figure below)
CE will bisect DB (as BCD is an isosceles right triangle)
i.e. DE = BE = \(\frac{3}{2}\)

Since, the diagonals in a square bisect at 90, point E is the centre of square
i.e. AE = BE = CE = DE = \(\frac{3}{2}\)

Also, DQ + QE = DE
i.e. 1 + QE = \(\frac{3}{2}\)
QE = \(\frac{1}{2}\)

Finally, Applying Pythagoras Theorem in triangle CEQ;
\(CE^2 + QE^2 = CQ^2\)
\((\frac{3}{2})^2 + (\frac{1}{2})^2 = x^2\)
\(x^2 = \frac{10}{4} \)
\(x = \frac{\sqrt{10}}{2}\)

Hence, option A