it is a little on the off side for the GRE.

However, in my opinion it is a reasonable question!

Regards

The diagonal of the rectangle will be the diameter of the circle.

Now we are told that \(x\) and \(y\) will take only certain values obeying the equality \(x+y=10\) and the inequality \(1<x<y\).

Now the list of values of \(x\) and \(y\) which obey these conditions are \(2\) and \(8\), \(3\) and \(7\) and \(4\) and \(6\).

Applying the Pythagoras theorem for each of these values of \(x\) and \(y\), which form the height and base of the right-angled triangle whose hypotenuse is the diagonal, we get the following values for the diagonal reduced to their simplest form.

\(\sqrt{2^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68} = \sqrt{4 \times 17} = 2\sqrt{17}\)

\(\sqrt{3^2 + 7^2} = \sqrt{9 + 49} = \sqrt{58}\)

\(\sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}\)

Thus the correct answers are Choices C, D and F.

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