Answer: Since square ABCD has a diagonal length of x, then we know that:
$x = s\sqrt{2}$
Let's consider the diagonals of the rhombus EFGH. Since EFGH is a rhombus, we know that its diagonals are perpendicular bisectors of each other and they bisect the angles of the rhombus. Let's call the length of the longer diagonal of EFGH D.
By the Pythagorean theorem, we know that the length of the shorter diagonal of EFGH is:
$\sqrt{(s/2)^2 + (s\sqrt{2}/2)^2} = s\sqrt{3/4}$
Since the diagonals of EFGH are perpendicular bisectors of each other, we know that the two diagonals of EFGH are congruent, so:
$D = s\sqrt{3}$
Comparing the lengths of the diagonals of ABCD and EFGH, we have:
$D/ x = (s\sqrt{3})/(s\sqrt{2}) = \sqrt{3/2}$
Therefore, the diagonal of EFGH is shorter than the diagonal of ABCD.
For a rhombus with side length s, the maximum length of a diagonal occurs when the angle between the two sides of the rhombus is 90 degrees (i.e., when the rhombus is a square). In this case, the diagonal length is:
$D = s\sqrt{2}$
So the maximum length of a diagonal of a rhombus with side length s is s times the square root of 2.
Explanation: