The area of a rhombus is 540 cm, and the length of one of its diagonals is 4.5 dm. What is the distance between the point of the intersection of the diagonals and the side of the rhombus?
Let $ABC$ be our rhombus, with diagonals intersecting at point $D$, as shown below - [asy]
pair A,B,C,D;
draw((0,0)--(1,0.8)--(2,0)--(1,-0.8)--cycle);
A=(0,0); B=(1,0.8); C=(2,0); D=(1,0);
draw(A--C--B--D--cycle, mediumgray);
draw(A--C,dashed);
label("$A$",A,SW);
label("$B$",B,N);
label("$C$",C,NE);
label("$D$",D,plane(1,5));
[/asy] The area of the rhombus is $\frac{1}{2} \times$ (product of diagonals) $= 540$. Thus, the product of the diagonals is $1080$. Given that one of the diagonals is $4.5$ decimeters or $45$ cm, we can solve \[45 \times d = 1080 \Rightarrow d = \frac{1080}{45} = 24.\] Thus, the length of both diagonals in equal before using the Pythagorean Theorem, and the length of a side of the rhombus is 24 cm. In $\triangle ADB$, by Pythagoras, $BD=\sqrt{AD^2 - AB^2}= \sqrt{24^2 - 22^2} = \sqrt{580} \approx \boxed{24.083}$ cm.