9 rectangles have the same area as 20 squares

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This is a fun little question that I encountered on a problem solving assessment:

Q) A small area is covered by $20$ identical square tiles or $9$ identical rectangular tiles. The length of the side of each square tile is a whole number, and this is $2 \text{ cm}$ shorter than the longer side of each rectangular tile. What is the length of the shorter side of the rectangular tile?

$$(A)\ 3\text{ cm} \quad\quad (B)\ 4\text{ cm}\quad\quad (C)\ 5\text{ cm}\quad\quad (D)\ 1\text{ cm}$$

Here's how I went about it:

Let the side of the square be $a$, the longer side of the rectangle be $l$ and the shorter side be $b$. From the given information, I deduce that $$20a^2 = 9lb$$ It is also given that $a = l - 2,$ $$\implies 20(l-2)^2 = 9lb$$ And well, I'm stuck with this quadratic equation. I need to get $b$ but I don't know how.

Got any ideas, folks?

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The area is $15\times 12$. You can cover this with nine $5\times 4$ tiles or with twenty $3\times 3$s.