The picture border!

With the area of the picture ( $P$ ) being $\dfrac{4}{9}$ of the total area ( $T$ ), the area of the frame ( $F$ ) is $\dfrac{5}{9}$ of the total area. $T - P = \dfrac{9}{9}T - \dfrac{4}{9}T = \dfrac{5}{9}T = F$

Thus, the difference between $F$ and $P$ : $F - P = \dfrac{5}{9}T - \dfrac{4}{9}T = \dfrac{1}{9}T$

Of the total area, the parts of the frame that do not have dimensions in common with the picture are the 4 corners ( $4C$ ). $C = 5 \times 5 = 25$ $4C = 4 \times 25 = 100$

Due to this observation, it's reasonable to assume that $4C$ is equal to the difference between $F$ and $P$ : $\dfrac{1}{9}T = 100$ Thus: $\dfrac{4}{9}T = \large\boxed{400}$

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To verify, each of the 4 edges of the frame that do not constitute as the corners would also have an area of $\dfrac{1}{9}T$ , with a width of 5cm, and a length of 20cm. $\dfrac{\frac{1}{9}T}{5} = \dfrac{100}{5} = 20$

$(4 \times 20) + (4 \times 25) = 500 = F$

The length of 20cm corresponds to the length of each edge of the square picture. $20^2 = \boxed{400} = P$

Which checks out with the total area: $T = F + P = 500 + 400 = 900$ $\dfrac{400}{900} = \dfrac{4}{9}$