4. In a planar convex quadrilateral with an area of $18 \mathrm{~cm}^{2}$, the sum of one pair of opposite sides and one diagonal is $12 \mathrm{~cm}$. Then the minimum value of the sum of the other diagonal and the other pair of sides is $\qquad$

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$4.6(\sqrt{2}+\sqrt{5}) \text{ cm}$.
Consider a plane convex quadrilateral $ABCD$, where the diagonal $AC$ and the sum of the lengths of one pair of opposite sides $AB$ and $CD$ is $12 \text{ cm}$. Let $AB = b$, $AC = a$, $CD = c$, $\angle BAC = \alpha$, and $\angle ACD = \beta$ (as shown in Figure 3). Then,
$$
a + b + c = 12,
$$
$S_{\text{quadrilateral } AEC D}$
$$
= S_{\triangle ABC} + S_{\triangle ACD},
$$

i.e., $18 = \frac{1}{2} a(b \sin \alpha +$
$$
\begin{array}{c}
c \sin \beta).
\end{array}
$$
Since $0 < \sin \alpha \leq 1$ and $0 < \sin \beta \leq 1$,
$$
18 \leq \frac{1}{2} a(b + c) = \frac{1}{2} a(12 - a),
$$
which simplifies to $(a - 6)^2 \leq 0$.
Since $(a - 6)^2 \geq 0$, it follows that $(a - 6)^2 = 0$. Solving this, we get $a = 6$.
Thus, $b + c = 6$, $\sin \alpha = \sin \beta = 1$, i.e., $\alpha = \beta = 90^\circ$.
Draw $DE \perp AB$ at $E$, then quadrilateral $ACDE$ is a rectangle. Therefore, $AE = CD = c$.
In $\triangle BDE$,
$$
\begin{array}{l}
BD = \sqrt{BE^2 + ED^2} = \sqrt{(AB + AE)^2 + AC^2} \\
= \sqrt{(b + c)^2 + a^2} = 6\sqrt{2} \text{ cm}.
\end{array}
$$
Also,
$$
\begin{array}{l}
AD + BC = \sqrt{AC^2 + CD^2} + \sqrt{AC^2 + AB^2} \\
= \sqrt{a^2 + c^2} + \sqrt{a^2 + b^2} \\
= \sqrt{36 + c^2} + \sqrt{36 + (6 - c)^2}.
\end{array}
$$
As shown in Figure 4, $PM = MN = NQ = 6 \text{ cm}$, $MP \perp MN$, $QN \perp MN$, and $R$ is on $MN$ such that $MR = c$. Then,
$$
\begin{array}{l}
AD + BC \\
= \sqrt{6^2 + c^2} \\
+ \sqrt{6^2 + (6 - c)^2} \\
= PR + RQ \geq PQ.
\end{array}
$$
The equality holds if and only if $P$, $R$, and $Q$ are collinear.
The minimum value of $AD + BC$ is
$PQ = 6\sqrt{5} \text{ cm} (PQ$ is calculated similarly to $BD)$.
Therefore, the required minimum value is $6(\sqrt{2} + \sqrt{5}) \text{ cm}$.