7. In $\triangle A B C$, it is known that $A B=10$, and the height from $A B$ is 3. If $A C \cdot B C$ is minimized, then $A C+B C$ $=$ . $\qquad$
7. $4 \sqrt{10}$.

From the formula for the area of a triangle, we have
$$
\frac{1}{2} A C \cdot B C \sin C=\frac{1}{2} \times 3 \times 10=15 \text {. }
$$

Therefore, $A C \cdot B C \geqslant 30$.
The equality holds if and only if $\angle C=90^{\circ}$.
Then, $A C^{2}+B C^{2}=100$.
Thus, $(A C+B C)^{2}=A C^{2}+B C^{2}+2 A C \cdot B C=160$,
which means $A C+B C=4 \sqrt{10}$.