7 In $\triangle A B C$, $A B=10, A B$ side's height is 3, when $A C \cdot B C$ is the smallest,
$A C+B C=$ $\qquad$
$74 \sqrt{10}$ Hint: From the area formula of a triangle, we get
$$
\frac{1}{2} \cdot A C \cdot B C \sin C=\frac{1}{2} \times 3 \times 10=15,
$$

so $A C \cdot B C \geqslant 30$. The equality holds if and only if $\angle C=90^{\circ}$, that is, $A C^{2}+$ $B C^{2}=100$, so
$$
(A C+B C)^{2}=A C^{2}+B C^{2}+2 A C \cdot B C=160,
$$

which means $A C+B C=4 \sqrt{10}$.