4. Let the three sides of $\triangle A B C$ be $B C=2, C A$ $=3, A B=4, h_{a}, h_{b}, h_{c}$ represent the altitudes to sides $B C, C A, A B$ respectively. Then
$$
\left(h_{a}+h_{b}+h_{c}\right)\left(\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}\right)=(\quad) .
$$
(A) $\frac{41}{6}$
(B) $\frac{39}{4}$
(C) $\frac{38}{5}$
(D) $\frac{38}{7}$
4. B.

Let the area of $\triangle A B C$ be $S$. Then
$$
\begin{array}{l}
h_{a}=\frac{2 S}{a}, h_{b}=\frac{2 S}{b}, h_{c}=\frac{2 S}{c} . \\
\text { Hence }\left(h_{a}+h_{b}+h_{c}\right)\left(\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}\right) \\
=\left(\frac{2 S}{a}+\frac{2 S}{b}+\frac{2 S}{c}\right)\left(\frac{a}{2 S}+\frac{b}{2 S}+\frac{c}{2 S}\right) \\
=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(a+b+c) \\
=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)(2+3+4)=\frac{39}{4} .
\end{array}
$$