4. The radii of the two bases of a frustum are $R$ and $r(R > r)$, respectively. A plane parallel to the bases divides the lateral surface of the frustum into two equal areas. Then, the radius of the cross-sectional circle is
(A) $\sqrt{R r}$
(B) $\sqrt{\frac{R^{2}+r^{2}}{2}}$
(C) $\frac{R+r}{2}$
(D) cannot be determined
4. (B)

Since the lateral areas of the two frustums are equal, and the angle formed by the slant height with the base and the section is equal, according to $S^{\prime}=S \cos \theta$, it can be concluded that the projection area of the upper frustum on the section is equal to the projection area of the lower frustum on the base.
Let the radius of the section be $x$, then we have
$\pi\left(x^{2}-r^{2}\right)=\pi\left(R^{2}-x^{2}\right)$. Solving for $x$ yields $x=\sqrt{\frac{R^{2}+r^{2}}{2}}$.