128 A cross-section of a parallelepiped is a pentagon, the ratio of any two sides of which is 1, 2, or 0.5, then this pentagon
A. has 4 angles of $120^{\circ}$, one angle of $60^{\circ}$
B. has 3 angles of $120^{\circ}$, two angles of $60^{\circ}$
C. all 5 angles are $108^{\circ}$
D. the size of the angles cannot be determined
128 A. The ratio of the smallest side $x$ to any other side is 1 or 0.5, so the lengths of the sides of the pentagon can only be $x$ and $2x$. Since there are 3 pairs of parallel faces in a parallelepiped, there are two pairs of parallel sides in the pentagon. Let the vertices of the pentagon be $A, B, C, D, E$, and $BC \parallel DE$, $CD \parallel AE$. Extend sides $AE$ and $BC$ to intersect at point $K$. Clearly, the parallelogram $CDEK$ contains the pentagon. Since $|AE| < |EK| = |CD|$, we know that $|AE| = x, |CD| = 2x$, and similarly, $|BC| = x, |DE| = 2x$.
Since $|AB| < |AK| + |KB| = 2x$, we know that $|AB| = |AK| = |KB| = x$, thus $\angle AKB = 60^\circ = \angle EDC$, $\angle AED = \angle BCD = 120^\circ = \angle EAB = \angle ABC$, meaning there are 4 angles of $120^\circ$ and one angle of $60^\circ$.