2009 AMC 10B Problem 24

Extend the non-parallel sides (legs) of every trapezoid downward. Since the trapezoids are congruent and glued leg to leg, all these lines pass through a single point $O$ on the ground line, the center of the arch. The arch spans a straight angle at $O$, and the nine trapezoids divide it evenly, so each trapezoid sits inside a wedge of
$$\frac{180^\circ}{9} = 20^\circ.$$

Each trapezoid is an isosceles triangle with apex $O$ and apex angle $20^\circ$, with its tip cut off by a line parallel to the base. The base angles of that triangle are $\frac{180^\circ - 20^\circ}{2} = 80^\circ$; these are the trapezoid's angles at its longer (outer) base.

The angles at the shorter (inner) base are supplementary to these, because the two bases are parallel: $180^\circ - 80^\circ = 100^\circ$. That is the larger interior angle.

The answer is $\boxed{\textbf{(A)}\ 100}$.

Congruent pieces arranged around a point share a center; extending their edges reveals it and converts the arch into equal central wedges. Each trapezoid is then a triangle minus a similar tip, and all its angles follow from the wedge angle. Check with the end trapezoid: its bottom is horizontal and its outer leg makes $80^\circ$ with the ground, consistent with the picture.

• Dividing 360 by 9 (a full circle) instead of 180 by 9, or reporting the smaller angle 80 or the wedge angle.
• Treating the trapezoids as pieces of a regular $18$-gon and mixing up interior and central angles.