18-88 Amy drew a square clock face as a dartboard, using the "clock positions" as boundary lines (as shown in the figure). If $t$ is the area of one of the 8 triangles (for example, the area between 12 o'clock and 1 o'clock), and $q$ is the area of one of the 4 quadrilaterals at the corners (for example, the area between 1 o'clock and 2 o'clock), then $\frac{q}{t}$ equals
(A) $2 \sqrt{3}-2$.
(B) $\frac{3}{2}$.
(C) $\frac{\sqrt{5}+1}{2}$.
(D) $\sqrt{3}$.
(E) 2.
[Solution] Let the center of the clock be $O$, and mark the triangle from 12 o'clock to 1 o'clock as $AOB$, the quadrilateral from 1 o'clock to 2 o'clock as $OBCE$, and $E$ represents the position of 3 o'clock (as shown in the figure), then $\triangle AOB \cong \triangle EOD$.
Let $AB=1$, since $\angle AOB=30^{\circ}$,

so $OA=\sqrt{3} . S_{OACE}=3, S_{\triangle AOB}=\frac{\sqrt{3}}{2}$,
thus $\frac{q}{t}=\frac{S_{OBCE}}{S_{\triangle AOB}}=\frac{S_{OACE}-2 S_{\triangle AOB}}{S_{\triangle AOB}}$
$=\frac{S_{OACE}}{S_{\triangle AOB}}-2=2 \sqrt{3}-2$.
Therefore, the answer is $(A)$.