8、As shown in the figure, in the regular hexagon $A B C D E F$, point $P$ is a point on $A B$. It is known that $S_{\triangle A F P}=8$, $S_{\triangle C D P}=42$, then $S_{\triangle E F P}=$ $\qquad$ _
【Answer】33
【Solution】As shown in the figure, $\mathrm{S} \triangle \mathrm{AFP}+\mathrm{S} \triangle \mathrm{CDP}=\frac{1}{2} S_{A C D F}=\frac{1}{2} \times \frac{2}{3} \times S_{A B C D E F}=\frac{1}{3} S_{A B C D E F}$; $S_{A B C D E F}=50 \times 3=150$, then similarly, $S_{\triangle E F P}+S_{\triangle B P C}=\frac{1}{3} S_{A B C D E F}=50$, connect $\mathrm{AC}$, $S_{\triangle B P C}=S_{\triangle A B C}-S_{\triangle A P C}=S_{\triangle A B C}-S_{\triangle A P F}=150 \div 6-8=17$; then $50-17=33$