Check the picture below.
[tex]\textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}asn ~~ \begin{cases} a=apothem\\ n=\stackrel{side's}{number}\\ s=\stackrel{side's}{length}\\[-0.5em] \hrulefill\\ a=7\\ s=8.1\\ n=6 \end{cases}\implies \stackrel{\textit{area of the hexagonal base}}{A=\cfrac{1}{2}(7)(8.1)(6)} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\textit{volume of a prism}\\\\ V=Bh ~~ \begin{cases} B=\stackrel{base's}{area}\\ h=height\\[-0.5em] \hrulefill\\ B=\frac{1}{2}(7)(8.1)(6)\\ V=3572.1 \end{cases}\implies 3572.1=\cfrac{1}{2}(7)(8.1)(6)h \\\\\\ 3572.1=170.1h\implies \cfrac{3572.1}{170.1}=h\implies \boxed{21=h}[/tex]
well, the length of the tunnel is "h", now two 8 meters cars, that's 8+8=16 meters plus a 3 meter connector between them, that's 16 + 3 = 19 meters, can those two cars connected like so fit inside the tunnel? sure thing, "h" can fit 19 meters just fine.
