Given a straight pyramid with a regular hexagonal base, we pass through the center of its base an alpha plane parallel to a side face. Given a straight pyramid with a regular hexagonal base, we pass through the center of its base an alpha plane parallel to a side face. Find the ratio between the area of ​​the obtained section and the side face.
I can't imagine this. What will be the polygon formed by this plane? Can someone draw me this situation?
Thanks for attention.


Here is a picture of how this will look (my awkward effort to show it). S is the center of the hexagonal base and PS is perpendicular to the base. The plane in question (say Q) is parallel to ABP side face, passes through CSD and cuts CFP, FEP and DEP sides. The cross section that you see is trapezoid CXYD parallel to ABP. As it is a regular hexagonal pyramid, a few things to note -
Say AB = a, then all 6 sides of the base is a; CS = SD = a.
Given plane Q is parallel to ABP, $\angle PMS = \angle TSN = \angle TNS$.
Also, $\triangle TSN$ and $\triangle PMN$ are similar. So,
Given $SN = \frac{MN}{2}$, $TS = \frac{PM}{2}$. So, $TN = \frac{PN}{2} = \frac{PM}{2}$
Also, $\triangle PXY$ is similar to $\triangle PFE$ and as $TN = \frac{PN}{2}$, $XY = \frac{FE}{2} = \frac{a}{2}$
$\triangle ABP = \frac{1}{2}.PM.AB = \frac{a}{2}PM$
Area of trapezoid $CXYD = \frac{1}{2}.(CD+XY).TS = \frac{1}{2}(2a+a/2)\frac{PM}{2}$
So, the ratio of area between trapezoid cross-section and side face ABP $= \frac{5}{4}$.
