3dgeometrypolygons
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.
A vectorial approach would be quite lean and effective.
Given two faces of the pyramid, sharing the common edge $V P_n$, and containing the contiguous base points $P_{n-1}$ and $P_{n+1}$, the dihedral angle between these two faces would be the angle made by the two vectors ($t_m, t_p$), normal to the common edge and lying on the respective face.
Clearly, that will be also the angle made by the normal vectors to the faces, provided that one is taken in the inward, and the other in the outward direction.
That is, by the right-hand rule, $$ {\bf n}_{\,m} = \mathop {P_{\,n} P_{\,n - 1} }\limits^ \to \; \times \;\mathop {P_{\,n} V}\limits^ \to \quad \quad {\bf n}_{\,p} = \mathop {P_{\,n} P_{\,n + 1} }\limits^ \to \; \times \;\mathop {P_{\,n} V}\limits^ \to $$
Then the dihedral angle $\alpha$ will be simply computed from the dot product $$ \cos \alpha = {{{\bf n}_{\,m} \; \cdot \;{\bf n}_{\,p} } \over {\left| {{\bf n}_{\,m} } \right|\;\;\left| {{\bf n}_{\,p} } \right|}} $$
Note that by that plane, the pyramid is cut into two tetrahedrons with base area 14 and sum of their heights equal to $s=3$. Since the volume of tetrahedron is given by $\displaystyle V=\frac {Sh} 3$ where $S$ is a base area and $h$ is a height from the base to apex, the volume of the pyramid is $$ \frac{14(h_1+h_2)}{3}=\frac{14\times 3}{3}=14. $$
Best Answer
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 $= \frac{5}{4}$.