Having problem on this one.
I drew it on GeoGebra, and managed to see some congruences, but couldn't prove it. I managed to do it using analytic geometry on $\mathbb{R}^3$, but I also tried using only synthetic geometry, and it got kinda big. Could someone try doing it via synthetic geometry?
Here's the question.
Let $VABCD$ be a regular pyramid whose lateral faces are equilateral triangles of side $1$. Let $P$ be an extension of segment $VA$ such that $A$ is on $VP$ and $AP = \frac{1}{2}$. Considering a plane $\pi$ determined by $P$ and the middle points of the segments $BC$ and $AD$, calculate the area of the intersection between the plane $\pi$ and the pyramid.
(a) $\frac{16\sqrt{11}}{7}$
(b) $\frac{7\sqrt{11}}{16}$
(c) $\frac{7\sqrt{11}}{64}$
(d) $\frac{64\sqrt{11}}{7}$
(e) $\frac{32\sqrt{11}}{7}$
For quick answer, say, in a tight test situation, one could observe that the intersection is a symmetric trapezoid, which has its one base line across the pyramid base along the middle and the other base line on one side of the pyramid at a moderate angle. So, its area is roughly half that of the pyramid base (which is 1). The closest choice then would be (C); all other answers are far off from 1/2.
Now, let's do it more carefully. First, the height of the pyramid is found to be $\sqrt{2}/2$ from the given of a unit square base and four equilateral sides. All other segments in the diagram are then derived and marked.
The long base length of the intersecting trapezoid is simply 1 and the short base length is 3/4, which can be obtained from similarly triangles of side ratio 1:3. The derivation of its height can be similarly carried out, from similar triangles of side ratio 1:2 and at $h=\sqrt{11}/8$.
So, its area is,
$$\frac{1}{2}\times \left( 1 + \frac{3}{4}\right) \times \frac{\sqrt{11}}{8}=\frac{7\sqrt{11}}{64}$$