4. The apothem of a regular hexagonal pyramid is $m$, and the angle between the base and the lateral face is $\alpha$. Find the area and volume of the pyramid.
Solution. Let $M$ and $N$ be the midpoints of sides $A F$ and $C D$ respectively (see diagram). Triangle $M N S$ is isosceles with leg $m=\overline{M S}=\overline{N S}$ and angle at the base $\alpha$. Therefore, $M N=2 m \cos \alpha$. If $a$ is the side of the base, then

$$
\overline{M N}=2 \overline{O M}=a \sqrt{3}
$$

Thus,

$$
a=\frac{2 \sqrt{3}}{3} m \cos \alpha
$$

For the area $P$ of the pyramid, we have

$$
P=6\left(\frac{a^{2} \sqrt{3}}{4}+\frac{a m}{2}\right)=2 m^{2} \sqrt{3}(1+\cos \alpha) \cos \alpha
$$

The height of the pyramid is $H=m \sin \alpha$, so for the volume $V$ we get:

$$
V=\frac{2 \sqrt{3}}{3} m^{2} \sin \alpha \cos ^{2} \alpha
$$

## Fourth Year

![](https://cdn.mathpix.com/cropped/2024_06_05_e3f82bd39eecc817d651g-5.jpg?height=553&width=459&top_left_y=257&top_left_x=991)