2.37. The base of the pyramid is a regular hexagon with a side equal to $a$. One of the lateral edges is perpendicular to the base plane and equals the side of the base. Determine the total surface area of the pyramid.
2.37. We have \( S_{\text {full }}=S_{\text {base }}+2 S_{\triangle A S F}+2 S_{\triangle F S E}+2 S_{\triangle E S D} \) (Fig. 2.32), where \( S_{\text{base}}=6 \cdot \frac{a^{2} \sqrt{3}}{4}=\frac{3 a^{2} \sqrt{3}}{2}, S_{\triangle A S F}=\frac{a^{2}}{2} \), since \( S A \perp A F \) and \( S A=a \). Further, since \( A E \perp E D \), then \( S E \perp E D \), which means \( S_{\triangle E S D}=\frac{1}{2} E D \cdot S E \).

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-476.jpg?height=467&width=500&top_left_y=109&top_left_x=350)

Fig. 2.32

But \( S E=\sqrt{S A^{2}+A E^{2}}=\sqrt{a^{2}+3 a^{2}}=2 a \), hence \( S_{\triangle E S D}=\frac{1}{2} a \cdot 2 a=a^{2} \). To calculate \( S_{\triangle F S E} \), draw \( S K \perp F E \); then \( A K \perp F E \). Considering that \( A K=\frac{a \sqrt{3}}{2} \left(\angle K A F=30^{\circ}\right), S K=\sqrt{S A^{2}+A K^{2}}=\sqrt{a^{2}+\frac{3 a^{2}}{4}}= \frac{a \sqrt{7}}{2} \), we find \( S_{\triangle F S E}=\frac{a^{2} \sqrt{7}}{4} \). Therefore,

\[
S_{\text {full }}=\frac{3 a^{2} \sqrt{3}}{2}+a^{2}+\frac{a^{2} \sqrt{7}}{2}+2 a^{2}=\frac{a^{2}(6+3 \sqrt{3}+\sqrt{7})}{2} .
\]

Answer: \( 0.5 a^{2}(6+3 \sqrt{3}+\sqrt{7}) \).