- Example 1 Sheet Layout Problem. There is a square cardboard with a side length of $a$, to make a regular triangular pyramid model, the base and sides cannot be spliced, and the volume should be maximized. How should the lines be drawn, and what are the volume and material utilization rate?
Solution: In a square with side length $a$, two isosceles triangles with base length $a$ and base angles of $15^{\circ}$ are removed. The figure 8 line is used as the unfolded diagram of the required regular triangular pyramid, which can be folded and assembled into the regular triangular pyramid $A-B C D$.
By the sine rule, we have
$$
\frac{C D}{\sin 30^{\circ}}=\frac{a}{\sin 75^{\circ}} \Rightarrow C D=\frac{\sqrt{6}-\sqrt{2}}{2} a \text {. }
$$

Draw $A O \perp(B C D)$ at $O$, then $O$ is on the height line $D E$.
$$
\begin{array}{l} 
A O=\sqrt{A D^{2}-O D^{2}} \\
=\sqrt{a^{2}-\left[\frac{2}{3} \times \frac{\sqrt{3}}{2}\left(\frac{\sqrt{6}-\sqrt{2}}{2}\right) a\right]^{2}} \\
=\sqrt{\frac{1+\sqrt{3}}{3}} a . \\
\therefore V_{A-B C D}=\frac{1}{3} \cdot A O \cdot S_{\triangle B D C} \\
=\frac{1}{3} \times \sqrt{\frac{1+\sqrt{3}}{3}} a \times \frac{\sqrt{3}}{4}\left(\frac{\sqrt{6}-\sqrt{2}}{2}\right)^{2} a^{2} \\
=\frac{1}{12} \cdot \sqrt{1+\sqrt{3}}(2-\sqrt{3}) a^{3} \\
\approx 0.04 a^{3} .
\end{array}
$$

Material utilization can be calculated in this way. As shown in figure 9, note that the sum of the areas of the two shaded triangles removed is equal to the area of the rectangle $B B_{2} N M$ cut out by drawing a line parallel to $B_{1} A$ through point $C$. The material area used is equal to the area of rectangle $M N A B_{1}$. Therefore, the material utilization rate is