As shown in the figure, a circular piece of paper has center $O$ and radius $5$ cm. An equilateral triangle $ABC$ on this paper has center at $O$. Points $D, E, F$ are on circle $O$. Triangles $DBC, ECA, FAB$ are isosceles triangles with $BC, CA, AB$ as their bases respectively. After cutting along the dashed lines and folding triangles $DBC, ECA, FAB$ along $BC, CA, AB$ respectively so that $D, E, F$ coincide, a triangular pyramid is formed. As the side length of $\triangle ABC$ varies, the maximum volume (in $\mathrm { cm } ^ { 3 }$) of the resulting triangular pyramid is \_\_\_\_

Points $A, B, C, D$ are on the surface of a sphere with radius 4. $\triangle ABC$ is an equilateral triangle with area $9 \sqrt { 3 }$. The maximum volume of the tetrahedron $D$-$ABC$ is
A. $12 \sqrt { 3 }$
B. $18 \sqrt { 3 }$
C. $24 \sqrt { 3 }$
D. $54 \sqrt { 3 }$

5. A right square frustum has upper and lower base edges of lengths 2 and 4 respectively, and lateral edge length 2. Its volume is ( )
A. $20 + 12 \sqrt { 3 }$
B. $28 \sqrt { 2 }$
C. $\frac { 56 } { 3 }$
D. $\frac { 28 \sqrt { 2 } } { 3 }$
【Answer】D 【Solution】 【Analysis】Using the geometric properties of the frustum, calculate the height of the solid and the areas of the upper and lower bases, then use the volume formula for a frustum.
【Detailed Solution】Draw a figure and connect the centers of the upper and lower bases of the right square frustum, as shown in the diagram. [Figure]
Since the upper and lower base edges of the frustum are 2 and 4 respectively, and the lateral edge length is 2, the height of the frustum is $h = \sqrt { 2 ^ { 2 } - ( 2 \sqrt { 2 } - \sqrt { 2 } ) ^ { 2 } } = \sqrt { 2 }$, the area of the lower base is $S _ { 1 } = 16$, and the area of the upper base is $S _ { 2 } = 4$, so the volume of the frustum is $V = \frac { 1 } { 3 } h \left( S _ { 1 } + S _ { 2 } + \sqrt { S _ { 1 } S _ { 2 } } \right) = \frac { 1 } { 3 } \times \sqrt { 2 } \times ( 16 + 4 + \sqrt { 64 } ) = \frac { 28 } { 3 } \sqrt { 2 }$ . Therefore, the answer is: D.

4. The South-to-North Water Diversion Project has alleviated water shortages in some northern regions, with part of the water stored in a certain reservoir. When the water level of the reservoir is at an elevation of 148.5 m, the corresponding water surface area is $140.0 \mathrm {~km} ^ { 2 }$; when the water level is at an elevation of 157.5 m, the corresponding water surface area is $180.0 \mathrm {~km} ^ { 2 }$. Treating the shape of the reservoir between these two water levels as a frustum, the volume of water added when the water level rises from an elevation of 148.5 m to 157.5 m is approximately ( $\sqrt { 7 } \approx 2.65$ )
A. $1.0 \times 10 ^ { 9 } \mathrm {~m} ^ { 3 }$
B. $1.2 \times 10 ^ { 9 } \mathrm {~m} ^ { 3 }$
C. $1.4 \times 10 ^ { 9 } \mathrm {~m} ^ { 3 }$
D. $1.6 \times 10 ^ { 9 } \mathrm {~m} ^ { 3 }$

8. A regular square pyramid has lateral edge length $l$, and all its vertices lie on the same sphere. If the volume of the sphere is $36 \pi$ and $3 \leqslant l \leqslant 3 \sqrt { 3 }$, then the range of the volume of the regular square pyramid is
A. $\left[ 18 , \frac { 81 } { 4 } \right]$
B. $\left[ \frac { 27 } { 4 } , \frac { 81 } { 4 } \right]$
C. $\left[ \frac { 27 } { 4 } , \frac { 64 } { 3 } \right]$
D. $[ 18,27 ]$
II. Multiple Choice Questions: This section contains 4 questions, each worth 5 points, for a total of 20 points. For each question, there may be multiple correct options. Full marks are awarded for selecting all correct options, 2 points for partially correct selections, and 0 points if any incorrect option is selected.

A cylinder and a cone have equal base radii and equal lateral surface areas, and both have height $\sqrt { 3 }$ . Then the volume of the cone is
A. $2 \sqrt { 3 } \pi$
B. $3 \sqrt { 3 } \pi$
C. $6 \sqrt { 3 } \pi$
D. $9 \sqrt { 3 } \pi$

As shown in the figure, consider a rectangular stone block with a vertex $A$ and a face containing point $A$. Let the midpoints of the edges of this face be $B , E , F , D$ respectively. Another face of the rectangular block containing point $B$ has its edge midpoints as $B , C , H , G$ respectively. Given that $\overline { B C } = 8$ and $\overline { B D } = \overline { D C } = 9$. The stone block is now cut to remove eight corners, such that the cutting plane for each corner passes through the midpoints of the three adjacent edges of that corner.
Find the length of $\overline { A D }$ and the volume of tetrahedron $A B C D$, and find the height from vertex $A$ to the base plane when $\triangle B C D$ is the base of the tetrahedron. (Volume of pyramid $= \frac { \text{Base area} \times \text{Height} } { 3 }$) (Non-multiple choice question, 8 points)