6. In a cube, the midpoints of the three edges passing through vertex $A$ are $E, F, G$ respectively. After cutting off the triangular pyramid $A-EFG$ from the cube, the front view of the resulting polyhedron is shown in the figure on the right. Then the corresponding left view is
A. [Figure]
B. [Figure]
C. [Figure]
D. [Figure]
[Figure]

7. In a cube, the midpoints of three edges passing through vertex $A$ are $E , F , G$ respectively. After cutting off the triangular pyramid $A - E F G$ from the cube, the front view of the orthogonal projection of the resulting polyhedron is shown in the figure on the right. Then the corresponding side view is ( [Figure]
A. [Figure]
B. [Figure]
C. [Figure]
D. [Figure]

11. Let $A, B, C$ be three points on the surface of a sphere $O$ with radius 1, and $AC \perp BC$, $AC = BC = 1$. Then the volume of the triangular pyramid $O-ABC$ is
A. $\frac{\sqrt{2}}{12}$
B. $\frac{\sqrt{3}}{12}$
C. $\frac{\sqrt{2}}{4}$
D. $\frac{\sqrt{3}}{4}$

14. A cone has a base radius of 6 and volume $30 \pi$. Then the lateral surface area of the cone is $\_\_\_\_$ .

19. (12 points) In a right triangular prism $ABC-A_1B_1C_1$, the lateral face $AA_1B_1B$ is a square, $AB = BC = 2$. $E, F$ are the midpoints of $AC$ and $CC_1$ respectively. $D$ is a point on edge $AB_1$. $BF \perp A_1B_1$.
(1) Prove that $BF \perp CE$;
(2) When $BD$ equals what value, is the sine of the dihedral angle between plane $BCC_1$ and plane $DFE$ minimized? [Figure]

20. (12 points) The parabola $C$ has its vertex at the origin $O$ and focus on the $x$-axis. The line $l: x =

As shown in the figure, the orthographic projection of a polyhedron is drawn on grid paper, where each small square has side length 1. Then the volume of the polyhedron is:
A. 8
B. 12
C. 16
D. 20

The BeiDou-3 Global Navigation Satellite System is an important achievement of China's space program. In satellite navigation systems, geostationary satellites orbit in the plane of Earth's equator at an orbital altitude of 36,000 km (orbital altitude is the distance from the satellite to Earth's surface). Treating Earth as a sphere with center $O$ and radius $r = 6400$ km, the latitude of a point $A$ on its surface is defined as the angle that $OA$ makes with the equatorial plane. The maximum latitude at which a geostationary satellite can be directly observed from Earth's surface is $a$. The surface area covered by the satellite signal on Earth's surface is $S = 2 \pi r ^ { 2 } ( 1 - \cos a )$ (in $\mathrm { km } ^ { 2 }$). Then $S$ accounts for approximately what percentage of Earth's surface area?
A. $26 \%$
B. $34 \%$
C. $42 \%$
D. $50 \%$

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. $1.6 \times 10 ^ { 9 } \mathrm {~m} ^ { 3 }$

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. $\frac { 28 \sqrt { 2 } } { 3 }$

As shown in the figure, $O$ is the center of a circle, $OA$ is the radius, arc $AB$ is part of the circle with center $O$ and radius $OA$, $C$ is the midpoint of chord $AB$, $D$ is on arc $AB$, and $CD \perp AB$. The formula for calculating the chord value $s$ is: $s = AB + \frac { CD ^ { 2 } } { OA }$. When $OA = 2$ and $\angle AOB = 60 ^ { \circ }$, then $s =$
A. $\frac { 11 - 3 \sqrt { 3 } } { 2 }$
B. $\frac { 11 - 4 \sqrt { 3 } } { 2 }$
C. $\frac { 9 - 3 \sqrt { 3 } } { 2 }$
D. $\frac { 9 - 4 \sqrt { 3 } } { 2 }$

Execute the program shown in the flowchart, the output $n =$
A. $3$
B. $4$
C. $5$
D. $6$

Execute the program flowchart on the right, the output $n =$
A. 3
B. 4
C. 5
D. 6

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.

Two cones A and B have equal slant heights. The sum of the central angles of their lateral surface developments is $2 \pi$. Let their lateral surface areas be $S _ { \text{A} }$ and $S _ { \text{B} }$, and their volumes be $V _ { \text{A} }$ and $V _ { \text{B} }$. If $\frac { S _ { \text{A} } } { S _ { \text{B} } } = 2$, then $\frac { V _ { \text{A} } } { V _ { \text{B} } } =$
A. $\sqrt { 5 }$
B. $2 \sqrt { 2 }$
C. $\sqrt { 10 }$
D. $\frac { 5 \sqrt { 10 } } { 4 }$

Two cones A and B have equal slant heights. The sum of the central angles of their lateral surface development diagrams is $2 \pi$. Let their lateral surface areas be $S_{\text{甲}}$ and $S_{\text{乙}}$, and their volumes be $V_{\text{甲}}$ and $V_{\text{乙}}$. If $\frac { S_{\text{甲}} } { S_{\text{乙}} } = 2$ , then $\frac { V_{\text{甲}} } { V_{\text{乙}} } =$
A. $\sqrt { 5 }$
B. $2 \sqrt { 2 }$
C. $\sqrt { 10 }$
D. $\frac { 5 \sqrt { 10 } } { 4 }$

In the tetrahedron $A B C D$ , $A D \perp C D , A D = C D$ , $\angle A D B = \angle B D C$ , and $E$ is the midpoint of $A C$.
(1) Prove: Plane $B E D \perp$ plane $A C D$ ;
(2) Given $A B = B D = 2 , \angle A C B = 60 ^ { \circ }$ , point $F$ is on $B D$ . When the area of $\triangle A F C$ is minimized, find the volume of the tetrahedron $F - A B C$ .

Xiaoming designed a closed packaging box as shown in the figure: the bottom face $ABCD$ is a square with side length 2. Triangles $\triangle E A B , \triangle F B C , \triangle G C D , \triangle H D A$ are all equilateral triangles, and the planes containing them are perpendicular to the bottom face.
(1) Prove that $E F \parallel$ plane $A B C D$ ;
(2) Find the volume of the packaging box (disregarding the thickness of the material).

(1) Find the equation of $C$;
(2) Let the lines $MD$ and $ND$ intersect $C$ at another point $A$ and $B$ respectively. Denote the inclination angles of lines $MN$ and $AB$ as $\alpha$ and $\beta$ respectively. When $\alpha - \beta$ attains its maximum value, find the equation of line $AB$.

[Elective 4-4: Coordinate Systems and Parametric Equations] In the rectangular coordinate system $x O y$, the parametric equation of curve $C _ { 1 }$ is $\left\{ \begin{array} { l } x = \frac { 2 + t } { 6 } \\ y = \sqrt { t } \end{array} \right.$ ($t$ is the parameter), and the parametric equation of curve $C _ { 2 }$ is $\left\{ \begin{array} { l } x = - \frac { 2 + s } { 6 } \\ y = - \sqrt { s } \end{array} \right.$ ($s$ is the parameter).
(1) Write the ordinary equation of $C _ { 1 }$ ;
(2) With the origin as the pole and the positive $x$-axis as the polar axis, the polar equation of curve $C _ { 3 }$ is $2 \cos \theta - \sin \theta = 0$ . Find the rectangular coordinates of the intersection points of $C _ { 3 }$ with $C _ { 1 }$, and the rectangular coordinates of the intersection points of $C _ { 3 }$ with $C _ { 2 }$ .

[Elective 4-4: Coordinate Systems and Parametric Equations] (10 points) In the rectangular coordinate system $xOy$, the parametric equation of curve $C _ { 1 }$ is $\left\{ \begin{array} { l } x = \frac { 2 + t } { 6 } \\ y = \sqrt { t } \end{array} \right.$ ($t$ is the parameter), and the parametric equation of curve $C _ { 2 }$ is $\left\{ \begin{array} { l } x = - \frac { 2 + s } { 6 } \\ y = - \sqrt { s } \end{array} \right.$ ($s$ is the parameter).
(1) Write the Cartesian equation of $C _ { 1 }$;
(2) With the origin as the pole and the positive $x$-axis as the polar axis, the polar equation of curve $C _ { 3 }$ is $2 \cos \theta - \sin \theta = 0$. Find the rectangular coordinates of the intersection points of $C _ { 3 }$ with $C _ { 1 }$, and the rectangular coordinates of the intersection points of $C _ { 3 }$ with $C _ { 2 }$.

Let $A=\{0,-a\}$, $B=\{1,a-2,2a-2\}$. If $A\subseteq B$, then $a=$
A. 2
B. 1
C. $\frac{2}{3}$

Execute the following flowchart, the output $B =$
A. $21$
B. $34$
C. $55$
D. $89$

As shown in the figure, the three-view drawing of a part drawn on grid paper, where the side length of each small square is 1. The volume of this part is
A. 24
B. 26
C. 28
D. 30

A cone with apex $P$ and base center $O$ has base radius $\sqrt { 3 }$. $PA$ and $PB$ are slant heights of the cone, $\angle A O B = 120 ^ { \circ }$. If the area of $\triangle P A B$ equals $\frac { 9 \sqrt { 3 } } { 4 }$, then the volume of the cone is
A. $\pi$
B. $\sqrt { 6 } \pi$
C. $3 \pi$
D. $3 \sqrt { 6 } \pi$