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Solid Geometry - Practice Questions (35)

Question 1: 1. If the radius of the base of a cylinder is 1 and its side is unfolded as a square, then the later...

1. If the radius of the base of a cylinder is 1 and its side is unfolded as a square, then the lateral area of the cylinder is ( )

  • A. A. $4 \pi ^ { 2 }$
  • B. B. $3 \pi ^ { 2 }$
  • C. C. $2 \pi ^ { 2 }$
  • D. D. $\pi ^ { 2 }$

Answer: A

Solution: The side length of the side expansion by the question is not $2 \pi \times 1 = 2 \pi$ and the area is $( 2 \pi ) ^ { 2 } = 4 \pi ^ { 2 }$.

Question 2: 2. The three views of a geometry are shown in the figure, then the volume of the geometry is ( ) ![]...

2. The three views of a geometry are shown in the figure, then the volume of the geometry is ( ) ![](/images/questions/solid-geometry/image-001.jpg) ![](/images/questions/solid-geometry/image-002.jpg) Side view ![](/images/questions/solid-geometry/image-003.jpg) Top view

  • A. A. $\frac { 2 } { 3 }$
  • B. B. $\frac { 4 } { 3 }$
  • C. C. 2
  • D. D. 4

Answer: A

Solution: Based on the three views of the geometry, the space diagram is shown below: ![](/images/questions/solid-geometry/image-004.jpg) $\therefore V = \frac { 1 } { 3 } \times \frac { 1 } { 2 } \times 2 \times 2 \times 1 = \frac { 2 } { 3 }$

Question 3: A triangular pheasant $P - A B C _ { \text {'s three lateral edges } } P A , P B , P C$ is perpendic...

A triangular pheasant $P - A B C _ { \text {'s three lateral edges } } P A , P B , P C$ is perpendicular to each other and has lengths $1 , \sqrt { 6 }$ and 3, respectively, then the surface area of the outer sphere of this triangular pheasant is B. $32 \pi$

  • A. A. $16 \pi$
  • B. B. A three-pronged pheasant $P - A B C _ { \text {'s three lateral edges } } P A , P B , P C$ two perpendicular to each other and of length $1 , \sqrt { 6 }$
  • C. C. $36 \pi$
  • D. D. $64 \pi$

Answer: A

Solution: The three lateral ribs $P A , P B , P C$ of the triangular pheasant $P - A B C$ are perpendicular to each other, and its outer sphere is the outer sphere of the rectangular body that takes $P A , P B , P C$ is the outer sphere of a rectangle of length, width, and height, respectively. The diagonal length of the rectangle is: $\sqrt { 1 + ( \sqrt { 6 } ) ^ { 2 } + 3 ^ { 2 } } = 4$ so the diameter of the sphere is 4, the radius is 2, and the surface area of the sphere is: $S = 4 \pi R ^ { 2 } = 16 \pi$.

Question 4: 4. The side expansion of a cylinder is a square, then the ratio of surface area to side area of the ...

4. The side expansion of a cylinder is a square, then the ratio of surface area to side area of the cylinder is ( )

  • A. A. $\frac { 1 + 2 \pi } { 4 \pi }$
  • B. B. $\frac { 1 + 2 \pi } { 2 \pi }$
  • C. C. $\frac { 1 + 2 \pi } { \pi }$
  • D. D. $\frac { 1 + 4 \pi } { 2 \pi }$

Answer: B

Solution: Let the radius of the base of the cylinder be $r$ and the height of the cylinder be $h$. $\because$ The side expansion of the cylinder is a square. $\therefore 2 \pi r = h$ $\therefore$ The side area of the cylinder is $2 \pi r h = 4 \pi ^ { 2 } r ^ { 2 }$ , and The area of the two bases of the cylinder is $2 \pi r ^ { 2 } , \therefore$ The surface area of the cylinder is $2 \pi r ^ { 2 } + 2 \pi r h = 2 \pi r ^ { 2 } + 4 \pi ^ { 2 } r ^ { 2 }$ , $2 \pi r h = 4 \pi ^ { 2 } r ^ { 2 }$ $\therefore$ The ratio of the surface area to the side area of the cylinder is $\frac { 2 \pi r ^ { 2 } + 4 \pi ^ { 2 } r ^ { 2 } } { 4 \pi ^ { 2 } r ^ { 2 } } = \frac { 1 + 2 \pi } { 2 \pi }$ , the

Question 5: 5. It is known that the point $A ( 3,0 , - 4 )$, the point $A$ has symmetry $B$ about the origin, an...

5. It is known that the point $A ( 3,0 , - 4 )$, the point $A$ has symmetry $B$ about the origin, and that $| A B | =$

  • A. A. 25
  • B. B. 12
  • C. C. 10
  • D. D. 5

Answer: C

Solution: Since the point ${ } ^ { A ( 3,0 , - 4 ) }$ has symmetry ${ } ^ { B }$ about the origin, $B ( - 3,0,4 )$, and hence $| A B | = \sqrt { ( - 3 - 3 ) ^ { 2 } + ( 0 - 0 ) ^ { 2 } + ( 4 + 4 ) ^ { 2 } } = 10$.

Question 6: 6. A prism whose sides are all rectangular must be

6. A prism whose sides are all rectangular must be

  • A. A. cuboid
  • B. B. triangular prism
  • C. C. rectangular parallelepiped
  • D. D. rectangular prism

Answer: D

Solution: A prism whose sides are all rectangular requires only that the lateral prongs be perpendicular to the base, i.e., a prism whose sides are all rectangular must be a rectilinear prism.

Question 7: 7. If the front, side, and top views of a spatial geometry are all circles with radius equal to 5, t...

7. If the front, side, and top views of a spatial geometry are all circles with radius equal to 5, then the surface area of the spatial geometry is equal to

  • A. A. $100 \pi$
  • B. B. $\frac { 100 \pi } { 3 }$
  • C. C. $25 \pi$
  • D. D. $\frac { 25 \pi } { 3 }$

Answer: A

Solution: It is easy to know that the geometry is a sphere with radius 5, and the surface area is $S = 4 \pi R ^ { 2 } = 100 \pi$.

Question 8: 8. The following statements are correct ( ). (1) A geometric body with triangles on all sides is a t...

8. The following statements are correct ( ). (1) A geometric body with triangles on all sides is a trigonometric pheasant; (2) When a plane is used to truncate a prismatic pheasant, the portion of the original prismatic pheasant between the base and the cross section is a prism; (3) All four faces of a trigonous pheasant can be right triangles; (4) The visualization of a trapezoid can be a parallelogram; (5) There are an infinite number of busbars through a point on the side of a circular platform; (6) All four sides of a quadrangular pheasant can be right triangles.

  • A. A. 0
  • B. B. 1
  • C. C. 2
  • D. D. 3

Answer: C

Solution: For (1), all 8 faces of the regular octahedron are triangles, and the regular octahedron is not a trigonal pheasant, (1) is wrong; For (2), when the cross-section is not parallel to the base of the prismatic pheasant, the portion between the base of the original prismatic pheasant and the cross-section is not a prism, (2) is incorrect; for (3), a triangular pyramid that passes through the acute vertices of the base right-angled triangles has lateral ribs perpendicular to the base, and all four faces are right-angled triangles, (3) is correct; For (4), the visualization of a trapezoid cannot be a parallelogram, (4) is incorrect; For (5), all the busbars of a circular table intersect at one point, so there is one busbars through a point on the side of the table, (5) is incorrect; For (6), a quadrangular pheasant with one lateral prong perpendicular to the base rectangle has right triangles on all four sides, (6) is correct, so the number of correct propositions is 2 .

Question 9: 9. The point $P ( 2 , - 31 )$ is symmetric about the origin.

9. The point $P ( 2 , - 31 )$ is symmetric about the origin.

  • A. A. $( - 2 , - 3 , - 1 )$
  • B. B. $( - 23 ; 1 )$
  • C. C. $( 2 , - 3 , - 1 )$
  • D. D. $( - 2,31 )$

Answer: B

Solution: Each coordinate of a point $P ( 2 , - 31 )$ symmetric about the coordinate origin becomes the opposite of the original, which is $( - 23 , - 1 )$. Choose B. This question examines the coordinates of the point of symmetry of a space point about the origin, familiar with the characteristics of the symmetry of the point about the point is the solution to the The key to solving the problem is to familiarize yourself with the characteristics of point symmetry.

Question 10: 10. The area of each of the three faces of a rectangle is $\sqrt { 2 } , \sqrt { 3 } , \sqrt { 6 }$,...

10. The area of each of the three faces of a rectangle is $\sqrt { 2 } , \sqrt { 3 } , \sqrt { 6 }$, then the volume of the rectangle is ()

  • A. A. 6
  • B. B. $\sqrt { 6 }$
  • C. C. 3
  • D. D. $2 ^ { \sqrt { 3 } }$

Answer: B

Solution: Let the length, width, and height of the rectangle be $x , y , z$, then $x y = \sqrt { 2 } , y z = \sqrt { 3 } , x z = \sqrt { 6 }$ $\therefore ( x y z ) ^ { 2 } = 6$ $\therefore V = x y z = \sqrt { 6 }$

Question 11: 11. The following proposition is false

11. The following proposition is false

  • A. A. The axial section of a cylinder is the one with the largest area among the sections crossing the busbar
  • B. B. The axial section of a cone has the largest area of all the sections passing through the vertices
  • C. C. All cross sections of a circular table parallel to the base are circles
  • D. D. All axial sections of a cone are isosceles triangles

Answer: B

Solution: For option A, in a cylinder, the cross sections over the mother line are all rectangles with equal heights, and the cross section area is maximized when the base is the diameter, i.e., the axial cross section of the cylinder is the largest of the cross sections over the mother line, so the proposition in option A is correct. For option B, the area of the axial section of a cone is $\frac { 1 } { 2 } \cdot l ^ { 2 } \cdot \sin \theta$, where ${ } _ { l }$ is the length of the base line, and $\theta$ is the maximum value of the angle between the two base lines of the cone, and it can be seen that the axial section has the largest area when $\theta \leq \frac { \pi } { 2 }$ is used. When $\frac { \pi } { 2 } < \theta < \pi$ exists, there must be a section of $\theta _ { 1 } = \frac { \pi } { 2 }$ that maximizes the area of the section, so the proposition of option B is wrong. According to the geometric properties of a circular table, all sections of the table parallel to the base are circles. According to the geometric properties of cones, all axial sections of cones are isosceles triangles, so the proposition of D is correct. To summarize, choose B. [Eyesight] This question is mainly examined the geometric properties of cylinders, cones and circular table, the examination of the formula for the area of the triangle and the maximum value of the area of the triangle, belonging to the basic questions.

Question 12: 12. If the radius of the tangent sphere of a cylinder is known to be 1, then the volume of the cylin...

12. If the radius of the tangent sphere of a cylinder is known to be 1, then the volume of the cylinder is ( )

  • A. A. $\pi$
  • B. B. $2 \pi$
  • C. C. $3 \pi$
  • D. D. $4 \pi$

Answer: B

Solution: From the meaning of the question, the radius of the circle at the base of the cylinder is 1, and the height of the cylinder is 2. So the volume of the cylinder is $\pi \times 1 ^ { 2 } \times 2 = 2 \pi$.

Question 13: 13. Expand the radius of the ball to ${ } ^ { \sqrt { 2 } }$ times its original size, then the volum...

13. Expand the radius of the ball to ${ } ^ { \sqrt { 2 } }$ times its original size, then the volume expands to ${ } ^ { \sqrt { 2 } }$ times its original size.

  • A. A. 2 times
  • B. B. $2 \sqrt { 2 }$ times
  • C. C. $\sqrt { 2 }$ times
  • D. D. $\sqrt [ 3 ] { 2 }$ times

Answer: B

Solution: Let the ball have radius ${ } _ { R }$ and volume $V _ { 1 } = \frac { 4 } { 3 } \pi R ^ { 3 }$. Expand the radius of the ball to $\sqrt { 2 }$ times its original radius, which is $\sqrt { 2 } R$, and the volume is $V _ { 2 } = \frac { 4 } { 3 } \pi ( \sqrt { 2 } R ) ^ { 3 } = \frac { 8 \sqrt { 2 } } { 3 } \pi R ^ { 3 }$. Then $\frac { V _ { 2 } } { V _ { 1 } } = 2 \sqrt { 2 }$, i.e., $V _ { 2 } = 2 \sqrt { 2 } V _ { 1 }$.

Question 14: 14. The height of a cylinder is 1, the perimeter of its base is 8, and its three views are shown in ...

14. The height of a cylinder is 1, the perimeter of its base is 8, and its three views are shown in the figure. The point $P$ on the surface of the cylinder corresponds to the point on the front view. The point $A$ on the surface of the cylinder is $A$, and the point $Q$ on the surface of the cylinder is $B$ in the left view, and the shortest path on the side of the cylinder is the one that goes from $P$ to $Q$, and the shortest path is the one that goes to $Q$. ]], the length of the shortest path is ( ) ![](/images/questions/solid-geometry/image-005.jpg) ![](/images/questions/solid-geometry/image-006.jpg) ![](/images/questions/solid-geometry/image-007.jpg)

  • A. A. $\sqrt { 17 }$
  • B. B. $\sqrt { 5 }$
  • C. C. $\frac { 3 } { 2 }$
  • D. D. 1

Answer: B

Solution:

Question 15: 15. Let the sides of a rectangle be $a , b ( a > b )$, and roll it in two ways to form a cylinder (w...

15. Let the sides of a rectangle be $a , b ( a > b )$, and roll it in two ways to form a cylinder (without a base) of height $a$ and $b$, whose volumes are $V ^ { a }$ and $V ^ { b }$, then the size relationship between $V ^ { a }$ and $V ^ { b }$ is $V ^ { b }$. FORMULA_4]], then the relationship between the sizes of $V ^ { a }$ and $V ^ { b }$ is

  • A. A. $V _ { a } > V _ { b }$
  • B. B. $V _ { a } = V _ { b }$
  • C. C. $V _ { a } < V _ { b }$
  • D. D. inconclusive

Answer: C

Solution: From the question, when rolled into a cylinder with height $a$, at this time, let the radius of the base of the cylinder is $r _ { 1 }$, then $2 \pi r _ { 1 } = b$, the Solve for $r _ { 1 } = \frac { b } { 2 \pi }$, then the volume of the cylinder is $V _ { a } = \pi r _ { 1 } ^ { 2 } a = \pi \times \left( \frac { b } { 2 \pi } \right) ^ { 2 } a = \frac { a b ^ { 2 } } { 4 \pi }$. When rolled into a cylinder with height ${ } ^ { b }$, then let the radius of the base of the cylinder be $r _ { 2 }$, then ${ } ^ { 2 \pi r _ { 2 } = a }$, and the volume of the cylinder is ${ } ^ { 2 \pi r _ { 2 } = a }$. Solve for $r _ { 2 } = \frac { a } { 2 \pi }$, then the volume of the cylinder is $V _ { b } = \pi r _ { 2 } ^ { 2 } b = \pi \times \left( \frac { a } { 2 \pi } \right) ^ { 2 } b = \frac { a ^ { 2 } b } { 4 \pi }$. By $a > b$, $\frac { a ^ { 2 } b } { 4 \pi } > \frac { a b ^ { 2 } } { 4 \pi }$ is $V _ { b } > V _ { a }$. [Eyesight] This question mainly examines the side of the cylindrical expansion diagram, as well as the calculation of the volume of the cylinder, in which the answer to the solution, according to the meaning of the question to solve the radius of the base circle of the two cylinders, the use of the formula for the volume of the cylinder, accurately solving the volume of the cylinder is the key to solving the answer, focusing on the test of the ability to reasoning and calculating, belong to the basic questions.

Question 16: 16. There is a tent, the lower part of which is in the shape of a square hexagonal prism with a heig...

16. There is a tent, the lower part of which is in the shape of a square hexagonal prism with a height of 1 m, and the upper part of which is in the shape of a square hexagonal pheasant with side angles of 3 m (as shown in the figure). When the volume of the tent is maximum, the distance from the apex $O$ to the center ${ } ^ { O } { } _ { 1 }$ of the bottom face of the tent is ![](/images/questions/solid-geometry/image-008.jpg)

  • A. A. 1 m
  • B. B. $\frac { 3 } { 2 } \mathrm {~m}$
  • C. C. 2 m
  • D. D. 3 m

Answer: C

Solution: Let ${ } ^ { O O _ { 1 } }$ be $^ { x \mathrm {~m} }$ and $1 < x < 4$, then $1 < x < 4$. Let the area of the base square hexagon be $S \mathrm {~m} ^ { 2 }$ and the volume of the tent be $V \mathrm {~m} ^ { 3 }$. Then the length of the bottom side of the square hexagon is $\sqrt { 3 ^ { 2 } - ( x - 1 ) ^ { 2 } } = \sqrt { 8 + 2 x - x ^ { 2 } } ( \mathrm {~m} )$ from the problem. and so $S = 6 \times \frac { \sqrt { 3 } } { 4 } \left( \sqrt { 8 + 2 x - x ^ { 2 } } \right) ^ { 2 } = \frac { 3 \sqrt { 3 } } { 2 } \left( 8 + 2 x - x ^ { 2 } \right)$ . So $V = \frac { 1 } { 3 } \times \frac { 3 \sqrt { 3 } } { 2 } \left( 8 + 2 x - x ^ { 2 } \right) ( x - 1 ) + \frac { 3 \sqrt { 3 } } { 2 } \times \left( 8 + 2 x - x ^ { 2 } \right)$ $= \frac { \sqrt { 3 } } { 2 } \left( 8 + 2 x - x ^ { 2 } \right) [ ( x - 1 ) + 3 ] = \frac { \sqrt { 3 } } { 2 } \left( 16 + 12 x - x ^ { 3 } \right) ( 1 < x < 4 )$, $V = \frac { 1 } { 3 } \times \frac { 3 \sqrt { 3 } } { 2 } \left( 8 + 2 x - x ^ { 2 } \right) ( x - 1 ) + \frac { 3 \sqrt { 3 } } { 2 } \times \left( 8 + 2 x - x ^ { 2 } \right)$ then $V ^ { \prime } = \frac { \sqrt { 3 } } { 2 } \left( 12 - 3 x ^ { 2 } \right)$. Let $V ^ { \prime } = 0$, solve for $x = 2$ or $x = - 2$ (rounded). When $1 < x < 2$, $V ^ { \prime } > 0 , ~ V$ is monotonically increasing; When $2 < x < 4$ is $V ^ { \prime } < 0 , V$ monotonically decreasing. Therefore, $x = 2 ( \mathrm {~m} )$ is largest when $V$ is $x = 2 ( \mathrm {~m} )$.

Question 17: 17. someone with the paper shown in the figure, folded along the crease and glued into a four-pronge...

17. someone with the paper shown in the figure, folded along the crease and glued into a four-pronged cone "lantern", square to do the bottom of the lamp, and there is a triangle written on the surface of the word "year", when the lamp rotates, just to see "! Happy New Year", then in (1), (2), (3) should be written in the order of ![](/images/questions/solid-geometry/image-009.jpg)

  • A. A. Fast, New, Fun
  • B. B. Happy, New, Fast
  • C. C. New, fun, fast
  • D. D. Happy, fast, new

Answer: A

Solution: According to the four-pronged pheasant graph, just to see the words "Happy New Year", can be seen in the order of (2) year (1) (3), so choose A . [Eyesight] This question examines the structural characteristics of the four-pronged pheasant, examining the students' knowledge of graphics, belonging to the basic questions.

Question 18: 18. The three views of a geometry are shown (in $m$), the volume of the geometry is () $m ^ { 3 }$ !...

18. The three views of a geometry are shown (in $m$), the volume of the geometry is () $m ^ { 3 }$ ![](/images/questions/solid-geometry/image-010.jpg) ![](/images/questions/solid-geometry/image-011.jpg)

  • A. A. $2 \pi$
  • B. B. $\frac { 8 \pi } { 3 }$
  • C. C. $3 \pi$
  • D. D. $\frac { 10 \pi } { 3 }$

Answer: B

Solution: From the three views, the geometry consists of a combination of two identical cones and a cylinder, with the base of the cones rounded in half. The radius of the base circle of the cone is 1, the height is 1, the length of the busbar of the cylinder is 2, and the radius of the base circle is 1, so the volume of the geometry is The volume of the geometry is $V = 2 \times \frac { 1 } { 3 } \times \pi \times 1 ^ { 2 } \times 1 + \pi \times 1 ^ { 2 } \times 2 = \frac { 8 } { 3 } \pi$.

Question 19: 20. Take any three vertices of a triangular prism as vertices to make triangles, from which any two ...

20. Take any three vertices of a triangular prism as vertices to make triangles, from which any two triangles, the two triangles are coplanar in the following cases ( )

  • A. A. 6 species
  • B. B. 12 species
  • C. C. 18 species
  • D. D. 30 species

Answer: C

Solution: If two triangles are coplanar, the two triangles must be in the same side, each side has 4 vertices and can be made into 4 triangles, there are $\mathrm { C } _ { 4 } ^ { 2 } = 6$ ways to choose any two triangles, and $6 \times 3 = 18$ pairs of coplanar triangles can be chosen from the three sides.

Question 20: 21. If the length, width, and height of a rectangle are $4 , \sqrt { 5 } , 2$ and each vertex of the...

21. If the length, width, and height of a rectangle are $4 , \sqrt { 5 } , 2$ and each vertex of the rectangle is on the sphere of the ball $O$, then the surface area of the ball $O$ is ( )

  • A. A. $18 \pi$
  • B. B. $20 \pi$
  • C. C. $24 \pi$
  • D. D. $25 \pi$

Answer: D

Solution: From the question, the intersection of the body diagonals of a rectangular body is at equal distances from each vertex, i.e., the center of the sphere $O$ is the point of intersection of the body diagonals. The radius is half of the diagonal of the body, i.e., the radius $O$ of the ball $r = \frac { \sqrt { 4 ^ { 2 } + ( \sqrt { 5 } ) ^ { 2 } + 2 ^ { 2 } } } { 2 } = \frac { 5 } { 2 }$. Then the surface area of the ball $O$ is $S = 4 \pi r ^ { 2 } = 25 \pi$.

Question 21: 23. As shown in the figure, the side length of the small square on the grid paper is 1. After a squa...

23. As shown in the figure, the side length of the small square on the grid paper is 1. After a square is partially truncated, the three views of the remaining geometry are shown in the figure, then the volume of the geometry is ( ) ![](/images/questions/solid-geometry/image-012.jpg)

  • A. A. 8
  • B. B. 16
  • C. C. $\frac { 16 } { 3 }$
  • D. D. $\frac { 8 \sqrt { 3 } } { 3 }$

Answer: B

Solution: From the question, the visualization of the geometry is shown in the figure, the geometry is a quadrangular pheasant $A - B C D E$, the base $B C D E$ is a right-angled trapezium, the length of the upper and lower base sides are $4 , 2$, the height is 4, the quadrangular pheasant, so the volume of the geometry is $\frac { 1 } { 3 } \times \frac { 1 } { 2 } \times ( 2 + 4 ) \times 4 \times 4 = 16$.

Question 22: 24. It is known that the triangular pheasant $B - P A C$ has equal lateral ribs, the midpoints of wh...

24. It is known that the triangular pheasant $B - P A C$ has equal lateral ribs, the midpoints of which are $D , E , F$, and the midpoints of the prongs $A C$ are $G , P B \perp$ The plane $^ { A B C }$ is $^ { A B C }$ and the midpoints of the prongs are $^ { A B C }$. ]. and $A B = 4 , \angle A B C = 120 ^ { \circ }$. If each vertex of a tetrahedron ${ } ^ { D E F G }$ is on the sphere ${ } ^ { O }$ of a ball ${ } ^ { O }$, then the total length of the lines of intersection of the sphere with the sides of the trigonal pheasant $B - P A C$ is ( )

  • A. A. $\frac { 7 \pi } { 3 }$
  • B. B. $\frac { 8 \pi } { 3 }$
  • C. C. $\frac { 10 \pi } { 3 }$
  • D. D. $\frac { 11 \pi } { 3 }$

Answer: C

Solution: As shown in the picture : ![](/images/questions/solid-geometry/image-013.jpg) The links $B G$ , $\angle A B C = 120 ^ { \circ }$ $\because A B = B C = B P = 4 , D , E , F , G$ are the midpoints of the prongs respectively, $\angle A B C = 120 ^ { \circ }$ $B D = B E = B F = B G = 2$ $\therefore \quad$ $\therefore$ point $B$ is the center of the ball $O$, $\because P B \perp$ $\because P B \perp$ the plane $A B C$ , the $A B C$ The total length of the lines of intersection of the $\therefore$ sphere with the sides of the triangular pheasant $B - P A C$ is $\frac { 120 + 90 + 90 } { 360 } \times \pi \times 2 ^ { 2 } = \frac { 10 \pi } { 3 }$ , and

Question 23: 26. TIANGONGKAIWU is a comprehensive scientific and technological work written by Song Yingxing, a s...

26. TIANGONGKAIWU is a comprehensive scientific and technological work written by Song Yingxing, a scientist of the Ming Dynasty in China, in which a method of making tiles is recorded. The senior class of a school plans to practice this method, and prepares clay and a cylindrical wooden drum with a diameter of 20 cm at the bottom and a height of 20 cm for the students to make tiles. First of all, in the outer side of the drum evenly covered with a layer of clay thickness of 2 cm, and then, along the direction of the drum bus will be divided into four equal parts of the clay layer (as shown in the figure), and so on the clay after drying, you can get the same size of the four pieces of tile. If each student makes four tiles, and there are 500 students in the class, which of the following figures is closest to the amount of clay needed to be prepared (not counting losses). (Reference data: $\pi \approx 3.14$) ![](/images/questions/solid-geometry/image-014.jpg) Split Line

  • A. A. $0.8 \mathrm {~m} ^ { 3 }$
  • B. B. $1.4 \mathrm {~m} ^ { 3 }$
  • C. C. $1.8 \mathrm {~m} ^ { 3 }$
  • D. D. $2.2 \mathrm {~m} ^ { 3 }$

Answer: B

Solution: Volume of four tiles from conditions $V = \pi \times 12 ^ { 2 } \times 20 - \pi \times 10 ^ { 2 } \times 20 = 880 \pi \left( \mathrm {~cm} ^ { 3 } \right)$ So the total volume of clay needed to make 4 tiles each for 500 students is $500 \times 880 \pi = 440000 \pi \quad \left( \mathrm {~cm} ^ { 3 } \right)$ , and $500 \times 880 \pi = 440000 \pi \quad \left( \mathrm {~cm} ^ { 3 } \right)$ and $\pi \approx 3.14$. So the total volume of clay needed is about $1.3816 \mathrm {~m} ^ { 3 }$ , and $1.3816 \mathrm {~m} ^ { 3 }$

Question 24: 27. A square $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$ is known to have prongs of length 2...

27. A square $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$ is known to have prongs of length 2, and the point $E$ is the midpoint of the prong $A B$, then the distance to the point $A _ { \text {到plane } } E B _ { 1 } C$ is

  • A. A. $\frac { \sqrt { 6 } } { 3 }$
  • B. B. $\frac { \sqrt { 3 } } { 4 }$
  • C. C. $\frac { \sqrt { 6 } } { 6 }$
  • D. D. $\frac { \sqrt { 5 } } { 5 }$

Answer: A

Solution: Since ${ } ^ { E }$ is the midpoint of $A B$, the distance from ${ } _ { \text {到plane } } E B _ { 1 } C$ is equal to the distance from ${ } ^ { B }$ to the plane ${ } ^ { E B } C$, and let this be the distance from $A B$. ![](/images/questions/solid-geometry/image-015.jpg) From the question $B _ { 1 } E = C E = \sqrt { 2 ^ { 2 } + 1 ^ { 2 } } = \sqrt { 5 } , B _ { 1 } C = 2 \sqrt { 2 }$, we can see that $B _ { 1 } E = C E = \sqrt { 2 ^ { 2 } + 1 ^ { 2 } } = \sqrt { 5 } , B _ { 1 } C = 2 \sqrt { 2 }$ So $S _ { \triangle B _ { 1 } C E } = \frac { 1 } { 2 } \times 2 \sqrt { 2 } \times \sqrt { ( \sqrt { 5 } ) ^ { 2 } - ( \sqrt { 2 } ) ^ { 2 } } = \sqrt { 6 }$ , . Since $V _ { B _ { 1 } - B C E } = V _ { B - B _ { 1 } C E }$ so $\frac { 1 } { 3 } \times \left( \frac { 1 } { 2 } \times 1 \times 2 \right) \times 2 = \frac { 1 } { 3 } \times \sqrt { 6 } \times h$ , and So $h = \frac { \sqrt { 6 } } { 3 }$.

Question 25: 28. It is known that the shape of a commodity is a circular table, the axial cross-section of the ta...

28. It is known that the shape of a commodity is a circular table, the axial cross-section of the table is an isosceles trapezoid with an upper base of 2, a lower base of 4 and a waist of 3, then the surface area of the circular table is ![](/images/questions/solid-geometry/image-016.jpg)

  • A. A. $\frac { 14 } { 3 } \pi$
  • B. B. $14 \pi$
  • C. C. $\frac { 7 } { 3 } \pi$
  • D. D. $7 \pi$

Answer: B

Solution: The surface area of this circular table $S = \pi \left( r _ { 1 } ^ { 2 } + r _ { 2 } ^ { 2 } + r _ { 1 } l + r _ { 2 } l \right) = \pi \left( 1 ^ { 2 } + 2 ^ { 2 } + 1 \times 3 + 2 \times 3 \right) = 14 \pi$.

Question 26: 29. The side area of a cone is twice the area of the base, and the central angle of the fan of the s...

29. The side area of a cone is twice the area of the base, and the central angle of the fan of the side view of the cone is

  • A. A. $60 ^ { \circ }$
  • B. B. $90 ^ { \circ }$
  • C. C. $120 ^ { \circ }$
  • D. D. $180 ^ { \circ }$

Answer: D

Solution: Let the radius of the base of the round pheasant be $r$ and the length of the busbar be $a$. Since the lateral area is twice the area of the base. So $\pi r a = 2 \pi r ^ { 2 }$ , the and $2 \pi r = \frac { n \pi a } { 180 ^ { \circ } }$, solving for $n = 180 ^ { \circ }$.

Question 27: 31. It is known that in a straight triangular prism $\square$ $A B C - A _ { 1 } B _ { 1 } C _ { 1 }...

31. It is known that in a straight triangular prism $\square$ $A B C - A _ { 1 } B _ { 1 } C _ { 1 }$, $A B \perp B C , A B = 6 , B C = 8$ and this triangular prism has an incisive sphere, the ratio of the surface area of the incisive sphere to that of the external sphere of this triangular prism is

  • A. A. $2 : 5$
  • B. B. $4 : 25$
  • C. C. $2 : \sqrt { 29 }$
  • D. D. $4 : 29$

Answer: D

Solution: From the question, the radius of the tangent circle of ${ } _ { V A B C }$ is $\frac { 6 + 8 - 10 } { 2 } = 2$, so for this triangular prism to have an internally tangent sphere, the height of this triangular prism ${ } ^ { A A _ { 1 } = 4 }$ is ${ } ^ { A A _ { 1 } = 4 }$, and so the radius of the internally tangent sphere ${ } ^ { r = 2 }$ is ${ } ^ { r = 2 }$; ![](/images/questions/solid-geometry/image-017.jpg) Take the midpoint of ${ } ^ { A C }$ to the midpoint of $D , { } ^ { A _ { 1 } C _ { 1 } }$ to the midpoint of ${ } ^ { D _ { 1 } }$ , then the midpoint of ${ } ^ { D D _ { 1 } }$ to the midpoint of $M$ is the center of the outer ball. So the radius of the outer sphere $R = \sqrt { 5 ^ { 2 } + 2 ^ { 2 } } = \sqrt { 29 }$, and hence the ratio of the surface area of the inner tangent to the outer sphere of the triangular prism is $4 \pi r ^ { 2 } : 4 \pi R ^ { 2 } = 4 : 29$

Question 28: 32. As shown in the figure, it is known that the radius of the circle on the base of the cylinder is...

32. As shown in the figure, it is known that the radius of the circle on the base of the cylinder is $\frac { 2 } { \pi }$, and the height is ${ } _ { 2 } , A B , C D$ which is the diameter of the two bases, and $A D B C$ is the base line. If a bug starts from point $A$ and crawls sideways to point $C$, find the shortest path for the bug to crawl on. ![](/images/questions/solid-geometry/image-018.jpg)

  • A. A. $\sqrt { 2 }$
  • B. B. $2 \sqrt { 2 }$
  • C. C. 2
  • D. D. $3 \sqrt { 2 }$

Answer: B

Solution: Unfold the side of the cylinder as shown in the figure. ![](/images/questions/solid-geometry/image-019.jpg) After unfolding, in the rectangle $A B C D ^ { \text {中 } } , A B = \pi \times \frac { 2 } { \pi } = 2 , B C = 2$ , the From the figure, the shortest length of the worm's crawling route is $A C = \sqrt { A B ^ { 2 } + B C ^ { 2 } } = \sqrt { 2 ^ { 2 } + 2 ^ { 2 } } = 2 \sqrt { 2 }$.

Question 29: 33. In the trapezoid $A B C D$, $\angle A B C = 90 ^ { \circ } , A D / / B C , B C = 2 A D = 2 A B =...

33. In the trapezoid $A B C D$, $\angle A B C = 90 ^ { \circ } , A D / / B C , B C = 2 A D = 2 A B = 2$. The volume of the geometry enclosed by the surface formed by rotating the trapezoid $A B C D$ about the line $A D$ is

  • A. A. $\frac { 2 \pi } { 3 }$
  • B. B. $\frac { 4 \pi } { 3 }$
  • C. C. $\frac { 5 \pi } { 3 }$
  • D. D. $2 \pi$

Answer: C

Solution: ![](/images/questions/solid-geometry/image-020.jpg) From the question, the rotated geometry is shown in the figure: Right-angled trapezoid ABCD rotated around the line where AD and the surface formed by the geometry is a base radius of 1, the length of the mother line is 2 cylinder digging a base radius of 1, the height of 1 cone obtained after the combination, so the The volume of the assemblage is $V = V _ { \text {圆柱 } } - V _ { \text {圆雉 } } = \pi \times 1 ^ { 2 } \times 2 - \frac { 1 } { 3 } \times \pi \times 1 ^ { 2 } \times 1 = \frac { 5 } { 3 } \pi$.

Question 30: 34. The three views of a geometry are shown in the figure, and the curved portion of its front view ...

34. The three views of a geometry are shown in the figure, and the curved portion of its front view is half an arc, then the surface area of the geometry is ( ). ![](/images/questions/solid-geometry/image-021.jpg) ![](/images/questions/solid-geometry/image-022.jpg) ![](/images/questions/solid-geometry/image-023.jpg) Make a visualization

  • A. A. $16 + 6 ^ { \sqrt { 2 } } + 4 \pi \mathrm {~cm} ^ { 2 }$
  • B. B. $16 + 6 ^ { \sqrt { 2 } } + 3 \pi \mathrm {~cm} ^ { 2 }$
  • C. C. $10 + 6 ^ { \sqrt { 2 } } + 4 \pi \mathrm {~cm} ^ { 2 }$
  • D. D. $10 + 6 ^ { \sqrt { 2 } } + 3 \pi \mathrm {~cm} ^ { 2 }$

Answer: C

Solution: The geometry is obtained from the three views and the surface area is found. From the three views, we can see that the geometry consists of a triangular prism and a half-cylinder, where the base of the triangular prism is an isosceles right triangle with a waist of 2 cm and lateral angles of 3 cm, the base of the half-cylinder has a radius of 1 cm, the base of the half-cylinder has a busbar of 3 cm, and the surface area is $10 + 6 ^ { \sqrt { 2 } } + 4 \pi \left( \mathrm {~cm} ^ { 2 } \right)$.

Question 31: 35. It is known that the three views of a simple geometry are shown in the figure, if the volume of ...

35. It is known that the three views of a simple geometry are shown in the figure, if the volume of the geometry is $24 \pi + 48$, then the surface area of the geometry is ![](/images/questions/solid-geometry/image-024.jpg) ![](/images/questions/solid-geometry/image-025.jpg) ![](/images/questions/solid-geometry/image-026.jpg)

  • A. A. $24 \pi + 48$
  • B. B. $24 \pi + 90 + 6 \sqrt { 41 }$
  • C. C. $48 \pi + 48$
  • D. D. $24 \pi + 66 + 6 \sqrt { 41 }$

Answer: D

Solution: The geometry is a combination of a prismatic pheasant and a quartered circular pheasant with a surface area of $V = \frac { 1 } { 3 } \left( \frac { 1 } { 4 } \pi ( 3 r ) ^ { 2 } + \frac { 1 } { 2 } \times 3 r \times 3 r \right) \times 4 r = 24 \pi + 48 , ~ r = 2$, and so $S = \frac { 1 } { 2 } \times 12 \times 8 + \frac { 1 } { 2 } \times 6 \times 6 + \frac { 1 } { 2 } \times \frac { 1 } { 2 } \pi \times 6 \times 10 + \frac { 1 } { 4 } \pi \times 6 ^ { 2 } + \frac { 1 } { 2 } \times 6 \sqrt { 2 } \times \sqrt { 100 - 18 } = 66 + 24 \pi + 6 \sqrt { 41 }$ , so choose D.

Question 32: 37. The three views of a geometry are shown in the figure, the length of the small square in the fig...

37. The three views of a geometry are shown in the figure, the length of the small square in the figure is 1, then the volume of the geometry is ![](/images/questions/solid-geometry/image-027.jpg)

  • A. A. 60
  • B. B. 48
  • C. C. 24
  • D. D. 20

Answer: C

Solution: ![](/images/questions/solid-geometry/image-028.jpg) From the three views: the geometry is a three-pronged column eliminating a three-pronged pheasant with the same base, as shown in the figure, the height of the three-pronged column is 5, the height of the eliminated three-pronged pheasant is 3, the three-pronged pheasant and the base of the three-pronged column are right triangles with right-angled sides of lengths of 3 and 4, respectively, $\therefore$ the geometry of the volume of the $V = \frac { 1 } { 2 } \times 3 \times 4 \times 5 - \frac { 1 } { 3 } \times \frac { 1 } { 2 } \times 3 \times 4 \times 3 = 30 - 6 = 24$, so choose C. [METHOD POINT] This question uses the three views of space geometry to focus on students' spatial imagination and abstract thinking ability and the volume formula of prisms and prisms, belonging to the difficult problem. Three views of the problem is to test students' spatial imagination ability is the most common type of questions, but also the college entrance examination hot spot. Observe the three views and its "translation" into a visual diagram is the key to solving the problem, not only to pay attention to the three elements of the three views "high flush, long to the right, wide and equal," but also pay special attention to the solid line and dotted line and the same figure of the different positions of the visualization of the geometry.

Question 33: 38. Nine chapters of arithmetic" is China's ancient extremely rich mathematical masterpieces, the bo...

38. Nine chapters of arithmetic" is China's ancient extremely rich mathematical masterpieces, the book has the following question: "Today there are commissioned beans according to the group within the corner, under the circumference of three zhang, seven feet high, asked the volume and geometry for beans? "The meaning is:" stacked against the wall of soybeans into a semiconical shape, soybean pile of arc length of the bottom surface of 3 zhang, 7 feet high, ask the volume of soybean pile and stacked soybeans have how many dendrobes? Known as 1 db of soybeans $= 2.43$ cubic feet, 1 zhang $= 10$ feet, the circumference of the circle is about 3, the estimated volume of soybeans stacked there are

  • A. A. 140 Dendrobium
  • B. B. 142 Dendrobium
  • C. C. 144 Dendrobium
  • D. D. 146 Dendrobium

Answer: C

Solution: According to the question, let the semicircular radius of the cone base be $r$, the length of the semicircular arc be $l$, then we have $\pi r = l$, that is, $r = \frac { l } { \pi }$, and the area of cone base is $S = \frac { 1 } { 2 } r l = \frac { l ^ { 2 } } { 2 \pi }$, then we have $l = 30$, and $h = 7$, that is, $V = \frac { 1 } { 3 } S h = \frac { 1 } { 3 } \cdot \frac { l ^ { 2 } } { 2 \pi } \cdot h \approx \frac { 1 } { 3 } \times \frac { 30 ^ { 2 } } { 2 \times 3 } \times 7 = 350$. 4]] And $l = 30$ feet, the height of the cone $h = 7$ feet, so we get the volume of the round pheasant is $V = \frac { 1 } { 3 } S h = \frac { 1 } { 3 } \cdot \frac { l ^ { 2 } } { 2 \pi } \cdot h \approx \frac { 1 } { 3 } \times \frac { 30 ^ { 2 } } { 2 \times 3 } \times 7 = 350$ in cubic feet. From $\frac { 350 } { 2.43 } \approx 144$ we get that there are 144 (cubic feet) of soybeans. So the pile of soybeans is about 144 (dwt).

Question 34: 39. It is known that all the vertices of the quadrangular pheasant $S - A B C D$ are on the same sph...

39. It is known that all the vertices of the quadrangular pheasant $S - A B C D$ are on the same sphere, and that the base $A B C D$ is square and the center $O$ is at the center of the sphere. In the same plane, when the volume of this quadrilateral is maximized, its surface area is equal to $4 + 4 \sqrt { 3 }$, then the volume of the ball $O$ is equal to

  • A. A. $\frac { 32 \sqrt { 2 } } { 3 } \pi$
  • B. B. $\frac { 16 \sqrt { 2 } } { 3 } \pi$
  • C. C. $\frac { 8 \sqrt { 2 } } { 3 } \pi$
  • D. D. $\frac { 4 \sqrt { 2 } } { 3 } \pi$

Answer: C

Solution: all vertices of the quadrangular pheasant $S - A B C D$ are on the same sphere, and the base $A B C D$ is square and in the same plane as the center $O$. So the center of the sphere $O$ is the center of the square $A B C D$. When the height of the quadrilateral is the radius of the ball, the volume of the quadrilateral is maximized. The quadrangle is then a square quadrangle. Let the radius of the ball $O$ be $R$, then $| A B | = \frac { \sqrt { 2 } } { 2 } | A C | = \sqrt { 2 } R$, $| S B | = \sqrt { | O B | ^ { 2 } + | S O | ^ { 2 } } = \sqrt { 2 } R$, $\triangle S B C$ and $S _ { \triangle S B C } = \frac { 1 } { 2 } | S B | ^ { 2 } \sin 60 ^ { \circ } = \frac { \sqrt { 3 } } { 2 } R ^ { 2 }$. $| S B | = \sqrt { | O B | ^ { 2 } + | S O | ^ { 2 } } = \sqrt { 2 } R$ $\triangle S B C$ is an equilateral triangle, then $S _ { \triangle S B C } = \frac { 1 } { 2 } | S B | ^ { 2 } \sin 60 ^ { \circ } = \frac { \sqrt { 3 } } { 2 } R ^ { 2 }$ So the surface area of this quadrangle is $4 S _ { \triangle S B C } + S _ { A B C D } = 2 \sqrt { 3 } R ^ { 2 } + 2 R ^ { 2 } = 4 + 4 \sqrt { 3 }$ So $R = \sqrt { 2 }$ and hence the volume of the ball $O$ is $V = \frac { 4 } { 3 } \pi R ^ { 3 } = \frac { 8 \sqrt { 2 } } { 3 } \pi$. ![](/images/questions/solid-geometry/image-029.jpg)

Question 35: 40. To make a cylindrical closed container with a volume of $216 \pi \left( \mathrm {~cm} ^ { 3 } \r...

40. To make a cylindrical closed container with a volume of $216 \pi \left( \mathrm {~cm} ^ { 3 } \right)$, the radius of the base of the cylinder is $216 \pi \left( \mathrm {~cm} ^ { 3 } \right)$ in order to minimize the amount of material used. High School Mathematics Assignment, October 29, 2025

  • A. A. 6 cm
  • B. B. $3 \sqrt [ 3 ] { 2 } \mathrm {~cm}$
  • C. C. $3 \sqrt [ 3 ] { 4 } \mathrm {~cm}$
  • D. D. $3 \sqrt { 3 } \mathrm {~cm}$

Answer: C

Solution:
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Solid Geometry

空间几何

35 Practice Questions

Practice with Chinese questions to prepare for the CSCA exam. You can toggle translations while practicing.

Topic Overview

Spatial geometry is the branch of mathematics that deals with the study of points, lines, surfaces, and bodies and their relationships in three-dimensional space. In the CSCA exam, this section often involves the calculation of surface area and volume of common three-dimensional (e.g., columns, cones, and spheres), the identification and reduction of three-view drawings, and the determination of spatial positional relationships. The questions usually incorporate graphs and data, requiring candidates to have strong spatial imagination and formula application skills.

Questions:35

Key Points

  • 1Calculating surface area and volume of common three-dimensional figures
  • 2Interconversion of three views and visualizations
  • 3Determination of parallel and perpendicular lines in space
  • 4External and internal tangent ball problems for simple geometries

Study Tips

More hands-on sketching of three-dimensional shapes, practicing spatial reductions in conjunction with three-view drawings, and proficient memorization of volume and surface area formulas for key geometries are recommended.

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