Teaching objectives

Three-dimensional graphics, the main test site focuses on the calculation of the surface area and volume of irregular bodies. Among them, there is a unique "dyeing problem" and a common "geometric olympiad problem"

In elementary school, besides plane graphics, I also learned some simple three-dimensional graphics, such as cuboid, cube, straight cylinder, straight cone and sphere. , and know their volume and surface area calculation formula, summarized as follows.

★★★★★ Cube: We can also call it a cube. It is a special cuboid, and its six faces are all squares. If its side length is, then we can get:

Surface area of cube:

Volume of the cube:

★★★★★ Cuboid: If the length, width and height of a cuboid are respectively, then:

Surface area of cuboid:

Volume of cuboid:

★★★★ Cylinder: As shown in the right figure, the bottom of the cylinder is a circle with radius: the side development of the cylinder is a rectangle, the width is equivalent to the height of the cylinder, and the length is equivalent to the circumference of the bottom of the cylinder;

Surface area of cylinder:

Cylinder volume:

★★★★ Cone: As shown in the figure on the right, the bottom of the cone is a circle with radius: the side development of the cone is a sector;

Volume of cone:

★★★★★ Sphere:

There are many interesting geometry problems in the math contest. The key to solving these interesting problems lies in ingenious conception and proper design, which combines image thinking with abstract thinking.

Teacher's version of the answer tip: As shown in the figure below, tilt the rectangular container so that the water at one end just reaches the edge A of the container mouth, and when the other end of the horizontal plane just reaches the edge B, the container is just half filled with water. If this is not the case, the water in the container is not half full. As shown in Figure ②, the water in the container is just half full, but not in Figures ① and ③. More than half in Figure ① and less than half in Figure ③.

Surface area of three-dimensional graphics

Example 1 A cube with a side length of 1 cm is stacked layer by layer, as shown in the figure. When it is stacked on the fifth floor, how many square centimeters is the surface area of this three-dimensional figure?

Analysis: The rule of the number of blocks contained in the diagram: the fourth layer 1, 1, 3,6, 10, 15, increasing by 2,3,4,5 … in turn. When overlapping to the fifth floor, the perspective view is up and down.

Example 2 has two cylindrical parts with a height of 10 cm and a bottom diameter of 6 cm. One end of this part has a cylindrical part with a height of 10 cm and a bottom diameter of 6 cm. One end of the part has a straight cylindrical hole, as shown in the figure. The diameter of the round hole is 4 cm and the depth of the hole is 5 cm. If the part in contact with air is painted with antirust paint, how many square centimeters should it be painted? ( )

Observation shows that the drawn part includes the outer surface of the cylinder and the inner surface of the circular hole.

The upper and lower bottom surfaces of the parts and the outer side surfaces of the parts;

The inside of the part is:, and the part coated with antirust paint is:.

On the right side of the picture is a hat. The top of the hat is cylindrical and made of black cloth; The brim part is a ring made of white cloth. If the radius, height and width of the crown are centimeters, which color of cloth is used more?

Analysis: as much. Black cloth: white cloth:.

Example 3 How many square centimeters of iron sheet does it take to make a workpiece as shown in the figure (both ends are not closed)? ( )

Analysis: The workpiece is neither a cylinder nor a cone, and it is not our common regular geometric figure. So we should consider how to transform this geometric figure into a familiar geometric figure. As shown in the figure below, if you take the same workpiece and put two workpieces together, you can make a right cylinder, then the lateral area of one workpiece is half of the lateral area of this cylinder. The height of the cylinder is:, and the side area of the cylinder is:. A workpiece needs iron sheet: (square centimeter). When solving the problem of irregular three-dimensional graphics, the key is to first convert it into regular three-dimensional graphics, and then use the formulas and properties we have to solve the problem. In fact, we have been exposed to this idea in the spring class.

Consolidation (related topics learned in the spring of fifth grade) (Hope Cup training in 2007) Saw off a cuboid block with a square bottom and turned it into a hexahedron ABCD-EFGH as shown on the right, in which the longest side DH = 8cm and the shortest side AB = BC = CD = DA = BF = 4cm. How many cubic centimeters is the volume of this hexahedron?

Analysis: 42. The volume of this hexahedron is half that of a cuboid with a length of 4 cm, a width of 4 cm and a height of 12 cm, that is, 4×4× 12÷2=96 (cubic cm).

Extension (Huajin Cup 2005), as shown in figure 1, is a curved surface development diagram of a straight triangular prism, in which the gray-black part is a square with a side length equal to 1. Q: What is the volume of this triangular prism?

Analysis: As shown in Figure 2, this triangular prism is a cube with the length of 1 cut diagonally. The volume of the cube is 1, and the volume of this straight triangular prism is half that of the cube.

Example 4 (Spring Cup Mathematics Invitational Tournament) The surface area of a cube is 54 square centimeters. What is the sum of the surface areas of two cuboids if one size fits all?

Analysis: It is known that the surface area of a square is 54 square centimeters, so the area of each side of the square is 54÷6=9 (square centimeters). After being cut into two cuboids, the sum of the surface areas of the two cuboids is 9×2= 18 (square centimeter) higher than the original square surface area. Therefore, the sum of the surface areas of two cuboids is 54+6548.

The shop in front is as shown on the right. The side length of a square ABCD is 6 cm. By drawing a straight line at any two points in the square, the square can be divided into nine small rectangles. What is the sum of the perimeters of these nine small rectangles?

Analysis: On the whole, when finding the sum of the perimeters of nine small rectangles, four sides of AB, BC, CD and AD are used 1 time, and the other four lines are used twice, so the sum of the perimeters of nine small rectangles is 4×6+4×2×6=72 (cm).

Qianpu (the topic of related ideas in fifth grade spring school) is a cube-shaped wood block with a side length of 1 m. Saw it into three pieces along the horizontal direction, each piece into four pieces according to any size, and each piece into five small pieces according to any size, * * * to get 60 cuboids of various sizes. What is the sum of the surface areas of these 60 cuboids?

The analysis shows that the original cube has six external surfaces, and the area of each surface is 1× 1 = 1 (square meter). No matter how many pieces are sawed later, the 6 square meters of these six outer surfaces are always included in the surface area of the later small pieces of wood. Considering that every time you saw it, you would get two sides of 1 square meter, and now it is a * *.

Example 5 (Tsinghua attached training questions in 2005) Divides a cuboid with a red surface area into several small cubes with a side length of 1cm, of which only three sides are not red. How many square centimeters is the surface area of the original cuboid?

Analysis: length: 3+1+1= 5 cm; Width:1+1+1= 3cm; Height:1+1+1= 3cm; So the surface area of the original cuboid is:

(3×5+3×5+3×3)3×2=78 square centimeters.

Champ (the topic of related ideas in fifth grade spring studies) is a 4×5×6 cube on the right. If its surface is painted red, how many cubes are painted red on one side, two sides and three sides?

Analysis: Eight cubes with only eight vertices are painted red on three sides;

Red painted on both sides, ***(4-2)×4+(5-2)×4+(6-2)×4=36 pieces;

The middle part of the surface, one side painted red:

(4-2) × (5-2 )× 2+(4-2 )× (6-2 )× 2+(5-2 )× (6-2 )× 2 = 52 blocks.

The squares not painted red are: (4-2)×(5-2)×(6-2)=24. Pay attention to helping children understand and then sum up the rules.

Extension (the topic of related ideas in fifth grade spring studies) The right picture is a 3×3×3 cube composed of 27 small cubes. If the surface of a cube is painted red, three of the eight cubes in the corner are red, while the cube in the center is not red at all. Of the remaining 18 cubes, 12 cubes are red on both sides and 6 cubes are red on one side. In this way, the number of small squares with red sides is twice that of small squares with red sides, and the number of small squares with red sides is eight times that of small squares without red at all. Q: How many cubes are there? When the surface is painted red, the opposite will happen, that is, there are twice as many cubes with one side red as those with both sides red, and there is no red at all.

Is a small square eight times as big as a small square with three sides red?

Analysis: For an n×n×n cube composed of n3 cubes, 8 cubes are painted red on three sides, 12× (n-2) cubes are painted red on both sides, 6× (n-2) cubes are painted red on one side, and (n-2) cubes are not painted. Given conditions, a small square with no red at all is eight times as big as a small square with red on three sides, that is, (n-2) 3 = 8× 8, and the solution is n = 6.

Volume of three-dimensional graphics

Example 6 (Huajin Cup in 2005) As shown in the figure, a conical container A and a hemispherical container B have the diameter of the circular opening and the height of the container. If container A is used to fill container B with water, how many times should it be filled at least?

Analysis: The volume of conical container A is: and the volume of hemispherical container B is:, so water should be injected at least 8 times.

A 24cm-high conical container filled with water. If water is poured into a cylindrical container with the same diameter as the bottom of the cone, how high is the water surface?

Analysis: let the bottom area be s and the water level in the cylinder be h, according to the meaning of the question:

As shown on the right, the water in the conical container is only its volume, and the height of the water surface is a fraction of the height of the container.

Analysis: If the height of the water surface is twice that of the container, then the radius of the water surface is twice that of the bottom of the container. According to the meaning of the question,

Example 8 The ball falls into a cylindrical barrel filled with water. The diameter of the ball is 15 cm, and the diameter of the barrel bottom is 60 cm. The ball has a submerged volume (see the right). After the ball fell into the water, how many centimeters did the water level in the bucket rise?

Analysis: the volume of the ball is: (cubic centimeter); The part of the ball immersed in water is: (cubic centimeter); The bottom area of the bucket is: (square centimeter); The height at which the water level rises is: (cm).

Example 9 (selected from the experimental class of Beijing No.5 Middle School in 2006) A cylindrical glass filled with water has a bottom area of 80 square centimeters and a water depth of 8 centimeters. Now, a cuboid iron block with a bottom area of 16 square centimeter is placed vertically in the water, and some iron blocks are still exposed. How many centimeters is the water depth now?

Analysis: According to the principle of equal product change, the volume of water divided by the bottom area of water is the height of water.

(Law1): 80× 8 ÷ (80-16) = 640 ÷ 64 =10 (cm);

Method 2: Let the water level rise by centimeters. According to the volume of the rising part = the volume of the iron block immersed in human body water, the equation is:, and the solution is:, 8+2= 10 (cm).

A cylindrical barrel with a radius of 20 cm at the bottom is consolidated, and a section of cylindrical steel with a radius of 5 cm in the barrel is immersed in water. After the steel was taken out of the bucket, the water in the bucket dropped by 6 cm. How long is this steel?

According to the meaning of the question, the volume of cylindrical steel is equal to the volume of water falling in the bucket, because the bottom radius of steel is the bottom radius of the bucket, that is, the bottom area of steel is the bottom area of the bucket. According to a certain volume, the bottom area of the cylinder is inversely proportional to the height, and it can be known that the length of steel is 16 times of the falling height of the water surface.

(Law 1): 6 ÷ () = 96 (cm)

Method 2: 3.14× 20× 6 ÷ (3.14× 5) = 96 (cm).

Spread out (the topic of fifth grade spring studies, I hope the teacher will take time to recall this topic as much as possible). The inner diameter of the bottom of a cylindrical container filled with water is 5 cm, the depth is 20 cm, and the water depth is 15 cm. Now put an iron cylinder with a bottom radius of 2 cm and a height of 18 cm vertically into the container. Find out how many centimeters the water depth of the container is at this time.

Analysis: There may be three situations in this question: ① The water depth is not as high as the iron tube; (2) the water depth is higher than the iron drum but does not overflow; ③ Water overflows. After being put into the iron drum, the cross section of water is annular around the iron drum, as shown in the figure, and the cross-sectional area is × 5× 5 —× 2 = 2 1. Income cylinder. 18cm. So the water depth of the container at this time is 17 cm.

[Comment] Please consider the situation that the water depth is 16 cm or 19 cm, and compare it with the result of this question.

Example 10 A three-dimensional figure is the same figure from top to bottom, from front to back, and from left to right. It is a square with a side length of 3 cm, divided into 9 small squares with equal areas (as shown in the right picture). Calculate the total surface area and volume of three-dimensional objects.

Analysis: According to the three views, it can be judged that the solid is a cube with a length of 3 cm, and a square with a bottom area of1cm 2 and a height of 3 cm is drawn in the center of each face, as shown in the right figure below.

Let the total surface area and total volume of solids be:

(square centimeter),

(cubic centimeter).

Make a hole in the center of each face of a cube with a side length of 4 cm. The hole is square, with side length 1 cm and hole depth 1 cm (as shown below). Find the surface area and volume of the block after digging.

Analysis: The side length of the big cube is 4cm, and the side length of the dug small cube is 1cm, which indicates that the big cube block has not been dug through. Therefore, every time a small cube is dug, the surface area of a large cube is increased by four lateral area in the "small hole". The sum of the newly added areas in the six small holes: 1× 1× 4× 6 = 24 (square centimeter), the original cube surface area: 42× 6 = 96 (square centimeter), the surface area after digging: 96+24 = 120 (square centimeter), and the volume.

As shown in the figure, a cuboid hole is made in the center of two pairs of side faces of a cube, and a cylindrical hole is made in the center of the upper and lower bottom faces. It is known that the side length of a cube is 10 cm, the holes on the side are squares with a side length of 4 cm, and the holes on the upper and lower sides are circles with a diameter of 4 cm. Find the surface area and volume of this three-dimensional figure.

Analysis: the lateral area is 6×10×10-4× 4-22× 2 = 536-8.

The internal surface area is16× 4× 3+2× (4× 4× 22)+2× 3 =192+32-8+24 = 224+16.

Total surface area = 224+16+536-8 = 760+8 = 785.12 (square centimeter).

When calculating the volume, take out the three-dimensional diagram of the hollowed-out part, as shown in the figure, as long as the volume of this geometry can be calculated. The volume of the mining geometry is:

4×4×4×3+4×4×4+2× ×22×3= 192+64+24 =256+24 .

The geometric volume is:1o×1o×1o-(256+24) = 668.64 (cubic centimeter).

[Comment] If this problem can be solved, the calculation of the surface area and volume of all irregular bodies is a cinch. It must be noted that the thinking should be clear, for example, the surface area is discussed from the outside to the inside, and the volume is directly the whole cut-off part. Details determine success or failure: first, when calculating the surface area, the square of the inner center MINUS the inscribed circle is easily ignored; Secondly, the side length of the large cube in this problem is 10 cm, which is a very nerve-racking number, which directly leads to the appearance of 3 in many places. Hehe, many people are embarrassed here.

Example 1 1 (03 Math TV Science Competition) As shown in figure 1, ABCD is a right-angled trapezoid (unit: cm).

(1) Take AB as the axis, and rotate the trapezoid around this axis to get a rotating body. What is its volume?

(2) If CD is the axis and the trapezoid rotates around this axis, what is the volume of the rotating body?

Analysis: (1) As shown in Figure 2, the required volume can be regarded as the sum of BCDE revolving body around AB and△ △AED revolving body around AB, namely (cubic centimeter).

(2) As shown in Figure 3, the required volume can be regarded as the difference between ABCE revolving around EC and△ △ADE revolving around EC, namely

(cubic centimeter).

Example 12 (the 7th Zu Chongzhi Cup Mathematics Invitational Tournament) has a rectangular piece of iron, with a length of 40cm and a width of 20cm. Please use it to make a 5 cm deep cuboid tin box without cover (the bigger the better). What is the volume of the tin box you made?

Analysis: Method 1: (1) As shown in the figure on the right, cut a square iron sheet with a length of 5 cm at the four corners of a 40×20 rectangular iron sheet, and then weld it into a rectangular open iron sheet box. The length of this tin box is 40-5-5 = 30 (cm) and the width is 20-5-5 = 60.

(2) As shown in the right figure, cut two squares (two pieces) with a side length of 5cm from the two corners on the left side of a 40×20 rectangular iron sheet, and weld them tightly in the middle of the right side, so that the length of the uncovered iron sheet box is 40-5 = 35 (cm), and the width is 20-5-5 =10.

(3) As shown in the figure on the right, cut a rectangular iron sheet (* * * two pieces) with a width of 5cm from the left and right sides of a 40×20 rectangular iron sheet, and weld them to the upper and lower parts in the middle, respectively, to make an uncovered iron sheet box with a length of 40-5-5-5 = 20 (cm) and a width of 20 (cm).

Method 2: If you want to maximize the volume, you should make full use of the iron piece in your hand. If you can use all the iron sheets, you can get the biggest iron box. As shown in the following figure (1), we cut four 5×20 cuboids from the original iron sheet, as shown in Figure (2), which can be welded into a 5 cm deep cuboid iron sheet box, so the maximum volume at this time is 20×20×5=2000 (cubic centimeter).

Theme outlook

If you want to know the content of the entrance exam, please pay attention to the content of the winter vacation class!

Exercise 3

1. Use a square with a side length of 1 cm to make a three-dimensional figure as shown in the figure below. What is the surface area of this figure? (Comparative Example 1)

Analysis: Looking at the area problem as a whole, the surface area of the upper and lower sides is always 3 × 3; Looking at the front, back, left and right, it is 2×3+ 1, so the total is 9×2+7×4= 18+28=46cm2.

2. How many square decimeters has the surface area increased by dividing a cube with a length of 6 decimeters into 27 cubes on average? (Comparative Example 4)

Analysis: In order to divide the cube into 27 cubes equally, it must be divided according to the graph. For each division, the surface area will increase the area of two squares with a side length of 6 decimeters. * * * needs to be divided into six places, increasing the area of 12 squares. 6 × 12=432 (square decimeter)

3. There are two conical containers filled with water, with a bottom radius of 10 cm and a height of 30 cm. Pour all the water into a cylindrical container with a bottom radius of 20 cm to find the water depth. (Comparative Example 7)

Analysis: (cm)

4. (Huajin Cup 2005) The three sides of a right triangle are 3, 4 and 5 respectively. If you rotate on the axis of side length 4, you will get a solid. Find the volume of this solid. (For example, 1 1)

Analysis: A solid rotating around a right-angled side with a length of 4 is also a cone with a bottom radius of 3, which is obtained from the volume formula of the cone:

5.(2005 National Primary School Mathematics Olympics) There is a cube block with a side length of 12 cm, and a square hole with a side length of 4 cm is drilled in its upper, front and left center (penetrating the cube block). What is the volume of the block after perforation? (Control Example 10)

Analysis: (cubic centimeter).

6. As shown in the figure, the conical container holds 3 liters of water, and the water level is exactly half the height of the cone. How much water can this container hold? (Comparative Example 6)

Analysis: If the bottom radius of conical container is, the water surface radius is, and the volume of container is,

The volume of water is:

It shows that the container can hold 8 portions of 3 liters of water, so it can also hold water: 3× (8- 1) = 2 1 (liter).

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