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1、高二数学-9.9棱柱与棱锥(二)(b版)（High school mathematics -9.9 prism and pyramid (two) (b Edition)）9.9 prisms and pyramids (two)Learning guidance1. to solve the problems in the pyramid, we should first clear the definition and related properties of pyramid, especially some intrinsic properties of the pyramidThe
2、three angles, the three distances, the area and the volume of the 2. pyramids are always through them. Through the proof and calculation of them, the true grasp of the pyramids is achievedKey points of knowledgeDefinition of the 1 pyramid of knowledgeOne of the faces is polygonal, and the rest of th
3、e faces are triangles with a common vertex. The closed geometry enclosed by these faces is called a pyramidThe definition of a regular pyramid: the bottom surface is a regular polygon, and the projection of the vertex at the bottom is the center of the base plane. The pyramid is called a regular pyr
4、amidThe properties of 2 pyramid of knowledge(1) high, and slant on the bottom surface of the slant pyramid projective form a triangle pyramid; high lateral and lateral ribs on the bottom surface of the projective form a right triangle; the side edge and the bottom edge of ofbending pyramid, (part of
5、) a right triangle in the bottom of the pyramid; slant projection in the side edge and the bottom, the bottom surface of the projection (part) to form a right triangle. (Figure 9 - 9 - 55: Rt VDH, Rt VAH, Rt VAD, Delta Delta Delta Rt HAD) calculation. That is often in the pyramid above four right an
6、gled triangle, is especially important.(2) using a frustum of a plane parallel to the base plane, the cross section is similar to the base, and some of the sides are proportional, and the area is proportional to the square of the corresponding edge(3) it is especially pyramid, each side edge is equa
7、l, each side are congruent isosceles triangleSpecial knowledge in Section 3 Pyramid: diagonal plane (section two is not adjacent to the side edge are parallel to the bottom surface of the section).Side area formula of 4 pyramid of knowledge pointThe drawing of horizontal horizontal drawing of 5 stra
8、ight pyramid of knowledge point: oblique two measuring methodRelative concepts and properties of knowledge points, 6 polyhedra and regular polyhedraEspecially regular polyhedron, it has and only five kinds (namely, regular tetrahedron, regular hexahedron, positive eight body, positive twelve body, p
9、ositive twenty body)Knowledge point 7 uses the properties of pyramids to study the relation between the lines and lines, lines and planes, planes and planes, and calculates the distances, the three angles and the area and volume of the pyramidsProblem solving methods and skills trainingThe proof and
10、 calculation of the relation between the line and the line in the 1 pyramidFigure 9956 in 1 cases, four pyramid PABCD, bottom ABCD is a right angled trapezoid, angle BAD = 90 degrees, AD = BC, AB = BC = a, AD = 2A, PA, ABCD and PD face the bottom surface, the bottom surface at an angle of 30 degrees
11、.(1) if the AE group PD, E group: BE PD pedal, confirmation;(2) the different surface lines AE and CD into the corner.analysis careful analysis is not difficult to find graphic features, PA, BA and DA 22 are perpendicular to each other, which is the establishment of the coordinate system, to provide
12、 the conditions of solid geometry problem is transformed into a vector, in order to find the easy way to give you one.proof (1) take A as the origin, and AB, AD and AP take the line as the coordinate axis, and establish the Cartesian coordinate systemPD group of AB and R, t AE PD dreams,PD BE. * t(2
13、) dreams of anomalous PA surface ABCD, PD and the bottom surface at an angle of 30 degrees,L / PDA = 30 E, EF t AD, AE pedal F = A / EAF = 60 deg,I (1) in solving the problem with vector, to choose the coordinate system, find out the coordinates of the point, write the vector coordinates. (2) in sol
14、ving the problems concerned with pyramid, should make full use of the definition and character of the pyramid, and if the pyramid on the surface of the point, line, space tend to figure the problem of plane.The proof and calculation of the relation between the line and the surface in the 2 pyramid2
15、cases of known: four PABCD pyramid, the bottom surface is a right angled trapezoid, which AB, CD, BA t AD, PAD t side bottom ABCD, (1): confirmation of anomalous PCD plane PAD plane; (2) if AB = 2, CD = 4, PBC is equal to the positive side of the long side of the triangle 10 the AC and PCD for the d
16、iagonal side angle sine value.(1) analysis it is proved that the plane and the plane are vertical. Generally, the judgment theorem of the line and the surface is verticalprove dreams quadrilateral ABCD right angled trapezoid, CD. AB saidAnd the dreams of anomalous PAD bottom ABCD surface, a bottom s
17、urface PAD ABCD = AD,AD group AB, AB group of PAD. star(2) analysis we should grasp the projection angle for line and plane anglesolution by (1), PAD group and PAD PCD, a plane surface PCD = PD. * PAD in the surface in A AH group PD. H. AH group of pedal PCD (theorem vertical surface). Then connect
18、to CH, then ACH for front line, face the angles (Figure 9 - 9 - 61).I line and plane angle grasping projective.Determine the projective points depends on nature theorem of vertical surface. The vertical line and surface vertical is often use each other and transform into each other.The proof and cal
19、culation of the relation between the plane and the surface in the 3 pyramid3 cases in figure 9966, ABCD AB t BCD tetrahedron, plane, BC = CD / BCD = 90 / ADB = 30 degrees, E degrees, F, AC, AD respectively is the midpoint.(1): confirmation of anomalous BEF plane ABC plane;(2) calculate the angle of
20、plane BEF and plane BCDanalysis to prove that the two planes are vertical, that is, to prove a plane through another plane of a vertical line, so you need to prove a straight line and the plane of the intersection of two vertical lines, and to prove that the two vertical line,(1) that established in
21、 Figure 9 - 9 - 67 shows the space coordinate system, A (0, 0, a), a / ADB = 30 deg,I use vector coordinate calculation in three-dimensional geometry, played a count generation card effect, reduce the degree of difficulty, embodies the new outline of the spirit.The calculation of the area and volume
22、 in the 4 pyramid4 cases of known plane anomalous ADE plane ABCD, Delta ADE is the side length of a ABCD is an equilateral triangle, rectangle, F is the midpoint of AB, EC and ABCD plane at an angle of 30 degrees, (1) for the four E AFCD pyramid volume; (2) E - CF - D for the dihedral angle size (;
23、3) for the D EFC point to surface distance.(1) solution in Figure 9 - 9 - 72, EAD group of ABCD dreams, and EAD is an equilateral triangle, triangular E projective point on the bottom of the ABCD on H, AD and AD in the midpoint of the link HC / ECH, then = 30 degrees.(2) analysis for E - CF - D dihe
24、dral angle, by definition of law (i.e. three vertical theorem as the angle, angle), you can also use the projective area method.Star delta HFC is a right angled triangle. And the anomalous HF / EFH as defined by FC. that the plane EFCD dihedral angle, in Rt Delta EFH,I volume conversion method is a
25、commonly used method of point to plane distance. Its idea is simple, easy to master.Easily mixed warningIn the understanding of the definition of pyramid attention must be paid to the conditions in the definition: the rest of the surface is a common vertex of the triangle. Some geometric properties
26、of the pyramid is to be straightened out, clear, also should be paid attention to and lateral side, the bottom of the pyramid location.5 cases have the following proposition of the bottom surface is polygonal pyramid is orthoprism; all side edge length equal is the pyramid is a pyramid pyramid; can
27、have two side edges perpendicular to the bottom surface; and a pyramid can have two sides perpendicular to the bottom surface, wherein, the correct proposition ()A.0 B.1C.2 D.3means choose A or C or D.analysis wrong due to the cause of the error is the concept and nature of the pyramid is not clear
28、enough on the line and line, line and surface, surface and surface properties, the mastery and application of theorem is not in place, in fact, as long as the definition of the proposition are familiar pyramid will not go wrong, the pyramid prisms in the same length of proposition at the bottom, can
29、 only determine the vertex projective point, but not sure it is positive if the pyramid pyramid, proposition two side edge is perpendicular to the bottom surface, the two side edge will be parallel, then it is not a pyramid, the proposition of a pyramid can have two sides perpendicular to the bottom
30、 surface, as long as a pyramid side edge is perpendicular to the bottom surface, it will have two sides and perpendicular to the bottom surface, so right. a. B. positive solutionIntegrated application innovationintegrated capabilities upgradingThe pyramid as a carrier, a comprehensive test of the re
31、lationship between the position of line and plane problem is another key problem in solid geometry. To solve these problems, a correct understanding of a variety of different positions in the pyramid lines; two to master the analysis and solve common problems in solid geometry.6 cases in Figure 9 -
32、9 - 75, P - ABCD in the four pyramid, anomalous PB bottom ABCD.CD t PD, ABCD bottom is a right trapezoid, AD BC AB, an BC, AB = AD = PB = 3, E PA and PE in the edge, = 2EA.(1) the different surface lines PA and CD into the corner;(2): PC / / EBD plane confirmation;(3) for the size of ABED dihedral a
33、ngle;(1) analysis according to the concept of the angle of the different straight line, in the bottom of the ABCD, the A can find the parallel line of DC, then we can make the angle of PA and CD, then we use the known condition to seek the characteristic of the triangle in itsolution in the bottom o
34、f ABCD A AF DC BC in F, PA and AF for acute angle or right angle.From the known, Delta DAB as an isosceles right triangle, the angle DBA = 45 deg,And / ABC = 90 deg / CBD, r = 45 deg,Dreams of anomalous PB plane ABCD, CD group PD, CD group of BD star,CDB is an isosceles right triangle star.Star delt
35、a PAF is an equilateral triangle.L / PAF = 60 degrees.Therefore, the angles of the PA and CD are 60 degrees(3) according to the analysis of anomalous DA plane PAB plane, so as to be composed of three vertical theorem A - BE - D dihedral angle.solution according to the meaning of problems, DA group o
36、f plane PAB, A AG BE G DG group, DG group of BE links.It is AGD / A - BE - D plane dihedral angle.In this case I (3), G in the extended line segment of BE (in order to simplify the original graphics, there is no picture). In the segment of AG long, alone in the plane PAB mapping (Figure 9 - 9 - 76),
37、In general, in space problems, if the whole graph is more complex, and solving the involved elements (or quantity) only in one part (usually graphic) is the part of get more intuitive graphic feature - named Sprite.application innovation capability upgradeFolding and unfolding of the thought, segmen
38、tation and compensation of thought, such as product (area and volume) application thought, the thought of the great transformation equation in this section in this section is the knowledge of knowledge in the application of the main problem solving ideas.analysis for the geometry on the surface of t
39、he distance between two points of the minimum value to the general geometry side. (Figure 9977, 9978)solution this is a pyramid of four P - ABCD PA along the cut, then it will start side link AA , AA, PB = E, AA AA PC = F, PD = G, then this is the four pyramid expansion graph reduction is beetles should be A - E - F - G - A (A) path crawling, through the shortest distance.The beetle in AEFGA path crawling through the shortest distance, and the shortest distance is 6.