TACN-group-12-1

Chủ đề : Xây dựng hệ thống tập hình học trường trung học phổ thơng Danh sách thành viên nhóm: Trần Thị Hồng Khánh Lưu Đức Mạnh Nguyễn Thị Thúy Nga Lê Doãn Thái Introduction Geometry is a very old subject It probably began in Babylonia and Egypt Geometry is the branch of mathematics which investigates the relations, properties and measurements of solids, surfaces, lines and angles Today geometry includes plane figures and solid figures In order to grasp the knowledge og geometry, we need to exersices with theoretical knowledge of the trade Plane figures in high school includes vectors, dot product of twovectors, coordinate system, relationships within a triangle and solution of triangles, geometric transformation More important is conventions for coordinates in the plane, because it’s the premise for solid figures.And it has linear equations and circle equations Plane geometry is pretty simple and easy to imagine Solid figures or figures with three dimentions It includes length, width, thickness, lines and planes in space, vectors in spce, distances, prism, pyramid, surfaces of revolution, spheres, the volume and the mrthod of coordinates in space Solid figures hard to imagine but beautiful drawing is also very easy In order to know more of geometry, we explore the topic: “Build exercises system of geometry in high school” PLANE GEOMETRY I A a Vector Brief theories Addition and subtraction of two vector Definition:Let then A b B Properties - Communtative property: - Associative property: - Zero vector property: - Given any three arbitrary point A,B and C, we always have: + 3-pointrule: + Subtraction rule: + Parallelogram rule: ABCD is a parallelogram if Scalar multiple of a vector C a Definition Let is a vector and a nonzero vector + if k > then is direction the same as + if k < then is oppsite tho that of + is vector with magnitude b Properties - Given any two vectors and and given any scalar h and k, we get: + + + + + - If , then two vectors and are parallel - Three separate point A, B and C are collinnear if and only if there is a nonzero scalar a so that - Given two vectors and that are not parallel Then any vector can be solely analysed according to two vectors and , that is to say there is only pair of scalar h, k so that Coordinate system - If  - Given two points A(xA ; yA) and B(xB ; yB).We get: = (xB – xA, yB – yA) - Coordinates of the midpoint of the line segment - Coordinates of the centroid of the triangle B Common mathematical formulas a - Prove that two vectors is congruent Method of solution: By definition, or b By parallelogram rule Example: Example 1: Given quadrilateral ABCD Let be point M, N, P and Q with midpoint of the sides AB, BC, CD and DA respectively Prove that: Prove: By assumption, we have : M A B Q D N P C (1) Hence MNPQ is parallelgram Therefore MN is parallel to PQ (2) From (1) and (2), we deduce that Example 2: Given any four point A, B, C and D, we always have Proof : In fact, given any point O, we have c Problems: Exercise 1: Prove that if and only if the midpoints of two line segments AD and BC coincide Exercise 2:Given triangle ABC, let D is midpoint of the segment BC and M, N, E and F are repectively midpoint of the segment AB, AC, CD and DB Prove that: Exercise 3: Given quadrilateral ABCD, let points M, N and P are respectively midpoint of the segment parallelogram if and only if Prove that three separate point A, B and C are collinear a Method of solution: Three separate point A, B and C are collinear if and only if : 1) There’s exist k is nonzero such that 2) There are exist α and β such that b Example: Example 1: Given ∆ABC with G as the centroid Let I be the midpoint of line segmemt AG and K be a point on side AB so that Put Prove that point C, I and K are collinear A K I G C B Solution: Let AD be the median of ∆ABC, we get Thus, From the above analysis, we get Thus three points C, I and K are collinear Example 2:Given four point O, A, B and C such that Prove that: A, B, C are collinear O A B C Prove: By subtraction rule, then: We have: By method of solution, we deduce that, three point A, B and C are collnear Example 3: Given triangle ABC On the segment BC, AC, AB, take point M, N and P segment respectively so that: and Prove that: M, N and P are collinear A P N M C B Solution By assumption: + , giving + , giving + , giving We have: = = = = = (since by 3-point rule) ( by subtraction rule ) Therefore, By method of solution, we deduce that M, N and P are collinear c Problems: Exercise 1: Given parallel ABCD On the ray AD, AB, take two points E and respectively such that Prove that: F, C and E are collinear Exercise 2: Given triangle ABC Take two point I, J such that: Prove that: I and J arecollinear Exercise 3: Given triangle ABC, take two point M, N and G is centroid of the triangle, such that: Prove that: M, G and N are collinear Prove that equation of vectors a Method of solution - Moved of two sides - Moved of everysides and used to formurlas: subtraction rule, 3-point rule, parallelogram rule, b Example Example 1: Given any four point A, B, C and D, we always have Proof : In fact, given any point O, we have Example : Point G is the center of triangle ABC if and only if A B G C I D Proof: The center G of ∆ABC is on the median AI Let D be symmetric to G with respect to I the BGCD is a parallelogram and G is the midpoint of line segment AD Thus and We get On the contrary, suppose Draw the parallelogram BGCD with I as the point of intersection between two diagonals Then , thus , so G is of the midpoint of line segment AD Therefore, three point A, G and I are on a straight line, with and point I Thus G is the centroid of triangle ABC c Problem Exercise 1: Given parallelogram ABCD Prove: Exercise 2: Let AM be the median of the triangle ABC and D be the midpoint of line segment AM Prove: a) b) , with O as an arbitrary point Exercise 3: Let M and N be the midpoints of sides AB and CD respectively of quadriteral ABCD Prove: Exercise 4: Given equilateral triangle ABC with O as the center and M as an arbitrary point in the triangle Let D, E and F be the feet of the penperdiculars drawn from M to BC, AC and AB Prove: A II A a P N M and applications The dot product of two vectors Brief theories C DotBproduct of two vectors Definition In the case at least one of two vectors and equals vector , we difine = b Properties and applications of dot product Given any three vectors , and any scalar k, = (a1; a2), = (b1; b2) We get: - Communtative property: = - Distributive property: - (k) = k( = = a1b1 + a2b2 Relationships within triangles and solution of triangles • Given triangle ABC: A A ma B - hb hc B C mb mc C Sides : BC = a, CA = b, AB = c The length of the medians from vertexs A, B, C: ma, mb, mc The length of the altitudes from vertexs A, B, C: ha, hb, hc The circumcircle and incircle: R, r The semi-perimeter: p The area of the triangle: S • The law of cosines a2 = b2 + c2 − 2bc.cos A b2 = c2 + a2 − 2ca.cosB ; ; c2 = a2 + b2 − 2ab.cosC a b c = = = 2R sin A sinB sinC The law of sines: • The length of a median of the triangle • ma2 mb2 mc2 2(b2 + c2) − a2 = 2(a2 + c2) − b2 = ; A A ; 2(a2 + b2) − c2 = hb ma B • hc C B mb mc C The area of the triangle S= 1 aha = bhb = chc 2 abc 4R pr = 1 bcsin A = casin B = absinC 2 p( p − a)( p − b)( p − c) = = = ( Heron’s Formula) • Relationships within triangles Given a right triangle ABC: = 90, AH is altitude of the triangle A C B H - BC = AB2 + AC AB2 = BC.BH AC = BC.CH , AH = BH CH (Pitago’s Theorem) , - AH = AB2 + AC AH BC = AB.AC b = a.sin B = a.cosC = c tanB = c cotC c = a.sinC = a.cosB = btanC = bcotC B Common mathematical formulas Dot product of two vectors a Method of solution: - By definition - By formulas b Example:Given points O(0; 0), M(-2; -1), N(3; -1) Compute the dot products ? - Solution: We have: = (-2; -1), = (3; -1) Therefore = (-2).3 + (-1).(-1) = -5 c Problems: Exercise 1: Given an isosceles right triangle ABC with AB = AC = a Compute the dot products and Exercise 2: Let three points O, A and B be collinear and OA = a and OB = b Compute the dot products in the cases: a) b) a b Point O is outside segment AB; Point O is inside segment AB Compute the angle between two vectors and the distance between two points Method of solution By from the angle between two vectors By from the distance between two points Example: b) c) : 2x + my +3z – = and : nx – 8y – 6z + = : 3x – 5y + mz - = and : 2x + ny – 3z + = II The distance between two subjects in 3D space 1: The distance between two points In three-dimensional space, points have three coordinates each To find the distance between A() and B(), use the formula: Example 1: Find the distance between the points A(2,-5,7) and B(3,4,5) From the distance formula we have: 2: The distance between a point and a plane Determining the distance between a point and a plane follows a similar strategy to determining the distance between a point and a plane Consider a plane defined by the equation : M ax+by+cz+d=0 and a point () in 3D space Then we use the formula: (P) Example 2: Calculate the distance between point M(1;-2;13) and plane (: 2x - 2y –z + = d( M,() = = 3: The distance between a point and a line Distance from a point to a line is equal to length of the perpendicular distance from the point to the line If M0(x0,y0,z0) point coordinates, =(m;n;p) directing vector of line, M1(x1,y1,z1) - coordinates of point on line, then distance between point M0(x0,y0,z0) and line be found using the following formula: Example : To find distance between point M(0, 2, 3) and line Solution From line equation find: - directing vector of line; M1(3; 1; -1) - coordinates of point on line Then M0M1={3 - 0; - 2; -1 - 3}={3; -1; -4} The distance between point M0(0, 2, 3) and line is: 4: The distance between two skew lines In Oxyz space, let skew lines and The line has the direction vector is , passing through point A The line has the direction vector is , passing through point B The distance between and is: d( ) = Another method of solution: The line and the plane (as you have noted) are parallel The distance from the plane to the line is the distance from the plane to any point on the line So just pick any point on the line and us "the formula" to find the distance from (P) Step 1: Write the equation of plane (P) contain the line and parallel to We have the normal vector of (P) is = Step 2: d( )= d(=d( Example :Calculate the distance between: ( ): and (): Solution: =(1;-2;-3); =(-4;1;5) Therefore, =(-7;7;-7) In this case, we have ==7 The point A(0;5;14) , the point B(9;3;-1) ( So = (9,-2;-15) = 28 It follows that d( ) = = = We also write the equation of plane (P) passing through A(0;5;14) and has normal vector = (-7;7;-7): (P): x – y + z – = d( )= d( = = 5: The distance between a line and a plane +) If the line and the plane have intersection then distance between them is equal to the +) If the line is parallel to the plane then the distance between them is the distance between point A and the plane, point A is on the line A ( d d( ) = d( A,()) Example 5: P2 Put (P1):x−z−3=0 (P2):x+2y+4z=6 (P3):3x+2y+2z−5=0 And Δ is intersection of two plane(P1) and (P2) P1 We have, a normal vector of the plane (P1) is = (1,0,−1), a normal vector of the plane (P2) is = (1,2,4) A direction vector of Δ is cross product of and , therefore= (3,2,2) Another way, M=(3,2,0) is on the line Δ and not belongs to (P3), thus Δ is parallel to the plane (P3) We have, the distance between the line Δ to the plane (P3) is the distance from the point M to the plane (P3) From here, we have the distance of and (P3) is From 5: The distance between two parallel planes If (P): Ax + By + Cz + = and (Q): Ax + By + Cz + = is a plane equation, then distance between planes can be found using the following formula: (Q) Example To find distance between planes 2x + 4y – 4z – = and x + 2y – 2z + = (P) Solution Let's check up, whether two planes are parallel, for this purpose we will multiply the equation of the second plane for So we have: 2x + 4y - 4z + 18 = As planes are parallel than for calculation distance between planes we use the formula: III Surface area and volume of solids The surface area is the area that describes the material that will be used to cover a geometric solid When we determine the surface areas of a geometric solid we take the sum of the area for each geometric form within the solid The volume is a measure of how much a figure can hold and is measured in cubic units The volume tells us something about the capacity of a figure Surface area of solids That is formula for calculating the surface area of soilds Example: Calculating the surface area of some solids: 1.1 Find the lateral and surface areas of the solid This is a triangular prism First thing we got to know is the perimeter of the triangle It's a2 + b2 = c2 102 + 102 = c2 200 = c2 c ≈ 14.1 yd That means we can calculate the perimeter of the triangle P = 10 yd + 10 yd + 14.1 yd = 34.1 yd Now that we know the perimeter and the height, here comes the lateral area L = Ph = (34.1 yd)(15 yd) L = 511.5 yd2 To finish off the entire surface area, we need to add the areas of the two triangular base SA = L + 2B SA = 511.5 yd2 + 2(½bh) SA = 511.5 yd2 + 2(½)(10 yd)(10 yd) = 611.5 yd2 1.2 Find the lateral area and the surface area of the cylinder Like always, we'll use our handy dandy surface area formula, SA = L + 2B In our case SA = 2πrh + 2( π r2 ) Substituting in the values we know, we get: SA = 2π(1.2 in)(7.3 in) + 2π(1.2 in)2 SA = 2π(8.76 in2) + 2π(1.44 in2) SA = 55.04 in2 + 9.05 in2 = 64.09 in2 We calculated the surface area (64.09 in2) 1.3 Find the lateral area and the surface area of the solid The first thing we should know is that since it's a rectangular prism, there could be di SA = L + 2B SA = Ph + 2B The perimeter of our base is + + + 2, or 18 cm The height is 13 cm Our rectangu SA = (18 cm)(13 cm) + 2(14 cm2) SA = 234 cm2 + 28 cm2 = 262 cm2 That means the surface area is 262 cm2, and the lateral area when the base is cm × c If the bases were the cm × 13 cm sides, our lateral surface area would be different Th One last base pair: the cm × 13 cm sides The perimeter would be 30 cm and the heig 1.4 Application of exercises: Exercise 1: In space, let ABCD be a square with side a Given that I and H are midp Compute the curved surface area of the cicular cylinder Exercise 2: Given the cube ABCD.A’B’C’D’ with side a Compute the curved surfac Exercise 3: A cylinder a given with the radius of r and the height h=r Compute the curved surface area and surface area of the cylinder Exercise 4: Cutting a cone by a plane containing its axis we get a cross-section whic Volume of solids Typically, we will calculate the volume of a solid by finding the length, width, thickness, and then replacing the formula Example 1:Given the pyramid S.ABCD with ABCD is a square AB = a, the triangle SAB is equilateral triangle.The plane ( SAB ) is perpendicular to the plane (ABCD ) Compute volume of pyramid S.ABCD Solution: Assume I is midpoint of side AB We have Therefore SI ⊥ ( ABCD ) or SI is thickness of pyramid S.ABCD We have SI is thickness of triangle SAB and the triangle SAB is the equilateral triangle whose length is a, then SI = The area of base ABCD is: S.ABCD is: S ABCD = a.a = a 1 a a3 VS ABCD = S ABCD SI = a = 3 Hence, the volume of pyramid We also use the method of coordinates in space in order to compute the volume of pyramid Compute volume of a solid by volume proportion method Example 2: Compute the volume of a pyramid by volume proportion method Let the pyramid S.ABC, A’ is point in SA, B’ is point in SB, C’ is point in SC Then, we have: Apply to specific exercises: Let the pyramid S.ABC, triangle ABC is right triangle SA=AB = BC = a SA is penpercular to the plane (ABC).Let H is midpoint of SB Calculated volume of S.AHC Solution: We have : AC= Then Hence: VSAHC = VSABC = a3 Example 3: Let ABC.A’B’C’ be a triangular prism Given that E and F are the midpoints of edges AA’ and BB’, respectively The straight line CE intersects the straight line C’A’ at E’ The straight line CF intersects the straight line C’B’ at F’ Let V be the volume of soild prism ABC.A’B’C’ a) Compute the volume of the soild pyramid C.ABFE according to V b) Let the soild polyedron (H) be the ramainer of the soild prism ABC.A’B’C’ after the soild pyramid C.AEFE was cut out Compute the ratio of the volume of (H) and the volume of soild pyramid C.C’E’F’ Solution: a) The pyramid C.A’B’C’ and the prism ABC.A’B’C’ have the base and the height which are congruent, so VC.A’B’C’ = V VC ABB ' A ' = V − V = V 3 From this we deduce Since EF is the midline of the parallelogram ABB’A’, the area of ABFE is equal toa half of the area of ABB’A’ Hence b) Applying the equations a) we receive 1 VC ABFE = VC ABB ' A ' = V CC ' Because EA’ is parallel and equal to ,according to Thales themrem, A’ is the midpoint of E’C’.similarly, B’ is the midpointof F’C’ So, the area of the triangle CE’F’ is four times as large as the area of the triangle A’B’C’ It follows from this, VC E ' F 'C ' = 4VC A ' B 'C ' = V VH VC E ' F 'C ' = Therefore Problems: Exercise 1:Let S.ABC be a regular triangular pyramid with AB = 5a, BC = 6a; CA = 7a The lateral SAB, SBC, SCA together with the base from an angle 60 Compute the volume of this soild pyramid Exercise 2: Given regular quadrilateral pyramid S.ABCD with a square with length of side a as its base and each for its lateral edges making the angle of 60 with the base Let M be the midpoint of SC The plane passing through AM and parallel to BD intersects SB at E and SD at F Compute the volume of the soild pyramid S.AEMF h=r Exercise 3: A cylinder a given with the radius of r and the height Compute the volume of the soild cylinder created by the given cylinder Exercise 4: A given cone has the height h = 20cm, the base radius of r = 25cm Compute the volume of the soild cone created by the cone CONCLUDING From the above essay, we get: - System of common mathematical formulas and application exercises in plane figures: + Vectors and the dot product of two vector and applications + Conventions for coordinates in the plane + Isometries and similarities in the plane - System of common mathematical formulas and application exerscises in solid figures (space): + Lines and planes in space – parallel relationships + Vectors and perpendicular relationships is space + Solid polyhedron: solid pyramid, solid prism + Conical surface of rerolution, surface of circular cylinder, sphere + The method of coordinates in space