resposta review 8 w8:resposta,review,2018,middle,sc:2018 middle school mathematics (Qinghai) review (test): Chapter 4 preliminary understanding of graphics and triangles, quadrilateral data download 2018 test mathematics (Qinghai) review (test): Chapter 4 graphics preliminary understanding a

resposta review 8 w8

2018 middle school mathematics (Qinghai) review (test): Chapter 4 preliminary understanding of graphics and triangles, quadrilateral data download 2018 test mathematics (Qinghai) review (test): Chapter 4 graphics preliminary understanding and triangle, quadrilateral The preliminary understanding of the four chapters of the figure and the triangle, the first section of the quadrilateral, the angle, the intersection line and the parallel line 1 (2017 Changde high school entrance examination), if an angle is 75, the degree of its residual angle is () A. 285°B. 105 ° C. 75°D. 15°2. (2017 from the point B on the line b and AB⊥BC=35 then ∠2=()A.45°B.50°C.55°D.60°, (2nd question)), (3rd) Title map)) 3 (2017 Yantai middle school entrance examination) The positional relationship of several roads in a city is as shown in the figure. The angle between AB∥CDAE and AB is 48. If the length of CF and EF is equal, the degree of ∠C is ()A. . 48°B. 40 ° C. 30°D. 24°4. As shown in the figure, ∠EACAD∥BC, ∠B=30, then the degree of ∠C is ()A. 50° . C. 30° . 5. In the following figures, ∠1=∠2 can get AB∥CD is (), A), C), 6. (2017 Chongqing exam paper) as shown in the figure EF ∥ GH point A in EF on GH at point B if ∠ FAC = 72 = 58 points D on the GH degree is 50 ° __., (6th figure)), (7th figure)) 7 (2017 Jingzhou senior high school entrance examination) A ruler and a triangular plate ABC (including 30 angles) placed in the position shown on the side of the ruler with the triangle two D, point E on the other side and the triangle The right-angled edges are respectively assigned to point F point A and ∠CDE=40, then the size of BAF is ()A. 40°B. 45 ° C. 50°D. 10°8. (2017 Shijiazhuang middle school simulation) A vertical square ruler is placed in different ways as shown in the figure. In the figure, ∠α and ∠β are equal (), A), B), C), D) 9. (2017 Ningbo Senior High School Entrance Test) It is known that the straight line m∥n will have a right angle triangle plate ABC with 30 angles as shown in the figure ABC=30), where A points on the straight line m respectively, if ∠1=20, then the degree of ∠2 is () A. 20°B. 30 ° C. 45°D. 50°, (9th title map), (10th title map)) 10 (2017 Texas Senior High School Entrance Examination) The method of using the ruler and the triangular plate to pass the parallel line of the known straight line l to the parallel line of the straight line l is based on the method. The same angle is equal to two straight lines. 11 (Kunming high school entrance examination) as shown in Figure CE at the point D = DF = 20, then the degree of ∠ B is 40 ° __. , (11th title map)), (12th title map)) 12 (Yibin middle school entrance examination) as shown by the line a∥b=45=30 then ∠P=75__°.13. (Anshun middle school test) as shown in the line m∥n is isosceles right triangle = 90 then ∠ 1 = 45__ °., (13th title map), (14th map)) 14 (2017 Guigang senior high school entrance examination) as shown in Figure E If the point F on the AB is on the CD, if ∠CFE:∠EFB=3:4=40 then the degree of BEF is 60°__. The second quarter three 1 (2017 Qingyang high school entrance examination) is known that a is the three sides of △ABC The result of long simplification |a+b-c|-|c-a-b| is ()A. 2a+2b-2c. +2b2. As shown in the graph paper, ΔABP is made with AB as one side and △ABC is equal. Find the point P that meets the condition from the four points of P. Then point P has ()A. One. One. One. 3 is known to be △ ABC and △ DEF congruent = ∠ D = 70 = 60, then the degree of ∠ F is (C) A. 50°B. 60 ° C. 50° or 60 or 704. (2017 Zhangjiajie High School Entrance Examination) is the midpoint on the side AB of △ABC, respectively. If the circumference of △ADE is 6, the circumference of △ABC is ()A. 6B. 12C. 18D. (4th title map)), (5th title map)) 5 (Dongguan senior high school entrance examination) The common point of the center line AD on the three sides of the figure is the point G. If S=12, the area of ??the shaded part in the figure is 4__. 6 (2017 Dazhou Senior High School Entrance Examination) △ ABC = 5 = 3 is the middle line of △ ABC. If the length of AD is m, the value of m is 1 < m < 47. (2017 Xinjiang senior high school entrance examination) as shown in the quadrilateral ABCD = AD = CD diagonal AC intersects at point O in the following conclusions: = ∠ ADC; 2AC and BD are equally divided; 3AC bisects the two sets of diagonal ABCD, respectively; 4 quadrilateral The area of ??ABCD is S=AC·BD. The correct one is 14__. (Fill in the serial number of all correct conclusions) 8 (2017 Wenzhou senior high school entrance examination) as shown in the pentagon ABCDE = ∠ EDC = 90 = ED = AD. (1) Proof: △ ABC ≌ △ AED; (2) When ∠ B = At 140 o'clock, ask for the degree of BAE. 1) ∵AC=AD∴∠ACD=∠ADC. ∵∠BCD=∠EDC=90∴∠ACB=∠ADE. △ABC≌△AED(SAS) in △ABC and △AED; (2)When ∠B=140=140 and ∵∠BCD=∠EDC=90∴Pentagon ABCDE ∠BAE=540-140-90=809. (2017 Huzhou senior high school test) As shown in Figure BP and CP, respectively, ∠ABC and ∠DCB pass point P and perpendicular to AB. If AD=8, the distance from point P to BC is ()A. 8B. 6C. 4D. 2, (9th title map), (10th title map)) 10 (Shaanxi senior high school entrance examination) as shown in △ ABC = 90 = 8 = 6. If DE is the median line of △ ABC extended DE △ ABC outer angle平ACM bisector at point F then the length of line segment DF is () A. 7B. 8C. 9D. 11. (2017 Daqing Senior High School Entrance Examination) As shown in the figure from 1∠1=∠2;2∠C=∠D;3∠A=∠F, two of the three conditions are selected as the known conditions and the other is the correct conclusion. The number of propositions is () A. 0B. 1C. 2D. 312. (2017 Wuhan senior high school entrance examination) as shown in Figure C on a straight line = ∠ BEA = BF = AE. Write the relationship between CD and AB and prove your conclusion. Solution: The relationship between CD and AB is: CD=AB and CD∥AB. Proof: ∵CE=BF=BE. △ΔCDF≌△BAE(SAS) in △CDF and △BAE. =AB=∠B 13. (2017 Zunyi middle school test simulation) As shown in the figure =90=2AB point D is the midpoint of AC. Place a right-angled triangle with an acute angle of 45 as shown in the figure so that the two end points of the hypotenuse of the triangle are respectively associated with A. Coincidentally connect BE to test the number and position of BE and EC and prove your conjecture. Solution: BE=EC reasons are as follows: ∵AC=2AB Point D is the midpoint of AC=AD=CD.=∠EDA=45∴∠EAB=∠EDC=135∵EA=ED∴∠AEB=∠DEC=EC. =∠AED=90∴BE=EC14. (Tai'an senior high school entrance examination) Figure = 90 on BC and AD = DE. Point F is the midpoint of AE and AB intersects with point M. (1) Prove: ∠ FMC = ∠ FCM; (2) AD and MC vertical ? And explain the reasons. Solution: (1) ∵AD⊥DE and AD=DE is the isosceles right triangle is the midpoint of AE ∴DF⊥AE, DF=AF=EF. and ∵∠ABC=90 both with ∠MAC mutual ∴∠DCF= ∠AMF. ∵∠DFC=∠AFM=90∴△DFC≌△AFM,∴CF=MF∴∠FMC=∠FCM; (2)AD⊥MC reason: extend AD to MC at point G by (1) ∠MFC=90=FE=FC=∠FMC=45=∠ADE=90 ie AD⊥ in quadrilateral ABCD=BC=CD=DAD=60 connected AC. (1) as shown in Figure 1 point E on side BC and BE=CF.Verification:; is an equilateral triangle; (2) If E has a point F on the straight line CD on the extension line of BC, make △AEF an equilateral triangle? Please prove your conclusion. (Figure 2 alternate) Solution: (1) 1 ∵ AB = BC = 60 ∴ △ ABC is an equilateral triangle. =AC. The same is also an equilateral triangle ∴∠ACF=∠B=60 and ∵BE=CF(SAS); ∴AE=AF=∠CAF.+∠CAE=60+∠CAE=60 ie ∠EAF=60∴△ AEF is an equilateral triangle; (2) there is a point F on the CD extension line so that CF=BE is connected to AE and (1)1 is similar. △ABE≌△ACF∴AE=AF=∠CAF.-∠CAE= ∠BAE-∠CAE.=∠BAC=60∴△AEF is equal (Note: Take F on CD extension line to make DF=CE also) 16 (2017 Chongqing Senior High School A) In △ABC=45 foot For M point C is a point connection AC on the BM extension line. (1) As shown in Fig. 1, if AB=3=5 find the length of AC; (2) as shown in Fig. 2, point D is a point on line AM = MC point E is △ ABC Outside point = AC connects ED and extends BC at point F and point F is the midpoint of line segment BC: ∠BDF=∠CEF. Solution: (1)∵∠ABM=45∴AM=BM=AB=3×= 3 then CM=BC-BM=5-3=2∴AC===; (2) EF to point G makes FG=EF connect BG. From DM=MC=∠AMC=AM∴△BMD≌ΔAMC(SAS ),∴AC=BD.CE=AC=CE.=FC=∠EFC=FE∴△BFG≌△CFE,∴BG=CE=∠E∴BD=BG=CE∴∠BDG=∠G=∠E Section III Isosceles Triangle and Right Triangle 4,5,6B. ,2,. 2,3,4D. 1,32. (2017 Nanchong Middle School) As shown in the figure, the side length of the △OAB is 2, and the coordinates of point B are ()A. (1) B. (, 1) C. (,) D. (1,) 3. (2017 Zunyi middle school test simulation) as shown in △ ABC = 45 points D on AB point E on BC. If AD=DB=DE=1, the length of AC is (). D., (3rd question map), (4th question map)) 4 as shown in △ABC = 30 vertical bisector line AB at point E foot is D bisector ∠ ACB BC = 2 then AC length for(). D. 5. (2017 Shaanxi Senior High School Entrance Examination) As shown in △ABC=36=AC is the angle bisector of △ABC. If BE=BC is connected to the edge AB, the middle waist triangle is shared ()A. 2 One. One. ((5th title map)), (6th title map)) 6 (2017 Guang'an senior high school entrance examination) The figure is the diameter of ⊙O and the midpoint H of the string CD is known ==5, then the length of OH is () . 7. (2017 Wuhan Senior High School) As shown in the figure =90, draw the isosceles triangle with the side of △ABC so that its third vertex can draw the most different isosceles triangles on the other side of △ABC. Is () A. 4B. 5C. 6D. 7, (7th title map)), (8th title map)) 8 (2017 Qingyang senior high school entrance examination) as shown in a triangle paper ABC = 90 = 8 = 6 now folded the paper so that point A coincides with point B Then the crease length is equal to __. (2017 绥化中考) In the isosceles △ABC, the straight line BC is at point D. If AD=BC, the apex angle of △ABC is 30° or 150 or 90.10. (Liaocheng middle school test) As shown in the figure △ ABC = 90 = 30 is the bisector of ∠ ABC, if AB = 6 then the distance from point D to AB is __. 11 (2017 Zunyi Hangzhong 2nd Mould) is known: the length of both sides of an isosceles triangle x satisfies the equations, and the circumference of this isosceles triangle is 5__. 12 (2017 Xuzhou senior high school entrance examination) As shown in the figure, AC⊥BC foot is C=4=3. The line segment AC is rotated by 60 in the counterclockwise direction to obtain the line segment AD connection DC(1) line segment DC=________; (2) The length of the line segment DB. Solution: (1) 4; (2) DE ⊥ BC at point E. Is an equilateral triangle ∴∠ ACD = 60 and ∵ AC ⊥ BC = ∠ ACB - ∠ ACD = 90-60 = 30 ∴ in = DC = 2=DC·=4×=2=BC-CE=3-=.∴中===.13(2017 Liaocheng Middle School) As shown in the figure, the endpoints of the large square segment AB consisting of 8 congruent rectangles are at If the point P of a small rectangle is the apex of a small rectangle connecting PA, then the number of points P of ΔABP is an isosceles right triangle is () A. 2B. 3C. 4D. 5, (13th title map)), (14th title map)) 14 (Jingmen middle school entrance examination) as shown in the figure = AC is the bisector of ∠BAC. It is known that AB=5=3, the length of BC is ()A. 5B. 6C. 8D. 1015. (2017 Shanxi Senior High School Entrance Examination) A set of triangular plates is arranged as shown in the figure to obtain △ABD and △BCD. ∠ADB=BCD=90A=60=45 is the midpoint of the point E is EF⊥CD at point F. If AD= 4 The length of EF is (+), (15th title map)), (16th title map)) 16 is a rectangular paper ABCD known AB=8=7 is AB a point=5 is now Cut an isosceles triangle paper (△AEP) so that the point P falls on one side of the rectangular ABCD, then the base length of the isosceles triangle AEP is 5 or 4 or 5.17. (2017 绥化中考) The midpoint of each side of the isosceles right triangle is obtained as the first small triangle, and the midpoint of each side of the small triangle obtained by the sequential connection is obtained to obtain the second small triangle. Thus, the area of ??the nth small triangle is __. 18 (2017 Changchun senior high school entrance examination) Figure 1 This pattern is a schematic diagram of this pattern. Figure 2, where the quadrilateral ABCD and the quadrilateral EFGH are squares are four congruent right triangles. If EF = 2 = 8, the length of AB is 10__. Figure 1 Figure 219 (2017 Zunyi middle school simulation) as shown in the isosceles is a right angle = BC put a 45-angle apex on both sides of the C and E to the E two points. (1) Rotate the obtained ΔACE with C as the center in the counterclockwise direction to △BCG to testify: △EFC≌△GFC; (2) If AB=10=3:4, find the length of EF. Solution: (1) Know by rotation: △BCG≌△ACE.=CE=∠=45∴∠BCG+∠BCF=45 ie ∠GCF=∠ECF=45 and CF is the common side (SAS); (2)Connect FG Known by △BCG≌△ACE: ∠CBG=∠A=45∴∠GBF=∠CBG+∠CBF=90 is known from △EFC≌△GFC: EF=GF set BG=AE=3x=4x in middle=5x∴ EF=GF=5x ??then 3x+5x+4x=10 solution x==.20 (2017 Jingmen Middle School) is known: as shown in the figure = 90 points D is the midpoint of AB E is the midpoint of CD over point C for CF∥AB The extension line of the AE is at point F. (1) to verify: △ ADE ≌ △ FCE; (2) If ∠ DCF = 120 = 2 to find the length of BC. Solution: (1) ∵E is the midpoint of CD = ∥CF, ∴∠BAF=∠AFC. △△≌≌FCE(AAS) in △ADE and △FCE; (2)(1) know=2DE .=2=4. At the midpoint D is the midpoint of AB∴AB=2CD=8=CD=AB.∴∠BDC=180-∠DCF=180-120=60∴∠DAC=∠ACD=∠BDC= ×60=30∴中中=AB=×8=4. The fourth section ruler is shown in Figure 1 (2017 Hebei Senior High School Entrance Examination). Figure C is made on the OB side of ∠AOB with a ruler. CN∥OA is made in the following In the trace of the figure is () A. An arc with a point C as the radius of the circle with the point C as the radius of the circle with the point E as the radius of the arc with the point E as the radius of the circle, (1st question)), (2nd question map) )) 2 (2017 four-city high school entrance examination) as shown in the figure >AC is the external angle of △ABC observation chart traces of the ruler drawing, the following conclusion is wrong () A. ∠DAE=∠B. =∠CCD. =∠EAC3. (2017 襄阳中考) As shown in Figure △ABC = 90 = 30 = 4 with point C as the center of the circle as the radius of the arc intersection AB at point D; then point B and point D as the center of the circle is larger than the radius of BD The arc two arcs intersect at point E for the ray CE intersection AB at point F, then the length of AF is () A. 5B. 6C. 7D. 84. (Zhangzhou senior high school entrance examination) The following ruler mapping can judge that AD is the height of △ABC (), A), B), C), D) 5. (2017 Tangshan Middle School) As shown in Fig. A, the center of the circle is circled with the same length as the radius, and the ray AM is intersected with the two points of B. Then the B is the center of the circle with the same length (greater than BC) as the arc. When the intersection is at point D and AD is connected, the following conclusion is wrong () A. AD ∠ ∠ MAN vertical halving BC = ∠ NCD quadrilateral ACDB must be diamond 6 known △ ABC

'Seven times Lang, how do you think about the past today? Did you guess that there is an apostle strike today?' Gecheng Mei Liqit looked at Shi Lei


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