First, multiple-choice questions (***8 small questions, 3 points for each small question, out of 24 points)
The square root of 1.49 is ()
A.7B. 7C﹣7D.49
Test center: square root.
Topic: Existential type.
Analysis: Answer according to the definition of square root.
Answer: Solution: ∫(7)2 = 49,
The square root of ∴49 is 7.
So choose B.
Comments: this question examines the definition of square root, that is, if the square of a number is equal to a, this number is called the square root of a, also called the square root of a.
2. The arithmetic square root of (-3) 2 is ()
3C。 ﹣3D.
Test center: arithmetic square root.
Special topic: calculation problems.
Analysis: From (-3) 2 = 9, the arithmetic square root of 9 =3.
Answer: Solution: ∵ (∵ 3) 2 = 9,
The arithmetic square root of 9 =3.
So choose a.
Comments: This question examines the definition of arithmetic square root: the positive square root of a positive number A is called the arithmetic square root of this number, and it is recorded as (a>0), and the arithmetic square root of 0 is designated as 0.
3. In real numbers -0,-π, 1.4 1, the irrational number is ().
1。
Test center: irrational number.
Analysis: According to the fact that irrational numbers are infinite acyclic decimals, we can get the answer.
Solution: π is an irrational number,
So choose: a.
Comments: This question examines irrational numbers, which are infinite acyclic decimals. Note that numbers with roots are not necessarily irrational numbers.
4. The points corresponding to 1 and 1 on the number axis are A and B respectively, and point B is the symmetrical point C of point A, then the real number represented by point C is ().
A.﹣ 1B. 1﹣C.2﹣D.﹣2
Test sites: real number and number axis.
Analysis: First, the length of the line segment AB is obtained according to the known conditions and the number axis, and then the result is obtained according to the symmetry.
Solution: ∵ The number axis represents 1, and the corresponding points are A and B respectively.
∴AB=﹣ 1，
Let the real number represented by the symmetric point C of point B relative to point A be x,
Then it is = 1,
The solution can be x=2﹣,
That is, the number corresponding to point C is 2.
So choose C.
Comments: This topic mainly investigates how to use the idea of combining numbers with shapes to find the distance between two points on the number axis, and also uses symmetry.
5. Prove the proposition by reduction to absurdity: "As shown in the figure, if AB∨CD and AB∨EF, then CD∨EF", and the first step of proof is ().
A. suppose CD∑EFB. AB∨EF is known.
C. suppose that CD and EFD are not parallel. Suppose AB and EF are not parallel.
Test center: reduce to absurdity.
Analysis: According to CD∑EF, it is directly assumed that CD and EF are not parallel.
Solution: Prove the proposition by reducing to absurdity: If AB∨CD and AB∨EF, then CD∨EF.
The first step of proof should be to start from the opposite side of the conclusion, so assume that CD is not parallel to EF.
So choose: C.
Comments: This question mainly examines the first step of reduction to absurdity, and the counterexample of the proposition conclusion is the key to solving the problem according to the meaning of the question.
6. As shown in the figure, the straight line L passes through the vertex B of the isosceles right triangle ABC, and the distances from two points A and C to the straight line L are 2 and 3 respectively, so the length of AB is ().
A.D. 5B
Test center: congruent triangles's judgment and nature; Pythagorean theorem; Isosceles right triangle.
Special topic: computational problems; The finale.
Analysis: From triangle ABC to isosceles right triangle, it can be concluded that AB=BC, ∠ABC is right angle, ∠ABD and ∠EBC are complementary. In right triangle ABD, two acute angles are complementary angles. Using the complementary angles of equal angles, one diagonal line is equal, and then a pair of right angles are equal, AB=BC. Triangle ABD and triangle BEC can be obtained by atomic absorption spectrometry.
Solution: Solution: As shown in the figure:
∵△ABC is an isosceles right triangle,
∴AB=BC,∠ABC=90，
∴∠ABD+∠CBE=90，
And AD⊥BD, ∴∠ ADB = 90,
∴∠DAB+∠ABD=90，
∴∠CBE=∠DAB，
At △ABD and △BCE,
,
∴△ABD≌△BCE，
∴BD=CE, and CE=3,
∴BD=3，
In Rt△ABD, AD=2, BD=3,
According to Pythagorean Theorem: AB==.
So choose d
Comments: This topic examines congruent triangles's judgment and nature, the nature of isosceles right triangle and Pythagorean theorem. The key to solve this problem is to flexibly use the mathematical thought of transformation and congruent triangles's judgment and nature.
7. As shown in the figure, in △ABC and △DEC, it is known that AB=DE, and two conditions need to be added to make △ ABC △ DEC, and the group of conditions that cannot be added is ().
A.BC=EC，∠B=∠EB。 BC=EC，AC=DCC。 BC=DC，∠A=∠DD。 ∠B=∠E，∠A=∠D
Test center: congruent triangles's judgment.
Analysis: According to congruent triangles's judgment method, they can be judged separately.
Solution: Solution: A, AB=DE is known, plus the conditions BC=EC, ∠B=∠E can be proved by SAS, so the options are irrelevant;
B, AB=DE is known, plus BC=EC and AC=DC can be proved by SSS, so the option is irrelevant;
C, AB=DE is known, plus BC=DC, ∠A=∠D can't prove △ ABC △ DEC, so the option meets the meaning of the question;
D, AB=DE is known, plus the conditions ∠B=∠E, ∠A=∠D can be proved by ASA, so the options are irrelevant;
So choose: C.
Comments: This question examines the judgment method of triangle congruence. The general methods to judge the coincidence of two triangles are SSS, SAS, ASA, AAS and HL.
Note: AAA and SSA cannot judge whether two triangles are congruent. When judging whether two triangles are congruent, there must be edges involved. If two angles are equal, the angle must be the included angle between two sides.
As shown in the picture, a 25-meter-long ladder stands obliquely on a vertical wall. At this time, the bottom of the ladder is 7 minutes away from the bottom of the wall. If the top of the ladder descends for 4 minutes, the smooth distance at the bottom of the ladder is ().
A.9 decimeter B. 15 decimeter C.5 decimeter D.8 decimeter
Test center: the application of Pythagorean theorem.
Analysis: In the right triangle AOC, the lengths of AC and OC are known, and the length of AO can be obtained according to Pythagorean theorem.
Solution: solution: AC = 25 decimeters, OC=7 decimeters,
∴AO==24 decimeters,
BO= 20 decimeters after a drop of 4 decimeters,
At this point, OD== 15 decimeter,
∴ CD =157 = 8 decimeters.
So choose D.
Comments: This topic examines the application of Pythagorean theorem in real life and the correct application of Pythagorean theorem in right triangle. Applying Pythagorean Theorem twice in this problem is the key to solving it.
Fill in the blanks (***6 small questions, 3 points for each small question, full score 18 points)
9. Calculation: =-2.
Test center: cube root.
Special topic: calculation problems.
Analysis: first transform to =, and then get the answer according to the concept of cube root.
Answer: Solution: = =-2.
So the answer is -2.
Comments: This topic examines the concept of cube root: If the cube of a number is equal to A, then this number is called the cube root of A, and it is recorded as.
10.Calculation:-﹣a2b？ 2ab2=﹣2a3b3.
Test center: single item multiplied by single item.
Analysis: According to the multiplication of monomials and monomials, multiply their coefficients respectively, add the powers of the same letters respectively, and the remaining letters and their exponents are unchanged as the factors of the product, and then calculate.
Answer: Solution: ﹣a2b? 2ab2=﹣2a3b3；
So the answer to is:-﹣2a3b3.
Comments: This question examines the multiplication operation of single item and single item, and mastering the algorithm is the key to solving the problem.
1 1. Calculation: (a2) 3 ÷ (-2a2) 2 = a2.
Test site: division in algebraic expressions.
Analysis: Calculated according to the power of the power sum product.
Solution: Solution: Original formula =a6÷4a4
=a2，
So the answer is a2.
Comments: This topic examines the division of algebraic expressions, and mastering the power sum product is the key to solving problems.
12. The picture shows the fan-shaped statistics of the number of people who participated in extracurricular interest groups in Class 7, 20 1 4/2015. If the number of people participating in the foreign language interest group is 12, then the number of people participating in the painting interest group is 5.
Test center: fan chart.
Special topic: calculation problems.
Analysis: According to the number of people who participate in foreign language interest groups is 12, accounting for 24%, calculate the total number of people, then subtract all known percentages from 1 to find the percentage of painting, and then multiply it by the total number of people to answer.
Answer: Solution: The number of people participating in the foreign language group is 12, accounting for 24% of the number of people participating in extracurricular interest groups.
∴ The number of people participating in extracurricular interest groups * * is: 12÷24%=50 (people)
The number of painting interest groups is 50× (1-14%-36%-16%-24%) = 5 (people).
So the answer is: 5.
Comments: This question examines the fan-shaped statistical chart, and finding relevant information from the chart is the key to solve this kind of problem.
13. As shown in the figure, in △ABC, the circumference of △ABD is 12 and AE=5, so the circumference of △ABC is 22.
Test site: the nature of the vertical line.
Analysis: the middle vertical line of AC passes through AC to E and BC to D. According to the nature of the middle vertical line, the two groups of line segments are equal. After the line segments are equivalently replaced, other known answers are combined.
Answer: solution: ∫DE is the middle vertical line of AC,
∴AD=DC,AE=EC=5，
Delta perimeter △ABD =AB+BD+AD= 12,
That is AB+BD+DC= 12, AB+BC= 12.
The circumference of ABC is AB+BC+AE+EC= 12+5+5=22.
The circumference of △ABC is 22.
Comments: This question mainly examines the geometric knowledge such as the nature of the perpendicular line of the line segment; Equivalent replacement of line segments is the key to correct solution.
14. As shown in the figure, in △ABC, ∠ c = 90, ∠ cab = 50. Draw according to the following steps: ① draw an arc with point A as the center and the length less than AC as the radius, and AB and AC intersect at point E and point F respectively; (2) Draw an arc with point E and point F as the center and the length greater than EF as the radius, and the two arcs intersect at point G; (3) If the ray AG passes through the BC side at point D, the degree of ∠ADC is 65.
Test center: congruent triangles's judgment and nature; Properties of right triangle; Drawing-complex drawing.
Analysis: According to the drawing steps in the known conditions, AG is the bisector of ∠CAB, which can be solved according to the properties of the angular bisector.
Solution: Solution 1: Connect EF.
∵ points e and f are the intersections of points a and AB with AC, and the radius is the center of the circle and the length less than AC.
∴af=ae；
∴△AEF is an isosceles triangle;
∵ Draw an arc with point E and point F as the center and the length greater than EF as the radius, and the two arcs intersect at point G;
∴AG is the perpendicular bisector of the line segment EF,
∴AG divides the taxi equally,
∫∠CAB = 50，
∴∠cad=25；
In delta △ADC, ∠ C = 90, ∠ CAD = 25,
∴∠ ADC = 65 (two acute angles in a right triangle are complementary);
Solution 2: According to the drawing steps in the known conditions, AG is the bisector of ∠CAB, and ∵∠ cab = 50.
∴∠cad=25；
In delta △ADC, ∠ C = 90, ∠ CAD = 25,
∴∠ ADC = 65 (two acute angles in a right triangle are complementary);
So the answer is: 65.
Comments: This topic comprehensively examines the nature of drawing-double drawing and right triangle. It is the key to solve this problem to infer that AG is -∠CAB bisector according to the drawing process.
Third, answer questions (***9 small questions, out of 78 points)
15. Decomposition factor: 3x2y+ 12xy2+ 12y3.
Test center: the comprehensive application of common factor method and formula method.
Analysis: extract the common factor from the original formula, and then decompose it with the complete square formula.
Solution: Solution: Original formula =3y(x2+4xy+4y2)
=3y(x+2y)2。
Comments: This topic examines the comprehensive application of common factor method and formula method, and mastering factorization is the key to solve this problem.
16. Simplify first, and then find 3a-2a2 (3a+4), where a =-2.
Test center: single multiplication polynomial.
Analysis: First, remove brackets according to the law of multiplication of single item and polynomial, then merge similar items, and finally substitute them into known values for calculation.
Solution: solution: 3a-2a2 (3a+4)
=6a3﹣ 12a2+9a﹣6a3﹣8a2
=﹣20a2+9a，
When a =-2, the original formula =-20× 4 × 9× 2 =-98.
Comments: This topic examines the simplification of algebraic expressions. Algebraic addition and subtraction is actually removing brackets and merging similar items, which is a common test site in the 20 15 senior high school entrance examination.
17. Given that a2﹣b2= 15 and a+b=5, find the value of A ﹣ B. 。
Test center: factorization-using formula method.
Special topic: calculation problems.
Analysis: It is known that the left side of the first equation is decomposed by the square difference formula, and a+b=5 is substituted to find the value of a-b. 。
Answer: Solution: A2-B2 = (a+b) (a-b) = 15, a+b=5.
Get a-b = 3.
Comments: This topic examines factorization-using formula method and mastering the square difference formula skillfully is the key to solve this problem.
18. As shown in the figure, it is known that in △ABC, AB=AC, M is the midpoint of BC, D and E are points on the sides of AB and AC respectively, and BD=CE. Verification: MD = me.
Test center: congruent triangles's judgment and nature; Properties of isosceles triangle.
Special topic: proving the problem.
Analysis: According to the nature of isosceles triangle, it can be proved that ∠DBM=∠ECM, △ BDM △ CEM, MD=ME, and the problem can be solved.
Answer: Proof: In △ABC,
AB = AC，
∴∠DBM=∠ECM，
∵M is the midpoint of BC,
∴BM=CM，
At △BDM and △CEM,
,
∴△BDM≌△CEM(SAS)，
∴MD=ME.
Comments: This question examines the nature of congruent triangles's judgment and congruent triangles's equivalence.
19. As shown in the figure, in the equilateral triangle ABC, points D and E are on the sides of BC and AC respectively, and de∨ab, the intersection point E is EF⊥DE, and the extension line passing through BC is at point F. 。
(1) The number of times to find ∠F;
If CD=2, find the length of DF.
Test center: the judgment and nature of equilateral triangle; A right triangle with an angle of 30 degrees.
Special topic: geometric problems.
Analysis: (1) According to the properties of parallel lines, we can get ∠ EDC = ∠ B = 60, which can be solved according to the theorem of triangle interior angle sum;
It is easy to prove that △EDC is an equilateral triangle, and then solve it according to the properties of right triangle.
Solution: Solution: (1)∫△ABC is an equilateral triangle.
∴∠B=60，
∫DE∨AB，
∴∠EDC=∠B=60，
∵EF⊥DE，
∴∠DEF=90，
∴∠f=90 ﹣∠edc=30；
∠∠ACB = 60，∠EDC=60，
△ EDC is an equilateral triangle.
∴ED=DC=2，
∠∠DEF = 90，∠F=30，
∴DF=2DE=4.
Comments: This question examines the judgment and nature of equilateral triangle and the nature of right triangle. A right-angled side with an acute angle of 30 degrees is equal to half of the hypotenuse.
20. As shown in the figure, CE⊥AB, BF⊥AC, BF and CE intersect at point D, and BD=CD.
(1) Verification: Point D is on the bisector of ∠BAC;
Is it true that the condition "BD=CD" is interchanged with the conclusion "Point D is on the bisector of ∠BAC"? Try to explain why.
Testing Center: congruent triangles's Judgment and Nature.
Analysis: (1) Deduce △ deb △ DFC according to AAS, calculate DE=DF according to the properties of congruent triangles, and get it according to the properties of angular bisector;
According to the properties of angular bisector, we can get DE=DF, according to ASA, we can get △ Deb △ DFC, according to the properties of congruent triangles.
Answer: (1) Proof: ∵CE⊥AB, BF⊥AC,
∴∠DEB=∠DFC=90，
In △DEB and △DFC,
,
∴△DEB∽△DFC(AAS)，
∴DE=DF，
∵CE⊥AB,BF⊥AC，
Point D is on the bisector of ∠BAC;
Solution: Established,
The reasons are: ∫ point D is on the bisector of ∠BAC, CE⊥AB, BF⊥AC,
∴DE=DF，
In △DEB and △DFC,
,
∴△DEB≌△DFC(ASA)，
∴BD=CD.
Comments: This topic examines the nature and judgment of congruent triangles and the application of the angle bisector. The key to solve this problem is to derive △ Deb △ DFC. Note: The distance from the point on the bisector of the angle to both sides of the angle is equal, and vice versa.
2 1. Let the comprehensive evaluation score of middle school students' physical health be X, with a full score of 100. It is specified that 85≤x≤ 100 is Grade A, 75≤x≤85 is Grade B, 60≤x≤75 is Grade C, and X.
(1) In this survey, a * * selected 50 students, α = 24%;
Complete the bar chart;
(3) The central angle corresponding to Grade C in the sector statistical chart is 72 degrees;
(4) If there are 2000 students in this school, please estimate how many D students there are in this school?
Test center: bar chart; Estimate the population with samples; Department statistical chart.
Topic: Chart types.
Analysis: (1) Calculate the total number of people drawn according to the number and percentage of people in grade B, and then divide the number of people in grade A by the total number of people to get a;
Subtract the number of people in A, B and D from the extracted total number of people to find the number of people in C, thus completing the statistical chart;
(3) Multiply 360 degrees by the percentage of Grade C to find the degree of the central angle corresponding to Grade C in the fan-shaped statistical chart;
(4) Multiply the percentage of D-level by the total number of students in this school to get the number of D-level students in this school.
Answer: Solution: (1) In this survey, the number of students selected by a certain * * =50 (people).
a =× 100% = 24%；
So the answer is: 50, 24;
The number of people in Grade C is: 50- 12-24-4 = 10 (person).
Supplementary pictures are as follows:
(3) The central angle corresponding to Grade C in the sector statistical chart is × 360 = 72;
So the answer is: 72;
(4) According to the meaning of the question: 2000×= 160 (person),
A: There are 160 D students in this school.
Comments: This topic examines the comprehensive application of bar chart and fan chart. Reading charts and getting the necessary information from different charts is the key to solving the problem. Bar charts can clearly show the data of each project. The fan-shaped statistical chart directly reflects the percentage of local parts in the whole.
22. The center of a typhoon is located in O, and the typhoon center moves to the northwest at a speed of 25 km/h, and it will be affected within a radius of 240 km. A city is 320 kilometers west of O city. Will it be affected by this typhoon? If it is affected, how many hours will it take?
Test center: the application of quadratic root; Pythagorean theorem
Analysis: Whether a city is affected depends on the minimum distance from the typhoon center to a city. If the vertical line passing through point A is ON, the vertical foot is H, and AH is the minimum value. Compared with the radius of 240 kilometers, it can be judged whether it is affected. Calculate the affected time. Draw an arc intersection line with a radius of 240 km between M and N with A as the center, then AM=AN=240 km, and the affected stage is from M to N. MH is calculated according to Pythagorean theorem, and the distance is calculated according to MN=2MH, and the affected time is: time = distance ÷ speed.
Solution: As shown in the figure, OA=320, ∠ AON = 45.
The vertical line that passes through point A and is ON, with H as the vertical foot and A as the center, has a radius of 240, and is in the range of m, n,
At Rt△OAH, ah = OASIN 45 = 160.
At Rt△AHM, MH===80.
∴MN= 160, the affected time is: 160÷25=6.4 hours.
A: City A was affected for 6.4 hours.
Comments: This topic examines the application of quadratic roots in solving practical problems. According to the meaning of the question, it is the key to construct a right triangle with Pythagorean theorem and calculate it.
23. Perception: As shown in Figure ①, point E is on the BC side of the square ABCD, point BF⊥AE is on point F, and point DG⊥AE is on point G, from which we can know that △ adg △ BAF. (No proof is required).
Extension: As shown in Figure 2, points B and C are on the AM and AN sides of ∠MAN, points E and F are on the ray AD inside ∠MAN, and ∠ 1 and ∠2 are the external angles of △ABE and △CAF, respectively. Known AB=AC, ∠ 6544; .
Application: As shown in Figure ③, in the isosceles triangle ABC, AB=AC, AB> BC. Point D is on the side of BC, CD=2BD, points E and F are on the line segment AD, ∠ 1=∠2=∠BAC. If the area of △ABC is 9, then the sum of the areas of △ABE and △CDF is 6.
Test center: congruent triangles's judgment and nature; The nature of isosceles triangle; The nature of a square.
Topic: the finale.
Analysis: expansion: using ∠ 1=∠2=∠BAC, and using the properties of the external angles of triangles, we get ∠4=∠ABE, and then prove △ Abe △ caf by AAS;
Application: firstly, according to the contour lines of △ABD and △ADC, and the bottom ratio is: 1: 2, then the area ratio of △ABD and △ADC is: 1: 2. Then it is proved that △ABE △ caf, and the sum of the areas of △ Abe and △CDF is the area of △ADC.
Answer: expand:
Proof: ∫≈ 1 =∠2,
∴∠BEA=∠AFC，
∠∠ 1 =∠ABE+∠3，∠3+∠4=∠BAC，∠ 1=∠BAC，
∴∠BAC=∠ABE+∠3，
∴∠4=∠ABE，
∴,
∴△ABE≌△CAF(AAS).
Application:
Solution: In the isosceles triangle ABC, AB=AC, CD=2BD,
∴△ABD is as high as△△ ADC, and the bottom ratio is 1: 2.
∴△ABD and△△ ADC area ratio: 1: 2
∫△ABC has an area of 9,
∴△ABD and△△ ADC area are 3 and 6 respectively;
∵∠ 1=∠2,
∴∠BEA=∠AFC，
∠∠ 1 =∠ABE+∠3，∠3+∠4=∠BAC，∠ 1=∠BAC，
∴∠BAC=∠ABE+∠3，
∴∠4=∠ABE，
∴,
∴△ABE≌△CAF(AAS)，
∴△ABE and△△△ Caf have the same area.
∴δthe sum of Δ Abe and Δ △CDF areas is the area of Δ △ADC.
∴△ABE and△△△ CDF have an area of 6.
So the answer is: 6.
Comments: This topic mainly examines the judgment and nature of triangle congruence and the solution of triangle area. According to the known results, ∠4=∠ABE and the area ratio of △ABD to △ADC: 1: 2 are the keys to solve the problem.
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