Mathematics examination questions in 2008

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In 2008, the entrance examination for secondary schools in Shandong Province.

Math test questions

Precautions:

1. This exam is divided into two parts: Volume I and Volume II. Volume 1 has four multiple-choice questions, with 24 points; Page 8 of Volume 2 is a multiple-choice question, with 96 points; Full volume *** 12 pages, full score 120 minutes, examination time 120 minutes.

Before answering the first volume, candidates must scribble their names, test numbers and test subjects on the answer sheet. At the end of the exam, the questions and answer sheets are taken back together.

3. After selecting the answers to each question in Book 1, you must use 2B pencil to blacken the answer label ABCD of the corresponding question on the answer sheet. If you need to change it, clean it with an eraser first, and then paint other answers.

4. Scientific calculators are not allowed to be used in the exam.

The first volume (multiple choice questions ***24 points)

1. Multiple choice question: This big question has ***8 small questions. Only one of the four options given in each question is correct. Please choose the correct option and score 3 points for each small question. Choose wrong, don't choose or choose multiple answers, zero.

1. You can't just embed the following graphics.

A. Triangle

B. quadrilateral

C. regular pentagon

D. regular hexagon

2. The following calculation results are correct.

B.=

3. In the plane rectangular coordinate system, if the point P (m-3, m+ 1) is in the second quadrant, the range of m is

A.- 1 0), the intersection point m is the ME⊥y axis, the intersection point n is the NF⊥x axis, and the vertical feet are e and f respectively.

Trial certificate: Mn ‖ ef.

(2) If other conditions in (1) remain unchanged, only point M and point N will be changed.

The position of is shown in Figure 3. Please judge whether MN and EF are parallel.

23. (The full mark of this question is 12)

At △ABC, ∠ A = 90, AB = 4, AC = 3, M is the moving point on AB (which does not coincide with A and B), the point passing through M is MN‖BC, AC is at N point, and MN is the diameter ⊙O, making a rectangle AM = X .. in ⊙.

The area s of (1)△MNP is expressed by an algebraic expression with x;

(2) When what is the value of x, ⊙O is tangent to the straight line BC?

(3) In the process of moving point M, remember that the overlapping area of △MNP and trapezoidal BCNM is y, try to find the functional expression of y about x, and find the value of x and the maximum value of y?

In 2008, the entrance examination for secondary schools in Shandong Province.

Reference answers and grading opinions of mathematics test questions

Tag description:

1. For each small question in multiple-choice questions and fill-in-the-blank questions, there are only two grading files, full marks and zero marks, and no intermediate marks are given.

2. The corresponding score in the answer to each small question refers to the cumulative score that the candidate should get if he answers this step correctly. Only one solution is given for each small question in this answer. Please refer to the grading opinions for other solutions.

3. If the candidate makes a calculation error in the middle of the answer, but does not change the nature and difficulty of the test question, the subsequent part will be given extra points as appropriate, but at most it will not exceed half of the correct answer score; If there is a serious logical error, the subsequent part will not be scored.

First, multiple-choice questions (this big question is ***8 small questions, each with 3 points and 24 points * * *)

Title 1 2 3 4 5 6 7 8

Answer C C A C B A B D

II. Fill in the blanks (5 small questions in this big question, 4 points for each small question, 20 points for * * *)

9.； 10. 120 ; 1 1.； 12.； 13.28 yuan; 14.； 15. 16.①②③⑤.

Third, answer the question (this big question is ***7 small questions, ***64 points):

17. (The full mark of this question is 6)

Solution: The original formula = ... 2 points.

= ... 3 points.

= ........................................................ 4 points.

When, when,

Original formula = ........................................ 6 points.

18. (The full mark of this question is 8)

Solution: (1) Suppose 6x individuals donate to 30 yuan, then 8x+6x = 42.

∴ x = 3 .................................................... 2 points.

∴ The number of donors * * * is: 3x+4x+5x+8x+6x=78 people, with 3 points for ..................

(2) According to the image, the mode is 25 yuan; Since the number of this set of data is 78, and both numbers in the middle position in size order are 25 (yuan), the median is 25 (yuan). ..............................................................................................................................

(3) donations from the whole school:

(9×10+12×15+15× 20+24× 25+18× 30) × = 34,200 (yuan) .......................................................................

19. (The full mark of this question is 8)

Solution: Set X sets of Olympic symbols and Y sets of Olympic mascots.

........................................, two points.

①× 2-②: 5x = 10000。

∴ x = 2000 ..................................................... 6 points.

Substituting x=2000 into ① gives: 5y = 12000.

∴ y=2400。

A: The factory can produce 2,000 sets of Olympic symbols and 2,400 sets of Olympic mascots. .....................................................................................................................................................

20. (The full mark of this question is 10)

It is proved that if point C is CF⊥AB, the vertical foot is f............... 1.

∫ AB‖CD, AB∶CD, ∠ A = 90 in the trapezoid,

∴ ∠D=∠A=∠CFA=90。

The quadrilateral AFCD is a rectangle.

Ad = cf, BF = ab-af =1.................................. 3 points.

In Rt△BCF,

CF2=BC2-BF2=8，

∴ CF=。

∴ AD = CF = ........................................................ 5 points.

E is the midpoint of AD,

∴ Germany = AE = AD = ........................................................ 6 points.

In Rt△ABE and Rt△DEC,

EB2=AE2+AB2=6，

EC2= DE2+CD2=3，

EB2+ EC2=9=BC2。

∴∠ CEB = 90 ......................................................... 9 points.

∴ EB ⊥ EC ............................................... 10.

2 1. (The full mark of this question is 10)

Solution: (1) As shown in the figure, from the meaning of the question, ∠ EAD = 45, ∠ FBD = 30.

∴ ∠EAC=∠EAD+∠DAC =45 + 15 =60。

∫AE‖BF‖CD，

∴ ∠FBC=∠EAC=60。

∴∠ DBC = 30 ...................................... 2 points.

And ∠∠DBC =∠DAB+∠ADB,

∴∠ ADB = 15.

∴ ∠DAB=∠ADB。 ∴ BD=AB=2。

In other words, the distance between B and D is 2 kilometers. .................................................................................................................................................................

(2) B is BO⊥DC, and its extension line is at point O,

In Rt△DBO, BD=2, ∠ dbo = 60.

∴ do = 2× sin60 = 2×, bo = 2× cos60 =1....................................................., 8 points.

At Rt△CBO, ∠ CBO = 30, Co = Botan 30 =,

∴ CD=DO-CO= (km).

That is to say, the distance between c and d is km. ..................................................................................................................................................................

22. (The full mark of this question is 10)

(1) proof: passing through point C and point D, CG⊥AB, DH⊥AB respectively,

If the vertical feet are G and H, then ∠ CGA = ∠ DHB = 90 ...1min.

∴ CG‖DH。

∫△ABC and△△△ Abd have the same area.

∴ CG = DH .......................................... 2 points.

∴ quadrilateral CGHD is a parallelogram.

∴ Record company ........................................ 3 points.

(2)① Proof: 4 points for connecting MF and NE .........................................................................................................................................

Let the coordinates of m point be (x 1, y 1) and the coordinates of n point be (x2, y2).

∵ point m, n is on the image of inverse proportional function (k > 0),

∴ , .

ME⊥y axis, NF⊥x axis,

∴ OE=y 1，OF=x2。

∴ △ EFM =, ... Yes ... Yes ... Yes ... Yes ... Yes.

S △ EFN = .......................... 6 points.

∴S△EFM =S△EFN。 ………………? 7 points

According to the conclusion in (1), Mn ‖ ef .............................................................................................................................................

(2) Mn ‖ ef ............................................10.

If students use other methods, as long as the solution is correct, they will all be given points.

23. (The full mark of this question is 12)

Solution: (1)∫Mn‖BC, ∴∠AMN=∠B, ∠ ANM = ∠ C.

∴ △AMN ∽ △ABC。

That is ∴.

∴ Ann = x ................................... 2 points.

∴ =. (0 < < 4) ............................ scored 3 points.

(2) As shown in Figure 2, if lines BC and ⊙O are tangent to point D and connect AO and OD, then AO =OD = MN. ..

In Rt△ABC, BC = = 5.

From (1), we know △ AMN ∽△ ABC.

That is ∴.