Grading Criteria and Reference Answers of Mathematics Examination Paper in May 2009
First, multiple-choice questions (***8 small questions, 4 points for each small question, ***32 points)
Title 1 2 3 4 5 6 7 8
Answer A C D D C B C B
Fill in the blanks (***4 small questions, 4 points for each small question, *** 16 points)
9.
10.
1 1.6
12.0 or 4 (2 points for a correct answer; On the basis of answering 0 or 4, only 2 points will be given for those who answer more. )
Iii. Answer questions (* * 13 small questions, ***72 points)
13. (5 points for this small question)
Solution: the original formula = ...........10 ……………………………………………………………………………………………………….
Five points.
14. (5 points for this small question)
Solution: 2 points.
......................................................., 3 points.
Solution ................................................... 4 points.
After checking, it is the solution of the original fractional equation with ................................................. score of 5.
15. (5 points for this small question)
Solution: The original formula = ... 3 points.
..........................................................., 4 points.
When, the original type of ................................................. 5 points.
16. (5 points for this small question)
Proof: ∫AD‖BC,
The Grand Canyon
AE = AB，
∴∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠873
∴∠ B = ∠ DAE ........................................................... 3 points.
And ad = BC,
∴△ ABC△ AED .................................................................... 4 points.
∴ Germany = AC ................................................................. 5 points.
17. (5 points for this small question)
Solution: replace, get.
∴ The coordinate of point A is (3, 1). ....................................................................................................................................................
If you substitute point A (3, 1), you will get 4 points.
∴ The analytical formula of inverse proportional function is .................................................................................................................................................................
18. (5 points for this small question)
Solution: (1) 31.6%; .............................................. 1 min.
(2) Complete the statistical chart; Four points.
(Note: the score of this question is ***3 points, ① Complete "surfing the Internet" and give 1 point; ② Complete the "fitness game" and give 2 points. )
(3) There are different answers, such as: appropriately reducing TV watching time and doing more exercise are good for health. (Give points if reasonable)
Five points.
19. (5 points for this small question)
Solution: in trapezoidal ABCD, AB‖CD,
∴∠ 1=∠2.
∠∠ACB =∠D = 90。
∴∠3=∠B.
∴ ............................................1min.
In Rt△ACD, CD = 4,
∴ .................................................. 2 points.
∴ ................................................ 3 points.
At Rt△ACB,
∴ .
..................................................., 4 points.
∴ ……………………………………………………………………………………………………………………………………………………………………………………………………………………………… 5 points.
20. (5 points for this short question)
Solution: Suppose 1978 there are X public libraries and Y museums in China, and there are ………………………………………………………………………………………………………………………………………………………………………………………………………………+0.
From the meaning of the question, get 3 points.
Solution 4 points.
Then,.
A: In 2008, there were 2,650 public libraries and 2,000 museums in China. .................................. scored 5 points.
2 1. (5 points for this small question)
Solution: From the meaning of the question, it can be concluded that △ABC and △BDC are right triangles.
In Rt△BDC, BD = 20, DBC = 30,
∴, ................................................. 2 points.
In Rt△ABC, ∠ ABC = 45,
∴ .................................................. 3 points.
................................................., 4 points.
∴ (Male) 5 points.
A: The height of the newly-built stairs will be increased by 7 meters.
22. (7 points for this small question)
Prove that (1) connects OC (as shown in Figure ①),
∵OA=OC,∴∠ 1=∠A.
∵OE⊥AC,∴∠A+∠AOE=90。
∴∠ 1+∠AOE=90。
∠ FCA =∠ AOE, Figure ①.
∴∠ 1+∠ FCA = 90。 That is ∠ OCF = 90 degrees.
∴FD is the tangent of ⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙.
(2) Connect BC (as shown in Figure ②),
∵OE⊥AC,∴AE=EC.
And ao = ob,
∴OE‖BC and .................... 3 points.
∴△OEG∽△CBG. Figure ②
∴ .
∵OG=2,∴CG=4.
∴ OC = 6 .................................................... 5 points.
That is, the ⊙O radius is 6.
(3)∫OE = 3, and BC = 2oe = 6 is obtained from (2).
∵ OB = OC = 6, ∴△OBC is an equilateral triangle. ∴∠ COB = 60 .......................................................................................................................
At Rt△OCD,
∴
....................................................., seven.
23. (5 points for this small question)
( 1)
..................................... 1 min.
Note: Just draw a crease.
(2)
................................................, two points.
(Note: Just draw a triangle that meets the conditions; The answer is not unique, as long as one side of the triangle is equal to the height of that side. )
(3) The length of one side of the triangle is equal to the height of the side .............................................................................. 3 points.
(4) Diagonal lines are perpendicular to each other. (Note: diamonds and squares are not given points) 5 points.
24. (7 points for this small question)
Solution: (1) Let the analytical formula of straight line AC be, and substitute it into A (- 1, 0) to get it.
∴ The analytical formula of linear communication is ...................................................................................................................................................................
According to the meaning of the question, the ordinate of point Q is -6.
Substitution, you can get ∴ point Q( 1,). ............................................................................................................................................
Point q is on the parabola axis of symmetry, which is a straight line.
Let the analytical expression of parabola be derived from the meaning, result and solution of the problem.
∴ The analytical formula of parabola is ...................................................................................................................................................................
(2) As shown in Figure ①, a vertical line passing through point C and AC intersects a parabola at point D,
Cross the x axis at point n, and then
∴ ,∴ .
∵ , ,∴ .
The coordinate of point n is (9,0)
The analytical expression of straight line CN can be obtained. Figure ①
From, it is found that the coordinates of point D are (,). .............................................................................................................................................................
(3) Let the symmetry axis of parabola intersect with the X axis at point E,
According to the meaning of the question,
∵ ,
Besides,
Here we go again.
Let P( 1, m), as shown in Figure ②.
(1) When point P is higher than point M, pm = m+4 = 3,
∴p( 1,- 1∴) ....................................................................................................................................
② When point P is lower than point M, pm =-4-m = 3,
∴ p (1,-7 ∴) .............................................................................. 7 points.
To sum up, the coordinates of point P are (1,-1), (1, -7).
25. (8 points for this short question)
(1) Proof: As shown in Figure ①, ∫∠ACB = 90°, AC=BC.
∴∠A=∠B=45。
Make ∠ ECF = ∠ ECB take CE as one side and CF as the other.
If CF=CB is truncated, then CF=CB=AC. Figure ①
If DF and EF are connected, △ CFE △ CBE ........................................................................................................................................1min.
∴FE=BE,∠ 1=∠B=45。
∠∠DCE =∠ECF+∠DCF = 45，
∴∠DCA+∠ECB=45。
∴∠DCF=∠DCA.
∴△ DCF△ DCA ....................................................................................... 2 points.
∴∠2=∠A=45。
∴∠DFE=∠2+∠ 1=90。
∴△DFE is a right triangle.
And AD=DF, EB=EF,
∴ Line segments DE, AD and EB can always form a right triangle. .............................................. 4 points.
(2) When AD=BE, the line segments DE, AD and EB can
Form an isosceles triangle.
As shown in Figure ②, similar to (1), with CE as one side, let
∠ECF=∠ECB, intercepting CF=CB on CF, you can get
△CFE≔△CBE，△DCF≔△DCA。
∴AD=DF。 Figure ②
∴∠ DFE = ∠1+∠ 2 = ∠ A+∠ B =120 .................................. 5 points.
If Δ Δ dfe is an isosceles triangle, it is exactly DF=EF, that is, AD=BE.
∴ When AD=BE, line segments DE, AD and EB can form an isosceles triangle. ................................................................................................................................................
And the vertex angle ∠DFE is 120.
(3) Proof: As shown in Figure ①,
∠∠ACE =∠ACD+∠DCE，∠CDB=∠ACD+∠A
And < DCE = < α = 45,
∴∠ACE=∠CDB.
And < a = < a = < b,
∴△ACE∽△BDC.
∴ .
∴ .
∵Rt△ACB, from, from.
.................................................., 8 points.
Note: Refer to the above criteria to give different correct answers to each question.