Current location - Training Enrollment Network - Mathematics courses - Answers to the big questions behind the mathematics of Hebei senior high school entrance examination in 2009.
Answers to the big questions behind the mathematics of Hebei senior high school entrance examination in 2009.
In 2009, Hebei junior high school graduates took part in the joint examination of cultural courses.

Reference answers to math test questions

First, multiple choice questions

The title is123455678911112.

Answer A A D C B B A B C C D C

Second, fill in the blanks

13.>; 14. 1.2 × 107; 15.36.4; 16. 1; 17.3; 18.20.

Third, answer questions.

19. Solution: Original formula =

= .

When a = 2,

Original formula = 2.

Note: If you directly replace the evaluation with this question, the correct result will get the corresponding score.

20. solution: (e point 1)∵OE⊥CD, CD=24,

∴ED = = 12。

In Rt△DOE,

∫sin∠DOE = =,

∴OD = 13 (m).

(2)OE=

= .

∴ Drainage requirements:

5/0.5 =10 (hour).

2 1. solution: (1) 30%;

② As shown in figure 1;

(3) ;

(4) As the average level of monthly sales is the same, the monthly sales of brand A show a downward trend, while the monthly sales of brand B show an upward trend.

Therefore, the store should distribute B brand TV sets.

22. Solution: (1)-3.

t =-6。

(2) Substituting (-4,0) and (-3,3) respectively to obtain

solve

Up.

(3)- 1 (the answer is not unique).

Note: writing t >-3, t≠0 or any of them will get extra points.

Solution: practical application

( 1)2; . ; .

(2) .

Extended association

The circumference of (1)∑△ABC is L, and ∴⊙O rotates in three planes.

The sum of the outer angles of a triangle is 360 degrees,

∴ At three vertices, ⊙O rotates (week).

∴⊙O*** rotated (+1) weeks.

(2) + 1.

24.( 1) It is proved that ∵ quadrilateral BCGF and CDHN are both squares.

In addition, point n coincides with point g, point m coincides with point c,

∴FB = BM = MG = MD = DH,∠FBM =∠MDH = 90。

∴△fbm?△mdh。

∴FM = MH。

∠∠fmb =∠DMH = 45° ,∴∠fmh = 90°。 ∴FM⊥HM.

(2) Proof: Connect MB and MD, as shown in Figure 2, and let FM and AC intersect at point P. 。

∫B, D and M are the midpoint of AC, CE and AE respectively.

∴MD‖BC, and md = bc = bf;; MB‖CD,

And MB = CD = DH.

Quadrilateral BCDM is a parallelogram.

∴ = Clean development mechanism.

And ∠FBP =∠HDC, ∴∠FBM =∠MDH. ..

∴△fbm?△mdh。

∴FM = MH,

And ∠ MFB = ∠ HMD.

∴∠fmh =∠FMD-∠HMD =∠APM-∠mfb =∠FBP = 90。

∴△FMH is an isosceles right triangle.

(3) Yes.

25. Solution: (1) 0,3.

(2) From the meaning of the question, you can get

, ∴ .

,∴ .

(3) From the meaning of the question, it is concluded that.

Tidy it up and bring it here.

From the meaning of the question, get

The solution is x ≤ 90.

Note: In fact, 0≤x≤90, and x is an integer multiple of 6.

According to the properties of linear functions, when x = 90, q is the minimum.

At this time, 90 sheets, 75 sheets and 0 sheets are cut respectively according to three cutting methods.

26. Solution: (1) 1,;

(2) QF⊥AC at point F, as shown in Figure 3, AQ = CP= t, ∴.

By △AQF∽△ABC

Yes ∴ 。

∴ ,

Namely.

(3) Yes.

① When DE‖QB, as shown in Figure 4.

∴pq⊥qb ∵de⊥pq, quadrilateral QBED is a right trapezoid.

At this time ∠ aqp = 90.

From △APQ ∽△ABC

Which is the solution.

② As shown in Figure 5, when PQ‖BC, DE⊥BC and quadrilateral QBED are right-angled trapezoid.

At this time ∠ apq = 90.

From △AQP ∽△ABC

Which is the solution.

(4) or.

Note: ① point p moves from c to a, and DE passes through point C.

Method 1: Connect QC and do QG⊥BC at G point, as shown in Figure 6.

, .

Gradually, gradually.

Method 2: from, from, and then from.

, um, ∴. ∴

② point p moves from a to c, and DE passes through point c, as shown in figure 7.

,