In 2007, Zibo senior high school entrance examination reviewed and guided a series of mathematical answers.

Review Guidance for Senior High School Entrance Examination in Zibo —— Mathematical Answers and Hints

The first review of basic knowledge

1. 1 real number and its operation

Synchronous training

I.1.a2.d3.c4.a5.c6.c7.d8.d.

2. 1.2 and -8 2. 3.7 4.0 or 2 or 4 5.3. 1, 3.2 6.

7. 1,48. Combination of numbers and shapes

Three. 1.( 1) 29 (2)

2.( 1) 18 o'clock (2) is not appropriate, because it is13 o'clock in the morning of October 27th (3) 165438+8 o'clock.

Test question

I.1.0.5 2.3.4.4 5.

6.602,

Second,/kloc-0 1.C2.D3.D4.b5.c6.a

3. 1.( 1) (2) 0 2. The integer part a =5 and the decimal part b= -5, so 3. -30.06 4.( 1) (2) (3)

5.( 1)33.5 yuan (2) The highest price per share this week is 35.5 yuan, and the lowest price is 30 yuan;

(3) (yuan)

1.2 Algebraic sum operation

Synchronous training

I. 1.2.3, omitting 4.5.6.

Second,/kloc-0 1.A2.C3.D4.A5.C6.C

Third, 1. ( 1); (2) .

2.4.3. 16.4.5.

Test question

I. 1. Omit 2. C4H 10 3。 B 4.5。 For example: 1030 10 6.30.

Second,/kloc-0 1.b2.d3.c4.c5.b6.a

Three. 1. Original formula =-9x+2.2. Same number. 3.33。

4.( 1) 343400 or (2) (3)

1.3 one-dimensional linear equation

Synchronous training

1. 1.x = 332.； 3. Move the project; 4.550; 5.5。

Second,1.d; 2.b； 3.d .

Third, 1. ( 1)x =- 16； (2)x =； 2. Xiaoming 1 1 year old; 3. The two digits are 56; 4.33 yuan.

Test question

1. 1.c； ; 2.b； 3.b； 4.b； 5.A

Second,1.1; 2.- ; 3.24; 4.4(4x+x)-2(x-3)= 8； 5. 10.

Third,1.1.9; 2.20; 3. Let the deposit principal be X yuan, then (1) x (1+3× 2.70%) 2 = 5000, and the solution is x=4278.77 yuan (2)x( 1+6×2.88%)=5000. 4. Assuming that X tons of product A are produced, then X tons of product B are produced, 1000x+900x = 530000, and x=200 tons.

1.4 Binary linear equations

Synchronous training

I. 1。 , 3; 2.2; 3.3,4; 4.3 km/h, 5 km/h; 5.2

Ii. 1. D2

Three. 1.( 1) (2) 2.(,)

3. Let's say that the 20-second broadcast is X times and the 40-second broadcast is Y times.

20x+40y= 180

X≥2 x=3 or x=5.

Y ≥ 2y = 3y = 2 (1) 3× 6000+3×10000 = 48000 yuan.

(2)5×6000+2× 10000=50000 yuan.

Test question

I. 1.a2.b3.b4.b5.b

Two. 1.2.( 1,3)

3.y= 2x+8 4. 15 12 5。 Y =- or y =

Three. 1.( 1) (2).

2. Solution: We buy X barrels of Class A purified water and Y barrels of Class B purified water. From the meaning of the problem:

A: omitted

3.720m, 7 minutes. 4. (1) A (-3,0 0) B (4 4,0) (2) Omit (3)

1.5 unary linear inequality (group)

Synchronous training

I) 1. x >-x > y； 7.4~5; 8.m≤2。

Second, 1. a； ; 2.c； 3.d； 4.b； 5.b； 6.a； 7.C。

3. 1 . x≤9； 2.x < 23.x < 2, abbreviated on the number axis; 4.< x≤。

4. 1.① x >，② x

Test question

I. 1. >-; 2 . x+5≤3； 3.> 13; 4.2; 5 . x ≥- 3； 6.5 < m < 327.-a < x 2； 2. 1; 3.32~36; 4. When the investment in the mall is higher than 20,000 yuan, it will be more profitable to sell it at the beginning of next month; When the investment in shopping malls is less than 20 thousand yuan, the profit from sales at the beginning of the month is more.

1.6 score

Synchronous training

1. 1, 3 2, m2-n23,4, such as. 5,-2 6,

Second,/kloc-0 1.c2.b3.d4.c

Third, 1. ( 1); (2) .2.( 1)②; (2) The denominator is missing; (3) .

3.a and B are opposites.

Test question

I. 1. - 1 2.3.5.2

Second,/kloc-0 1.d2.d3.c4.c5.d

Third, 1. ( 1) 1; (2).2.3.( 1) Let Mr. Wang's riding speed be. After checking, it is the solution of the original equation, which accords with the meaning of the question. (If you don't write, you won't be penalized) Teacher Wang's riding speed is15 km/h. 。

(2) Answer: I can get to school before 8 o'clock.

Let's assume that Mr. Wang spent several hours meeting Xiao Gang and went to school after the meeting, then we can get a solution from the meaning of the problem:

Can get to school before 8 o'clock.

4.( 1) < ( > >0)

Proof: ∫-= < 0 (provided that > > 0)

∴ >0, >0)

(3) Let the original floor space and window area be, and the increased area be,

From (2), we can see that: >, so the lighting conditions of the house have improved.

1.7 quadratic root

Synchronous training

I. 1.2.3. 1 4.0

Second,/kloc-0 1.d2.c3.d4.a

Three. 1.( 1) (2) 2.

Test question

I. 1.b2.b3.c4.c5.d

2. 1.2.3.4. (The answer is not unique) 5.

Three. 1.( 1) (2) 2.

3. Then, it might as well be established, because

So it must form a triangle.

4. Integer (2) of (1)

Proof: left = = right.

So, the equation holds.

1.8 unary quadratic equation

Synchronous training

I./kloc-0 1.C2.C3.B4.C

2. 1., 2., 3.-4 4.

Third, 1. (1) It is proved that the original equation has two unequal real roots.

(2)k = 1 ^ 2。 The solution of (1) is from departure to t seconds, cm2, PB= 16-3t, CQ=2t, ×BC=, when cm2, that is, at that time, the solution is: t=5, which meets the meaning of the question after inspection.

Test question

I./kloc-0 1.b2.c3.d4.c5.a

Two. 1.，2.3.3m4.5.-3 1

3. 1.( 1) If the original equation is a quadratic equation, then the solution is:, that is, or 2, and the original equation is a quadratic equation. (2) If the original equation is an unary quadratic equation, then sum, then the solution is k= 1. When k= 1, the original equation.

2. Suppose that each shirt should be reduced in price by RMB, and the solution is: because the inventory should be reduced as soon as possible, it should be scrapped, so each shirt should be reduced in price by 20 yuan.

Determination of 1.9 position

Synchronous training

I. 1.b2.a3.d4.a5.b

2. 1.(2, 1), (2, 2), (2, 3) 2.60, 3km, 30, 6km, due south, 4km3.804. The abscissa is unchanged, and the ordinate is multiplied by-1 respectively.

Three, 1. (1) Sports training base, tennis court; Still need distance; (2) Baihua Garden; In the direction of 30 southwest of the school, there is also the Huanghai Hotel, which is different from the school. (3) bearing and distance.

2.( 1)2,3,4,0; (2) Axisymmetric; (3) Points (0,0) and (4,2); (0,2) point and (4,0) point

3. If 0 < a ≤ 3, the point P 1 is on the line OM. pp2 = PP 1+p 1p 2 = 2op 1+2p 1m = 2(op 1+p 1m)。 If a > 3, point P 1 is to the right of point m, and pp2 = pp1-p/p2 = 2op1-2p1m = 2 (op1-p/kl.

Test question

I. 1.c2.b3.c4.c5.d

Second, 1. (4,3) 2.3 3.5 4. 1,-3 5.(,0).

Three. Area 1. A4→A3 →A2 →B2 →C2 →c 1 →d 1 →D2.

2.( 1) The left picture and the right picture are symmetrical about the Y axis, so the coordinates of the left and right flowers in the left picture are (-3,4) and (-2,5) respectively. (2) The vertical coordinates of the left and right flowers are unchanged, and the horizontal coordinates are increased by 2 units respectively. (3) The left and right flowers in the pattern move upward 1.

3.( 1) sketch, A (0 0,40), B (80 80,0); (2) Because our ship is close to O Island and the speed is the same, the interception point should be the intersection of the perpendicular line of AB and OB. Let this intersection be M and OM=x, then BM = AM = 80-X. In Rt△AOM, AM2=OA2+OM2 is obtained by Pythagorean theorem, so (80-X) 2 = 402.

1. 10 linear function

Synchronous training

I. 1.d2.b3.c

2. 1.y = 100+0.2x，x，y ^ 2。 Omit 3.4.y =-2x+3 5. - 1,2.

Third, 1. ( 1)(2)m = 9； Let the straight line intersect the Y axis at point C, then s △ AOB = s △ AOC+s △ COB =× 4×1+× 4× 9 = 20.

2. Solution: (1)y= 150-x (2) According to the meaning of the question, y≥2x. ∴ 150-x≥2x，x≤50。

∵ x≥0， 150-x≥0，∴ 0≤x≤50，∴p = 600 x+ 1000( 150-x)=-400 x+65438。

∵ p decreases with the increase of x, and 0≤x≤50, ∴-400× 50+150000 ≤ p ≤-400× 0+150000, that is,130000 ≤

3.( 1)A(—n，0)，B( 1，0)，P()。

(2)P(，)； y = x+ 1； Connect OP, s quadrilateral PQOB =

Test question

I. 1.d2.b3.d4.c5.b

Second, 1.2 2.y =-2x+6 3. () 4.20 5.Sa < sb。

3. 1.p(，2)。 Tip: If the PD⊥y axis is at point D, the ordinate of point P can be 2 first.

2. (1) y =10x-1000, y = 15x-2500 (2) 234 (3) 0 sold tickets and compensated1000 yuan (4) sold. If less than 100 tickets are sold, there will be losses.

(1) According to the meaning of the question, y = 20x+ 15 (7-x), that is, y=5x+ 105.

(2) 50x+35 (7-x) ≤ 300, and the solution is x≤.

The function value of ∵ y=5x+ 105 increases with the increase of x, which is a natural number. ∴ When x=3, the maximum value of y is 3×5+ 105= 120 (ten thousand people) .7-x = 4.

A: TV stations broadcast 3 episodes of A and 4 episodes of B every week, so as to maximize the total number of viewers per week, that is, 654.38+0.2 million.

1. 1 1 inverse proportional function

Synchronous training

I. 1. C 2。 A 3。 A 4。 A

Second, 1. For example, y=-2. Decrease by 3.4.5. -2 < x < 0 or x > 3.

Three. 1.( 1)A( 1+， 1-) B ( 1-， 1+) (2) 2。 ( 1) b= 2。

Test question

I. 1.a2.b3.d4.d

Second, 1. The answer is not unique. For example, the relationship written can satisfy that the xy value is positive. 2.3. parallelogram 4.3.6 5.20 (hint: x 1 =-x2, y 1 =-y2).

Third, 1. ( 1); (2)R =20 (Europe) 2. ( 1)(x≥5)； (2)25 minutes

3.( 1)m=-3，k = 9； (2) By knowing that -x-6 =,

That is x2+6x+k=0. To make the images of two functions have two different intersections, it is necessary to make the equation x2+6x+k=0 have two unequal real roots. ∴ 62-4k = 36-4k > 0。 The solution is k < 9, k≠0.

When k < 9, k≠0, the images of these two functions have two different intersections.

(3) When k =-2 and -2 are within the acceptable range of k, the image of the function y =- is in the second and fourth quadrants, so its two intersections with the straight line y =-x-6, A and B, should be in the second and fourth quadrants respectively, and at this time ∠AOB is an obtuse angle.

1. 12 quadratic function

Synchronous training

I. 1.a2.b3.c4.d

Second, 1. ①; ② ; ③ ; ④ ; ⑤ ; ⑥ .

2. (1,2) 3. (-2,5) or (4,5) 4.

Three. Sketch of 1 ( 1); (2) According to the image, it can be estimated as a parabola, y = 0.002x2+0.01x; ; (3) 150km/h overspeed.

(1), (2) Let the resolution function of this parabola be, and substitute O (0,0) for the solution, ∴, that is.

(3) Let A, then OB=m and AB=DC=. According to the axisymmetry of parabola, we can get: ∴ that is, AD = 12-2m.

∴ =AB+AD+DC=

= = .

When m=3, that is, OB = 3m, the maximum sum of the lengths of the three wooden poles is 15m.

Test question

I. 1. B 2。 D 3。 A 4。 B

2. 1.( 1,0) 2. Given the image passing point of quadratic function y=x2+bx+c (1, 0), and the symmetry axis is a straight line x=- 1, find its analytical formula 3.34. ( 1).(2)③④

Three. 1.( 1) (- 1,0), (3,0), (0,6), 16; (2) x 3。

2.( 1)y =-0.2 x2+3.5； (2)0.2 meters.

3.( 1)P( 1，-4)，A(- 1，0)； (2) It is easy to find that the parabola is y = x2-2x–3, and C(0, -3) and B(3, 0) are obtained. The analytical formula of the straight line BM can be found as y=-x +3, the abscissa of the point M is a, and there is-a+3 = A2-2A–3, thus m (-2,0) can be found.

1. 13 algebra synthesis problem

I.1.d2.b3.b4.c5.d6.c7.c8.c9.a10.d1.b12.c.

2. 1.2.4 3.4.5. Total number of small circles: 1, 7, 19, 37, 61; Relationship:

3. 1.2. Because the solution is: x= 1, and substitute it into the original formula y = = 2007.

3. The original formula is changed to: naming; Understandable;

Test: After test, it is the solution of the original equation.

4.( 1) >0 (2) ; ,

(3) two intersection points; (2,0); ( ,0)

5.( 1) ; (2) Invest 70,000 yuan in Class A commodities and 30,000 yuan in Class B commodities, with a maximum profit of 58,000 yuan.

6.( 1) (2), the vertex coordinate is (65, 1950), the figure is omitted; Therefore, when 65 yuan is designated, the daily average profit is the most, and the highest daily average profit is 1950 yuan.

1. 14 data collection and processing

Synchronous training

1. 1.c； ; 2.d； 3.A。

Second,1.87,88; 2.7 ; 3.7.

Third, 1. Solution: (1) The PE achievement rate of Class 2 in Grade 3 is (1-0.02) × 100% = 98%.

The achievement rate of other categories is 1- 12.5% = 87.5%.

Answer: The rate of reaching the standard of physical education in Class 2 and other classes in Grade 3 is 98% and 87.5% respectively.

(2) There are X students in the whole school. The question means:

50×98%+(x-50)×87.5%≥90%，

Solution: x≤2 10

A: The number of people in the whole segment shall not exceed 2 10.

2. Solution: (1), (2) The median height falls within158 ≤ X.

Because the sample size is 60 data arranged from small to large, the median of the sample should be the average of the 30th and 3 1 data, and at158 ≤ X.

(3) The height should be155 ≤ x.

Because there are 4 1 students in this range, and the height is relatively close, 40 students are selected from them to participate in the competition, and the team is relatively neat.

3. Solution: (1) 33; (2) See the figure below; (3) As can be seen from the chart, the demand of textile workers, turners, electronic component manufacturers and welders is greater than the number of job seekers; The number of job seekers such as administrative staff, accountants, secretaries and typists far exceeds the demand.

Test questions:

1. 1 . d； 2.d； 3.c； 4.C。

Second,1.64; 2.8; 3.b； 4.2.

Three. 1. Solution: (1)

The mean, median and mode variance are higher than 85 points.

One piece 84 84 84 14.4 0.3

B 84 84 90 34 0.5

(2) The mode of A score is 84, and the mode of B score is 90. Judging from their scoring pattern, B scored better.

The variance of grade A is 14.4, and the variance of grade B is 34. Judging from the variance of grades, Grade A is relatively stable.

The median and average of Grade A and Grade B are both 84, but from the frequency above 85, Grade B is better.

2. Solution: (1) The number of tourists in B tourist spots increased the fastest in 2004 compared with the previous year.

(2) (ten thousand yuan) (ten thousand yuan)

From 200 1 to 2005, both tourist attractions A and B received an average of 30,000 tourists per year, but the fluctuation of tourist volume in scenic spot A was greater than that in scenic spot B. 。

(3) Get a solution of 5-≤ 4 from the meaning of the question, and x ≥ 100 100-80 = 20.

A: Tickets for tourist attraction A should be increased by at least 20 yuan.

3. Solution: (1) 1000, 0.25, 100, 0.05; 2000; (2) ellipsis; (3)500; (4) ellipsis.

4. Solution: (1) According to the figure, the accommodation consumption is 34,382,400 yuan, accounting for 22.62% of the tourism consumption.

∴ Tourism consumption * * * 3438.24 ÷ 22.62% = 15200 (ten thousand yuan) = 1.52 (one hundred million yuan).

Traffic consumption accounts for 17.56% of tourism consumption, and ∴ traffic consumption is15200×17.56% = 2669.12 (ten thousand yuan).

∴ This year's "May 1" Golden Week, the median of all kinds of consumption in our city's tourism consumption is

(3438.24+2669.12) ÷ 2 = 3053.68 (ten thousand yuan).

(2) Solution: Assume that the average annual growth rate of tourism consumption from 2005 to 2007 is X.

, solution

Because the growth rate cannot be negative, it is discarded. ∴ x = 0.5 = 50%。

A: The average annual growth rate of tourism consumption from 2006 to 2008 was 50%.

1. 15 probability

Synchronous training

1. 1.a； ; 2.a； 3.a；

Second, 1. 2.a； 3.； 4.3 ,2 , 1.

Third, 1. Solution: Xiaohua is Party B's reason: Let A 1 stand for the first black ball, A2 for the second black ball, B 1 for the first white ball, and B2 for the second white ball. There are 24 possible results (which can be illustrated by tree diagram or table), and 8 are arranged in black and white. Therefore, Party A,

2. Solution: (1)∫ The number of white balls is 50- 1-2- 10 = 37.

What is the probability of missing the prize?

(2) There is only one possibility for winning 10 yuan, that is, touching two yellow balls at the same time.

The probability of winning 10 yuan prize is =

Test question

1. 1.d； ; 2.a； 3.a； 4.B。

Second,1.3; 2. Uncertainty; 3.30%; 4.25%; 5.6 species.

Three. 1.30 2. Black: × = white: × =+=

3. Fairness, 1.2 Possible situations:/kloc-0 /×1=1.2×1= 2; 1×2=2 2×2=4; 1×3=3 2×3=6

4.( 1) Form:12; 25%; (2) ellipsis; (3) 25%.

1. 16 comprehensive questions of statistics and probability

1. 1.b； ; 2.d； 3.a； 4.b； 5.A

Second,1.5; 2.30.0; 30.0; 32.0; 3.2; 4.7 and 8.

Third, 1. Solution: (1) 50; (2)0.22; (3)3≤t < 4； (4) .

2. Solution: (1)132,48,60; (2) 4,6

3. Solution: (1) A = 50× 0.24 = 12, B = 8 ÷ 50 = 0.16;

⑵ (5+6+3)÷50=28%; ⑶ The median of the third group is between 46 and 48.

Solution: (1) This game is unfair to both parties.

∵ ; ; ; ,

∴ Huayang scored every time on average (points);

Ji Hong's average score is 100.

This game is unfair to both sides.

(2) It is revised as: Huayang gets 3 points when the figure is a villain or an electric light, and Ji Hong gets 2 points when the figure is a hill or a house. If other rules remain the same, the game is fair to both sides. (The answer is not unique)

1. 17 positional relationship of the plan.

Synchronous training

I. 1，D 2，B 3，C 4，A 5，B。

2. 1, 22 2, 1, 2,1,12 or 3,4,9, 10 or 5,6,7,8 or 3,3/kloc.

AD，BE

3. 1, ( 1)65.(2)45。

2、

3、

Test question

I. 1，D 2，D 3，B 4，B 5，D。

2. 1, 133 2, 5 3, 35.4. The greater the angle, the smaller the length of the line segment.

Three. 1. As shown in the figure, the straight line AE is a drawn straight line obtained from the characteristics of the grid.

2、∵∠ADE=∠B,∴DE//BC ∴∠ 1=∠DCB

∵∠ 1=∠2,∴∠2=∠DCB ∴FG//CD

∵FG⊥ AB,∴∠FGB=∠CDB=90。 ∴CD⊥ company

3、( 1)90。 (2) unchanged

1. 18 triangle

Synchronous training

1. 1, B 2, B 3, B 4, B 5, a

2. 1, ∠A=∠F or BC = ed 2,8 3,4,30. 5、5

Third,

Test question

I. 1, B 2, B 3, B 4, C 5, c

Second, 1, 2, 1233, 8 4, isosceles triangle 5, 1

Third,

(3) Can spell out the graph that proves Pythagorean theorem, and the graph is rough.

Translation, Rotation and Symmetry of 1. 19 Graph

Synchronous training

I. 1. B 2。 A 3。 C

Second, 1. Omit, 2. (1) as shown in figure (2) 34 (3) Pythagorean Theorem 3. As shown in the figure.

Test question

A 1。 B 2。 D 3。 B 4。 C

2 1.60 2.(2,2) 3.60

Three 1. As shown in figure 4.

2.( 1) The position of the rotation center point p is shown in the figure, and the coordinate of the point p is (0, 1).

(2) The rotated triangle ④ is as shown in the figure.

3.( 1) as shown in figure (b)

(2) ;

(3) As shown in Figure (c), it holds.

That is: (or by rotation)

Extend the delivery time and deliver it to

,

When the rotation angle is large, the conclusion still holds.

1.20 quadrilateral

Synchronous training

I./kloc-0 1.C2.A3.B4.A

Second, 1. 6 2.3.

Three. 1.( 1) (omitted) (2) 90 2.

3.( 1)90 .

(2) Solution: A is AE⊥BC, and the vertical foot is E. 。

Then AE line divides quadrilateral ABCD into two parts: △ABE and quadrilateral AECD.

Rotate △ Abe 90 counterclockwise with A as the rotation center, and the divided two parts will be recombined into a square.

The extension of AF‖BC cross CD is longer than that of f, ∫∠ABC+∠ADC = 180,

∴∠ABC=∠ADF.

AD=AB，∠ AEC = ∠ AFD = 90，∴△ABE≌△ADF.

∴AE=AF.∴ Quadrilateral AECF is a square.

Test question

I. 1. B 2。 B 3。 D 4。 B

2.1.122.163.24.8 The shadow area of any square piece of paper is 8 cm2.

3. 1.( 1) ∵ Quadrilateral ABCD is a parallelogram, ∴∠EBC =∠ADF, BC=AD.

* daf = * BCE，∴△ADF≌△CBE.

(2) In \an \u BC, \ anb = \ nbc.bn \ ABC, \ ABC = 60,

∴∠ NBC =∠ ABN = 30。 It is also obtained by (1): ∠ DAF = ∠ ECB = 20.

∴∠amn= 180-30-20 = 130

2.( 1) vertical bisection,

The quadrilateral is a diamond.

(2) When the diamond is a square.

Diamonds are squares.

3. The point P drawn by (1) is on AC, not the midpoint and endpoint of AC;

(2) Lian P 1A, P 1D, P 1B, P 1C, and.

∠AP 1D=∠AP 1B，∠DP 1C =∠BP 1C∠AP 1B+∠BP 1C = 180

P 1 is on AC, and P2 is also on AC.

In △DP 1P2 and △BP 1P2.

∠ dp2p 1 = ∠ bp2p 1, ∠ dp 1p2 = ∠ bp 1p2 Male * * *

△DP 1p 2?△BP 1p 2

DP 1=BP 1 DP2=BP2, so B and D are symmetrical about AC.

Let p be any point of P 1P2 and connect PD and PB, which can be obtained from symmetry.

∠DPA=∠BPA，∠DPC=∠BPC

So point p is a semi-equiangular point of a quadrilateral.

(3) Draw the symmetrical point B 1 of point B about AC, and extend the intersection of DB 1 and AC at point P, which is the demand.

1.2 1 similar shape

Synchronous training

I. 1.c2.c3.b

Second, 1.2. △ Abe ∽△ ACD ∽△ Ade3.30

Third, 1. Solution: AD//BC ∽ In trapezoidal ABCD,

AD= 10，BC=20

∵ ,

We still need 200× 10=2000 yuan, and the remaining funds are 2000-500 = 1500 < 2000.

So there is not enough money.

2.

3. According to the meaning of the question: AB⊥BH, CD⊥BH, FG⊥BH

Rt△ABE and Rt△CDE in cd⊥bh. ∵ab⊥bh

∴CD//AB, it can be proved that: △ Abe △ CDE.

∴ ① Similarly: ②

And CD = fg = 1.7m, which can be obtained from ① and ②:

The solution is: BD = 7.5m. Substituting BD = 7.5 into ① is: AB=5.95m≈6m.

Answer: The height of pole AB is about 6m.

Test question

I. 1.c2.d3.b

2. 1.∠ACD=∠B (the answer is not unique) 2. (4,-3) 3. (4,0) or (3,2) 4.2.

Three. 1. Similarly, when AP=2

2.( 1) ellipsis (2) bit similarity ratio is 1:2 (3) ellipsis.

3. yes. According to the meaning of the question, ∠B+∠ACB=∠E+∠DFE, ∠B∠E, ∠B≦DFE.

Let ∠ b < ∠DFE be ∠EFG=∠B, G be on DE, ∠BCH=∠E ∠ E, and H be on AB (as shown in the figure). Then you can get △AHC∽△DGF, △ HBc ∽△ GFE.

1.22 Solving Right Triangle

Synchronous training

I. 1. A 2。 B 3。 D 4。 B

Second, 1. B 2。

3. 1. As shown in the figure, after a little, it is over.

Because,

So the quadrilateral is a parallelogram.

So ...

By,

Yes

In,,, by,

Get it. So ... Yes,

Get it.

2.( 1) , .

In ..

(2) Fill in the form in the following order:

(or), (or), (or)

(3) The whole roof truss has 18 auxiliary columns,

The shortest auxiliary column on the right is, and the penultimate column is,

(meter)

Answer: The shortest auxiliary column length is about1.3m..

Test question

I./kloc-0 1.A2.C3.C4.B5.D

Two. 1.45 2.8.2 3.7.3

Three. 1. As shown in the figure, the intersection point B is BE⊥AD, AD is in E, and AC is in F.

BF=2，DE=BC= CD=4 EF=2

In Rt△AEF, tan α = ∴ ∠ α = 30.

Answer: The minimum included angle α between the elevator and the ground on the first floor is 30.

2. the intersection point b is perpendicular to CD and AC, and the vertical feet are e and f respectively.

∵∠ BAC = 30，AB = 1500m ∴ BF = EC = 750m。

AF=7503 let fc = x m≈DBE = 60∴de = 3 x m m.

∫∠DAC =45∴∴AC = CD, that is, 7503+X = 750+3 X.

X=750 ∴CD=(750+7503) meters.

A: The CD of the mountain height is (750+7503) meters.

3. the intersection point c is perpendicular to the straight line AB, and the vertical foot is D.

Let the tractor running route CF intersect with AD at point E, AC = 300, ∠ ACD = 45,

∴CD=AD=300 ÷ =300。 DE=CD？ tan30 =300× = 170。

∴BE=300-36- 170=94.

If point B is BH⊥CF and vertical foot is H, then ∠ EBH = 30.

∴BH=BE？ Cos 30 = 94× ≈ 80。 ∵ 80 < 100, ∴ B classroom is affected by tractor noise.

Make an arc with a radius of 100 with point B as the center, and cross CF at point M and point N, then MN=2 =2×60= 120.

The time for classroom B to be affected by noise is: 120÷8= 15 (seconds).

Let ah' ⊥ cf and h' be vertical feet, then ∠ eah' = 30, while AE = 36+94 =130, and ∴ ah' = AE? cos 30 = 130×= 1 1 1。

∫11>100, ∴ A classroom is not affected by tractor noise.

1.23 circle

Synchronous training

I. 1, C 2, C 3, C 4, B 5, B 6, a

2. 1, 1.8cm or 22cm 2, r=3 or r >；; 6 3、 4、

Third, 1, link OE to prove OE⊥AE.

2. Solution: OE=OF, connecting OA and OB, then OA=OB, ∴∠A=∠B∴△OAE≌△OBF∴OE=OF.

3. hint: extend CB to e, make BE=BA, and connect ME, ma, MB and MC. It is proved that ∠MBE=∠ABM, and BM=BM, BE=BA, ∴△MBA≌△MBE, ∴ME=MA=MC, ∴ED=CD, that is, AB+BD=CD, ab =

4. Solution: (1) A vertical line connecting O O as O E and O F; BC; Make o g ⊥o e o e cross to g, as shown in the figure, ∵⊙O is inscribed with rectangular ABCD, AB= 16, ∴R=8, and in RT △ o o o g, we can get by pythagorean theorem.

The solutions are r =32- 16, r =32+ 16 (excluding ∴R=8), and r=32- 16.

(2) No, because 2r >；; AB, so you can't cut such a small round iron piece.

Test question

I. 1，B 2，D 3，B 4，C 5，B

Second, 1 75 or152,453,4, inscribed or circumscribed 5,

Iii. 1, (1) Omit (2)

2. It is proved that the connections AD, CF, FD, BD, EF, ∫△ACB are isosceles right triangles, ∴∠ CDF = ∠ CAF = 45.

∴DF bisects∠∠ CDE, in △CFD, ∠∠ DCF =135-∠ CFD = ∠135-∠ CFB. In △BCF,

∠ BCF = 135-∠ CFB, ∴∠DCF=∠BCF, that is, CF divided by ∠DCB, ∴F is the heart of △CDE.

3. Solution: When (1) crosses point O, let ⊥ AM be in F, and when = r = 2, O is tangent to AM, and OA=4cm, so

X=AD=2cm。 (2) When crossing point O, OG⊥AM is in G, ob = oc = 2, ∠ BOC = 90, ∴ BC = 2, ∫og⊥BC.

∴og= ∴bg=cg=，∫≈a = 30，∴OA=2 ,∴x=AD=2 -2。

4.( 1) If ∠ EBC = ∠ CDA = 72, ∠ BFD = ∠ C, the quadrilateral BCDF is a parallelogram and BC=CD, it can be concluded that (2) it comes from ∠ Ade = ∠ AEB.

∠∠ 1 =∠2, ∴AE=ED, ∠∠3 =∠4, ∴DE=DF, ∴FD=AE, and the conclusion can be proved.

1.24 views and projections

Synchronous training

I. 1. B 2。 C 3。 C 4 explosive D

Second, 1. Parallel to the same line. 2. The circle is getting smaller and smaller. 3.24.20 5. Bright and clear. 6. The bigger it is.

Third, 1. Solution: facing the sun (south-east); Xiao Qiang is very tall, because the sun rays are parallel rays, and his height is directly proportional to the growth of the shadow.

Step 2: Solutions