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Who knows which questions are more difficult for the final exam in 2009 in various provinces and cities? Introduce it! Introduce it!
In 2009, the senior high school entrance examination mathematics finale problem assembly (including part of the problem-solving process)

Beijing, 2009. As shown in the figure, in the plane rectangular coordinate system, the coordinates of three kinds of mechanical warfare are as follows

Extend AC to point D, make CD=, pass point D, and make the extension line of DE‖AB to BC of point E.

(1) Find the coordinates of point D;

(2) Make the symmetrical point F of point C about the straight line DE, and connect DF and EF respectively. If the straight line passing through point B divides the quadrangle CDFE into two quadrangles with equal perimeter, the analytical formula of this straight line is determined;

(3) Let G be a point on the Y axis, and point P starts from the intersection of the straight line and the Y axis, then reaches point G along the Y axis, and then reaches point A along GA. If the moving speed of point P on the Y axis is twice as fast as that on the straight line GA, try to determine the position of point G, so that the time for point P to reach point A is the shortest. (Requirements: Briefly describe the method of determining the position of G-point, but do not require proof)

(Chongqing, 2009) 26. As shown in the figure, in the plane rectangular coordinate system, the edge OA of the right-angle OABC is on the positive semi-axis of the Y axis and OC is on the positive semi-axis of the X axis, with OA=2 and OC = 3. The bisector passing through the origin O is ∠AOC passes through AB at point D, connecting DC, and points D pass through DE⊥DC and OA.

(1) Find the analytical formula of parabola passing through points E, D and C;

(2) After rotating ∠EDC clockwise around point D, one side of the angle intersects with the positive semi-axis of the Y axis at point F, and the other side intersects with the line segment OC at point G. If DF intersects with the parabola in (1) at another point M, and the abscissa of point M is, is EF=2GO true? If yes, please give proof; If not, please explain the reasons;

(3) For point G in (2), is there a point Q on the parabola located in the first quadrant, so that the △PCG formed by the intersection point P of straight lines GQ and AB and points C and G is an isosceles triangle? If it exists, request the coordinates of point q; If it does not exist, please explain why.

26. The solution: (1) is given by the known, and is obtained.

,

.

. ( 1)

The analytical formula of parabola passing through this point is.

Substitute the coordinates of this point and you get it.

Substitute the coordinates of the sum points respectively, and get

(2 points)

To solve this system of equations, you must

Therefore, the analytical formula of parabola is. (3 points)

(2) established. (4 points)

The point is on a parabola, and its abscissa is,

The ordinate of this point is. (5 points)

Let's assume that the analytical formula is,

Substitute the coordinates of each point respectively to obtain

solve

The analytical formula of is. (6 points)

,. (7 points)

Do something excessive,

Then.

,

.

Say it again,

.

.

. (8 points)

.

(3) Click,, and then set.

, , .

(1) if, then,

The solution is. At this point, the point coincides with the point.

. (9 points)

② If, then,

Solution, this axis.

The abscissa of the intersection with the parabola in the first quadrant is 1,

The ordinate of this point is.

. (10)

③ If, then,

At this point, the solution is an isosceles right triangle.

Passing through this point is the axis of this point,

Then, suppose,

.

.

Solve (give up)

. (12)

To sum up, there are three things that meet the requirements.

That's or or or or.

Qijiang County, Chongqing, 2009. (1 1) As shown in the figure, it is known that a parabola passes through a point, and the vertex of the parabola is a ray. A straight line passing through the vertex and parallel to the axis intersects the ray at this point, and it is connected to the positive semi-axis of the axis.

(1) Find the analytical formula of parabola;

(2) If the moving point starts from this point and moves along the ray at the speed of 1 length unit per second, the moving time of this point is. When is the value? Are quadrilaterals parallelograms respectively? Right trapezoid? Isosceles trapezoid?

(3) If the moving point and the moving point start from point and point at the same time, they move at the speed of 1 length unit and 2 length units per second respectively. When one of the points stops moving, the other point stops moving. Let the time of their movement be connection. When what value is taken, the area of the quadrilateral is the smallest? And find the minimum value and length at this time.

* 26. Solution: (1) parabola passing point,

1 point

The analytic formula of quadratic function is: 3 points.

(2) If the vertex of the parabola is overstressed, then,

4 points

When a quadrilateral is a parallelogram.

5 points

When a quadrilateral is a right trapezoid.

If you go too far, then

(If not found, you can find it)

6 points

When, the quadrilateral is an isosceles trapezoid.

To sum up: at 0, 5 and 4, the corresponding quadrangles are parallelogram, right-angled trapezoid and isosceles trapezoid respectively. Seven points

(3) According to (2), it is an equilateral triangle.

rule

If you go too far, you will get 8 points.

= 9 points

, the minimum area is 10 point.

now

1 1 min

(Hebei Province, 2009). (The full score of this small question is 12)

As shown in figure 16, in Rt△ABC, ∠ C = 90, AC = 3, AB = 5..P Point P starts from point C, moves along CA to point A at a constant speed of 1 unit per second, and immediately returns to point A at the original speed; Point Q starts from point A and moves along AB to point B at a constant speed of 1 unit per second. With the movement of P and Q, DE divides PQ vertically, and intersects PQ at point D. The intersection line QB-BC-CP starts from point E, and points P and Q move at the same time. When point Q reaches point B, it stops moving, and point P also stops moving. The moving time of points P and Q is set as t seconds.

(1) When t = 2, AP =, and the distance from Q to AC is;

(2) In the process of moving point P from C to A, find the sum of the area s of △APQ.

Functional relationship of t; (Don't write the range of T)

(3) In the process of moving point E from point B to point C, can quadrilateral QBED become?

Is it a right-angled trapezoid? If yes, find the value of t, if not, please explain the reason.

(4) When DE passes through point C, please write the value of t directly.

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.

,

(Henan Province, 2009) 23. (1 1) As shown in the figure, in the plane rectangular coordinate system, the three vertices of rectangular ABCD are known as B (4 4,0), C (8 8,0) and D (8 8,8). Parabola y=ax2+bx passes through point A and point C. 。

(1) Write the coordinates of point A directly and find the analytical formula of parabola;

(2) The moving point P starts from the point A and moves along the line AB to the end point B, while the point Q starts from the point C and moves along the line CD.

Move to the end D. The speed is 1 unit length per second, and the moving time is t seconds. Passing through point p is PE⊥AB, and AC is at point E.

① If you cross point E, do EF⊥AD at point F, and pass parabola at point G. When t is the value, what is the longest segment of eg?

② Connect EQ. During the movement of points P and Q, how many moments make △CEQ an isosceles triangle?

Please write the corresponding t value directly.

Solution. (1) The coordinate of point A is (4,8) .................1min.

Substitute the coordinates of A (4 4,8) and c (8 8,0) into y=ax2+bx respectively.

8= 16a+4b

get

0=64a+8b

The solutions are a =- and b = 4.

The analytical formula of parabola is: y =-x2+4x. ....................................................................................................................................................

(2)① In Rt△APE and Rt△ABC, tan∠PAE= =, that is =

∴PE= AP= t.PB=8-t

The coordinate of point e is (4+ t, 8-t).

The ordinate of point G is: -(4+t) 2+4 (4+t) =-t2+8. ............................................................................................................................

∴EG=- t2+8-(8-t)

=- t2+t。

∫-< 0, ∴ When t=4, the longest line segment eg is 2. ...........................................................................................................................................

(2) * * * there are three moments .......................... 8 points.

T 1=, t2=, T3 = ...............11min.

Shanxi Province, 2009, 26. (Question 14) As shown in the figure, it is known that straight lines intersect at points and intersect at two points respectively. The vertices of a rectangle are on a straight line, the vertices are all on the axis, and the points coincide with each other.

(1) area;

(2) Find the length of the sum of the sides of the rectangle;

(3) If the rectangle starts from the origin and translates in the opposite direction of the axis at the speed of/kloc-0 per unit length per second, let.

The moving time is seconds, the area of rectangle and overlapping part is, and the gap is found.

Function relation of, and write the corresponding value range.

26.( 1) solution: the coordinates of the obtained point are

The coordinates of the obtained point are

∴ (2 points)

According to the solution, the coordinate of point ∴ is (3 points).

∴ (4 points)

(2) Solution: ∵ Point is on, and

The coordinate of the point is (5 points)

Point over and over again.

The coordinate of the point is (6 points)

∴ (7 points)

(3) Solution 1: When, as shown in figure 1, the rectangle and the overlapping part are pentagons (when, they are quadrangles). If you go too far, then

That's VII.

Namely (10)

(Taiyuan, Shanxi Province, 2009) 29. (The full score of this small question is 12)

problem solving

As shown in figure (1), fold the square paper in half, so that the point falls on the edge (does not coincide with the point), and flatten it to get the crease. When, evaluate.

Analogical induction

In the diagram (1), the value of if is equal to; If the value is equal to; otherwise; If (is an integer), the value of is equal to. (expressed by inclusion)

Contact tuoguang

As shown in Figure (2), fold the rectangular piece of paper so that the point falls on the edge (does not coincide with the point), and the crease rule value obtained after flattening is equal to. (expressed by inclusion).

29. Problem solving

Solution: Method 1: Connect as shown in the figure (1- 1).

Quadrilateral and quadrilateral are symmetrical about a straight line.

Vertical division. 1min

∴, quadrilateral is a square

Set rules

Yes,.

∴ solution, namely 3 points.

One in and one out,

,

,

5 points

Set rule seven

The solution is 6 points.

7 points

Method 2: Same as Method 1, 3 points.

As shown in figure (1-2), make intersections and connections at points.

A quadrilateral is a parallelogram.

Similarly, quadrangles are parallelograms.

live in harmony with

5 points

6 points

7 points

Analogical induction

(or); ; 10 point

Contact tuoguang

12 point

Scoring description: 1. If your correct answer is different from the reference answer provided above, you can refer to the scoring instructions for scoring.

2. If the answer consists of multiple questions, and the answer of the previous question is wrong or unanswered, it will not affect the answer of the following question, and the score can be made according to the reference answer and scoring instructions.

Anhui Province (2009) 23. It is known that the functional relationship between the wholesale unit price and the wholesale quantity of a certain fruit is shown in the figure (1).

(1) Please explain the practical significance of the two function images in the figure.

solve

(2) Write down the wholesale fund amount of this fruit between W (yuan) and M (kg).

Functional relationship; Draw a function image in the following coordinate system; Indicate the amount.

Within a certain range, this kind of fruit can be wholesale in large quantities with the same funds.

solve

(3) After investigation, a letter between the highest daily sales volume and the retail price of this kind of fruit sold by a dealer.

As shown in Figure (2), the dealer plans to sell more than 60 kilograms of this fruit every day.

And the retail price of the day remains unchanged, please help the dealer design the purchase and sale plan.

Maximize the profits made on that day.

solve

23.( 1) Solution: Figure ① shows the fruit with a wholesale volume of not less than 20kg and not more than 60kg.

Wholesale according to 5 yuan/kg; ..... 3 points

Figure ② shows that the wholesale volume of this kind of fruit is higher than 60kg, and it can be wholesale according to 4 yuan /kg.

Three points

(2) solution: Get: from the meaning of the question, the function image is shown in the figure.

Seven points

As can be seen from the figure, when the amount of funds meets 240 < w ≤ 300, the same amount of funds can be used.

Wholesale this fruit in large quantities. .................................................... scores 8 points.

(3) Solution 1:

Assuming that the retail price of the day is X yuan, the maximum daily sales can be obtained from the figure.

When m > 60, x < 6.5.

Judging from the meaning of the question, the sales profit is

.......................... 12.

When x = 6, m = 80.

That is, the dealer will wholesale 80kg of this fruit, and the daily retail price will be set at 6 yuan/kg.

Maximum profit of the day 160 yuan ..........................................................................................................................................................

Solution 2:

The maximum daily sales volume is xkg (x > 60).

Then the daily retail price p in Figure ② satisfies:, so

Sales profit ..................... 12 points.

When x = 80, p = 6.

That is, the dealer will wholesale 80kg of this fruit, and the daily retail price will be set at 6 yuan/kg.

Maximum profit of the day 160 yuan ..........................................................................................................................................................

(Jiangxi Province, 2009) 25. As shown in figure 1, in the isosceles trapezoid,, is the midpoint and the intersection point is the intersection point.

(1) Find the distance from point to point;

(2) A point is a moving point on a line segment, passing through it, crossing the line at that point, connecting and setting.

① When the point is on the line segment (as shown in Figure 2), does the shape of the point change? If it does not change, the calculated perimeter; If yes, please explain the reasons;

② When the point is on the line segment (as shown in Figure 3), is there a point that makes it an isosceles triangle? If it exists, request all the values that meet the requirements; If it does not exist, please explain why.

25.( 1) As shown in figure 1, it is marked as a point 1 by a dot.

Yes, the midpoint,

At, ∴ 2 points

That is, the distance from point to point is 3 o'clock.

(2)① When a point moves on a line segment, its shape remains unchanged.

∵ ∴

∵ ∴ ,

Similarly, 4 points.

As shown in fig. 2, this point has ended.

rule

Yes,

∴ circumference = 6 points

② When a point moves on a line segment, its shape will change, but it is still an equilateral triangle.

As shown in Figure 3, when it is completed in, then

Similar to 1,

7 points

∵ is an equilateral triangle, ∴

At this point, 8 o'clock.

As shown in fig. 4, at this time

At this moment,

As shown in fig. 5, when,

Zeyou

Therefore, this point coincides with a right triangle.

At this moment,

To sum up, when or 4 or, is an isosceles triangle. 10.

(Guangzhou, Guangdong, 2009) 25. (The full score of this small question is 14)

As shown in figure 13, the image of quadratic function intersects with X axis at points A and B, and intersects with Y axis at point C (0,-1), and the area of δδABC is.

(1) Find the relation of quadratic function;

(2) Take a point M(0, m) on the Y-axis as the vertical line of the Y-axis, and if the vertical line and the circumscribed circle of ABC have a common point, find the value range of m;

(3) Is there a point D on the image of quadratic function, which makes the quadrilateral ABCD a right trapezoid? If it exists, find the coordinates of point D; If it does not exist, please explain why.

25. (The full score of this short question is 14)

Solution: (1)OC= 1, so q=- 1, from 0.5OC×AB=, AB=,

Let A(a, 0), B(b, 0)AB=b? A= =, the solution is p=, but p

So the analytical formula is:

(2) Let y=0 and solve the equation, so A (0) and B (2 2,0) are in the right triangle AOC.

AC= available, BC= available. Obviously, AC2+BC2=AB2, and triangle ABC is a right triangle. ab blood type

Is the hypotenuse, so the diameter of the circumscribed circle is AB=, so.

(3) Existence, AC⊥BC,① If AC is the base, then BD//AC, the analytical formula of AC is easy to find, that is, y=-2x- 1, and the analytical formula of BD can be set.

For y=-2x+b, when b (2,0) is substituted, the analytical formula of BD is y=-2x+4, and d (0 0,9) is obtained by solving the equations.

② If BC is taken as the radix, then BC//AD, the analytic formula of easy-to-find BC is y=0.5x- 1, and the analytic formula of AD can be set to y=0.5x+b, then

The analytical formula of replacing AD with A (0 0,0) is y=0.5x+0.25, and the equations are solved with D ().

To sum up, there are two points: (9) or ().

(Zhongshan, Guangdong Province, 2009) 22. (The full mark of this question is 9) The side length of the square ABCD is 4, and M and N are two moving points on BC and CD respectively. When m moves on BC, AM and MN remain vertical.

(1) proof: rt △ ABM ∽ rt △ MCN;

(2) Let BM=x and the area of trapezoidal ABCN be y, and find the functional relationship between y and x; When the m point moves to what position, the area of the quadrilateral ABCN is the largest, and the maximum area is calculated;

(3) When the point M moves to what position, Rt△ABM∽Rt△AMN, and find the value of X at this time.

(Harbin, 2009) No.28. (This question is 10)

As shown in figure 1, in the plane rectangular coordinate system, point O is the coordinate origin, quadrilateral ABCO is a diamond, and the coordinate of point A is (-3,4).

Point C is on the positive semi-axis of X-axis, straight line AC intersects Y-axis at point M, and AB intersects Y-axis at point H. 。

(1) Find the analytical formula of straight line AC;

(2) Connect BM, as shown in Figure 2. The moving point P starts from point A and moves to the end point C at a constant speed of 2 units/second along the dotted line ABC. Let the area of △PMB be S(S≠0) and the moving time of point P be t seconds, and find the functional relationship between S and T (the range of independent variable T is required);

(3) Under the condition of (2), when t is what value, ∠MPB and ∠BCO are complementary angles, find the tangent of the acute angle between the straight line OP and the straight line AC at this time.

(Tai 'an City, Shandong Province, 2009) 26 (Full score for this small question 10)

As shown in the figure, in the right-angled trapezoidal ABCD, ∠ ABC = 90, AD‖BC, AB=BC, E is the midpoint of AB, and CE⊥BD.

(1) verification: Be = AD

(2) Verification: AC is the middle vertical line of line segment ED;

(3) Is △ DBC an isosceles triangle? And explain why.

26. (Full score for this small question 10)

Proof: (1) ∠ ABC = 90, BD⊥EC.

∴∠ 1 and ∠3 are complementary, ∠2 and ∠3 are complementary,

∴∠1= ∠ 2 ...........................................1min.

∠∠ABC =∠DAB = 90 °, AB = AC

∴△ bad△ CBE ................................................................... 2 points.

∴ ad = be ... 3 points.

(2)E is the midpoint of AB,

∴EB=EA

The score is from (1) ad = be: AE = ad ... 5 points.

700 BC

∴∠7=∠ACB=45

∵∠6=45

∴∠6=∠7

From the properties of isosceles triangle, we can get: EM=MD, AM⊥DE.

That is, AC is the middle vertical line of line segment ED. Seven points

(3)△DBC is an isosceles triangle (CD = BD) with .............................. score of 8.

The reason for this is the following:

From (2): CD=CE

From ( 1): CE=BD

∴CD=BD

△ DBC is an isosceles triangle. .......................... 10.

Weihai City (2009) 25. (12)

The image of a linear function intersects the axis and the axis respectively, and intersects the image of an inverse proportional function respectively. The intersection points are axis, axis and vertical foot respectively, and they are intersection points and connecting lines.

(1) If this point is on the same branch of the inverse proportional function image, as shown in figure 1, try to prove that:

① ;

② .

(2) If the points are on different branches of the inverse proportional function image, as shown in Figure 2, is it equal to? Try to prove your conclusion.

25. (The full score of this short question is 12)

Solution: (1)① axis, axis,

A quadrilateral is a rectangle.

Shaft, shaft,

A quadrilateral is a rectangle.

Shaft, shaft,

Quadrilaterals are all rectangles. 1 min.

,

,

.

.

,

,

.2 points

② Known by (1).

.

.4 points

,

.5 points

.

.6 points

Axis,

A quadrilateral is a parallelogram.

.7 points

In the same way.

.8 points

(2) still equal to 0.9 points.

,

,

Say it again,

. 10 point

.

.

,

.

.

. 1 1.

Axis,

A quadrilateral is a parallelogram.

.

In the same way.

. 12 point

(Yantai, 2009) No.26. (The full mark of this question is 14)

As shown in the figure, the parabola intersects the axis at two points, intersects the axis at point C, and passes through that point. The axis of symmetry is a straight line, and the vertex is.

(1) Find the function expression corresponding to parabola;

(2) Is there such a point on the parabola that a straight line intersects the axis through two points, so that the quadrilateral with this point as the vertex becomes a parallelogram? If it exists, request the coordinates of the point; If it does not exist, please explain the reason;

(3) Let the intersection of the straight line and the Y axis be, take any point on the line segment (not coincident with it), and try to judge the shape through the intersection of the three-point circle with the straight line at this point, and explain the reasons;

(4) When it is any point on a straight line, does the conclusion in (3) hold? Please write the conclusion directly.

26. (The full mark of this question is 14)

Solution: (1) According to the meaning of the question, score 2 points.

solve

The function expression corresponding to parabola is .3 points.

(2) existence.

In the middle, make, get.

Make, get.

, , .

Say vertex again. 5 points

The expression that is easy to find a straight line is.

In the middle, make, get.

, .6 points

In the middle, make, get.

.

The quadrilateral is a parallelogram, at this point .8 points.

(3) It is an isosceles right triangle.

Reason: middle, make, get, make, get.

The intersection of a straight line and a coordinate axis is.

, .9 points

More, ... 10 points

As shown in the figure. 1 1.

And ... It is an isosceles right triangle. 12 points.

(4) When a point is any point on a straight line, the conclusion in (3) holds. 14 point.

Rizhao, Shandong, 2009. (The full mark of this question is 10)

In the known square ABCD, e is a point on the diagonal BD, if it passes through e, it is EF⊥BD, BC and F, connecting d F, and G is the midpoint of DF, connecting EG and CG.

(1) verification: eg = CG

(2) Rotate △BEF in Figure ① counterclockwise around point B by 45? As shown in Figure ②, take the G point in DF and connect EG and CG. Is the conclusion in (1) still valid? If yes, please give proof; If not, please explain why.

(3) Rotate the △BEF in Figure ① at any angle around point B, as shown in Figure ③, and then connect the corresponding line segments, and ask whether the conclusion in (1) still holds? What conclusion can you draw from the observation? (No proof required)