(1) and similar? Take figure 1 as an example to explain the reason;
② Li Ruomi.
① Find the moving speed of the moving point;
② Let the area be (square centimeter), and find the functional relationship;
(3) Explore the quantitative relationship among the three, and take figure 1 as an example to explain the reasons.
The answer is: (1) The reasons are as follows: As shown in Figure 1,
.
(2) cm.
Vertical score, cm.
=4cm。
① The moving speed of the set point is cm/s. 。
As shown in figure 1, when, by (1).
that is
As shown in Figure 2, it is easy to know when it is appropriate.
To sum up, the point velocity is1cm/s.
②
As shown in figure 1, when,
.
As shown in Figure 2, when,,,
.
All in all,
(? ) ?
The reason for this is the following：
As shown? Extend to, manufacture, connect,?
, bisect each other, quadrilateral is parallelogram,.
, , .
Vertical division,.
Similar triangles's judgment of the test center.
The analysis (1) is obtained in the following ways.
therefore
(2) (1) Because the point starts from the point and moves along the ray at the speed of centimeters per second, the time for the point to reach the point from the point along the ray is 4 seconds, which should be discussed in two cases respectively. (2) divided into two situations and.
(3) to explore the quantitative relationship between the three, we should put them in a triangle and pretend that the auxiliary line extends to, make, connect and get, so that in,
5、(20 1 1? Lianyungang, Jiangsu) as shown in the figure, at Rt△ABC, ∠ C = 90, AC=8, BC=6, point P is on AB, AP=2, point E and point F start from point P at the same time, and move at a constant speed along PA and PB to point A and point B at a speed of 1 unit length per second respectively. After point E arrives at point A, it will move at a constant speed immediately. Make a square EFgh with ef as its side, so that it is on the same side as the line segment AB of △ABC. Let the moving time of e and f be t/ s (t > 0), and the overlapping area of square EFGH and △ABC is S. 。
(1) When t= 1, the side length of the square EFGH is 1. When t=3, the side length of the square EFGH is 4.
(2) When 0 < t ≤ 2, find the functional relationship between S and T;
(3) Direct answer: In the whole exercise process, when t is what value, S is the largest? What is the maximum area?
Inspection center: the judgment and nature of similar triangles; Maximum value of quadratic function; Pythagorean theorem; The nature of a square.
Special topic: computational problems; Geometric moving point problem; Classified discussion.
Analysis: (1) When t= 1, it can be obtained that EP= 1, PF= 1, and EF=2 is the side length of a square EFGH; When t=3, PE= 1, PF=3, that is, EF = 4；;
(2) The overlapping parts of square EFGH and △ABC are square, pentagonal and trapezoidal in turn; The answer can be divided into three paragraphs: ① When 0 < t ≤; ② When < t ≤; ③ When < t ≤ 2; Find the functional relationship between s and t in turn;
(3) When 3)t = 5, the area is the largest;
Solution: Solution: (1) When t= 1, PE= 1, PF= 1,
The side length of square EFGH is 2;
When t=3, PE= 1, PF=3,
The side length of square EFGH is 4;
(2): ① When 0 < t ≤,
The functional relationship between s and t is y = 2t× 2t = 4t2;
② When < t ≤,
The functional relationship between s and t is:
y=4t2﹣ [2t﹣ (2﹣t)]× [2t﹣ (2﹣t)]，
=﹣t2+ 1 1t﹣3；
③ When < t ≤ 2;
The functional relationship between s and t is:
y= (t+2)× (t+2)﹣ (2﹣t)(2﹣t)、
= 3t
(3) When t=5, the maximum area is:
s= 16﹣××=；
Comments: This topic examines the problem of moving point function, in which the properties of similarity, square, Pythagorean theorem and so on are applied, which exercises students' ability to solve problems by using comprehensive knowledge.
6.(20 1 1? Huai 'an, Jiangsu) A research team made a special study on the graphic area, and they found the following conclusions:
(1) The ratio of the areas of two equilateral triangles is equal to the ratio of the corresponding heights of the sides;
(2) The ratio of the areas of two triangles corresponding to an angle is equal to the ratio of the products of two sides sandwiching the angle;
…
Now please continue to discuss the following questions, and the above conclusions can be directly applied to the exploration process. (s stands for area)
Question 1: As shown in figure 1, there is a triangular cardboard with ABC, P 1, P2, AB, R 1, R2 and AC on it.
Upon investigation, it is known that = 13 s △ ABC, please prove it.
Question 2: If there is another triangular cardboard, it can be combined with the cardboard in question 1 to form a quadrilateral ABCD, as shown in Figure 2, Q 1, Q2 trisecting DC. Please discuss the quantitative relationship with S quadrilateral ABCD.
Question 3: As shown in Figure 3, P 1, P2, P3, P4 equally divide AB, Q 1, Q2, Q3, Q4 equally divide DC. if
S quadrilateral ABCD = 1, find.
Question 4: As shown in Figure 4, quadrants AB, Q 1, Q2, Q3, DC, P 1Q 1, P2Q2, P3Q3.
The quadrilateral ABCD is divided into four parts with areas of S 1, S2, S3 and S4. Please write the equation containing S 1, S2, S3 and S4 directly.
Answer: Question 1: ∵ P 1, P2 is three points AB, R 1, R2 is three points AC,
∴p 1r 1∨p2r 2∨BC。 ∴△ AP1r1∽△ AP2R2 ∽△ ABC, the area ratio is 1: 4: 9.
∴ =4- 19 seconds △ABC= 13 seconds △ABC
Question 2: Connect Q 1R 1, Q2R2, as shown in the figure. As can be seen from the conclusion of the question 1
∴ = 13 S△ABC，= 13 S△ACD
∴+= 13s quadrilateral ABCD
By: p1,P2 shares AB, R 1, R2 shares AC, Q 1, Q2 shares DC,
We can get p1r1:p2r2 = q2r2: q1r1:2, and p 1r 1∑p2r 2, q2r2.
∴∠p 1r 1a=∠p2r2a,∠q 1r 1a=∠q2r2a.∴∠p 1r 1q 1=∠p2r2 Q2。
It can be seen from conclusion (2) that =.
∴ =+= 13 s quadrilateral ABCD ..
Question 3: Let = A, = B, = C,
From the conclusion of question 2, we can know that A = 13 and B = 13.
A+B = 13 (S quadrilateral ABCD+C) = 13 (1+C).
∫C = 13(A+B+C), that is, c =13 [1+c)+c].
C = 15, that is = 15.
Question 4: S 1+S4 = S2+S3.
Determination of parallelism of test sites, determination and nature of similar triangles, and equivalent replacement.
Analysis problem 1: judging from parallelism and similar triangles, and the property that similar triangles area ratio is the square of the ratio of corresponding sides.
Question 2: From the result of the question 1 and the conclusion (2) given, it can be concluded that the ratio of the area of an angle corresponding to two equal triangles is equal to the ratio of the product of two sides holding this angle.
Question 3: The result of question 2 can be obtained by equivalent substitution.
Question 4: Question 2 shows S 1+S4 = S2+S3 =.
7.(20 1 1? Nantong, Jiangsu) As shown in the figure, it is known that the straight line L passes through point A (1, 0) and the hyperbola Y = m x.
(x > 0) passes through point b (2, 1). The intersection P(p, P- 1) (P > 1) is the plane of the x axis.
The straight line intersects the hyperbola y = m x (x > 0) and y =-m x (x < 0) at m and n points respectively.
(1) Find the value of m and the analytical formula of straight line L;
(2) If the point P is on the straight line Y = 2, it is verified that: △ PMB ∽△ PNA;
(3) Is there a real number p that makes s △ AMN = 4s △ amp? If it exists, request all values of p that meet the conditions; if
Does not exist, please explain why.
Solution: (1) From point B (2, 1) on y = m x, there is 2 =, that is, m = 2.
Let the analytical formula of the straight line L be, from point A (1, 0) and point B (2, 1), we get
Solve it and get it.
The analytical formula of ∴ straight line L is.
(2) point P(p, p- 1) is on the straight line y = 2, and ∴P is on the straight line l, which is the intersection of the straight line y = 2 and l, as shown in figure (1).
According to the conditions, the coordinates of each point are n (- 1, 2), m (1, 2) and p (3, 2).
∴np=3-(- 1)=4,mp=3- 1=2,ap=，
Blood pressure =
∴, ∠ MPB = ∠ NPA, in △PMB and △PNA.
∴△PMB∽△PNA。
(3)S△AMN=. The following points are discussed:
? When 1 < P < 3, extend MP across the x axis to q, as shown in Figure (2). Let the straight line MP be
solve
Then the straight MP is
When y = 0, x =, that is, the coordinate of point Q is (,0).
Then,
If there is 2 = 4, solve, p = 3 (disagree, give up), p =.
? When p = 3, see figure (1) s △ amp = = s △ AMN. This is beside the point.
? When p> is at 3 o'clock, extend the intersection of PM and X axis to Q, as shown in Figure (3).
At this point, S△AMP is greater than? The triangle area S△AMN when p = 3. So there is no real number p, so s △ AMN = 4s △ amp.
To sum up, when p =, s △ AMN = 4s △ amp.
Inverse proportional function of test center, linear function, undetermined coefficient method, binary linear equations, Pythagorean theorem, similar triangles unary quadratic equation.
Analytic (1) Substitute the coordinates of point B (2, 1) into y = m x to get the value of m, and use the undetermined coefficient method to solve the binary linear equations to get the analytical expression of straight line L.
(2) The point P(p, p- 1) is on the straight line Y = 2, which actually means that the point is the intersection of the straight line Y = 2 and L, so it is necessary to prove that △PMB∽△PNA is only proportional to the corresponding line segment.
(3) The position of point P should be considered first. In fact, when P = 3, it is easy to find out S△AMP = S△AMN. When p>3, note that when p = 3, S △ amp is larger than the triangle area, so it is larger than S △ AMN. So as long as we mainly study the situation when 1 < P < 3. After making the necessary auxiliary lines, first find the equation of the straight line MP, then find the coordinates of each point (expressed by P), then find the expression of the area, and then find the value of P after substituting S △ AMN = 4S △ AMP.
8.(20 1 1? Suzhou, Jiangsu) It is known that the image of quadratic function intersects with X axis at point A and point B, and intersects with Y axis at point C, and point D is the vertex of parabola.
(1) As shown in Figure ①, connect AC and fold △OAC along AC straight line. If the corresponding point O' of point O just falls on the axis of symmetry of parabola, find the value of number A;
(2) As shown in Figure ②, in the square EFGH, the coordinates of points E and F are (4,4) and (4,3) respectively, and the edge HG is located on the right side of the edge EF. After exploration, Kobayashi found a correct proposition: "If point P is any point on edge EH or edge HG, then the four line segments of PA, PB, PC and PD cannot be connected with the four sides of any parallelogram. Please actively explore and write the exploration process;
(3) As shown in Figure ②, when point P is on the parabola axis of symmetry, let the ordinate t of point P be a constant greater than 3. Is there a positive number A that makes four line segments PA, PB, PC and PD correspond to four sides of a parallelogram (that is, these four line segments can form a parallelogram)? Please explain the reason.