(1) as shown in the figure, verification: △ ade ∽△ AEP;
(2) Let OA=x and AP=y, find the resolution function of Y about X, and write its domain;
(3) When BF= 1, find the length of the line segment AP.
Solution: (1) Connect OD,
Known conditions prove that OD⊥AB,
∴∠oda=∠dep ∵ep⊥ed
∫od = OE ∴∠ode=∠oed
∴∠ADE=∠AEP
∠∠a=∠∴△ade∠△AEP。
(2)∫∠ABC = 900，AB=4，BC=3，∴AC=5
∵OA=x, it is easy to get OE=OD=, AD= ∴AE=x+
∫△ade∽△AEP∴ That is to say,
∴ 。
As can be seen from this example, the second resolution function uses the conclusion of the first term "△ADE∽△AEP".
Example 2,2001finale of Shanghai senior high school entrance examination paper: It is known that in trapezoidal ABCD, AD‖BC, AD < BC, and AD = 5, AB = DC = 2.
(1) As shown in the figure, p is a point on AD, which satisfies ∠ BPC = ∠ a. 。
1 verification; △ABP∽△DPC
② Find the length of AP.
(2) If point P moves on the edge of AD (point P does not coincide with points A and D), and it satisfies ∠ BPE = ∠ A, the intersection BC of PE is at point E, and the intersection DC is at point Q, then
① When Q is on the extension line of line DC, let AP = X and CQ = Y, find the resolution function of Y about X, and write the definition domain of the function;
② When CE = 1, write the length of AP (there is no need to write the problem-solving process).
Solution: (1)① Prove: ∫∠ABCD = 180-∠A-∠APB, ∠DPC = 180-∠BPC-∞.
② solution: let AP = x, then DP = 5-x, from △ABP∽△DPC, that is, the solution is X 1 = 1, and X2 = 4, then the length of AP is 1 or 4.
(2)① solution: similar to (1)①, it is easy to get △ABP∽△DPQ.
That is to say,
Get ( 1 < x < 4)
② AP = 2 or AP = 3-.
Because the domain of the function in item (2) of this example is difficult, if there is no conclusion that the length of AP in item (1) is 1 or 4, the domain answer can easily get 0 < x < 5. Therefore, the conclusion of item (1) plays a foreshadowing and suggestive role.
The foreshadowing function of the idea of solving the first small problem
Example 3, the last question of the 2003 Shanghai senior high school entrance examination paper: as shown in figure 1, in the square ABCD, AB= 1, arc AC is the arc of a circle with point B as the center and length AB as the radius. Point e is any point on the edge AD (point e does not coincide with points a and d). Passing E is the tangent of the circle where arc AC is located, the intersection DC is at point F, and G is the tangent point.
(1) When ∠DEF=450, the verification point g is the midpoint of the line segment EF;
(2) Let AE=x and FC=y, find the resolution function of Y about X, and write the definition domain of the function;
(3) △D 1EF is obtained by folding △DEF along the straight line EF, as shown in Figure 2. When EF=, discuss whether △AD 1D and △ED 1F are similar. If they are similar, please prove them. If they are not similar, just write a conclusion, not a reason.
Solution, (1) ∠ def = 450 ∴ de = df ∴ ad = DC ∴ AE = fc.
It is easy to prove that the tangent circle B of AD and CD is at point A and point C. According to the tangent length theorem, AE=EG and FC=GF can be obtained.
∴EG=GF, that is, point G is the midpoint of line segment EF.
(2)∵EG=AE=x，FG=CF=y ∴ED= 1-x，FD= 1-y，EF=x+y
In Rt△DEF, (1-x) 2+( 1-y) 2 is derived from ED2+FD2=EF2? = (x+y)2
∴y= (0； 0), then N(R+ 1, r),
Substituting into the expression of parabola, the solution is
② When the straight line MN is below the X axis, let the radius of the circle be r (r >; 0),
Then N(r+ 1, -r),
Substitute the parabola expression and you get ... 7 points.
The radius of a circle is 7 points or ........................
(4) When the Y axis intersects with AG at point Q, the parallel line passing through point P,
G(2, -3) is easily obtained, and the straight line AG is ........................................................................................................................................................
Let P(x,), then Q(x, -x- 1), pq.
Nine points
When △APG has the largest area.
At this point, the coordinates of point p are, ...................................................................................................................................................................
5(08 Guiyang, Guizhou) 25. (The full mark of this question is 12) (There is no answer to this question)
The hotel housekeeping department has 60 rooms for tourists to live in. When the price of each room is 200 yuan per day, the room will be full. Each room will be increased by RMB 65,438+00 per day, and one room will be given as a gift. There are rooms for tourists, and the hotel needs to pay various fees for each room in 20 yuan every day.
Let the daily price of each room increase by RMB. Q:
(1) The daily occupancy of the room (room) is a function of (yuan). (3 points)
(2) The daily room rate of the hotel (yuan) is a function of (yuan). (3 points)
(3) the functional relationship between the daily profit (yuan) and (yuan) of the housekeeping department of this hotel; When the price of each room is several yuan per day, there is a maximum. What is the maximum value? (6 points)