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

II. Fill in the blanks (5 small questions in this big question, 4 points for each small question, 20 points for * * *):

13.7; 14.① and ③ (or ② and ④); 15.60; 16.； 17. Or

Third, answer the question (this big question is ***8 small questions, ***64 points):

18. (The full mark of this question is 6)

Solution:

= ... 2 points.

= ... 3 points.

= ... 5 points.

= ... 6 points.

19. (The full mark of this question is 6)

Proof: (1) In a word,

∵,

,

, 2 points.

∴△ABC≌△BAD, 3 points.

............................................., 4 points.

(2)∫∠AOC =∠BOD，，，

∴△ AOC△ BOD ................................................... 6 points.

20. (The full mark of this question is 8)

Solution: (1) mode is 5 and mode is.

The median is 5; Four points.

(2) Bar chart; Six points.

(3) ............................................... scored 8 points.

2 1. (The full mark of this question is 8)

Solution: Let one side of the rectangle be xcm and the other side be (18-x) cm .....................................1min.

According to the meaning of the question, the circumference of an isosceles triangle is 36cm, and if the waist length is known as 13cm, its base length is 10cm, and the height on the base of the isosceles triangle is (cm). ............................................................................................

If the area of a rectangle is equal to the area of an isosceles triangle, and x (18-x) =, then

The solution is x =

When x =,18-x =; When x =, 18-x =.

Answer: One side of the rectangle is () cm long, and the other side is () cm long. ......................... is 8 points.

22. (The full mark of this question is 8)

Solution: (1) Because AB is the diameter of ⊙O, ∠ ACB = 90? ................... 1 min.

Because ABC = AQC = 30? , AC=6, AB =12 ........................ 3 points.

(2) From (1), it is known that ∠BAC=60? , AO=6, since AQ is the bisector of ∠BAC,

So ∠ CAQ = ∠ BAQ = 30? , there is ∠ baq = ∠ ABC = 30? ,

So △APB is an isosceles triangle.

Connect PO, then PO is the distance from point P to AB, which is 5 minutes from ....................................................

In Rt△AOP, po = ao tan 30? =2.

Therefore, the distance from point P to AB is 2 ................................................. 6 minutes.

(3) Because ∠ bcq = ∠ baq = 30? , so ∠AQC=∠BCQ, then PQ = CP.

Since AP is the bisector of ∠BAC, ACP = AOP = 90?

Therefore, CP=PO=2 and PQ = 2. ..........................................................................................................................................................

23. (The full mark of this question is 8)

Solution: (1) It is known that warehouse A will transport x tons to place A, then warehouse A will transport (80-x) tons to place B, warehouse B will transport (50-x) tons to place A, and transport (x- 10) tons to place B. .....................

Therefore, y =10x+40 (80-x)+20 (50-x)+30 (x-10) = 3900-20x. ...........................................................................................

(2) According to the known knowledge, 10≤x≤50, 10≤x≤50.

Therefore, when x=50, the total freight is the least, which is 2900 yuan; Seven points

When x= 10, the total freight is the largest, which is 3700 yuan. ..............................................................................................................................................

24. (The full mark of this question is 10)

Solution: (1) It is known that the original parabola passes through the origin o and point a,

Therefore, the expression of the original parabola can be set as ..................................................................................................................................................................

If the formula is obtained, the coordinate of its vertex B is ...................................................................... 2 points.

Because vertex B is on a straight line, we can get ... 3 points by substituting it.

According to the meaning of the question, the coordinates of the parabola vertex obtained after translation are, that is.

Because this point is still on a straight line,

Finishing, because, so, 4 points.

So, the original expression of parabola is. (or) ... 5 points

(2) Method 1: According to (1), the coordinates of this point are,

Judging from the meaning of the question, the coordinate of the point is 6 o'clock.

If it is perpendicular to the y axis in c, if it is perpendicular to the y axis in d, 7 o'clock.

Because, so the area of the delta

= triangle area+trapezoid area-triangle area

=+-

=, ........................8 points.

, the solution is ........................................................... 10.

Method 2: According to (1), the coordinates of this point are,

Judging from the meaning of the question, the coordinate of the point is 6 o'clock.

So, let the expression of a straight line

Solution ... 7 points.

Then the expression of the straight line is.

Let the intersection of a straight line and the X axis be C, then the coordinate of point C is 8 o'clock.

Because, so > 0, > 0, > 0,

therefore

Solution ............................. 10.

25. (The full mark of this question is 10)

Solution: (1) Since the quadrilateral ABCD is a parallelogram, OA = oc.

△AOB and△ △BOC have the same height on the OA and OC sides, so S 1 = S2.

Similarly, S2 = S3, S3 = S4, S4 = S 1, so S 1 = S2 = S3 = S4. ....................................................................................................................

(2) Because the vertical foot of AC⊥BD is O, S 1 = OA ob, S2 = ob oc,

S3 = OC OD, S4 = OD OA, so s1S3 = s2s4 ............................................................. 4 points.

(3) Let the distance from point B to line segment AC straight line be h 1, and the distance from point D to line segment AC straight line be h2.

Then s 1 = OA h 1, S2 = oc h 1, S3 = oc H2, S4 = OA H2,

So there is S1S3 = S2 S4 ............................................................................................ 6 points.

(4) Solution 1:

Given that ∠BAC=∠BDC and ∠AOB=∠DOC, ∠ DCA = ∠ Abd.

When AB and CD are not parallel, they must intersect at one point.

Let's assume that the extension of line segment BA and CD intersect at point E. 。

Given that AC=BD and ∠AEC=∠DEB, so △ AEC △ Deb,

Ae = de, ce = be, so ab = DC, so △ AOB △ doc, then S 1 = S3.

Since S 1S3=S2S4, S 12=S2S4,

Then s = s1+S2+S3+S4 = 2s1+S2+S4 = S2+S4+2 (or = (+) 2); ..................................................................., eight.

When AB is parallel to CD, the base heights of △ABD and △BAC are the same, with S 1+S2=S 1+S4.

Then S2 = S4, because S 1S3 = S2=S4.

So S22= S 1S3, s = s1+S3+2s2 = s1+S3+2 (or = (+) 2). ....................................................................................................

Solution 2:

In △AOB and △DOC, ∠BAC=∠BDC and ∠AOB=∠DOC, then ∠DCA=∠ABD,

So △AOB∽△DOC, then =

Let h3 be the distance from point A to the straight line of line segment BD, and h4 be the distance from point B to the straight line of line segment AC.

The distance from point C to BD line is h5, and the distance from point D to AC line is H6.

S4 = H6 = H3, so =.

Similarly =, then =.

It is known that AC = BD, so =, that is =,

According to the arrangement, S1S2+S3 S4 = S2S3+S1S4, so S2 (S1-S3) = S4 (S1-S3), ……………………………………………………………………….

When S 1≠S3, S2 = S4. Because S 1S3 = S2=S4. ，S22 = S 1S3，

Then s = s 1+S3+2 (or = (+) 2).

When S 1 = S3, similarly, there is S = S2+S4+2 (or = (+) 2). .............................................................................................................................

I don't understand:/tsearch/resource/servlet/portalbar/showresattribute? Community = 00049 & Ampresid = 20080903140438530 #