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The final exam questions of the first volume of mathematics in the second day of junior high school
First, multiple-choice questions (4 points per question)

1. Among the following four patterns, () is an axisymmetric pattern.

1。

2. In the following operations, the correct result is ().

A.B.

C.D.

3. It is known that the relationship between,,, then,, and is ().

A.& gt& gtb . >; & gtC. & ltd . >; & gt

4. As shown in the figure, A, B, C, D, E and F are six points on the plane, so the degree of ∠A+∠B+∠C+∠D+∠E+∠F is ().

180

5. The following groups of line segments with different lengths can form a triangle ()

A. 1.5 cm, 3.9 cm, 2.3 cm B.3.5 cm, 7. 1 cm, 3.6 cm.

C.6 cm, 1 cm, 6 cm D.4 cm, 10 cm, 4 cm.

6. As shown in the figure, the bisector of ∠BAC and ∠CBE intersects at point P, BE=BC, PB and CE intersect at point H, PG∨AD intersects at point BC and AB intersects at point G, and the following conclusions are drawn: ① GA = GP; ② ; ③BP bisects CE vertically; ④FP = FC; The correct judgment is ()

A. only 12B. Only 34C. Only 134D. 12344.

7. After △ABC ∠A=∠D △DEF, AB=DE, ∠A=∠D. is added with the following conditions, it cannot be judged that △ ABC △ def is ().

A.BC=EF B.∠B=∠E C.∠C=∠F D.AC=DF

8. If is, the value of is ()

A.B. C. D。

9. If both x and y in the score are enlarged by 2 times, the value of the score is ().

A. Invariant B. Zoom in 2 times C. Zoom in 4 times D. Zoom out 2 times

10. Students in Class A and Class B plant trees. It is known that Class A plants 5 more trees every day than Class B. The number of days for Class A to plant 80 trees is equal to the number of days for Class B to plant 70 trees. If Class A plants X trees every day, the equation listed according to the meaning of the question is ().

A.B. C. D。

1 1. The Pythagorean school in ancient Greece called the numbers 1, 3,6, 10 … as "triangular numbers" and the numbers 1, 4,9, 16 … as "square numbers".

a . 20 = 6+ 14 b . 25 = 9+ 16 c . 36 = 16+20d . 49 = 2 1+28

12. As shown in the figure, points P and Q are the moving points on the AB side and BC side of the equilateral △ABC with a side length of 4cm (where P and Q are not coincident with the end points). Point P starts from vertex A at the same time, and point Q starts from vertex B at the same time, and their velocities are all1cm/s. If the connecting line AQ and CP intersect at point M, points P and Q will simultaneously block at △ abq △ (3) The number of ∠ CMQ is always equal to 60; (4) When second or second, △PBQ is a right triangle. The correct conclusion is ()

1。

Two. Fill in the blanks (4 points for each question)

13. In a right triangle, one acute angle is 50 and the other acute angle is 50.

14. As shown in the figure, in △ABC, ∠ c = 90, and the bisector BD of △ ABC intersects with AC at point D. If BD= 10cm and BC=8cm, the distance from point D to straight line AB is _ _ _ _ _ _ _ _.

15. If x+y=-4 and x-y=8, then the value of the algebraic expression x2-y2 is.

16. Observe the following equations:,,, …, and calculate according to the law you found: = _ _ _ _ _ (n is a positive integer).

17. As shown in the figure, in the plane rectangular coordinate system, the coordinates of vertices A, B and C of right-angle ABCD are (0, 0), (20, 0) and (20, 10) respectively, and there are moving points M and N on line segments AC and AB respectively, so when BM+MN is minimum, then

18. If the score is equal to 0, the value of is _ _.

Three. Calculation questions (7 points for each question)

19. Calculation: (﹣)

20. Solve the equation:

Iv. Answer questions (2 1-24 10 for each question, 25-26 12 for each question)

2 1. Simplify first and then evaluate.

22. As shown in the figure, in △ABC, AB=AC, points D, E and F are next to AB, BC and AC respectively, and BE=CF and BD=CE.

(1) proves that △DEF is an isosceles triangle;

(2) When ∠ a = 40, find the degree of ∠DEF;

23. As shown in the figure, △ACB and △ECD are isosceles right triangles, ∠ ACB = ∠ ECD = 90, and point D is a point on the side of AB. If AB= 17 and BD= 12,

(1) verification: △ BCD △ ace;

(2) Find the length of DE.

24. As shown in the figure, in △ABC, AB=BC, and point D is on the extension line of AB.

(1) Draw with a ruler as required, and mark the corresponding letters on the drawing (keep drawing marks, don't write).

① Make the bisector of ∠CBD;

② The middle vertical line of BC side intersects BC side at point E, connecting AE and extending bisector of ∠CBD at point F. 。

(2) From (1): The positional relationship between BF and side AC is.

25. A company plans to build a Hope Primary School for poor mountainous areas. Two engineering teams, A and B, submitted their bidding plans. If the project is completed independently, the time spent by Team A is 1.5 times that of Team B; If Team A and Team B cooperate to complete this project, it will take 72 days.

(1) How many days does it take for Team A and Team B to complete the school-building project alone?

(2) If the construction is carried out by Team A alone, the average daily cost is 8,000 yuan. In order to shorten the construction period, the company chose team B, but the total construction cost should not exceed that of team A. What is the maximum daily construction cost of team B?

26. 1. problem situation: place a pair of right triangles (Rt△ABC and Rt△DEF) as shown in figure 1, where ∠ ACB = 90, CA=CB, ∠ FDE = 90, and O is the midpoint DF⊥AC and point D.

Exploration and demonstration: Xiaoyu showed the following correct answers:

Solution: OM=ON, which is proved as follows:

Connect CO, then CO is the center line on the side of AB,

Ca = cb, ∴CO is the angular bisector of ∞∠ACB. (According to 1)

∵OM⊥AC, ON⊥BC, ∴OM=ON. (radix 2)

Reflective communication:

(1) The "basis 1" and "basis 2" in the above proof process refer to:

According to1:;

Foundation 2:.

(2) Do you think differently from Xiaoyu? Please write down your proof process.

Extension extension:

(3) Translate Rt△DEF in figure 1 to the positiON shown in figure 2 along the direction of ray BA, so that point D falls on the extension line of BA, the extension line of FD and the extension line of CA intersect at point M vertically, and the extension line of BC intersects at point N vertically with DE, connecting OM and ON. Try to judge the quantitative relationship and positional relationship between line segment OM and ON, and write the proof process.

Reference answer

1.C。

2.D。

3.A。

4.B

5.C

6.D。

7.A

8.B

9.A。

10.D

1 1.D

12.A。

13.40 .

14.6 cm

15.-32.

16.。

17.( 12,6).

18.6.

19.x﹣ 1

20.,

2 1.-3.

22.( 1)∵AB=AC

∴∠B=∠C

BE=CF,BD=CE。

∴DE=FE

△ def is an isosceles triangle.

(2)∵

∴∠BDE=∠CEF

∠∠A = 40

∴∠B =∠C =70

∴∠BDE+∠BED= 1 10

∴∠CEF+∠BED= 1 10

∴ .

23.( 1) It is proved that △ACB and △ECD are isosceles right triangles.

∴AC=BC,EC=DC.

∵∠ace=∠dce﹣∠dca,∠bcd=∠acb﹣∠dca,∠acb=∠ecd=90,

∴∠ACE=∠BCD.

In △ACE and △BCD,

∴△ace≌△bcd(sas);

(2) 13.

24. The positional relationship between BF and side AC is parallel.

25.( 1) It takes 180 days for Party A to complete the school building project alone, and 120 days for Party B to complete the school building project alone;

(2) The average daily construction cost of Team B is as high as 65,438 yuan+200,000 yuan.

26.( 1) Solution: Therefore, the answer is: the three lines of an isosceles triangle are unified (or the bisector of the top corner of an isosceles triangle, the midline on the bottom edge and the height on the bottom edge coincide), and the distance from the point on the bisector of the corner is equal to both sides of the corner.

(2) prove: CA = CB,

∴∠A=∠B,

O is the midpoint of AB,

∴OA=OB.

∵DF⊥AC,DE⊥BC,

∴∠AMO=∠BNO=90,

* in △OMA and △ONB

,

∴△OMA≌△ONB(AAS),

∴OM=ON.

(3) Solution: OM=ON, OM⊥ON. The reasons are as follows:

Connect OC,

∠∠ACB =∠DNB,∠B=∠B,

∴△BCA∽△BND,

∴ = ,

AC = BC,

∴DN=NB.

∫∠ACB = 90 degrees,

∴∠NCM=90 =∠DNC,

∴MC∥DN,

And ∵DF⊥AC,

∴∠DMC=90,

That is ∠ DMC =∠ MCN =∠ DNC = 90,

∴ Quadrilateral DMCN is a rectangle,

∴DN=MC,

∠∠B = 45,∠DNB=90,

∴∠3=∠B=45,

∴DN=NB,

∴MC=NB,

∫∠ACB = 90, O is the midpoint of AB, AC=BC,

∴∠ 1 = ∠ 2 = 45 = ∠ b, OC=OB (the center line of the hypotenuse is equal to half of the hypotenuse).

In △MOC and △NOB.

∴△MOC≌△NOB(SAS),

∴OM=ON,∠MOC=∠NOB,

∴∠MOC﹣∠CON=∠NOB﹣∠CON,

That is ∠ mon = ∠ BOC = 90,

∴OM⊥ON.