People's education printing plate eighth grade mathematics final examination questions
A, multiple-choice questions * * * 3 points for each small question, ***30 points * * *
1. Among the following symbols, * * * * can be regarded as an axisymmetric figure.
A.B. C. D。
2. The following calculation is correct * * * *
a.a﹣ 1÷a﹣3=a2 b . * * * * * 0 = 0 c . * * * a2 * * * 3 = a5d . * * * ***﹣2=
3. If the two sides of an isosceles triangle are 3 and 7 respectively, then its circumference is * * * * * *.
A. 17b.15c.13d.13 or17.
4. As shown in the figure, in △ABC, point D is on BC, AB=AD=DC and ∠ B = 80, then the degree of ∠C is * * * *.
30 BC to 40 BC
5. As shown in the figure, in △ABC and △DEF, AB=DE and ∠B=∠DEF, which of the following conditions cannot prove △ ABC △ def * * * *
A.AC∥DF B.∠A=∠D C.AC=DF D.∠ACB=∠F
6. Given that the polynomial x2+kx+ is completely flat, the value of k is * * * * *.
A.65438+b.﹣ 1
7. As shown in the figure: △ABC, ∠ C = 90, AC=BC, AD bisector ∠CAB passes through BC in D, and DE⊥AB in E, AB=6cm, then the circumference of △DEB is * * * *.
A.6 cm B.4 cm C. 10 cm D. None of the above are correct.
8. The result of simplification is * * * * * *
Fax number of b.x-1c.x.
9. A factory now produces 50 more machines a day than originally planned, and it takes the same time to produce 600 machines as originally planned to produce 450 machines. Assuming that the original plan produces X machines on average every day, according to the meaning of the question, the following equation is correct.
A.= B. = C. = D. =
10. As shown in the figure, in △ABC, AQ=PQ, PR=PS, PR⊥AB in R, PS⊥AC in S, then three conclusions ① As = AR; ②QP∑AR； ③ * * * * in △ BPR △ QSP
A. All correct B. Only ① and ② are correct C. Only ① is correct D. Only ① and ③ are correct.
Second, fill in the blanks * * * 4 points for each small question, *** 16 points * * *
1 1. Decomposition coefficient: ax4-9ay2 =.
12. As shown in the figure, in Rt△ABC, D and E are two points on the hypotenuse AB, and BD=BC and AE=AC, then the size of ∠DCE is * * * *.
13. As shown in the figure, ∠ E = ∠ F = 90, ∠B=∠C, AE=AF. The following conclusions are given: ① ∠1= ∠ 2; ②BE = CF； ③△ACN?△ABM； (4) CD = DN。 The correct conclusion is. * * * Fill in the serial numbers of all conclusions you think are correct * * *
14. As shown in the figure, the symmetry points of point P about OA and OB are C and D respectively. Connect CD, cross OA to m, cross OB to n, if CD= 18cm, then the circumference of △PMN is cm.
Three. Answer the question * * * * 74 points * * *
15. Decomposition factor: * * x ﹣1* * * * x ﹣ 3 * *+1.
16. Solve the equation: =.
17. Simplify first, then evaluate: * * * *, and select an appropriate one from the four numbers-2,0, 1, 2 for evaluation.
18. As shown in the figure, EF∨BC, AC shares ∠BAF, ∠ B = 80. Find the degree of ∠ C.
19. As shown in the figure, in a small square grid with a side length of 1 unit length, the vertex of △ABC*** is the intersection of grid lines * * *.
* * *1* * Please draw △A 1B 1C 1 where △ABC is symmetrical about the straight line L;
* * * 2 * * Translate the line segment AC to the left by 3 units, and then by 5 units to draw the translated line segment A2C2, and make a grid △A2B2C2 with it as one side, so that A2B2=C2B2.
20. As shown in the figure, in Rt△ABC, ∠ ABC = 90, point D is on the side of AB, so that DB=BC, crossing point D is EF⊥AC, AC is sent to point E respectively, and the extension line of CB is at point F. 。
Proof: AB=BF.
2 1. From Guangzhou to a city, you can take the ordinary train or the high-speed train. It is known that the driving distance of high-speed rail is 400 kilometers, and the driving distance of ordinary trains is 1.3 times that of high-speed rail.
* * *1* * Find the running distance of the ordinary train;
***2*** If the average speed of the high-speed train * * * km/h is 2.5 times that of the ordinary train * * * km/h, and the time required to take the high-speed train is 3 hours less than that of the ordinary train, find the average speed of the high-speed train.
22. As shown in the figure, point D is on the AB side of △ABC, ∠ ACD = ∠ A. 。
* * *1* * is the bisector DE of ∠BDC ∠, draw it with a ruler from BC to E***, and keep the traces of drawing, without requiring the writing method * * *;
***2*** Under the condition of * *1* * *, it is not necessary to prove the positional relationship between the straight line DE and the straight line AC * * *.
23. As shown in the figure, in △ABC, D is the midpoint of BC, the straight line GF passing through D intersects with AC at F, the parallel line BG intersects with AC at G, DE⊥DF, and AB intersects with E, connecting EG and EF.
*** 1*** Verification: BG = CF
* * * 2 * * Please judge the relationship between BE+CF and EF and explain the reasons.
Reference answer
A, multiple-choice questions * * * 3 points for each small question, ***30 points * * *
1. Among the following symbols, * * * * can be regarded as an axisymmetric figure.
A.B. C. D。
Axisymmetric figure of test point.
According to the concepts of axisymmetric figure and central symmetric figure, the analysis problem is solved.
Solution: a, it is not an axisymmetric figure, but a centrally symmetric figure, which does not meet the meaning of the question;
B, it is not an axisymmetric figure, but a centrally symmetric figure, which does not conform to the meaning of the question;
C, not an axisymmetric figure, but a centrally symmetric figure, which does not conform to the meaning of the question;
D, an axisymmetric figure, meets the meaning of the question.
Therefore, choose: d.
2. The following calculation is correct * * * *
a.a﹣ 1÷a﹣3=a2 b . * * * * * 0 = 0 c . * * * a2 * * * 3 = a5d . * * * ***﹣2=
The negative integer exponential power of the test point; The power of power and the power of products; Zero exponential power.
The analysis can be calculated according to the arithmetic rules of negative integer exponential power and 0 exponential power respectively.
Solution: A, the original formula = A * *- 1+3 = A2, so this option is correct;
B, * * * * 0 =1,so this option is wrong;
C, ***a2***3=a6, so this option is wrong;
D, * * * *-2 = 4, so this option is wrong.
So choose a.
3. If the two sides of an isosceles triangle are 3 and 7 respectively, then its circumference is * * * * * *.
A. 17b.15c.13d.13 or17.
Test the nature of the central isosceles triangle; The trilateral relationship of a triangle.
Because the analysis does not clearly indicate which side is the waist and which side is the bottom, it needs to be divided into: * *1* * when the waist of the isosceles triangle is 3 hours; ***2*** When the waist of the isosceles triangle is 7; Discuss these two cases and get its perimeter.
Solution: ① When the waist of an isosceles triangle is 3 and the base is 7, 3+3.
② When the waist of an isosceles triangle is 7 and the base is 3, the circumference is 3+7+7= 17.
So the circumference of this isosceles triangle is 17.
So choose: a.
4. As shown in the figure, in △ABC, point D is on BC, AB=AD=DC and ∠ B = 80, then the degree of ∠C is * * * *.
30 BC to 40 BC
Properties of isosceles triangle.
Firstly, the number of ∠ADB is obtained according to the nature of isosceles triangle, then the number of ∠ADC is obtained according to the definition of right angle, and a conclusion can be drawn according to the nature of isosceles triangle.
Solution: ∫△ABD, AB=AD, ∠ B = 80,
∴∠B=∠ADB=80，
∴∠ADC= 180 ﹣∠ADB= 100，
AD = CD，
∴∠C= = =40。
Therefore, choose: B.
5. As shown in the figure, in △ABC and △DEF, AB=DE and ∠B=∠DEF, which of the following conditions cannot prove △ ABC △ def * * * *
A.AC∥DF B.∠A=∠D C.AC=DF D.∠ACB=∠F
Congruent triangles's judgment of the test center.
According to congruent triangles's decision theorem, we can get an answer.
Solution: AB = DE, ∠B=∠DEF,
∴ Add AC∨df to get ∠ACB=∠F, which proves △ ABC △ def, so A and D are correct;
When ∠A=∠D is added, the △ ABC △ def can also be proved according to ASA, so B is correct;
But when AC=DF is added, there is no SSA theorem and △ ABC △ def cannot be proved, so C is incorrect;
So choose: C.
6. Given that the polynomial x2+kx+ is completely flat, the value of k is * * * * *.
A.65438+b.﹣ 1
The test center is completely flat.
The first and last terms here are x and the square of these two numbers, so the middle term is twice the product of x plus or minus.
Solution: ∫ Polynomial x2+kx+ is completely flat.
∴x2+kx+ =***x ***2，
∴k= 1，
So choose a.
7. As shown in the figure: △ABC, ∠ C = 90, AC=BC, AD bisector ∠CAB passes through BC in D, and DE⊥AB in E, AB=6cm, then the circumference of △DEB is * * * *.
A.6 cm B.4 cm C. 10 cm D. None of the above are correct.