Life System of Teaching and Research Section of Qingshan District Education Bureau 20 10, 10

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17

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20

2 1

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25

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Multiple choice answer sheet

L title123455678910112 score.

I answered

Choose one first (this big question * * 12 small questions, 3 points for each small question, 36 points for * * *).

The following questions are accompanied by four alternative answers, one and only one of which is correct. Please fill in the code of the correct answer in the corresponding question number on the answer sheet above.

1, real number -2, 0.3,,, and the number of irrational numbers is ().

A.2 B.3 C.4 D.5

2, the following "QQ expression" belongs to the axisymmetric graphics is ().

3. As shown in the figure, △ ABC △ EFD, ∠B and ∠F are corresponding angles, then ()

A.AB=DE，AC=EF，BC=DF B. AB=DF，AC=DE，BC=EF

C.AB=EF，AC=DE，BC=DF D.AB=EF，AC=DF，BC=DE

4. The coordinate of point P(2, -3) about the axis symmetry point of Y is ().

A.(2，3) B.(-2，-3) C.(-2，3) D.(-3，2)

5. If the formula is meaningful in the real number range, the value range of x is ().

A.x & gt-5 b . x & lt； -5c . x≦-5d . x ≥- 5

6, the following four conditions, can prove that two right triangle congruence is ().

A. the two acute angles are equal. A right-angled side is equal.

C. the hypotenuse is equal. D. Two right-angled sides are equal.

7, the following properties, isosceles triangle and right triangle does not necessarily have is ().

A. the sum of two sides is greater than the third side. The bisector of an angle is perpendicular to the opposite side of the angle.

C. the sum of two acute angles is equal to 90 D, and the sum of internal angles is equal to 180.

8. As shown in the figure, in △ABC, AB=AC, ∠ A = 36, BD and CE are bisectors of △ABC and △BCD respectively, then the isosceles triangle in the figure has ().

A.5 B.4

C.3 d.2

9. The following patterns are made by splicing isosceles right triangles with equal hypotenuse according to certain rules.

Yes According to this rule, there are () triangles in the ninth pattern that can be congruent with the triangles in the first pattern. A 49 B.64 C.8 1。 D. 100

10, as shown in the figure, there is a little d in △ABC, and DA=DB=DC, if ∠ dab = 20.

∠ DAC = 30, then∠∠ BDC is ()

A. 100

1 1. As shown in the figure, AE and AF belong to BC and CD respectively in quadrilateral ABCD.

The vertical line, ∠ EAF = 80, ∠ CBD = 30, then ∠ADC degree is ().

A.45 B.60

c . 80d 100

12, as shown in the figure, it is known that △ABE is an equilateral triangle, and BC divides ∠ Gbe and DF∨AB equally. The conclusions are as follows: ①△BGC is an equilateral triangle; ②BO+OC = GO； ③ Boping ∠ AOG; ④AF-EF=BF, which holds ()

A.①②③④ B.①②④

C.①②③ D.①③

Two, fill in a fill (3 points for each question, *** 12 points)

13、 =_____, =____, =____

14. As shown in the figure, the symmetrical points of point P relative to OA and OB are point C and point D respectively, which connect CD and intersect OA and OB at points M and N respectively. If the circumference of △PMN is 8 cm, the length of CD is _ _ _ _ _ cm.

15, as shown in the figure, AB=AC, and the condition to be added proves △ ADB △ ADC.

It cannot be _ _ _ _ _ _ (only one of them is written).

16, as shown in the figure, in △ABC, the coordinates of point A are (o, 1) and the coordinates of point B are

(3, 1), and the coordinates of point C are (4,3). If △ABD and △ABC are complete,

Wait, so the coordinates of point D are _ _ _ _ _.

Third, solve the following problems (this big question has 9 small questions, ***72 points)

17. (6 points for this question) Calculation: (+)-3

18 (6 points in this question) As shown in the figure, C is the midpoint of the AB line, CD bisects ∠ACE, CE bisects ∠BCD, and CD = CE.

Verification: ACD?△BCE

19 (6 points in this question) If m=-+4x, find the arithmetic square root of m. ..

20. (7 points in this question) As shown in the figure, CD⊥AB, vertical foot is D, ∠ ACB = 90, ∠ A = 30.

Verification: BD= AB

2 1, (7 points in this question) As shown in the figure, the vertex coordinates of △ABC are known as: A (-5,4), B (-3, 1), C (- 1, 3).

(1) Draw a graph △A'B'C', in which △ABC is symmetrical about the straight line x=2 (denoted as Ⅲ);

(2) the coordinates of the symmetry point. A is _ _ _ _ relative to the straight line M, and the coordinate of the symmetry point of point B relative to the X axis is _ _ _ _ _ _ _ _;

(3) the area of △ a' b' c' is _ _ _ _ _ _ _ _

22. (8 points in this question) As shown in the figure, it is known that AC = AE, FC = Fe, ∠ ABC = ∠ ade = 90, BC and DE intersect at point F, connecting CD and EB.

(1) verification: △ ABC △ Abe;

(2) verification: af ⊥ BD.

23.( 10) Students in Class 8 (1) have math activity classes, and divide a corner with a square (as shown in the figure). The following scheme is designed:

(I)∠AOB is any angle. Place the right-angled vertex P of the square between rays OA and OB, move the square so that the same scales on both sides of the square coincide with M and N, that is, PM=PN, and the ray OP passing through the vertex P of the square is the bisector of ∠AOB.

(ii)AOB is any angle. Let OM = be on the sides OA and OB respectively, let P be the right-angled vertex of the square between rays OA and OB, move the square so that the same scales on both sides of the square coincide with M and N, that is, PM=PN, and the ray OP passing through the vertex P of the square is the bisector of ∠ AOB.

(1) Are both Scheme (1) and Scheme (2) feasible? Please prove the feasible scheme;

(2) In the case of scheme (i) where PM = PN, continue to move the square, and at the same time make PM⊥OA and PN⊥OB. Is this scheme feasible? Please explain the reason. (0

24. (This question 10) As shown in the figure, it is known that in the equilateral triangle ABC, points D, E and F are the midpoint of three sides AB, AC and BC respectively.

M is the moving point on the straight line BC, and △DMN is an equilateral triangle (when the position of point M changes, △DMN moves as a whole).

(1) As shown in Figure ①, when point M is on the left of point B, the quantitative relationship between en and MF is _ _ _ _ _ _ _ _;

(2) As shown in Figure ②, when point M is on BC, other conditions remain unchanged. (1) Is the quantitative relationship between en and MF still valid? If yes, please use Figure ② to prove it; If not, please explain the reasons;

(3) If point M is on the right side of point C, please draw the corresponding figure in Figure ③ to judge the relationship between en and (1).

Is the quantitative relationship of MF still valid? Please write the conclusion directly, without proof.

25. (This question 12 points) As shown in the figure, in the plane rectangular coordinate system, △AOB is an isosceles right triangle, OA-AB.

(1) As shown in the figure, draw an axisymmetric graph of △AOB about BO △A 1OB. If A (-3, 1), find the coordinates of point A 1:

(2) When △AOB rotates around the origin O to the position shown in the figure, the AB and Y axes intersect at point E, and AE = be. AF ⊥ Y axis intersects BO at F, connects EF, and makes AG//EF intersect Y axis at G. Try to judge the shape of △AGE and explain the reason;

(3) When △AOB rotates around the origin O to the position shown in the figure, if A (3), C is a point on the X axis, OC=OA, ∠ BOC = 15, and P is a point on the Y axis, if P passes, PN⊥AC is in N, and PM ⊥.

Qingshan district 2010-2011the first semester midterm of grade eight.

Grading standard of mathematics test paper

Choose one first (this big question * * 12 small questions, 3 points for each small question, 36 points for * * *).

The title is123455678911112.

Answer a c c b b b b b b b c b c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c

Two, fill in a fill (3 points for each question, *** 12 points)

13,4,-2,314,815, ∠B=∠C or ∠ADB=∠ADC (only one is required).

16, (4,-1) or (-1, 3) or (-1,-1)

Third, solve the following problems (this big question has 9 small questions, ***72 points)

17, (subject 6 points) calculation:

= 3+ 1+...4 points

=...6 points

18, which proves that:

∫C is the midpoint of line AB.

∴ AC = BC... 1 point

∫CD aliquot ∠ACE, CE aliquot ∠BCD

∴∠ACD=∠ECD, BCE = ECD...3 points.

∴∠ ACD = ∠ BCE...4 points.

In △ACD and △BCE

∴△ Automatic Call Distributor△ BCE (SAS) ... 6 points.

19, solution: ∫x- 1≥0, 1- x≥0.

∴ x ≥ 1, x ≤ 1...2 points.

∴ x = 1...3 points.

= 4 ...4 points

The arithmetic square root of ∴ is 2...6 points.

20. Proof: ∫∠A = 30, ∠ ACB = 90.

∴∠ B = 60, BC = AB...3 points.

You are ∵CD⊥AB

∴∠BDC=90

∴∠ BCD = 30...5 points.

BC ∴BD=

= × AB

= ab...7 points

2 1, solution:

(1) as shown in the figure: △ a ′ b ′ c ′ is the drawing.

..... 2 points

(2)(9, 4), (7,- 1) ...5 points.

(3) 5 ...7 points

22、

(1) Proof: In △ACF and △AEF,

∴△ACF≌△AEF(SSS)

∴∠ ACB = ∠ AEF...2 points.

In △ABC and △ABE,

∴△ ABC△ Abe (AAS) ... 4 points.

(2) from ( 1): CF=EF，△ABC?△ABE。

∴BC=DE

∴ Blast furnace = DF...5 points.

In Rt△ABF and Rt△ADF.

∴Rt△ABF and RT△ ADF (HL)...6 points.

∴∠ Air base = ∠ AFD...7 points.

∴AF⊥BD (combination of three lines at the bottom of an isosceles triangle) ... 8 points.

23. Solution: (1) Scheme (Ⅰ) is not feasible, and Scheme (Ⅱ) is feasible ... 1 score.

Proved as follows:

At △OPM and △OPN,

∴△ OPM△ OPN (SSS) ... 4 points

∴∠AOP=∠BOP (the angles corresponding to congruent triangles are equal) ... 5 points.

(2) When ∠AOB is a right angle, the scheme is feasible ..... 6 points.

At this time ∠ AOB = ∠ ∠AOB =∠PMO =∠MPO =∠PNO = 90°, and the sum of the internal angles of the quadrilateral is equal to 360 ... 7 minutes.

And PM⊥OA,PN⊥OB,

PM=PN。

∴OP is the bisector of∠ ∠AOB. (The points with equal distance to both sides of the angle are on the bisector of this angle) ... 8 points.

When ∠AOB is not a right angle, this scheme is not feasible, at this time ∠ AOB+∠ PMO+∠ MPO+∠ PNO < 360.

..... 10 point

24 、( 1)EN = MF； ..... 2 points

(2) established. The evidence is as follows:

Link DE...3 points.

∫△ABC is an equilateral triangle,

∴AB=AC=BC,∠A=60

AD = AB，AE= AC。

∴AD=AE

Delta ade is an equilateral triangle.

∴ DE = AD = BD ①...4 points.

∠ADE=60

It can also be proved that ∠ BDF = 60.

∴∠MDF+∠BDM=60

△DMN is an equilateral triangle.

∴dn = 2 marks ... 5 points

∠MDN=60

∴∠EDN+∠BDM=60

∴∠ EDN =∠ MDF ③...6 points.

From ① ② ③ :△ DNE△ DMF (SAS)

∴ en = MF...7 points.

(3) Drawing correctly (connecting line segment ne)...9 points.

MF=NE still holds ... 10 points.

25、

(1) solution: as shown in the figure: △a 10b is the drawn axisymmetric figure ... 1 min.

Make the AC⊥x axis in C and the A 1D⊥x axis in D through A.

∫A(-3， 1)

∴AC= 1,OC=3

OA = AB，∠BAO=90

∴∠BOA=45

∴∠BOA 1=45

∴∠AOA 1=90

∴∠AOC+∠A 1OD=90

∠∠AOC+∠OAC = 180-∠ACO = 90。

∴∠CAO=∠A 1OD

∠∠ACO =∠ODA 1 = 90。

AO=A 1O

∴△ ACO△ ODA1... 3 points

∴AC=OD= 1,OC=A 1D=3

∴ A 1, (1, 3)...4 points.

(2)△AEG is an isosceles triangle ... 5 points

Proof: the extension line from B to BH⊥AB to B to AF is in H.

∠∠OAE =∠ABH = 90

∠AOE =∠ Ba =90 -∠OAH

OA=AB

∴△AEO?△ BHA ... 6 points.

∴AE=BH=BE,∠AEO=∠BHA

∠∠EBF =∠HBF = 45。

BF=BF

∴△BEF≌△BHF(SAS)

∴∠ BHF = ∠ BEF...7 points.

∫AG∨EF

∴∠EAG=∠BEF

∴∠EAG=∠AEG

∴AG=EG

That is, △AEG is an isosceles triangle ... 8 points.

(3)PO+PN-PM=3.

Solution: Make the AL⊥x axis pass through A at L to connect AP and PC...9 points.

∫A(，3)

∴ Al = 3... 10 point

∫∠AOC = 45+ 15 = 60

OC=OA

△ AOC is an equilateral triangle.

∴ ao = co = AC... 1 1.

∵

It's also VIII

∴

∴ Po+PN-PM = Al = 3... 12 points.