Test questions and answers in chapter 1-2 of the first volume of the eighth grade mathematics of the new education edition.

Chapter 1 1 of the eighth grade mathematics triangle test questions.

Fill in the blanks.

1. Among the three external angles of the triangle, there are at most _ _ _ obtuse angles and at most _ _ _ acute angles.

2. When building houses, triangular structures are often used. Mathematically, _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

3. Form a triangle with three line segments _ _ _ _ _, with the lengths of 200px, 225px and 250px respectively. (Fill in "Can" or "Can't")

4. In order to prevent the pentagonal wooden frame from deforming, at least _ _ _ _ _ pieces of wood should be nailed.

5. It is known that in △ABC, ∠ A = 40, ∠ B-∠ C = 40, then ∠ B = _ _ _ _ _ _.

6. As shown in figure 1, AB∥CD, ∠ A = 45, ∠ C = 29, then ∠ E = _ _ _ _ _

( 1) (2) (3)

7. As shown in Figure 2, ∠ α = _ _ _ _.

8. The sum of the internal angles of a regular decagon is equal to _ _ _ _ _, and each internal angle is equal to _ _ _ _ _.

9. If the sum of the inner angles of a polygon is half of the sum of the outer angles, then its number of sides is _ _ _ _ _.

10. If you combine a regular triangle with a square with the same side length, if you use two squares, you need _ _ _ regular triangles to splice.

1 1. The circumference of an isosceles triangle is 500px, and the length of one side is 150px, so the length of the bottom side is _ _ _ _ _.

12. If the sum of the internal angles of a polygon is 1260, then a vertex of this polygon has _ _ _ _ diagonal lines.

13. As shown in Figure 3, * * * has _ _ _ _ triangles, among which _ _ _ _ _ _ _ _ _ _ _ _.

14. As shown in Figure 4, ≈ A+≈ B+≈ C+≈ D+≈ E = _ _ _ _ _.

(4) (5) (6)

Second, multiple choice questions.

15. The following statement is wrong ().

A. Three high lines, three middle lines and three bisectors of an acute triangle intersect at one point respectively.

B. An obtuse triangle has two high lines outside the triangle.

C. A right triangle has only one high line.

Any triangle has three high lines, three middle lines and three angular bisectors.

16. Among the following regular polygon materials, the one that cannot be used for paving the ground alone is ().

A. regular triangle B. regular quadrangle C. regular pentagon D. regular hexagon

17. As shown in Figure 5, in △ABC, D is on AC, connected to BD, and ∠ABC=∠C=∠ 1, ∠A=∠3, then the degree of ∠A is ().

A.30 B.36 C.45 D.72

18.d is a point in △ABC, so the following conclusion is wrong ().

A.BD+CD & gt； BC B.∠BDC >∠A C.BD >CD D . a b+ AC & gt； BD+CD

19. A regular polygon with an internal angle equal to 144 is a regular () polygon.

a . 8 b . 9 c . 10d . 1 1

20. As shown in Figure 6, BO and CO are bisectors of ∠ABC and ∠ACB, respectively, and ∠ A = 100, then the degree of ∠BOC is ().

120 D. 140

2 1. If the sum of the inner angles of a polygon is k times the sum of the outer angles, then the number of sides of the polygon is ().

a . k . b . 2k+ 1 c . 2k+2d . 2k-2

22. As shown in the figure, there is a parallelogram in a rectangle with a length of 125px and a width of 75px, so the area of the parallelogram is ().

a . 175 px2 b . 200 px2 c . 225 px2 d . 250 px2

Third, answer the question.

23. As shown in the figure, in △ABC:

(1) Draw high AD and medium AE beside BC.

(2) If ∠ b = 30 and ∠ ACB = 130, find the degrees of ∠BAD and ∠CAD.

24. As shown in the figure, BE bisects ∠ABD, DE bisects ∠CDB, and the point E where BE and DE intersect on AC. If ∠ Bed = 90, please explain AB∨CD.

25. As shown in the figure, the straight line AD and BC intersect at O, AB∥CD, ∠AOC = 95, ∠ B = 50, and find ∠ A and ∠ D. 。

26.( 1) If the sum of the internal angles of a polygon is 2340, find the number of sides of the polygon.

(2) Every outer corner of a polygon is equal. If the ratio of the degrees of its inner angle to its outer angle is 13: 12, find the number of sides of the polygon.

Fourth, the proof questions

27.(4 18) As shown in the figure, in △ABC, the bisector of ABAC, ∠ABC and the bisector of external angle ∠ACF intersect at point P, PD∨BC, D is on AB, and PD is on AC and E is proved: DE = BD-CE.

28.(279) As shown in the figure, E is a point on the extension line of CA, F is a point on AB, and D is a point on the extension line of BC. Please explain: ∠ 1

V. Answering questions

29.(462) Xiaoming is known to have two pieces of wood, the lengths of which are 50px and150px respectively; ; Xiao Wang has two pieces of wood, the lengths of which are 100px and150px respectively. ; Xiao Zhang has two pieces of wood, the lengths of which are 75px and 7cm respectively. How many triangles can each person take?

30.(5 1 13) As shown, in △ABC, ∠ A = 60? ，∠B=70？ The bisector of ∠ACB intersects with AB in D and the bisector of ∨ BC intersects with AC in E, so we can find ∠BDC and ∠ EDC.

3 1.(356) As shown in the figure, e is a point on the extension line of AC side in △ABC, and the bisector of ∠BCE intersects the extension line of AB at point D. If ∠ cab = 40 and ∠ CBD = 68, find the degree of ∠CDB.

32.(238) As shown in the figure, in △ABC, ∠ B = 60, ∠BAC = 50, and AD divides ∠BAC equally, with point D on BC. Find the degree of ∠ 1 and ∠2.

Answer: Xkb 1.com

I. 1.3

2. Stability instability of triangle

3. ability 4. 2 5.90 50 6. 16.

7.75 8. 1440 144 9.3 10.3

1 1.200px or 150px 12.6

13.3 △ABD，△ABC △ACD，△ACB

14. 180

2.15.c16.c17.b18.c19.C20.d21.c22.a.

Three. 23.( 1) As shown in the answer sheet.

②∠BAD = 60，∠CAD=40。

24. Proof: In △BDE,

∠∠BED = 90°，

∠BED+∠EBD+∠EDB= 180，

∴∠ebd+∠edb= 180-∠bed = 180-90 = 90。

And ∵ average ∠ABD, de average ∠CDB,

∴∠ABD=2∠EBD,∠CDB=2∠EDB，

∴∠abd+∠cdb=2(∠ebd+∠edb)=2×90 = 180，

∴AB∥CD.

25. Solution: ∫∠AOC is the outer corner of △AOB.

∴∠AOC=∠A+∠B (one outer angle of a triangle is equal to the sum of two non-adjacent inner angles).

∠∠AOC = 95，∠B=50，

∴∠A=∠AOC-∠B=95 -50 =45。

∫AB∨CD，

∴∠

∴∠D=45。

26. Solution: (1) Let the number of edges be n, then

(n-2) 180 =2340，n= 15。

A: The number of sides is 15.

(2) The degree of each external angle is 180× = 24.

The number of sides of a polygon = 15.

A: The number of sides is 15.

27. solution: BD AC extends to point e, ∠ CDB = 90+32+21=143, so it is unqualified.

28. Yes, as shown on the answer sheet.

Four. 29.( 1) aa aa aa

(2) Description: According to the sum of the internal angles of the triangle is equal to 180, which is the first net of the new curriculum standard.

You can get ∠ ABC+∠ ACB = 180-∠ A,

According to the meaning of angle bisector, there are

∠6+∠8 =(∠ABC+∠ACB)=( 180-∠A)= 90-∠A，

So ∠ BIC = 180-(∠ 6+∠ 8)

= 180 -(90 -∠A)

=90 +∠A，xkb 1.com

That is ∠ BIC = 90+∠ A.

(3) Complementarity. . xkb 1.com

Verb (abbreviation of verb) 30.( 1) R2 (2) R2 (3) R2 (4) R2

Eighth grade mathematics Chapter 12 congruent triangles examination questions (new curriculum standard)

(Time limit: 100 minute, total score: 100 minute)

1. Multiple choice questions: Fill in the code options for the correct answers to the following questions in the table below. 2 points for each small question, 24 points for * * *.

Title number

three

four

five

six

seven

eight

nine

1 1

answer

1. The following statement is wrong ()

A. The corresponding sides of congruent triangles are equal. The angles of congruent triangles are equal.

C. The circumference of congruent triangles is equal. Congruent triangles is equal in area.

2. Point O is a point within △ABC, the distances from point O to three sides are equal, and ∠ BAC = 60, then ∠BOC has a degree ().

A.60 B.90 C. 120 D. 150

3. As shown in the figure, it is known that △ABC and △DEF are congruent triangles, so the isometric line in the figure is ().

A. 1 b group 2 c group 3d group 4 group

4. As shown in the figure, △ ABC △ def, AC∨DF, then the angle corresponding to ∠C is ().

A. France B. France C. AEF D.

5. As shown in the figure, in △AB=AC, AB=AC, D and E are on BC, AD = AE, BD = CE.

If ∠ bad = 30 and ∠ DAE = 50, ∠BAC is ().

A. 130 b . 120 c . 1 10d . 100

6. As shown in the figure, in △ABD and △ACE, AB = AC and AD = AE. To prove △ Abd △ ace, the supplementary condition is ().

A.∠B =∠C B .∠D =∠E C .∠DAE =∠BAC D .∠CAD =∠DAC

7. As shown in the figure, AD and BC intersect at point O, and it is known that ∠ A = ∠A=∠C, according to "ASA"

To prove △ AOB △ COD, one more condition needs to be added: ()

A.AB=CD B. AO=CO C.BO=DO D.∠ABO=∠CDO

8. As shown in the figure, given AB = AD, it is still uncertain whether △ ABC△ ADC is () after adding one of the following conditions.

A.CB = CD B .∠BAC =∠DAC c .∠BCA =∠DCA D .∠B =∠D = 90

9. As shown in the figure, the intersection o of AC and BD, and OA = OC, OB = OD, then the congruent triangles logarithm in the figure has ().

A.2 to B.3 to C.4 to D.6.

10. As shown in the figure, in △ABC, ∠ A = 36, ∠ C = 72, and BD is the bisector of ∠ABC.

Then the degree of ∠BDC is ()

A.36 B. 48 C. 60 D. 72

1 1. As shown in the figure, P is the point of the bisector of ∠BAC, PM⊥AB is in M, and PN⊥AC is in N, so the following conclusions are obtained:

⑴PM = PN； ⑵AM = AN； ⑶ The areas of △ APM and △APN are equal; (4) ∠ pan+∠ APM = 90。 Among them, the number of correct conclusions is ()

A. 1 .

12. Draw the following conclusions: ① An acute angle and a hypotenuse correspond to the congruence of two right-angled triangles; ② Two isosceles triangles with equal top and bottom angles are congruent; ③ Two isosceles triangles with equal top and bottom angles are congruent; ④ A triangle with two equal angles is congruent. The correct number is ().

1。

2. Fill in the blanks: This big question is ***8 small questions, each with 3 points and * * * 24 points.

13. In △ABC, ∠ b = ∠ c, if one of the triangles congruent with △ABC has an angle of 92,

Then the degrees of the three angles of △ABC are ∠ A = degrees respectively; ∠B =； ∠C=。

14. Given △ABC △ dEF, BC = EF = 6, the area of △ ABC is 18, and the height of the ef side is.

15. As shown in the figure, at △ABC, ∠ BAC = 90, AB = AC, F is the upper point of BC, the extension line of BD⊥AF passing through AF is at D, CE⊥AF is at E, given CE = 5 and BD = 2, then ED =.

16. As shown in the figure, it is known that ∠ DC = EC = ∠ A = 90, BE⊥AC is at point B, DC = EC, and Be = 200px.

So ab+ad =

17. As shown in the figure, in the right triangle ABC, ∠ BAC = 90, AB = AC, and B and C are perpendicular to the straight line passing through point A respectively. If BD = 75px, CE = 100px, then DE =.

On 18. △ABC, ∠ C = 90, and AD divides ∠BAC equally. Given BC = 200px, BD = 125px, the distance from point D to AB is.

19. Various conditions for judging the congruence of two right-angled triangles: (1) an acute angle and an edge; (2) Both parties are equal; (3) The two acute angles are equal. The condition that two right triangles are congruent is.

20. Rotate the right triangle ABC clockwise around the right vertex C to the position of △ dec, if point E is on the side of AB and ∠ DCB = 160, ∠ AED =.

Iii. Answering questions: (This big question ***52 points)

2 1. (2 points for each small question, ***8 points) A graphic is the same before and after translation, folding and rotation. Write the corresponding edges and corners according to the following congruent triangles.

(1) △ ABC △ CDA corresponds to an edge of and a corresponding angle of;

(2) △ AOB △ Doc, corresponding edge is, corresponding angle is;

(3) △ AOC △ BOD, the corresponding edge is, and the corresponding angle is;

(4) △ ace △ BDF, the corresponding edge is and the corresponding angle is.

22. (5 points for this short question)

As shown in the figure, in △AB=DC and △DCB, AC and BD intersect at point O, AB=DC and AC = BD.

Proof: △ ABC △ DCB.

23. (This little question is 10)

It is known that AB = AD, AC = AE, ∠ 1 = ∠ 2,

Verification: (1) △ ABC △ ade ∠ 2 ∠ b = ∠ d.

24. (This small question is 10) As shown in the figure, point C and point B are respectively taken as the center line AD on the BC side of △ABC.

The vertical line and vertical foot of the extension line are E and F respectively.

Verification: ① BF = CE; ②AE+AF=2AD。

25. (9 points in this small question) It is known that figure ∠ ab = BC = 90, AB = BC, D is a point above AC, and CE⊥BD is in.

E, AF⊥ straight BD in F. verification: ef = ce-af.

26. (this small question is 10) as shown in figure 1A, e, f and c are in a straight line, AE = CF

DE⊥AC in E and BF⊥AC in F.

(1) If AB = CD, verify: GE = GF.

⑵ Move the edge EC of △DEC to Figure ② along the AC direction, and other conditions remain unchanged. Is the above conclusion valid? Please explain the reason.

Reference answer

First, multiple-choice questions: 1. b； ; 2.c； 3.d； 4.a； 5.c； 6.c； 7.b； 8.c； 9.c； 10.d； 1 1.d； 12.b；

2. Fill in the blanks: 13.92, 44, 44; 14.6; 15.3; 16.200px 17. 175 px； 18.3; 19.⑴⑵; 20.70 ;

Third, answer questions:

2 1. omitted; 22. Omission; 23. Omission;

24. Omission;

25. Proof: It can be proved by conditions: △ ABF △ BEC.

∴AF=BE,FB=EC

∫BF = ef+be again

∴EC=EF+AF

∴EF=CE-AF.

Twenty-six proofs. ∫AE = CF AE+EF = CF+EF