Name: Time: 90 minutes, full mark: 100, score:

A, multiple-choice questions (this topic is entitled *** 10 small questions, 3 points for each small question, 30 points for each small question. Only one of the four options given in each small question meets the requirements of the topic.

1. With the following groups of line segments as edges, what can form a triangle is ().

A.2cm，3cm，5cm B.5cm，6cm， 10cm

C. 1cm， 1cm，3cm D.3cm，4cm，9cm

2. If one side of an isosceles triangle is equal to 4 and the other side is equal to 9, then its circumference is ().

A. 17b.22c. 17 or 22d. 13

3. The △ABC of the applicable condition ∠A= ∠B= ∠C is ().

A. acute triangle B. right triangle C. obtuse triangle D. equilateral triangle

4. It is known that one angle of an isosceles triangle is 75 and its vertex angle is ().

A.30 b.75 c. 105 d.30 or 75.

5. The sum of the inner angles of a polygon is 180, which is more than twice the sum of its outer angles, and the number of sides of this polygon is ().

A.5 B.6 C.7 D.8

6. An external angle of a triangle is an acute angle, so the shape of the triangle is ().

A. acute triangle B. obtuse triangle C. right triangle D. uncertainty

7. The following proposition is true ()

The bisector, midline and height of a triangle are all within the triangle.

At least one inside angle of the triangle is not less than 60 degrees.

C. A right triangle has only one height.

The height of the hypotenuse of a right triangle is higher than half of the hypotenuse.

8. Can constitute the basic graphics as shown in the figure is ()

(A) (B) (C) (D)

9. It is known that the bottom of the isosceles waist BC△ABC = 8cm, │AC-BC│=2cm, then the length of waist AC is ().

A.10cm or 6cm B. 10/0cm C.6cm D.8cm or 6cm.

10. As shown in figure 1, fold the △ABC paper along the DE. When point A falls within the quadrilateral BCDE, there is a quantitative relationship between ∠A and ∠ 1+∠2. Please try to find this rule, and the rule you find is ().

A.∠A =∠ 1+∠2 b . 2∠A =∠ 1+∠2 c . 3∠A = 2∠ 1+∠2d . 3∠A = 2(∠ 1+∠2)

( 1) (2) (3)

Fill in the blanks (this big question is ***8 small questions, each with 3 points and ***24 points. Fill in the answers on the lines in the questions)

1+2x 0 1。 The lengths of the three sides of the triangle are 5, 1+2x, 8 respectively, so the value range of x is _ _ _ _ _.

12. The lengths of the four line segments are 5cm, 6cm, 8cm and 13cm respectively, and any three of them can form _ _ _ _ _ triangles.

13. The following figure 2: ∠ A+∠ B+∠ C+∠ D+∠ E+∠ F equals _ _ _ _ _ _ _ _.

14. If the sum of the internal angles of a regular polygon is 900, then this regular polygon is a regular polygon.

15.n Every outer angle of the polygon is equal to 45, so n = _ _ _ _ _ _

16. Starting from Station A by train, you can go through three stations along the way to get to Mile Mile, so you need to arrange _ _ _ _ different tickets between Station A and bilibili.

17. Fold a piece of regular hexagonal paper in half and completely overlap it, then the obtained figure is a _ _ _ _ _ polygon, and the sum of its internal angles (calculated by one layer) is _ _ _ _ _ _ _.

18. As shown in Figure 3, given that ∠ 1 = 20, ∠ 2 = 25 and ∠ A = 55, the degree of ∠BOC is _ _ _ _.

Third, the solution (this big question ***6 small questions, ***46 points, the answer should be written to explain the process or calculus steps)

19.(6 points) As shown in the figure, divide BD by ∠ABC, DA⊥AB, ∠ 1 = 60, ∠ BDC = 80, and find the degree of ∠ C. 。

20.(8 points) As shown in the figure:

(1) Draw the outer corner ∠BCD of △ABC, and then draw the bisector CE of ∠BCD.

(2) If ∠A=∠B, please complete the following certificates:

It is known that in △ABC, ∠A=∠B, and CE is the bisector of the external angle ∠BCD.

Proof: ce∨ab.

2 1.(8 points) (1) As shown in Figure 4, there is a right-angled triangle XYZ placed on △ABC, and it happens that two right-angled sides XY and XZ of the triangle XYZ pass through △ ABC at point B and point C respectively, and ∠ A = 30, then ∠ ABC+∠ ACB.

(4) (5)

(2) As shown in Figure 5, if the position of the right-angled triangle XYZ is changed so that the two right-angled sides XY and xz of the triangle XYZ still pass through B and C respectively, will the size of ∠ABX+∠ACX change? If yes, please give examples; If not, request the size of ∠ABX+∠ACX.

22.(8 points) Fascinating collocation problems are familiar to adults and children. This picture is a diagram made up of matches. Take four matches and leave five squares. Each match left is an edge or part of a square. Please give two schemes, which are drawn in Figure (1) and (2) respectively.

23.(8 points) In a plane, with three, five or six matches connected end to end, what shape triangle can you spell? By trying, the list is as follows:

Q: (1) Can four matches make a triangle?

(2) How many triangles can eight matches and 12 matches make? And draw their schematic diagrams.

24.(8 points) As shown in the figure, BC⊥CD, ∠ 1=∠2=∠3, ∠ 4 = 60, ∠ 5 = ∠ 6.

Is (1)CO the height of △BCD? Why?

(2) What is the degree of ∠ 5?

(3) Find the degree of each internal angle of quadrilateral ABCD.

Answer:

1.B

2. measure b: according to the meaning of the question, the three sides of the triangle may be 4, 4, 9 or 4, 9, 9. But 4+4.

3.B nudge: let ∠ A = X, then ∠ B = 2x, ∠ C = 3x, from the triangle interior angle theorem, x+2x+3x = 180. The solution is x = 30. ∴ 3x = 3x。

4.d inching: Discuss the top angle of 75 and the bottom angle of 75.

C-dial: According to the meaning of the question, get (n-2)? 180 = 2× 360+ 180. The solution is n = 7. So I chose C.

6.B

7.B CuO: If all three internal angles in a triangle are less than 60, the sum of the three internal angles is less than 180, which contradicts the theorem of internal angle sum. Therefore, at least one inner angle in the triangle is not less than 60.

8.B

9. Push gently: ∵BC=8cm, │AC-BC│=2cm, ∴AC= 10cm or 6cm. After inspection, we can use 10 cm, 10cm, 8cm or 6 cm, 6cm and 8cm as side lengths to form a triangle.

10.b nudge: it can be proved according to the theorem of the sum of internal angles of triangles and quadrilaterals.

1 1. 1 & lt； Dial at x<6 o'clock: 8-5

12.2 nudge: A triangle can be formed with side lengths of 5cm, 6cm, 8cm or 6cm, 8cm and 13cm respectively.

13.360 nudge: ∵ There are exactly two triangles in the figure: △AEC, △BDF, ∴ ∠ A+∠ B+∠ C+∠ D+∠ E+∠ F = 30.

14.7

15.8 dial: n = = 8.

16. 10

17. iv; 360

18. 100 nudge: Connect AO and expand it, and it is easy to know that ∠ BOC = ∠ BAC+∠1+∠ 2 = 55+20+25 =100.

19. solution: in △ABD, ∫∠A = 90, ∠ 1 = 60,

∴∠ABD=90 -∠ 1=30。

∫BD shares ∠ ∠ABC, ∴∠ CBD =∠ Abd = 30.

In △BDC, ∠ c =180-(∠ BDC+∠ CBD) =180-(80+30) = 70.

20.( 1) As shown in the figure.

(2) prove that:

∠∠A =∠B, ∠BCD is the outer corner of △ABC,

∴∠BCD=∠A+∠B=2∠B，

∫CE is the bisector of the external angle∠∠ BCD,

∴∠BCE= ∠BCD= ×2∠B=∠B，

∴ce∨ab (internal dislocation angles are equal and two straight lines are parallel)

Hugging: As shown in the answer sheet, to prove that two straight lines are parallel, just prove that the inner angle ∠B=∠BCE.

2 1.( 1) 150 ; 90

(2) No change.

∫∠A = 30，

∴∠ABC+∠ACB= 150，

∫∠X = 90，

∴∠XBC+∠XCB=90，

∴∠abx+∠acx=(∠abc-∠xbc)+(∠acb-∠xcb)

=(∠ABC+∠ACB)-(∠XBC+∠XCB)= 150-90 = 60。

Hugging: A holistic approach to this problem.

22. As shown in Figure 7-2.

23. Solution: (1) Four matches cannot form a triangle;

(2) Eight matches can form a triangle (3, 3, 2);

12 matching can make three different triangles (4, 4, 4; 5,5,2; 3, 4, 5). Sketch.

24. solution: (1)CO is the height of △BCD.

Reason: in △BDC, bcd = 90, ∠ 1=∠2, ∴∠ 1 = ∠ 2 = 90 ÷ 2 = 45.

∵∠ 1=∠3, ∴∠ 3 = 45.

∴∠doc= 180-(∠ 1+∠3)= 180-2×45 = 90，

∴CO⊥DB.

∴CO is the height of △BCD.

(2)∠5=90 -∠4=90 -60 =30 .

(3)∠CDA =∠ 1+∠4 = 45+60 = 105，∠DCB=90，

∠DAB=∠5+∠6=30 +30 =60，

∠ABC= 105。