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The second unit of junior one mathematics examination questions.
Chapter VII Comprehensive Test of Triangle

Editor-in-chief is the honor guard (90 minutes, full score 100)

Fill in the blanks. (2 points for each small question, ***28 points)

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 lengths of 8cm, 9cm and10cm 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 20cm, and the length of one side is 6cm, so the length of the bottom 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: (3 points for each small question, ***24 points)

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 5cm and a width of 3cm, so the area of the parallelogram is ().

a . 7 cm2 b . 8 cm2 c . 9 cm2 d . 10 cm2

Three. Answer: (***48 points)

23. As shown in the figure, in △ABC:

(1) Draw high AD and midline AE on the side of BC. (3 points)

(2) If ∠ B = 30 and ∠ ACB = 130, find ∠BAD and ∠CAD degrees. (5 points)

24.(5 points) 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.(5 points) 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. (4 points)

(2) Every outer corner of a polygon is equal. If the ratio of the degrees of the inner angle to the outer angle is 13: 12, find the number of sides of the polygon. (4 points)

27.(5 points) The shape of the parts is as shown in the figure. According to the regulations, ∠A should be equal to 90, ∠B and ∠ C should be 32 and 2 1 respectively. If the inspector measures ∠ BDC = 148, the part is judged unqualified and tried out.

28.(5 points) The gardener collected a large amount of marble scraps from the land, and prepared to use them to pave the aisle floor of public green space. The shape of the corner material is shown in the figure. Can you use these wastes to pave the floor? If possible, please design at least two schemes.

Fourth, thinking expansion questions: (***6 points)

29. Please complete the following instructions:

(1) As shown in Figure ①, the bisector of the external angle of △ABC intersects with G. Try to explain that ∠ BGC = 90-∠ A. 。

Description: According to the sum of the angles in the triangle is equal to 180, we can know that ∠ ABC+∠ ACB =180-∞ _ _ _.

According to the angle180 ∠ Abe+∠ ACF =180× 2 = 360,

So ∠ EBC+∠ FCB = 360-(∠ ABC+∠ ACB) = 360-(180-∞ _ _ _ _) =180+∞.

(2) As shown in Figure ②, if the bisector of the internal angle of △ABC intersects with point I, try to explain that ∠ BIC = 90+∠ A. 。

(3) With the conclusions of (1) and (2), can you tell the relationship between ∠BGC and ∠BIC?

① ②

Verb (abbreviation of verb) cooperative inquiry: (***6 points)

30. As shown in the figure, build a fan-shaped lawn with a radius of r at each corner of a triangle, a quadrilateral and a pentagon (the shaded part in the figure).

(1) The lawn area in Figure ① is _ _ _ _ _; (2) The lawn area in Figure ② is _ _ _ _ _;

(3) The lawn area in Figure ③ is _ _ _ _ _;

(4) If the number of sides of a polygon is n and other conditions remain unchanged, then you think the lawn area is _ _ _ _.

Answer:

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

11.8cm or 6cm 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 fact that the sum of the internal angles of the triangle is equal to 180,

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,

That is ∠ BIC = 90+∠ A.

(3) Complementarity.

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