Eighth grade triangle test questions
First, multiple-choice questions (3 points for each small question, 30 points for * * *)
1. With the following groups of line segments as edges, what can form a triangle is ().
A. 1 cm, 2 cm, 4 cm B.8 cm, 6 cm, 4 cm.
C. 12 cm, 5 cm, 6 cm D.2 cm, 3 cm, 6 cm
2. The sides of an isosceles triangle are 5cm and10cm respectively, so the perimeter of the triangle is ().
A.15cm b.20cm d.20cm or 25cm
3. As shown in the figure, after the window is opened, it can be fixed with a window hook.
The geometric principle used here is ()
A. the stability of triangles
B. The line segment between two points is the shortest
C. Two points determine a straight line
D. The shortest vertical line segment
4. It is known that the bisector of ∠ABC and ∠ACB in △ABC intersects at point O, then ∠BOC must be ().
A. less than right angle B. equal to right angle C. greater than right angle D. uncertain
5. The following statement is true ()
Triangle can be divided into oblique triangle, right triangle and acute triangle.
Any internal angle of an isosceles triangle may be obtuse or right.
C. the outer angle of a triangle must be obtuse.
△ D in △ABC, if ∠AB∠C, ∠ A60, ∠ C60.
6.(20 14? The sum of the internal angles of a Pentagon is ()
180
7. The line segment that is not necessarily within the triangle is ().
A. Angle bisector of triangle B. Center line of triangle
C. The height of the triangle D. None of the above is correct.
8. It is known that at △ABC, the perimeter is 12, then b is ().
a3 b . 4 c . 5d . 6
9. As shown in the figure, in △ABC, point D is on BC, AB=AD=DC, ∠ B = 80, then
The number of times ∠C is ()
30 BC to 40 BC
10. The degree of the angle formed by the intersection of two acute bisectors of a right triangle is ()
A.45 B. 135 C.45 or 135 D none of the above answers are correct.
Fill in the blanks (3 points for each small question, 24 points for * * *)
1 1.(20 14? In middle school, the degree of the external angle was known to be 0.
12. As shown in the figure, a piece of paper with a right triangle is cut off to get a 4.
A polygon, then ∠ 1+∠ 2 =.
13. If the number of sides of a polygon increases by 1 times, the sum of its internal angles will increase by _ _ _ _ _ _.
14.(20 14? The included angle between the height of one waist and the height of the other waist of an isosceles triangle is 36, so the degree of the bottom angle of the isosceles triangle is _ _.
15. Let it be the length of three sides of △ABC, then.
16. As shown in the figure, AB=29, BC= 19, AD=20, CD= 16. If AC=, the value range of is.
17. As shown in the figure, AD is a diagonal of the regular pentagon ABCDE, so ∠ bad = _ _ _ _ _ _.
18. If every external angle of a polygon is 36, then the diagonal of the polygon has _ _ _ _ _.
Iii. Answering questions (***46 points)
19.(6 points) For convex polygons, the sum of all internal angles except one is 2 750. Find the number of sides of this polygon.
20.(6 points) As shown in the figure, in △ABC, AB=AC, and the midline on the side of AC divides the perimeter of the triangle into two parts, 24 cm and 30 cm, and finds the length of each side of the triangle.
2 1.(6 points) Some people say that they have big steps and can walk more than four meters in one step. Can you believe it? Use what you have learned in mathematics to explain why.
22.(6 points) It is known that two sides of a triangle are equal in length and the third side is equal in length. If all sides of a triangle are integers, try to judge the shape of the triangle.
23.(6 points) As shown in the figure, there are three stations A, B and C in Wuhan, which form a triangle, and a bus goes from mile to mile.
Station C.
(1) When the car moves to point D, it happens that BD=CD. What is the line segment connecting AD and AD? △ How many such line segments are there in △ABC? Is there a triangle with equal area at this time?
(2) The car keeps moving forward. When moving to point E, I found ∠BAE=∠CAE, so what is the line segment of AE? △ How many such line segments are there in △ABC?
(3) The car keeps moving forward. When it moves to point F, it is found that ∠ AFB = ∠ AFC = 90, so what line segment is AF? How many such line segments are there?
24.(8 points) Known: as shown in the figure, DG⊥BC, AC⊥BC, EF⊥AB, ∠ 1=∠2, verified: CD⊥AB.
25.(8 points) It is stipulated that all sides of (1) are not equal and are integers, and (2) the ratio of the height of the shortest side to the height of the longest side is an integer k. Such a triangle is called a high-ratio triangle, where k is called a high-ratio coefficient. Answer the following questions according to the regulations:
(1) Find the ratio coefficient k of a ratio triangle with a perimeter of 13.
(2) Write the perimeter of an aspect ratio triangle with only four aspect ratio coefficients.
1.B analysis: According to the fact that the sum of any two sides in a triangle is greater than the third side, only B can form a triangle, so B is chosen.
2.c analysis: Because the sum of any two sides in a triangle is greater than the third side, the waist can only be 10 cm, so the perimeter of this triangle is 10+ 10+5=25(cm). So, C.
3.a analysis: This topic mainly examines the application of triangle stability in life.
4.c analysis: Because in △ABC, ∠ ABC+∠ABC+∠ACB 180,
therefore
So ∠ BOC 90. So choose C.
5.d analysis: A. Triangle includes right triangle and oblique triangle, and oblique triangle includes acute triangle and obtuse triangle, so A is wrong;
B. The isosceles triangle has only an obtuse angle or a right angle, so B is wrong;
C the outer angle of a triangle may be obtuse, acute or right, so c is wrong;
D. Because in △ABC, ∠A∠B∠C, if ∠ A ≤ 60 or ∠ C ≥ 60, the sum of the interior angles of the triangle is 180, so the original conclusion is correct, so D is chosen.
6.c analysis: the formula of the inner angle of a polygon is, when,.
7.c analysis: Because the midline and bisector of the triangle are both inside the triangle, and part of the height of the obtuse triangle is outside the triangle, the answer is C.
8.b analysis: Because, so.
Again, so I chose B.
9.b analysis:
.
10.c analysis: as shown in the figure: ∵ AE and BD are bisectors of two acute angles in a right triangle.
∴ ∠OAB+∠OBA=90 ÷2=45。
The bisector consists of two angles: ∠BOE and ∠EOD.
According to the triangle exterior angle sum theorem ∠ BOE = ∠ OAB+∠ Oba = 45,
EOD =180-45 =135, so c
1 1. 140 Analysis: According to the theorem that ∠C = 40, the outer angle of ∠C is.
12.270 analysis: as shown in the figure, according to the meaning of the question ∠ 5 = 90,
∴ ∠3+∠4=90 ,
∴ ∠ 1+∠2= 180 + 180 -(∠3+∠4)=360 -90 =270 .
13. analysis: calculation by using the theorem of polygon internal angle sum.
Because the sum of polygon and polygon's internal angles is sum,
So the inner angle has increased again.
14.27 or 63 analysis: When the isosceles triangle is an obtuse triangle, as shown in Figure ①,
.
Answers to the question 14
When the isosceles triangle is an acute triangle, as shown in Figure ②:
.
15. Analysis: Because it is the length of three sides of △ABC,
So,,,
So the original formula =
16. 10 & lt； & lt36 analysis: in △ABC, AB-BCACAB+BC, so1048;
In △ADC，AD-DCACAD+DC，so 436。 So 1036.
17.72 analysis: every internal angle of the regular pentagon ABCDE is = 108, △AED is an isosceles triangle, ∠ EAD = (180-108) = 36, so ∠ DAB.
18.35 Analysis: Let the number of sides of this polygon be, so this polygon is a decagon. Because the total number of diagonals of an edge is, the number of diagonals of this polygon is.
19. Analysis: Because the removed internal angle is greater than 0 and less than 180, there are two unknowns in the problem, but there is only one equivalent relationship. This problem often appears in some competition questions and needs to be evaluated according to the special meaning of two unknowns in the conditions.
Solution: Let the number of sides of this polygon be (natural number) and the removed inner angle be (0.
According to the meaning of the question, you must
∵ ∴
∴ ,∴ .
Directions: In this question, after the equation is obtained by using the formula of the inner angle of a polygon, the number of sides of the polygon is obtained by solving the inequality with the help of the range of the angle. This is also a common method to solve the problem of the sum of inner and outer angles of polygons.
20. Analysis: Because BD is the midline, AD=DC. The reason why the two parts are not equal is that the waist and the bottom are not equal, so we should discuss them in different situations.
Solution: let AB=AC=2, then AD=CD=,
(1) When AB+AD=30 and BC+CD=24, there is 2=30.
∴ = 10, 2 =20, BC =24- 10= 14.
The three sides are 20 cm, 20 cm, 14 cm respectively.
(2) When AB+AD=24 and BC+CD=30, there is =24,
= 8, BC =30-8=22. The three sides are16cm,16cm and 22cm respectively.
2 1. analysis: human legs can be regarded as two line segments, and walking steps can also be regarded as line segments. Then these three line segments just form three sides of a triangle, so they should satisfy the trilateral relation theorem.
Solution:No..
If this person can walk more than four meters in one step, it is concluded from the relationship between the three sides of the triangle that the sum of his two legs is more than four meters, which is inconsistent with the actual situation.
So he can't walk more than four meters in one step.
22. Analysis: Given the length of three sides of a triangle, list the inequalities according to the relationship between the three sides of the triangle, and then solve them.
Solution: According to the trilateral relationship of the triangle, we can get
& lt& lt,
0 & lt& lt6-，0 & lt& lt。
Because 2,3-x is a positive integer, it = 1.
So the three sides of a triangle are 2, 2 and 2 respectively.
Therefore, a triangle is an equilateral triangle.
23. Analysis: (1) Because BD=CD, point D is the midpoint of BC and AD is the midline. The midline of the triangle divides the triangle into two triangles with equal areas;
(2) AE is the bisector of a triangle, because ∠ BAE = ∠ CAE;
(3) Since ∠ AFB = ∠ AFC = 90, AF is the high line of a triangle.
Solution: (1)AD is the midline of BC in △ABC, and there are three midlines in the triangle. At this time, the areas of △ABD and △ADC are equal.
(2)AE is the bisector of ∠BAC in △ABC, and the triangle has three bisectors.
(3)AF is the high line on the side of BC in △ABC, sometimes outside the triangle, the triangle has three high lines.
24. Analysis: Use the definition of verticality flexibly, and pay attention to the angle of 90 from verticality and verticality from 90. Combined with the judgment and properties of parallel lines, CD ∠ AB can be obtained as long as ADC = 90 is proved.
Proof: ∫dg⊥bc, AC⊥BC (known),
∴∠dgb =∠ACB = 90° (vertical definition),
∴DG∑AC (same angle, two straight lines are parallel).
∴ ∠2=∠ACD (two straight lines are parallel and the internal dislocation angles are equal).
∫∠ 1 =∠2 (known),
∴ ∠ 1=∠ACD (equivalent substitution),
∴ef∑CD (same angle, two straight lines are parallel).
∴ ∠AEF=∠ADC (two lines are parallel and have the same angle).
∫ef⊥ab (known), ∴∠ AEF = 90 (vertical definition),
∴∠ ADC = 90 (equivalent substitution).
∴ CD⊥AB (vertical definition).
25. Analysis: (1) According to the definition, the triangular relationship is "the sum of any two sides >; The third side, the difference between any two sides