Examination paper for the first monthly exam of seventh grade mathematics (2)

8. The following statement is true ()

A. Two disjoint line segments are parallel lines.

B. Two disjoint straight lines are parallel lines.

C. Two disjoint rays are parallel lines.

D on the same plane, two disjoint straight lines are parallel lines.

Test center parallel lines.

Analysis can be answered according to the definition of parallel lines.

Solution: According to the definition of parallel lines, two disjoint lines are parallel lines in the same plane.

A, b and c errors; D is correct;

Therefore, choose: d.

9. As we all know, as shown in the figure, AB∨CD, then, the relationship between and is ().

A.++=360? B.﹣+= 180？

C.+﹣= 180? D.++= 180？

The nature of parallel lines in test sites.

The analysis can be solved according to the fact that two straight lines are parallel, the internal angles on the same side are complementary and the internal angles are equal. In the process of solving this problem, auxiliary lines need to be added.

Solution: If point E is EF∨AB, then EF∨CD.

∫EF∨AB∨CD，

? +? AEF= 180？ ,? FED=，

? += 180? +,

That is +-= 180? .

So choose C.

10. The condition that two straight lines cannot be judged to be parallel is ()

A. Equal waist angle B. Equal internal dislocation angle

C. the internal angles on the same side are equal. D. They are all parallel to the third line.

Determination of parallel lines of test points.

By analyzing and judging that two straight lines are parallel, we have learned two methods: ① the inference of parallel axiom, ② the judgment axiom of parallel lines and the judgment theorem of two parallel lines.

Solution: the isosceles angles are equal and the two straight lines are parallel;

Internal dislocation angles are equal and two straight lines are parallel;

The internal angles on the same side are complementary and the internal dislocation angles are equal;

Parallel to three lines and parallel to two lines.

So choose C.

1 1. A student is practicing driving in the square. After two turns, the driving direction is the same as the original direction. The angle of these two turns may be ().

A. Turn left 30 for the first time? Turn right 30 for the second time?

B.turn right 50 for the first time? Second left turn 130?

C. Turn left 50 for the first time? Second right turn 130?

D. Turn left 50 for the first time? Turn left the second time 130.

The nature of parallel lines in test sites.

First, draw a schematic diagram of each option according to the meaning of the question, observe the figure, and get the answer according to the fact that the same angle is equal and the two straight lines are parallel.

Solution: As shown in the figure:

So choose: a.

12. As shown in the figure, CD? AB, the vertical foot is d, AC? BC, the vertical foot is C. The length of the line segment in the figure can represent the distance from the point to the straight line (or line segment).

A. 1 Article B.3, Article C.5, Article D.7

Distance from test center to straight line.

There are six line segments in the graph of this topic, namely AC, BC, CD, AD, BD and AB, of which the two endpoints of line segment AB are not vertical, which can not represent the distance from a point to a straight line, and everything else can be done.

Solution: The line segment representing the distance from point C to straight line AB is CD.

The line segment representing the distance from point b to straight line AC is BC,

The line segment representing the distance from point a to line BC is AC,

The line segment representing the distance from point a to straight line DC is AD,

The line segment representing the distance from point b to straight line DC is BD,

* * * Five articles.

So choose C.

Fill in the blanks (notes)

13. As shown in the figure, suppose AB∑CD, and the cross section EF intersects AB and CD at two points, M and N, respectively. Would you please choose two angles that you think are equal? 1=? 5 .

The nature of parallel lines in test sites.

Analyzing AB∑CD, these two parallel lines are cut by straight line EF; The congruent angles formed are equal, and the internal dislocation angles are equal.

Solution: ∫AB∨CD, 1=? 5 (the answer is not unique).

14. As shown in the figure, in order to translate △ABC, get △A? b？ c？ , you can first move △ABC to the right by 5 squares, and then move it up by 3 squares.

Changes in coordinates and graphics of test sites-translation.

The analysis can be directly solved by using the variation law of the translation midpoint.

The change law of translation midpoint is: the abscissa moves to the right and decreases to the left; The ordinate increases and decreases.

Solution: Starting from point A, move 5 squares to the right and 3 squares up to get a? Then the whole character moves like this. Therefore, fill in two blanks: 5, 3.

15. As shown in the figure, AE∨BD,? 1= 120? ,? 2=40? And then what? What's the degree of C 20? .

The nature of parallel lines in test sites.

This analysis is based on the property that two straight lines are parallel and the internal dislocation angles are equal. The number of AEC, and then the sum of the internal angles of the triangle is equal to 180? The solution can be calculated in the form of columns.

Solution: ∫AE∨BD, 2=40? ,

AEC=？ 2=40? ,

∵? 1= 120? ,

C= 180？ ﹣? 1﹣? AEC= 180？ ﹣ 120? ﹣40? =20? .

So the answer is: 20? .

16. As shown in the figure, if AB∨CD is known, then? 1 and? 2,? What does it matter? 1=? 2+? 3 .

Determine the parallel lines of the test site; Theorem of sum of interior angles of triangle.

According to the analysis, the sum of the internal angles of the triangle is equal to 180? The two straight lines are parallel and complementary.

Solution: ∫AB∨CD,

1+? C= 180？ ,

Again? C+？ 2+? 3= 180? ,

1=? +? 3.

17. as shown in the figure, AB∨CD,? B=68？ ,? E=20？ And then what? The degree of d is 48 degrees

Test the external angle properties of the central triangle; The essence of parallel lines.

According to the properties of parallel lines? BFD=？ B=68？ Then according to the triangle, an outer angle is equal to the sum of two non-adjacent inner angles. What? D=？ BFD﹣？ E, what can be obtained? D.

Solution: ∫AB∨CD, B=68？ ,

BFD=？ B=68？ ,

And then what? D=？ BFD﹣？ E=68？ ﹣20? =48? .

So the answer is: 48.

18. As shown in the figure, do straight lines DE cross? The edge of ABC is at point d, if DE∨BC,? B=70？ And then what? The degree of ADE is 70 degrees.

The nature of parallel lines in test sites.

The basis of analysis is that two straight lines are parallel and the same angle is equal.

Solution: ∫DE∨BC, B=70？ ,

ADE=？ B=70？ .

So the answer is: 70.

Third, answer questions (notes)

19. as shown in the figure, AB∨DE∨GF,? 1:? d:？ B = 2: 3: 4, beg? Degree 1?

The nature of parallel lines in test sites.

Analyze the first set? 1=2x？ ,? D=3x？ ,? B=4x？ According to the fact that two straight lines are parallel and the internal angles on the same side are complementary, it can be expressed as? GCB 、? The degree of FCD, and then according to? GCB 、? 1、? Is FCD 180? You can get the value of x, and then you can get? 1 degree.

Solution: ∵? 1:? d:？ B=2:3:4，

? Settings? 1=2x？ ,? D=3x？ ,? B=4x？ ,

∫AB∑DE，

GCB=？ ,

∫DE∨GF，

FCD=？ ,

∵? 1+? GCB+？ FCD= 180？ ,

? 180﹣4x+x+ 180﹣3x= 180，

The solution is x=30,

1=60? .

20. known: as shown in the figure, 1=? 2,? 3=? B, AC∑DE, and B, C and D are in a straight line. Verification: AE∨BD.

Determination and properties of parallel lines in test site.

Is the analysis based on the properties of parallel lines? 2=? 4. discovery? 1=? 4. According to the judgment of parallel lines, get AB∨CE, and according to the properties of parallel lines? B+？ BCE= 180？ , get to know it? 3+? BCE= 180？ , according to the judgment of parallel lines.

The answer proves: ∫AC∨DE,

2=? 4.

∵? 1=? 2,

1=? 4,

? AB∨CE，

B+？ BCE= 180？ ,

∵? B=？ 3,

3+? BCE= 180？ ,

? AE∨BD。

2 1. As shown in the figure, it is known that DE∑BC and EF are equally divided. AED, ef? AB，CD？ AB, please explain that the CD is equally divided. ACB。

Determination and properties of parallel lines in test site.

According to the properties of parallel lines, EF∨CD is obtained. AEF=？ ACD，？ EDC=？ BCD, according to the definition of angular bisector? AEF=？ Federal Reserve, launch? ACD=？ BCD, you can get the answer.

Solution: ∫DE∨BC,

EDC=？ BCD，

∵EF split equally? AED，

AEF=？ Federal reserve,

∵EF？ AB，CD？ AB，

? EF∑CD，

AEF=？ ACD，

ACD=？ BCD，

? CD split equally? ACB。

22. As shown in the picture, you know? DAB+？ D= 180？ , communication split? DAB, what else? CAD=25？ ,? B=95？

(1) Q? Degree of DCA;

(2) Q? Degree of DCE.

Determination and properties of parallel lines in test site.

By using the definition of angular bisector, it can be solved (1)? The degree of DAB, and then according to? DAB+？ D= 180？ Beg? The degree of d can be obtained by using the theorem of the sum of internal angles of triangles in △ACD. Degree of DCA;

(2) According to (1), it can be proved that AB∨DC can be solved by using the property theorem of parallel lines.

Solution: (1)∵ AC splitting? DAB，

CAB=？ DAC=25？ ,

DAB=50？ ,

∵? DAB+？ D= 180？ ,

D= 180？ ﹣50? = 130? ,

∫in∫ACD，？ D+？ DAC+？ DCA= 180？ ,

DCA= 180？ ﹣ 130? ﹣25? =25? .

(2)∵? DAC=25？ ,? DCA=25？ ,

DAC=？ DCA，

? AB∨DC，

DCE=？ B=95？ .

23. As shown in the picture, you know? 1+? 2= 180? ,? 3=? B, explain? AED=？ ACB。

Determination and properties of parallel lines in test site.

Analyze and judge first? AED and? ACB is a pair of isosceles angles, and then DE∨BC is deduced according to known conditions, and the two angles are equal.

The answer proves: ∵? 1+? 4= 180? (boxer definition),? 1+? 2= 180? (known),

2=? 4,

? EF∨AB (internal dislocation angles are equal and two straight lines are parallel),

3=? ADE (two straight lines are parallel and the internal dislocation angles are equal),

∵? 3=? B (known),

B=？ ADE (equivalent substitution),

? DE∑BC (same angle, two straight lines are parallel),

AED=？ ACB (two straight lines are parallel and have the same angle).

24. As shown in the picture, you know? 1=? 2. Split communication? DAB, please explain DAB.

Determination of parallel lines of test points.

Can it be analyzed according to the properties of the angular bisector? 1=? Taxi, plus conditions? 1=? 2. Available? 2=? CAB, and then according to the internal angles are equal and the two straight lines are parallel, CAB can be obtained.

The answer proves: ∫ communication split? DAB，

1=? Taxi,

∵? 1=? 2,

2=? Taxi,

? CD∨AB。

25. known? Age =? DHF？ 1=? 2. How many pairs of parallel lines are there in the picture? What is the difference? Why?

Determination of parallel lines of test points.

Analyze first? Age =? DHF obtains AB∨CD according to the fact that the isosceles angles are equal and two straight lines are parallel, and then what can be obtained according to the fact that the isosceles angles are equal? AGF=？ Swiss francs again? 1=? 2. According to the definition of right angle? MGF=？ NHF, according to the same angle, two straight lines are flat to get GM∨HN.

Solution: There are two pairs of parallel lines in the diagram, namely AB∨CD and GM∨HN.

∵? Age =? DHF，

? AB∨CD，

AGF=？ Swiss franc,

∵? MGF+？ AGF+？ 1= 180?

? NHF+？ CHF+？ 2= 180? ,

Again? 1=? 2,

MGF=？ NHF，

? General motors ∨HN.

26. Given a straight line A∨b, b∨c, c∨D, what is the relationship between A and D, and why?

Parallel axiom and test center reasoning.

Analysis is easy to draw conclusions from the transitivity of parallel lines.