People's education printing plate eighth grade mathematics volume 1 unit 2 examination paper 1, multiple-choice questions
1. The side length △ABC of a regular triangle is 3. Take points A 1, B 1 and C 1 on the sides of AB, BC and CA in turn, so that AA1= BB1= CC1= 65438.
A.B. C. D。
2. In Rt△ABC,? C=90？ ，AB= 10。 If the circle with point C as the center and CB as the radius just passes through the midpoint D of AB, then AC= ().
The fifth century BC
3. Place a pair of right-angle triangular rulers as shown in the figure, if? AOD=20？ And then what? The scale of Bank of China is ()
A. 140？ B. 160？ C. 170？ D. 150？
4. As shown in the figure, in △ABC,? C=90？ ,? B=30？ , the perpendicular line DE of AB side intersects AB at point E and BC at point D, and CD=3, then the length of BC is ().
a6 b . 6 c . 9d . 3
5. As shown in the figure, in Rt△ABC,? B=90？ ,? A=30？ , DE bisects the hypotenuse AC vertically, AB intersects D, E stands for vertical foot, and connects CD. If BD= 1, the length of AC is ().
A.2 B.2 C.4 D.4
6. As shown in the figure, in △ABC,? B=30？ The perpendicular line of BC intersects AB at point E, and the vertical foot is D, and CE is equally divided. ACB。 If BE=2, the length of AE is ().
A.B. 1
7. As shown in the figure, AC Road and BC Road are perpendicular to each other, and there is a lake between the midpoint M and point C of AB Road. If the measured length of AM is 1.2km, the distance between M and C is ().
A. 0.5 km b. 0.6 km c. 0.9 km d. 1.2 km
8. As shown in the figure, a rectangular piece of paper is cut off to get a triangle. So? 1+? The degree of 2 is ()
.30 caliber? B.60？ C.90？ D. 120？
9. As shown in the figure, in △ABC,? A=45？ ,? B=30？ ，CD？ AB, the vertical foot is D, CD= 1, then the length of AB is ().
The second century BC.
10. In a right triangle, there is an acute angle equal to 60? Then the degree of the other acute angle is ()
A. 120？ B.90？ C.60？ D.30？
1 1. Connect four thin strips of equal length end to end, nail them into a quadrilateral ABCD, and rotate the quadrilateral to change the shape. B=90？ , as shown in figure 1, measured AC=2, when? B=60？ As shown in figure 2, AC= ()
A.B.2 C. D.2
12. Will it be a 45? The right-angled vertex of the triangle is placed on the edge of a paper tape with a Zhang Kuan of 3 cm. The other vertex is on the other side of the paper tape. Measure that one side of the triangle is 30? As shown in the figure, the length of the largest side of the triangle is ()
A. 3 cm b. 6 cm c.cm d.cm
13. As shown in the figure, in △ABC,? ACB=90？ , split equally? ABC, ed? AB. if in d? A=30？ , AE=6cm, then CE is equal to ()
A. 2 cm long, 3 cm wide and 4 cm deep
14. As shown in the figure, you know? AOB=60？ , point p is on the edge OA, OP= 12, points m, n are on the edge OB, PM=PN, if MN=2, OM= ().
a3 b . 4 c . 5d . 6
15. As shown in the figure, in △ABC,? C=90？ ,? B=30？ , AD split equally? CAB intersects BC at point D, E is a point above AB, and DE is connected, so the following statement is wrong ().
A.？ CAD=30？ AD=BD C.BD=2CD D.CD=ED
Second, fill in the blanks
16. Because the wooden clothes hanger is inflexible, it is not easy to operate when hanging clothes. Xiao Min designed a kind of clothes hanger, which can be easily folded when used, and released after being put into clothes. As shown in figure 1, the hanger rod OA=OB= 18cm, and if the hanger is folded, AOB=60? As shown in Figure 2, the distance between point A and point B is cm.
17. In △ABC,? B=30？ , AB= 12, AC=6, then BC=.
18. As shown in the figure, in △ABC,? C=90？ ,? B=30？ , AD split equally? CAB, crossing BC at point D, if CD= 1, BD=.
19. As shown in the figure, it is known that the side length of square ABCD is 4, diagonal AC and BD intersect at point O, and diagonal E is on the extension line of DC side. CAE= 15？ ，AE=。
20. In the rectangular ABCD, diagonal AC and BD intersect at point O, if? AOB=60？ , AC= 10, then AB=.
Chapter 2 Special Triangle
People's Education Press Grade 8 Mathematics Volume 1 Unit 2 Reference Answers and Test Paper Analysis I. Multiple choice questions (*** 15 small questions)
1. The side length △ABC of a regular triangle is 3. Take points A 1, B 1 and C 1 on the sides of AB, BC and CA in turn, so that AA1= BB1= CC1= 65438.
A.B. C. D。
Determination and properties of equilateral triangle in test sites.
Theme finale.
Draw a picture according to the analysis of the meaning of the question. The intersection point A 1 is a 1D∑BC, and the intersection point D is a small regular triangle with a side length of 1 △AA 1D; AC 1=2, AD= 1, and point d is the midpoint of AC1,so we get s △ aa1c1= 2s △ aa1d =; In the same way, we can get s △ cc1b1= s △ bb1a1=; Finally, it is composed of s △ a1b1c1= s △ ABC-s △ aa1c1-s △ cc1b1-s △ bb655.
Solution: Draw a picture according to the meaning of the question, as shown below:
The crossing point A 1 is a 1D∑BC, and the crossing point D is AC, so it is easy to know that △AA 1D is an equilateral triangle with a side length of 1.
AC 1 = AC-cc 1 = 3- 1 = 2，AD= 1，
? Point d is the midpoint of AC 1,
? s△aa 1c 1 = 2S△aa 1D = 2 12 =；
Similarly, s △ cc1b1= s △ bb1a1=,
? s△a 1b 1c 1=s△abc﹣s△aa 1c 1﹣s△cc 1b 1﹣s△bb 1a 1=？ 32﹣3? = .
So choose B.
This question is not difficult to examine the judgment and nature of equilateral triangle. The entrance to this problem is wide and there are many solutions. Students can try different ways to solve it.
2. In Rt△ABC,? C=90？ ，AB= 10。 If the circle with point C as the center and CB as the radius just passes through the midpoint D of AB, then AC= ().
The fifth century BC
The determination and properties of the equilateral triangle of the test center: a right triangle with an angle of 30 degrees; Pythagorean theorem
Thematic calculation problems; The finale.
By analyzing the properties of the connecting line CD on the hypotenuse of right triangle and the midline, it is obtained that CD=DA=DB, and CD=CB=DB by using equal radius. It can be judged that △CDB is an equilateral triangle, then? B=60？ , so? A=30？ Then according to the relationship between the three sides of a 30-degree right triangle, BC is calculated first, and then AC is calculated.
Solution: link the CD, as shown in the figure.
∵? C=90？ D is the midpoint of AB,
? CD=DA=DB，
And CD=CB,
? CD=CB=DB，
? △CDB is an equilateral triangle,
B=60？ ,
A=30？ ,
? BC= AB=？ 10=5,
? AC= BC=5。
So choose C.
This topic reviews the judgment and properties of equilateral triangles: a triangle with three equal sides is an equilateral triangle; All three internal angles of an equilateral triangle are equal to 60? The properties of the hypotenuse midline of the right triangle and the relationship between the three sides of the 30-degree right triangle are also studied.
3. Place a pair of right-angle triangular rulers as shown in the figure, if? AOD=20？ And then what? The scale of Bank of China is ()
A. 140？ B. 160？ C. 170？ D. 150？
The nature of right triangle in test center.
Analyze and utilize the nature of right triangle and the relationship of mutual redundancy, and then get? The degree of COA, you can get the answer.
Solution: ∵ Place a pair of right-angle triangular rulers as shown in the figure. AOD=20？ ,
COA=90？ ﹣20? =70? ,
BOC=90？ +70? = 160? .
Therefore, choose: B.
The comment on this topic mainly investigates the nature of right triangle and draws a conclusion? The degree of COA is the key to solve the problem.
4. As shown in the figure, in △ABC,? C=90？ ,? B=30？ , the perpendicular line DE of AB side intersects AB at point E and BC at point D, and CD=3, then the length of BC is ().
a6 b . 6 c . 9d . 3
The test center contains a right triangle with an angle of 30 degrees; Properties of the vertical line in the line segment.
According to the analysis, the distance between the point on the vertical line in the line segment and the two ends of the line segment is equal, and AD=BD can be obtained. DAE=30？ Is it easy to get? ADC=60？ ,? CAD=30？ What is the advertisement? The angular bisector of BAC is obtained from the property of angular bisector DE=CD=3, and then according to the right triangle 30? BD=2DE is obtained when the right angle of the angle is equal to half of the hypotenuse, and the result is obtained.
Solution: ∫DE is the median vertical line of AB,
? AD=BD，
DAE=？ B=30？ ,
ADC=60？ ,
CAD=30？ ,
? AD is? Backing angle bisector,
∵? C=90？ , Germany? AB，
? DE=CD=3，
∵? B=30？ ,
? BD=2DE=6，
? BC=9，
So choose C.
The comment on this topic mainly investigates the nature of perpendicular bisector, the nature that the distance between the point on the bisector of the angle and both sides of the angle is equal, and the right triangle is 30? The right-angled side of an angle is equal to half of the hypotenuse. Remember that all properties are the key to solving the problem.
5. As shown in the figure, in Rt△ABC,? B=90？ ,? A=30？ , DE bisects the hypotenuse AC vertically, AB intersects D, E stands for vertical foot, and connects CD. If BD= 1, the length of AC is ().
A.2 B.2 C.4 D.4
The test center contains a right triangle with an angle of 30 degrees; The nature of the vertical line in the line segment; Pythagorean theorem
Analysis and calculation? ACB, according to the nature of the midline of the line segment, find AD=CD, and deduce? ACD=？ A=30？ , get to know it? You can get DCB, BD, BC, according to the content of 30? The right triangle property of an angle can be calculated as AC.
Solution: ∫ In Rt△ABC,? B=90？ ,? A=30？ ,
ACB=60？ ,
∫DE bisects the hypotenuse AC vertically,
? AD=CD，
ACD=？ A=30？ ,
DCB=60？ ﹣30? =30? ,
In Rt△DBC,? B=90？ ,? DCB=30？ ，BD= 1，
? CD=2BD=2，
From Pythagorean Theorem: BC= =,
In Rt△ABC,? B=90？ ,? A=30？ ，BC=，
? AC=2BC=2，
So choose a.
Comment on this topic, and investigate the application of the sum theorem of internal angles of triangles, the properties of isosceles triangles, the pythagorean theorem and the properties of right-angled triangles with angles of 30 degrees. The key to solve this problem is to find the length of BC. Note: in a right triangle, if an angle equals 30? Then the right-angled side it faces is equal to half of the hypotenuse.
6. As shown in the figure, in △ABC,? B=30？ The perpendicular line of BC intersects AB at point E, and the vertical foot is D, and CE is equally divided. ACB。 If BE=2, the length of AE is ().
A.B. 1
The test center contains a right triangle with an angle of 30 degrees; The nature of the angular bisector; Properties of the vertical line in the line segment.
First, according to the nature of the median vertical line, it is concluded that BE=CE=2, so it can be concluded that? B=？ DCE=30？ And then defined by the angular bisector? ACB=2？ DCE=60？ ,? ACE=？ DCE=30？ , using the triangle interior angle sum theorem? A= 180？ ﹣? B﹣？ ACB=90？ , and then press 30 in Rt△CAE? If the right angle side opposite to the angle is equal to half of the hypotenuse, AE= CE= 1.
Solution: ∫In△ABC,? B=30？ The vertical line of BC passes through AB in E, and BE=2.
? BE=CE=2，
B=？ DCE=30？ ,
∵CE split equally? ACB，
ACB=2？ DCE=60？ ,? ACE=？ DCE=30？ ,
A= 180？ ﹣? B﹣？ ACB=90？ .
In Rt△CAE, ∫? A=90？ ,? ACE=30？ ，CE=2，
? AE= CE= 1。
So choose B.
Comments This topic examines the properties of a right triangle with an angle of 30 degrees, the properties of the midline of the line segment, the properties of an isosceles triangle, the definition of the bisector of the angle, and the theorem of the sum of the internal angles of the triangle. How to find it? A=90？ Is the key to solve this problem.
7. As shown in the figure, roads AC and BC are perpendicular to each other, and there is a lake between the middle point M and point C of road AB. If the measured length of AM is 1.2km, the distance between M and C is ().
A. 0.5 km b. 0.6 km c. 0.9 km d. 1.2 km
Test the center line on the hypotenuse of the center right triangle.
Special application problems.
According to the analysis, the median line on the hypotenuse of a right triangle is equal to half of the hypotenuse, and MC=AM= 1.2km can be obtained.
Solution: ∫ In Rt△ABC,? ACB=90？ , m is the midpoint of AB,
? MC= AB=AM= 1.2km。
So choose D.
This question examines the nature of the midline on the hypotenuse of a right triangle: in a right triangle, the midline on the hypotenuse is equal to half of the hypotenuse. It is the key to solve the problem to understand the meaning of the problem and turn the actual problem into a mathematical problem.
8. As shown in the figure, a rectangular piece of paper is cut off to get a triangle. So? 1+? The degree of 2 is ()
.30 caliber? B.60？ C.90？ D. 120？
The nature of right triangle in test center.
Thematic standard questions.
This analysis is based on the complementary solutions of two acute angles of a right triangle.
Solution: According to the meaning of the question, the remaining triangle is a right triangle.
So? 1+? 2=90? .
So choose: C.
This question examines the nature that the two acute angles of a right triangle complement each other, and memorizing the nature is the key to solving the problem.
9. As shown in the figure, in △ABC,? A=45？ ,? B=30？ ，CD？ AB, the vertical foot is D, CD= 1, then the length of AB is ().
The second century BC.
The test center contains a right triangle with an angle of 30 degrees; Pythagorean theorem; Isosceles right triangle.
AD in Rt△ACD, BD in Rt△CDB are obtained by analysis, and then AB is obtained.
Solution: in Rt△ACD,? A=45？ ，CD= 1，
Then AD=CD= 1,
In Rt△CDB, B=30？ ，CD= 1，
Then BD=,
So AB=AD+BD=+1.
So choose D.
This topic reviews the isosceles right triangle and 30? The properties of right-angled triangles require us to master the properties of these two special right-angled triangles.
10.(20 14? In a right triangle, there is an acute angle equal to 60? Then the degree of the other acute angle is ()
A. 120？ B.90？ C.60？ D.30？
The nature of right triangle in test center.
The analysis can be obtained by calculating the reciprocal formula of two acute angles of a right triangle.
Solution: In a right triangle, an acute angle is equal to 60? ,
? The degree of the other acute angle =90? ﹣60? =30? .
Therefore, choose: d.
This question examines the nature that the two acute angles of a right triangle complement each other, and memorizing the nature is the key to solving the problem.
1 1. Connect four thin strips of equal length end to end, nail them into a quadrilateral ABCD, and rotate the quadrilateral to change the shape. B=90？ , as shown in figure 1, measured AC=2, when? B=60？ As shown in figure 2, AC= ()
A.B.2 C. D.2
Determination and properties of equilateral triangle in test center: application of Pythagorean theorem: properties of square.
By analyzing the Pythagorean theorem in graph 1, the side length of a square can be obtained. Figure 2 shows that there is a 60 angle? An isosceles triangle is an equilateral triangle.
Solution: As shown in Figure 1,
AB = BC = CD = DA，？ B=90？ ,
? The quadrilateral ABCD is a square,
Connect AC, then AB2+BC2=AC2,
? AB=BC= = =，
As shown in figure 2,? B=60？ , connect AC,
? △ABC is an equilateral triangle,
? AC=AB=BC=。
This topic examines the properties of squares, Pythagorean theorem and the judgment and properties of equilateral triangles. Using Pythagorean theorem, it is concluded that the side length of a square is the key.
12. Will it be a 45? The right-angled vertex of the triangle is placed on the edge of a paper tape with a Zhang Kuan of 3 cm. The other vertex is on the other side of the paper tape. Measure that one side of the triangle is 30? As shown in the figure, the length of the largest side of the triangle is ()
A. 3 cm b. 6 cm c.cm d.cm
The test center contains a right triangle with an angle of 30 degrees; Isosceles right triangle.
After analyzing that the other vertex C is the vertical line CD, we can get a right triangle according to 30? The opposite side of the angle is equal to half of the hypotenuse, and it can be found that there are 45? The right-angled side of a triangle, and then get the largest side from the isosceles right-angled triangle.
Solution: Make a CD after C? AD，？ CD=3，
In a right-angled triangular ADC,
∵? CAD=30？ ,
? AC=2CD=2？ 3=6,
Also, does the triangle have 45? Angle triangle,
? AB=AC=6，
? BC2=AB2+AC2=62+62=72，
? BC=6，
Therefore, choose: d.
The comment on the knowledge points examined in this question is 30? The key to the problem of right triangle and isosceles right triangle is to find the right side first, and then find the maximum side with Pythagorean theorem.
13. As shown in the figure, in △ABC,? ACB=90？ , split equally? ABC, ed? AB. if in d? A=30？ , AE=6cm, then CE is equal to ()
A. 2 cm long, 3 cm wide and 4 cm deep
The test center contains a right triangle with an angle of 30 degrees.
Thematic standard questions.
According to the analysis, in a right-angled triangle, the right-angled side opposite to 30 degrees is equal to half of the hypotenuse, and AE=2ED is obtained. According to the equal distanCE from the bisector of the angle to both sides, ED=CE is obtained, and the value of CE can be obtained.
Solution: ∫ED? AB，？ A=30？ ,
? AE=2ED，
AE = 6cm，
? ED=3cm，
∵? ACB=90？ , split equally? ABC，
? ED=CE，
? CE = 3cm
So choose: C.
Comment on this question, including 30? The knowledge point of right triangle is the basic property that the right angle side of 30 degrees is equal to half of the hypotenuse and the bisector of the angle in right triangle. The key is to find ED=CE.
14. As shown in the figure, you know? AOB=60？ , point p is on the edge OA, OP= 12, points m, n are on the edge OB, PM=PN, if MN=2, OM= ().
a3 b . 4 c . 5d . 6
The test center contains a right triangle with an angle of 30 degrees; Properties of isosceles triangle.
Special calculation problems.
Analysis p is PD? OB intersects with OB at point D. In a right triangle POD, the length of OD is determined by the definition of acute trigonometric function, so PM=PN, D is the midpoint of MN by the combination of three lines, the length of OD﹣MD is determined by MN, and the length of OM can be determined by OD ﹣ MD. 。
Solution: PD after P? OB, cross OB at point d,
In Rt△OPD, cos60? = =，OP= 12，
? OD=6，
PM = PN，PD？ MN，MN=2，
? MD=ND= MN= 1，
? OM=OD﹣MD=6﹣ 1=5.
So choose: C.
This topic reviews the properties of 30-degree right triangle and isosceles triangle. Mastering the properties of right triangle is the key to solve this problem.
15. As shown in the figure, in △ABC,? C=90？ ,? B=30？ , AD split equally? CAB intersects BC at point D, E is a point above AB, and DE is connected, so the following statement is wrong ().
A.？ CAD=30？ AD=BD C.BD=2CD D.CD=ED
The test center contains a right triangle with an angle of 30 degrees; Properties of angular bisector; Determination and properties of isosceles triangle.
Thematic geometry problems.
Is the analysis based on the triangle interior angle sum theorem? Taxi, did you find it? CAD=？ Bad =? B, only introduce AD=BD and AD=2CD.
Solution: ∫In△ABC,? C=90？ ,? B=30？ ,
CAB=60？ ,
∵ advertising split? Taxi,
CAD=？ Bad =30? ,
CAD=？ Bad =? b，
? AD=BD，AD=2CD，
? BD=2CD，
As we all know, CD=DE cannot be deduced.
That is, only D is wrong, and the answers to options A, B and C are all right;
Therefore, choose: d.
This topic examines the internal angles and theorems of triangles, the determination of isosceles triangles, and the application of the properties of right-angled triangles with angles of 30 degrees. Note: in a right triangle, if an angle equals 30? Then the right-angled side it faces is equal to half of the hypotenuse.
Second, fill in the blanks
16. Because the wooden clothes hanger is inflexible, it is not easy to operate when hanging clothes. Xiao Min designed a kind of clothes hanger, which can be easily folded when used, and released after being put into clothes. As shown in figure 1, the hanger rod OA=OB= 18cm, and if the hanger is folded, AOB=60? As shown in Figure 2, the distance between point A and point B is 18 cm.
Determination and properties of equilateral triangle in test sites.
Special application problems.
According to analysis, an angle is 60? The equilateral triangle of isosceles triangle can be solved
Solution: OA = OB, AOB=60？ ,
? △AOB is an equilateral triangle,
? AB=OA=OB= 18cm，
So the answer is: 18.
Comment on this topic and investigate the equilateral triangle problem. The key is that an angle is 60? The equilateral triangle of isosceles triangle is analyzed.
17. In △ABC,? B=30？ , AB= 12, AC=6, then BC= 6.
The test center contains a right triangle with an angle of 30 degrees; Pythagorean theorem
Analysis basis? B=30？ , AB= 12, AC=6, with 30? The right-angled side is equal to half of the hypotenuse. It is easy to get that △ABC is a right triangle, and the length of BC is calculated by Pythagorean theorem.
Solution: ∵? B=30？ ，AB= 12，AC=6，
? △ABC is a right triangle,
? BC= = =6，
So the answer is: 6.
Comment on this question, including 30? The nature of right triangle and Pythagorean theorem are the key to solve this problem.
18. As shown in the figure, in △ABC,? C=90？ ,? B=30？ , AD split equally? CAB, cross BC at point D, if CD= 1, BD= 2.
The test center contains a right triangle with an angle of 30 degrees; Properties of angular bisector.
Is the analysis based on the properties of the angular bisector? Bad degree, according to the nature of a right triangle with a 30-degree angle, BD can be obtained by finding AD.
Solution: ∵? C=90？ ,? B=30？ ,
CAB=60？ ,
Advertising split? Taxi,
Bad =30? ,
? BD=AD=2CD=2，
So the answer is 2.
In this paper, the properties of 30-degree right triangle and the application of angle bisector are investigated, and it is found that the length of AD is the key to solve this problem.
19. As shown in the figure, it is known that the side length of square ABCD is 4, diagonal AC and BD intersect at point O, and diagonal E is on the extension line of DC side. CAE= 15？ ，AE= 8。
The test center contains a right triangle with an angle of 30 degrees; The nature of a square.
Analysis can be obtained from the properties of the square first? BAC=45？ ，AB∨DC，？ ADC=90？ By who? CAE= 15？ According to the nature of parallel lines and the sum and difference of angles? E=？ BAE=？ BAC﹣？ CAE=30？ Then in Rt△ADE, according to 30? When the right angle of the angle is equal to half of the hypotenuse, AE=2AD=8 can be obtained.
Solution: the side length of the square ABCD is 4, and the diagonal AC and BD intersect at point O.
BAC=45？ ，AB∨DC，？ ADC=90？ ,
∵? CAE= 15？ ,
E=？ BAE=？ BAC﹣？ CAE=45？ ﹣ 15? =30? .
∫ in Rt△ADE,? ADE=90？ ,? E=30？ ,
? AE=2AD=8。
So the answer is 8.
Comment on the nature of a right triangle with an angle of 30 degrees: in a right triangle, 30? The right angle of an angle is equal to half of the hypotenuse. The properties of squares and parallel lines are also studied. What is the solution? E=30？ Is the key to solving the problem.
20. In the rectangular ABCD, diagonal AC and BD intersect at point O, if? AOB=60？ , AC= 10, then AB= 5.
The test center contains a right triangle with an angle of 30 degrees; The properties of rectangles.
According to the properties of rectangle, it can be concluded that △AOB is an equilateral triangle, so the length of OA can be found, and then the length of AB can be found.
Solution: ∵ Quadrilateral ABCD is a rectangle,
? OA=OB
Again? AOB=60？
? △AOB is an equilateral triangle.
? AB=OA= AC=5，
So the answer is: 5.