Eight grades mathematics first volume 1 chapter Pythagorean theorem unit examination paper.
First, multiple choice questions
At 1. △ABC？ A: B, is it? The opposite sides of C are A, B and C respectively, and the pseudo-proposition in the following proposition is ()
A. if? C﹣？ B=？ A, then △ABC is a right triangle.
B. If C2 = B2-A2, then △ABC is a right triangle, and? C=90？
C if (c+a) (c-a) = B2, then △ABC is a right triangle.
D. if? A:? b:？ C = 5: 2: 3, then △ABC is a right triangle.
2. The following three numbers in each group can be used as three sides to form a right triangle ()
a . 1.2，3 B.32，42，52 C .，D.0.3，0.4，0.5
3. Pythagorean theorem is an important theorem in geometry. China's ancient mathematical work "The Book of Changes suan Jing" recorded that "if three or four strands are hooked, then five strings". As shown in figure 1, it consists of a small square and an equilateral right triangle, and its area relationship can be used to verify the pythagorean theorem. Put the figure 1 into a rectangle to get Figure 2. BAC=90？ , AB=3, AC=4, and points D, E, F, G, H, I are all on the sides of the rectangle KLMJ, so the area of the rectangle KLMJ is ().
a . 90 b . 100 c . 1 10d . 12 1
4. In RT △ ABC, the hypotenuse length BC=3, and the value of AB2+AC2+BC2 is ().
A.18b.9c.6d. cannot be calculated.
5. In Rt△ABC, where A, B and C are three sides of △ABC, the following relationship is correct ().
A.a2+b2=c2 B.a2+c2=b2
C.b2+c2 = a2d。 All the above relationships are possible.
6. In △ABC, AB= 15, AC= 13 and AD= 12, then the circumference of △ ABC is ().
A.42 B.32 C.42 or 32 D.37 or 33
Step 2 fill in the blanks
7. It is known that A, B and C are the lengths of the two right angles and the length of the hypotenuse of Rt△ABC, respectively. a+b= 14 and c= 10, then S△ABC=.
8. Xiao Qiang walked 200 meters east on the playground, then walked 150 meters, and then walked 250 meters back to his original place. Xiao Qiang walked 200 meters to the east on the playground, and then walked in the following direction 150 meters.
9. As shown in the figure, it is known that in Rt△ABC, ACB=90? , AB=4, make semicircles with diameters of AC and BC respectively, and the areas are marked as S 1 and S2 respectively, then S 1+S2 is equal to.
Three. solve problems
10. As shown in the figure, AC? CE, AD=BE= 13, BC=5, DE=7, find AC.
1 1. As shown in the figure, there is a rectangular yard ABCD, where AB=9m and AD= 12m. There is a telephone pole at B and an electric lamp at E, which is 8m above the ground. What is the distance from point D to lamp E?
12. As shown in the figure, a parallel beam of sunlight enters the classroom window. Xiao Qiang measured BC =1m.
NC= m, BN= m, AC=4.5m, MC=6m. Find the length of MA.
13. As shown in the figure, the rectangular body length 15, width 10, height 20, and the distance from point B to point C is 5. What is the shortest distance an ant needs to crawl from point A to point B along the surface of a cuboid?
14. As shown in the figure, in the rectangular paper ABCD, AB= 18, fold the rectangular paper in half along the straight line AC, with point B at point E and point AE at point F. If AF= 13, find the length of AD.
15. As shown in the figure, for any eligible right triangle BAC, rotate 90 counterclockwise around its acute vertex? Get △DAE, so? BAE=90？ The quadrilateral ACFD is a square, and its area is equal to that of quadrilateral ABFE, which is equal to the sum of the areas of Rt△BAE and Rt△BFE. Write a method to prove Pythagorean theorem according to the diagram.
Unit 20 1 6 of the first volume of eighth grade mathematics in Beijing Normal University (chapter1).
Reference answers and analysis of test questions
First, multiple choice questions
At 1. △ABC？ A: B, is it? The opposite sides of C are A, B and C respectively, and the pseudo-proposition in the following proposition is ()
A. if? C﹣？ B=？ A, then △ABC is a right triangle.
B. If C2 = B2-A2, then △ABC is a right triangle, and? C=90？
C if (c+a) (c-a) = B2, then △ABC is a right triangle.
D. if? A:? b:？ C = 5: 2: 3, then △ABC is a right triangle.
Test site KS: the inverse theorem of Pythagorean theorem; K7: Theorem of Sum of Interior Angles of Triangle.
The judgment methods for analyzing right-angled triangles are as follows: ① Find an angle of 90? ② Using the inverse theorem of Pythagorean theorem.
Solution: a, according to the theorem of triangle internal angle sum, angle c can be found to be 90 degrees, so it is correct;
B, the solution should be? B=90 degrees, so it is wrong;
C, after simplification, c2=a2+b2. According to Pythagorean theorem, △ABC is a right triangle, so it is correct;
D. Let the triangles be 5x, 3x and 2x respectively. According to the theorem of the sum of internal angles of triangles, we can get that the three external angles are 90 degrees, 36 degrees and 54 degrees respectively, so △ABC is a right triangle, so it is correct.
So choose B.
This topic reviews the judgment of examining right-angled triangles.
2. The following three numbers in each group can be used as three sides to form a right triangle ()
a . 1.2，3 B.32，42，52 C .，D.0.3，0.4，0.5
Test site KS: the inverse theorem of Pythagorean theorem.
Analysis can be judged according to the inverse theorem of Pythagorean theorem.
Solution: ∫0.32+0.42 = 0.25, 0.52=0.25,
? 0.32+0.42=0.52,
? 0.3, 0.4 and 0.5 can form three sides of a right triangle.
So choose D.
This topic examines the inverse theorem of Pythagorean theorem. The key to solving problems is to remember the solution format of the inverse theorem of Pythagorean theorem, which belongs to the senior high school entrance examination.
3. Pythagorean theorem is an important theorem in geometry. China's ancient mathematical work "The Book of Changes suan Jing" recorded that "if three or four strands are hooked, then five strings". As shown in figure 1, it consists of a small square and an equilateral right triangle, and its area relationship can be used to verify the pythagorean theorem. Put the figure 1 into a rectangle to get Figure 2. BAC=90？ , AB=3, AC=4, and points D, E, F, G, H, I are all on the sides of the rectangle KLMJ, so the area of the rectangle KLMJ is ().
a . 90 b . 100 c . 1 10d . 12 1
KR: proof of pythagorean theorem.
Topic 1: general questions; 16: the finale.
By analyzing the extension of AB intersection KF at O point and AC intersection GM at P point, it can be concluded that quadrilateral AOLP is a square, then the side length of the square, then the length and width of rectangle KLMJ, and then it can be solved according to the area formula of the rectangle.
Solution: As shown in the figure, extend AB intersection KF to point O and AC intersection GM to point P.
The quadrilateral AOLP is a square,
Side length AO=AB+AC=3+4=7,
So KL=3+7= 10, LM=4+7= 1 1,
So the area of the rectangle KLMJ is 10? 1 1= 1 10.
So choose: C.
This question examines the proof of Pythagorean theorem, making auxiliary lines and constructing squares are the key to solving problems.
4. In RT △ ABC, the hypotenuse length BC=3, and the value of AB2+AC2+BC2 is ().
A.18b.9c.6d. cannot be calculated.
Test site KQ: Pythagorean theorem.
Using Pythagorean Theorem, AB2+AC2 is transformed into BC2, and then it is evaluated.
Solution: ∵Rt△ABC, BC is the hypotenuse,
? AB2+AC2=BC2，
? AB2+AC2+BC2=2BC2=2？ 32= 18.
So choose a.
This topic examines Pythagorean theorem. It is the key to correctly judge the right and hypotenuse of right triangle by Pythagorean theorem and get the equation.
5. In Rt△ABC, where A, B and C are three sides of △ABC, the following relationship is correct ().
A.a2+b2=c2 B.a2+c2=b2
C.b2+c2 = a2d。 All the above relationships are possible.
Test site KQ: Pythagorean theorem.
According to Pythagorean theorem analysis, points? C is a right angle. B is a right angle. A is a right angle, and the relationship between A, B and C can be obtained by discussing three situations.
Solution: In Rt△ABC, A, B and C are three sides of △ABC.
? C is a right angle, then a2+B2 = C2；;
? B is a right angle, then a2+C2 = B2；;
? A is a right angle, then b2+c2=a2.
Therefore, choose: d.
Examine Pythagorean Theorem: In any right triangle, the sum of squares of two right-angled sides must be equal to the square of hypotenuse.
6. In △ABC, AB= 15, AC= 13 and AD= 12, then the circumference of △ ABC is ().
A.42 B.32 C.42 or 32 D.37 or 33
Test site KQ: Pythagorean theorem.
Analysis of this topic should be discussed in two situations:
(1) When △ABC is an acute triangle, in Rt△ABD and Rt△ACD, the lengths of BD and CD can be obtained by pythagorean theorem, and the length of BC can be obtained by adding them, so that the circumference of △ABC can be obtained;
(2) When △ABC is an obtuse triangle, in Rt△ABD and Rt△ACD, the length of BD and CD can be obtained by pythagorean theorem, and the length of BC can be obtained by subtracting them, so the circumference of △ABC can be obtained.
Solution: This problem should be explained from two aspects:
(1) When △ABC is an acute triangle, in Rt△ABD,
BD= = =9，
In Rt△ACD,
CD= = =5
? BC=5+9= 14
? The circumference of △ABC is:15+13+14 = 42;
(2) When △ABC is an obtuse triangle,
In Rt△ABD, BD= = =9,
In Rt△ACD, CD= = =5,
? BC=9﹣5=4.
? The circumference of △ABC is: 15+ 13+4=32.
? When △ABC is an acute triangle, the circumference of △ABC is 42; When △ABC is an obtuse triangle, the circumference of △ABC is 32.
So choose C.
This topic reviews the knowledge of Pythagorean theorem and right triangle solution. When solving this problem, we should discuss it in two situations. An easy mistake is to leave out the solution. It is difficult for students to think comprehensively.
Step 2 fill in the blanks
7. It is known that A, B and C are the lengths of the two right angles and the length of the hypotenuse of Rt△ABC, respectively. a+b= 14 and c= 10, then S△ABC= 24.
Test site KQ: Pythagorean theorem; K3: the area of triangle.
The analysis directly uses Pythagorean theorem to get the equation about B, and then gets the answer.
Solution: ∫a, B and C are the lengths of two right angles and the hypotenuse length of Rt△ABC, respectively, with a+b= 14 and c= 10.
? A = 14-b, then (14-b) 2+B2 = C2.
Therefore, (14-b) 2+B2 = 102,
Solution: b 1=6, b2=8,
Then a 1=8, a2=6,
That is S△ABC= ab=? 6? 8=24.
So the answer is: 24.
This paper mainly investigates Pythagorean theorem and the solution of triangle area, and correctly concludes that the right angle side length is the key to solving the problem.
8. Xiao Qiang walked 200 meters east on the playground, then walked 150 meters, and then walked 250 meters back to his original place. Xiao Qiang walked 200 meters east on the playground, and then walked north or south 150 meters.
Library: Application of Pythagorean Theorem.
According to the meaning of the question, the figure is analyzed, and the right triangle is judged by the inverse theorem of Pythagorean theorem, so that the answer can be determined.
Solution: As shown in the figure, AB = 200m, BC = BD = 150m, AC = AD = 250m,
According to 2002+ 1502=2502:? ABC=？ ABD=90？ ,
? Xiao Qiang walked 200 meters east on the playground, and then walked north or south 150 meters.
So the answer is: North or South.
So the answer is north or south.
This topic examines the application of Pythagorean theorem. The key to solving the problem is to make a graph according to the meaning of the problem, which is moderately difficult.
9. As shown in the figure, it is known that in Rt△ABC, ACB=90? , AB=4, make semicircles with diameters of AC and BC respectively, and record the areas as S 1 and S2 respectively, then S 1+S2 equals 2? .
Test site KQ: Pythagorean theorem.
Topic 1 1: calculation problem.
According to the formula of semicircle area and Pythagorean theorem, it is known that S 1+S2 is equal to the semicircle area with the hypotenuse as the diameter.
Solution: S 1=? ( )2= ? S2 AC2 =？ BC2，
So S 1+S2=? (AC2+BC2)=？ AB2=2？ .
So the answer is: 2? .
According to the area formula of semicircle and pythagorean theorem, it is proved that the sum of the areas of two right-angled semicircles of a right triangle is equal to the area of the semicircle of the hypotenuse diameter, and the emphasis is on verifying pythagorean theorem.
Three. solve problems
10. As shown in the figure, AC? CE, AD=BE= 13, BC=5, DE=7, find AC.
Test site KQ: Pythagorean theorem.
The pythagorean theorem can be used to find the length of EC, then the length of CD, and then the length of AC.
Solution: ∫AC? CE，AD=BE= 13，BC=5，DE=7，
? EC= = 12，
DE = 7，
? CD=5，
? AC= = 12。
Comment on this question to examine students' application of the nature of right triangle and Pythagorean theorem.
1 1. As shown in the figure, there is a rectangular yard ABCD, where AB=9m and AD= 12m. There is a telephone pole at B and an electric lamp at E, which is 8m above the ground. What is the distance from point D to lamp E?
Library: Application of Pythagorean Theorem.
Analyze BD in Rt△ABD, and then use Pythagorean theorem in Rt△EBD to find the length of DE.
Solution: in Rt△BAD,? Bad =90? , meters,
In Rt△EBD, EBD=90？ , meters.
Therefore, the distance from point D to lamp E is17m.
This topic comments on the application of Pythagorean theorem, which belongs to the basic topic. The key to solve this problem is to master the expression of Pythagorean theorem skillfully.
12. As shown in the figure, a parallel beam of sunlight enters the classroom window. Xiao Qiang measured BC =1m.
NC= m, BN= m, AC=4.5m, MC=6m. Find the length of MA.
Library: Application of Pythagorean Theorem.
Firstly, the shape of △BCN is judged according to the inverse theorem of Pythagorean theorem, and then the conclusion is drawn from Pythagorean theorem.
Solution: ∫BC = 1m, NC= m, BN= m,
? BC2= 1，NC2=，BN2=，
? BC2+NC2=BN2，
? AC？ MC。
In Rt△ACM,
AC = 4.5m，MC=6m，MA2=AC2+CM2=56.25，
? MA=7.5 meters.
Comment on the application of Pythagorean theorem in this topic. First, judge AC according to the meaning of the question. MC is the key to solve this problem.
13. As shown in the figure, the rectangular body length 15, width 10, height 20, and the distance from point B to point C is 5. What is the shortest distance an ant needs to crawl from point A to point B along the surface of a cuboid?
Test site KV: plane expansion-shortest path problem.
The most direct way to analyze the shortest path between two points of a cuboid is to expand the edges of the cuboid and then use the shortest solution of the line segment between the two points.
Solution: Just cut the right side of the cuboid and form a rectangle with the plane where the front side is located, as shown in figure 1:
∫ The width of the cuboid is 10, the height is 20, and the distance from point B to point C is 5.
? BD=CD+BC= 10+5= 15，AD=20，
In the right triangle ABD, according to Pythagorean theorem:
? AB = = = 25
Just cut the right side of the cuboid and form a rectangle with the plane where the upper side is located, as shown in Figure 2:
∫ The width of the cuboid is 10, the height is 20, and the distance from point B to point C is 5.
? BD=CD+BC=20+5=25，AD= 10，
In the right triangle ABD, according to Pythagorean theorem:
? AB = = = 5；
Just cut the upper surface of the cuboid to form a rectangle with the plane where the back surface is located, as shown in Figure 3:
∫ The width of the cuboid is 10, the height is 20, and the distance from point B to point C is 5.
? AC=CD+AD=20+ 10=30，
In the right triangle ABC, according to Pythagorean theorem:
? AB = = = 5；
∵25 & lt； 5 ,
? The shortest distance for ants to crawl is 25.
Comment on this question mainly examines the shortest line segment between two points.
14. As shown in the figure, in the rectangular paper ABCD, AB= 18, fold the rectangular paper in half along the straight line AC, with point B at point E and point AE at point F. If AF= 13, find the length of AD.
Test site PB: Folding transformation (folding problem).
Through folding analysis:? EAC=？ BAC, AE=AB= 18, according to the properties of parallel lines, AF=FC= 13, then EF=5, and the length of EC, that is, the length of AD, is calculated by Pythagorean theorem.
Solution: from folding:? EAC=？ BAC，AE=AB= 18，
∵ quadrilateral ABCD is a rectangle,
? DC∨AB，
DCA=？ BAC，
EAC=？ DCA，
? FC=AF= 13，
∫AB = 18，AF= 13，
? EF= 18﹣ 13=5，
∵? E=？ B=90？ ,
? EC= = 12，
AD = BC = EC，
? AD= 12。
The comment on this question is a folding question. It is not difficult to examine the nature of rectangle and folding. It belongs to the conventional test type and is familiar with the situation that the two corresponding angles are equal before and after folding; When the inner angle is equal to the parallel line, an isosceles triangle is obtained, and the side length is calculated according to the equilateral angle, and the problem is solved by Pythagorean theorem.
15. As shown in the figure, for any eligible right triangle BAC, rotate 90 counterclockwise around its acute vertex? Get △DAE, so? BAE=90？ The quadrilateral ACFD is a square, and its area is equal to that of quadrilateral ABFE, which is equal to the sum of the areas of Rt△BAE and Rt△BFE. Write a method to prove Pythagorean theorem according to the diagram.
KR: proof of pythagorean theorem.
When analyzing and proving Pythagorean theorem, several congruent right triangles are used to form a positive figure, and then the area of the simplified quadrilateral ABFE is equal to the sum of the areas of Rt△BAE and Rt△BFE, and the Pythagorean theorem is obtained.
Solution: As can be seen from the figure:
The area of square ACFD = the area of quadrilateral ABFE = the sum of the areas of =Rt△BAE and Rt△BFE,
That is, s squared ACFD=S△BAE+S△BFE.
? b2= c2+，
Ordered: a2+b2=c2.
This review mainly examines the proof of Pythagorean theorem. There are many ways to prove Pythagorean theorem, generally using jigsaw puzzles. When solving problems, pay attention to: first use the puzzle method, and then use the area equality to prove the Pythagorean theorem.