Mathematical Pythagorean Theorem The first part of mathematical thought is the essence of mathematical knowledge, and it is also a bridge to transform knowledge into ability. Flexible use of mathematical ideas can effectively improve the ability to analyze and solve problems and enhance the awareness of applying mathematical knowledge. In this chapter of Pythagorean Theorem, there are many important mathematical ideas. Here are some examples.
First of all, the concept of equation
In graphs with right triangles, Pythagorean theorem is often used to find the length of line segments. If Pythagorean theorem cannot be directly used for calculation, it needs to be solved by column equation.
Second, the idea of conversion.
The idea of transformation is to change the form of the problem that needs to be solved into another problem that has been solved or easily solved through certain methods or channels, so that the original problem can be solved.
Example 3, as shown in Figure 3, has a rectangular body length 15cm, a width 10cm and a height of 20cm. The distance between point B and point C is 5 cm. If a snail wants to climb from point A to point B along the surface of a cuboid, what is the shortest crawling distance?
Analysis: Because the snail crawls along the surface of the cuboid, it is necessary to expand the cuboid into a plane figure. According to the shortest line segment between two points, there are two possibilities for snails to crawl a short distance, as shown in Figures 4 and 5. It is easy to find the length of AB in two figures by Pythagorean theorem. After comparison, it can be found that the shortest distance for snails to crawl is 25cm.
Note: here, through the expansion diagram of a cuboid, the three-dimensional figure is transformed into a plane figure, and the shortest path problem of snail crawling is reduced to the problem of finding the distance between two points by pythagorean theorem.
Example 4 is a quadrilateral grass ABCD as shown in Figure 6, in which? A = 60O，？ B =？ D = 90O, AB = 20m, CD = 10m, find the length of AD and BC (accurate to 0. 1m,? 1.732).
(Tianjin Senior High School Entrance Examination in 2004)
Analysis: What if there is no right triangle in the picture? Imagine a right triangle with an angle of 30O, thus extending the intersection of AD and BC at point E, then? E = 30O，AE = 2AB = 40m，CE = 2CD = 20m。 According to Pythagorean theorem, DE == m, BE == m, so AD = 40? 22.7m，BC = 20？ 14.6m.
Description: This problem makes full use of the characteristics of known graphs and skillfully transforms quadrilateral problems into right triangle problems by constructing new graphs.
Third, the combination of numbers and shapes.
The combination of numbers and shapes is to grasp the essential relationship between numbers and shapes, combine abstract mathematical language with intuitive graphics, and pass? Use form to help numbers? Or? Solve the shape by number? Simplify complex problems and concretize abstract problems, so as to achieve the goal of solving problems quickly.
Example 5 There are two monkeys in a tree 10m high. One of them climbed down the tree and went straight to the pond 20 meters away from the tree, and the other climbed to the top of the tree and went straight to the pond. If two monkeys pass the same distance, how tall is the tree? (Longyan City, Fujian Province, 2005)
Analysis: Draw Chart 7 according to the meaning of the question, where D is the top of the tree, AB = 10m, C is the pond, and AC = 20m. Let BD = (m), then the height of the tree is AD = (+10) m) m. Because AC+AB = BD+DC, DC = (30) m. Equation 202+(+ 10)2 = (30)2 can be obtained from Pythagorean theorem.
Explanation: Pythagorean theorem itself is a model combining numbers and shapes, which makes a right triangle have a right angle. Shape? Features, into three sides? Count? The key to solving practical problems by Pythagorean theorem is to transform practical problems into right triangle models through the combination of numbers and shapes, and then solve them by equations.
Fourth, the idea of classified discussion.
In the process of solving problems, when the conditions or conclusions are uncertain or not unique, there are often several possible situations, which requires classifying the problems according to certain standards and then solving them separately for different situations. Finally, the conclusion of the whole problem is drawn by synthesizing various results. Classification discussion is essentially a kind of. Break the whole into parts, one by one, and then add up to the whole? The mathematical method of.
Example 6 If the two sides of a right triangle are 3cm and 4cm respectively, then the length of the third side is _ _ _ _ _.
Analysis: In this question, it is known that both sides of a right triangle are long, but it is not specified whether it is a right side or a hypotenuse. Therefore, it is necessary to discuss the classification, and the answer is 5cm or cm.
Example 7? Shuguang Middle School? There is a triangular flower bed ABC, which can be measured directly now. A = 30O，AC = 40，BC = 25。 Please find out the area of this flower bed.
Analysis: Because the title does not clearly tell us the shape of △ABC, it needs to be discussed in two situations.
In fig. 8, s △ ABC =10 (20+15) m2;
In fig. 9, S△ABC= 10(20 15) m2.
Note: Because there are no numbers in the title, such questions often need to be discussed in categories, and it is easy to miss the answer because of poor consideration. I hope the students will understand it carefully.
V. Overall thinking
For some mathematical problems, it is difficult to solve them by sticking to the rules and starting from the local; If you think about a certain part or parts of a question as a whole, you can broaden your mind and answer the question quickly.
Example 8 It is known that the circumference of a right triangle is 30cm and the length of the hypotenuse is 13cm, so the area of this triangle is _ _ _ _ _.
Analysis: Let the length of two right-angled sides of this right-angled triangle be and the hypotenuse be, then = 30 13 = 17, so (+) 2 = 2+2 = 172 = 289, which is 65438+ from the Pythagorean theorem.
Note: what we require is the area, that is, we don't need to find the value of sum separately, just the whole one.
Example 9: As shown in figure 10, seven squares are placed on a straight line in turn. It is known that the areas of three squares placed obliquely are 1, 2, 3, and the areas of four squares placed sequentially are S 1, S2, S3, S4, so S 1+S2+S3+S4 = _ _.
Analysis: according to the known conditions, AC = EC,? ABC =？ CDE = 90O, which can be easily proved by the complementary relationship of angles? ACB =？ CED, so we can get △ ABC △ CDE, so BC = ED. In Rt△ABC, we can get AC2 = AB2+BC2 = AB2+DE2 by Pythagorean theorem. From S 1 = AB2, S2 = DE2, AC2 = 1, there is s.
Note: This problem does not directly solve S 1, S2, S3, S4, but solves S 1+S2, S3 +S4 with the help of Pythagorean theorem, which reflects the flexible application of the whole idea in solving problems.
The second part of mathematical Pythagorean Theorem, mathematical thinking method, is the guiding ideology and universally applicable method with specific mathematical content as the carrier and higher than specific mathematical content. It can make people understand the true meaning of mathematics and play a guiding and regulating role in people's thinking activities of learning and applying mathematical knowledge to solve problems. Mi Shan Kunsan, a Japanese mathematics educator, believes that if students have no chance to apply mathematics after entering the society, then mathematics, as knowledge, will usually be forgotten after leaving school for less than a year or two. However, no matter what industry they are engaged in, the mathematical spirit and mathematical thinking method engraved in their minds will play an important role in their lives and work for a long time. Flexible use of mathematical thinking methods to solve problems can often turn difficult into easy, turn decadent into magic, and get twice the result with half the effort. The following is an example of mathematical thought permeated in Pythagorean theorem.
First, the idea of classification.
Example 1. (20 13) If two sides of a right triangle are 3 and 4 respectively, then the length of the third side is ().
Comment: What is the error-prone point of this question? Hook three strands, four strings and five? As a result, the side with a side length of 4 is directly regarded as a right-angled side, which leads to the wrong choice of A and the mistake of not considering the problem comprehensively.
Second, the idea of equation.
Example 2. (JINAN 20 13) As shown in Figure 1, Xiao Liang pulled the flag-raising rope to the bottom of the flagpole, and the rope end just touched the ground, and then pulled the rope end to a position 8m away from the flagpole. It is found that the end of the rope is 2m away from the ground, so the height of the flagpole (the part above the pulley is ignored) is ().
a . 12mb . 13mc . 16md . 17m
Analysis: Observing the figure, when the end of the rope is pulled to a position 8 meters away from the flagpole, a vertical line can be drawn on the flagpole through the end of the rope, so that a right triangle can be obtained, and then the height of the flagpole is set as an unknown, and then the equation can be solved by Pythagorean theorem.
Solution: As shown in Figure 2, if the flagpole height is X, then AC=AD=x, AB=x-2, BC=8.
In Rt△ABC, (x-2)2+82=x2 is obtained by Pythagorean theorem.
The solution is x= 17m, that is, the flagpole height is 17m, and the answer is d 。
Third, the general idea.
Example 3. (Yangzhou, Jiangsu, 20 13) If the difference between two adjacent sides of a rectangle is 2 and the diagonal length is 4, the area of the rectangle is _ _ _ _ _ _ _.
Analysis: Let two adjacent sides of a rectangle be A and B (A >; B), then according to the meaning of the question, a-b=2, a2+b2= 16. The area of the rectangle is equal to ab. The key is to find a way to transform the two equations into a formula containing ab.
Solution: Let two adjacent sides of a rectangle be A and B (A >; B), then a-b=2.
Fifth, the idea of combining numbers with shapes.
Example 5. (Zhangjiajie, Hunan, 20 13) As shown in Figure 4, in the plane rectangular coordinate system, the coordinates of vertices A and C of right-angled OABC are (10,0) and (0,4) respectively, point D is the midpoint of OA, and point P moves on BC. When △ODP is an isosceles triangle with a waist length of 5, the coordinates of point P are.
Analysis: It is easy to know that OD=5. To make △ODP an isosceles triangle with a waist length of 5, we can make a circle with point O as the center and OD as the radius. You can also make a circle with point D as the center and point OD as the radius.
Solution: From C (10/0,0), OD=5.
(1) Make a circle intersection with point O as the center and OD as the radius.
Examples of intransitive verb structure thinking 6. The same example 3
Analysis: According to the known conditions, the Pythagorean theorem is proved by combining the string diagram. This example has the following ingenious solutions.
Mathematical Pythagorean Theorem Part III Correct mathematical thinking is the key to successful problem solving. When using Pythagorean theorem to solve problems, if we can correctly grasp mathematical thinking, we can broaden our thinking and make the method simple and fast. The following are some mathematical ideas often used in the application of Pythagorean theorem for your reference.
First of all, the concept of equation
◆ Example 1 As shown in figure 1, there is a piece of paper with a right triangle, and its two right-angled sides AC=6cm and BC=8cm. Now the right-angle side AC is folded along the straight line AD, so that it falls on the hypotenuse AB, and the point C falls on the point E, then CD is equal to ().
A.2cm, A.2cm, A.2cm, a.2cm.
Analysis: from the meaning of the question, ACD and? AED is symmetrical about linear advertising, so there is? ACD？ AED。 Further, AE=AC=6cm, CD=ED, DE? AB。 Let CD=ED=xcm, then in? In DEB, we can get DE2+BE2=BD2 from Pythagorean theorem. In ABC, AB2=AC2+BC2=62+82= 100, AB= 10. So we have x2+( 10- 6) 2=(8- x)2, and the solution is x=3. So we choose B.
Second, change ideas.
Example 2 As shown in Figure 2, a cuboid is 3cm high, with a square bottom and 2cm sides. At present, a bug starts from A, crawls along the surface of a cuboid, and reaches C. What is the shortest distance that the bug walks?
Analysis: To find the shortest distance on the geometric surface, the geometric surface can usually be expanded and the three-dimensional figure can be transformed into a plane figure. For this problem, you can fold the right surface of a cuboid to the front surface, so that points A and C are * * * planes, and the length of line segment AC is the shortest distance (as shown in Figure 3). According to Pythagorean Theorem, AC2=32+42=52, that is, the shortest distance that a bug has traveled.
Third, the idea of classified discussion.
Where is Example 3? In ABC, AB= 15, AC=20, and the height of the side of BC = 12. Try to find out the length of BC.
Analysis: The height of one side of a triangle can be inside or outside the triangle, so the problem should be considered in two cases. When is the high AD near BC? The internal time of ABC is shown in Figure 4, and BD2=AB2-AD2 and BD = 9； are obtained from Pythagoras theorem; CD2=AC2-AD2 and CD= 16, then BC = BD+CD = 9+16 = 25; Where are the high advertisements in BC? The outer time of ABC, as shown in Figure 5, can also be obtained by Pythagorean theorem CD= 16 and BD=9. At this time BC=CD-BD= 16- 9=7, then the length of BC is 25 or 7.
Fourth, the combination of numbers and shapes.