Lesson Plan by: Stephanie Miller

Transcription

1 Lesson: Pythagorean Theorem and Distance Formula Length: 45 minutes Grade: Geometry Academic Standards: MA.G Find the lengths and midpoints of line segments in one- or two-dimensional coordinate systems. MA.G Prove and use the Pythagorean Theorem. Performance Objectives: 1) Given thirteen problems, the geometry students will apply the Pythagorean Theorem to determine the length of the side of a right triangle at least 11 times correctly. 2) Given 5 problems, the geometry students will apply the Distance Formula to determine the length of a line segment at least 4 times correctly. Assessment: Students will be assigned 18 problems (pp #2-36 evens) which will be solved correctly with at least 83% accuracy. Advance Preparation by Teacher: 1) Print off note taking guide for each student. 2) Open PowerPoint labeled Pythagorean Theorem Procedure: Introduction: Previously, we have been working on isosceles and equilateral triangles and their many properties. Today we will be moving on to right triangles. What we are about to learn today is probably one of the most important concepts in mathematics. In fact, you benefit from this concept every day because it plays a big part in making sure buildings and bridges do not collapse from lack of support. Step by Step Plan: 1. Hand out note taking guide for section 4.4 to the students. 2. [Second slide] We will begin today by going over a few terms that you need to know in order to better understand the lesson. 3. [Third slide] Since we are talking more today about right triangles, it is important to know the difference between a leg and a hypotenuse. 4. Call on student to read the definition of leg aloud. 5. Call on a different student to read the definition of hypotenuse aloud. 6. [Press enter] Now that we know the definition of a leg and hypotenuse, can anyone tell me which side is the hypotenuse? (Bloom s Knowledge, Answer: side labeled c ) 7. Correct, side c is the hypotenuse because it is opposite the right angle. Sides a and b are legs because they help form the right angle. 8. [Fourth slide] Okay, now you have an understanding of the terminology, we are moving on to learn about the Pythagorean Theorem. 9. [Fifth slide] This is the type of story problem we will be working on solving today. 10. Ask a student to read the problem aloud. 11. Now, let s set this problem up. 12. [Sixth slide] Here, we have a boat on a lake.

2 13. [Press enter] We drop the over the side of the boat and allow a little bit of slack. Notice there is some extra chain length. Because of this extra chain length, the boat is able to drift slightly 14. [Press enter] Eventually though, the chain is pulled taut, creating a straight line. Can anyone see the shape we have created? (Answer: right triangle) 15. [Press enter] Exactly! We have created a right triangle. The dotted lines represent the legs of the triangle. 16. How long is this leg that goes straight down? (Answer: 19 feet) Note: Many will probably answer 16 feet because that is how deep the lake is, but the chain went over the edge of the boat, which is 3 feet above the surface of the lake. 17. [Press enter] What is the name of this long side of the triangle? (Bloom s Knowledge, Answer: hypotenuse) 18. And how long is the hypotenuse? (Answer: 21 feet) Note: Student may struggle to answer this question. 19. [Press enter] Yes, this hypotenuse actually represents the total amount of chain that was let out, which means it includes the 19 feet of chain needed to hit the bottom and the additional two feet of slack. 20. [Press enter] What we don t know, but want to find out is how far we drifted. The distance traveled is represented by this top leg of the triangle, which we will label b 21. [Seventh slide] This is the same triangle we just created, without the background pictures. We will be able to use what we learn today to determine the length of b, or how far the boat drifted. As you listen to the rest of the lesson, try to figure out what you could do to solve this problem because we will come back to it later. 22. [Eighth slide] Everyone get out a sheet of graph paper and a straightedge. If you do not have a straight edge, use the side of your paper. Draw a right triangle with legs of 3 units each. For each side, draw a square where one side of the square shares a side with the triangle. Note: This may have to be demonstrated for one side. (Gardener: Visual-Spatial) 23. Once you have done that, see if you can find the relationship between the areas of the squares that you drew. Then, using a different sized right triangle, test to see if the relationship holds true for that triangle as well. 24. Allow time for students to complete this activity. 25. What did you find? (Bloom s Analysis, Answers will vary, but should be along the lines that the area of the squares of the legs sum up to the area of the square of the hypotenuse) 26. [Ninth slide, press enter] So when you add the size of the two smaller squares, you get the size of the larger square. 27. [Press enter] Looking at this picture, the area of square with side length a has an area of a squared and the area of square with side length b has an area of b squared. Adding them together gives you the area of square with side length c which is c squared. 28. [Press enter twice] Thus, we have the relationship a squared plus b squared equals c squared. 29. [Tenth slide] With this activity, we have actually just defined the Pythagorean Theorem. 30. Ask student to read the theorem aloud. 31. [Eleventh slide] Now let s use this theorem we just learned. 32. [Twelfth slide] The easiest type of problem is to use the theorem to determine the length of the hypotenuse. 33. [Press enter] The first step when using this formula is to determine where the hypotenuse is. The theorem stated it only applied to right triangles. For what reason do you think there is a rule that we can only use right triangles? (Bloom s Synthesis, Answer: Hypotenuses only exist on right triangles) Note: Students attention may need to be reverted back to the first step in order to answer this question. 34. [Press enter] The next step is to set up the formula. It might be easier for you, while you are still learning the formula, to first write it down then substitute in the numbers for the problem

3 35. [Press enter] The third step is to solve for the unknown variable. This will sometimes include some algebra. 36. [Press enter] The last step is the conclusion. For the conclusion simply write the answer in the context of the problem and include the proper unit. 37. [Thirteenth slide] Alright, let s try it. Find the length of the hypotenuse for the given triangle. 38. [Press enter] Step 1: Determine which side is the hypotenuse. So which side is the hypotenuse? (Answer: side labeled x ) 39. [Press enter twice] Step 2: Set up the formula. So we are going to start with a squared plus b squared equals c squared. 40. [Press enter] We know that c is the hypotenuse which we just determined is the side with length x, so we can replace c with x. For the legs, we will let 4 represent a and 3 represent b. Keep in mind that you could switch it and make 4 be b and 3 be a. 41. [Press enter] Step 3: Solve for the unknown, which is x in this problem. 42. [Press enter] Here s our formula. We are going to square 4 and [Press enter] Now we have 16+9=x squared. Add these two numbers together. 44. [Press enter] 25=x squared. Now we need to get rid of the squared, so we take the square root 45. [Press enter twice] and we get x= [Press enter] Step 4: Conclusion. So we say 47. [Press enter] The length of the hypotenuse is 5 units long 48. [Fourteenth slide] Now let s move on to another type of problem, find the length of the leg of a triangle. We will use the same steps that we just used. 49. [Press enter] Which side is the hypotenuse? (Answer: side with length 6 ) 50. [Press enter twice] So if we are going to set up the formula, what would we put as our c? (Answer: 6) 51. What would we put for our a? (Answer: x or 5 ) 52. [Press enter] And we would make the other side be our b. So we have x squared plus 5 squared equals 6 squared. Note: If students said 5 in step 53, they may have a different order. This does not matter. 53. [Press enter twice] So now we solve for the unknown. Let s square 5 and [Press enter] Now subtract 25 from both sides. 55. [Press enter twice] And to get rid of the square we take the square root of both sides. 56. [Press enter twice] So we have x=the square root of 11 which is approximately [Press enter] And our conclusion is 58. [Press enter] The length of the unknown side is approximately 3.32 units. 59. [Fifteenth slide] You need to be careful because there are a few common mistakes students make when using this formula. First, students apply the formula when it is not a right triangle. Second, sometimes students let their unknown always be c instead of reserving c for the hypotenuse. Third, sometimes students forget to get rid of the square to get the unknown by itself. Doing any of these will give you a wrong answer, so be careful. 60. [Sixteenth slide] Now we are going to talk about the Distance Formula. 61. [Seventeenth slide] If you have a point A on the coordinate plane at x sub one, y sub one, and you have a point B at x sub 2, y sub 2, and you want to find the distance between them you use the following formula: The length of line segment AB is equal to the square root of the quantity x sub 2 minus x sub 1 squared minus the quantity y sub 2 minus y sub 1 squared. 62. [Eighteenth slide] Here I will show you what that means geometrically. We have a coordinate plane with two points, A and B, and we want to find the length of the line segment that connects them, which is also the distance between them. What we do is make this line the hypotenuse of a right triangle.

4 63. [Press enter] We can easily find to length of these legs. To find the bottom leg, we simply find the difference between the x coordinates. To find the length of the side leg, we simply find the difference of the y coordinates. 64. [Press enter] How would you solve for this unknown side using what we have learned today? (Bloom s Application, Answer: Pythagorean Theorem) 65. [Press enter] Now that we have a triangle, and the length of the two legs, we can solve for the length of the hypotenuse using the Pythagorean Theorem. 66. [Press enter] So we have a squared, which is x sub 2 minus x sub 1 and b squared, which is y sub 2 minus y sub 1 equals AB squared. 67. Can anyone see how this equation is related to the Distance Formula? (Bloom s Analysis, Answer: They are the same once we get rid of the exponent on the left side) 68. [Press enter] Exactly, taking the square root of both sides gives us the Distance Formula. 69. [Nineteenth slide] Alright, so let s do a quick example. 70. [Twentieth slide] Find the difference between points F and G. 71. [Press enter] First, plot the points and the line between them. 72. [Press enter] So we have point F in red at (0,1) and point G in blue at (3,3) and a black line that connects them. 73. [Press enter] Now add legs to make a right triangle. 74. [Press enter] So we added this red line and this blue line from the points F and G to create a right triangle. 75. [Press enter] Finally, use the Distance Formula. 76. [Press enter] We have line segment FG is equal the square root of the quantity x sub 2 which is the x coordinate of our G point minus x sub 1 which is the x coordinate of our F point squared plus the quantity of y sub 2 which is the y coordinate of G minus y sub 1 which is the y coordinate of F. Now let s simplify. 77. [Press enter] We subtracted 3 minus 1 and 3 minus zero. Now square those numbers. 78. [Press enter] So we have 2 squared plus 3 squared under the square root. Square them. 79. [Press enter] Now we have under the square root 4 plus [Press enter] This gives us the square root of thirteen which is approximately 81. [Press enter] [Twenty-first slide] Ok, so now you should have a pretty good idea of how to use the Pythagorean Theorem and Distance Formula, so let s go back to our original problem statement, which is on your note guide. 83. [Twenty-second slide] Every find a partner and see if you can figure out the distance the boat drifted. When you figure it out, show your work on the board and take a seat. (Gardener: Interpersonal and Logical-Mathematical) 84. Allow time for students to solve the problem. Check each groups answer. (Answer: square root of 80 which is approximately 8.94) Ask them either how they arrived at the correct answer or where they found difficulties. 85. [Twenty-third slide] Are there any questions? If not, here is the assignment due tomorrow. Please get started on it. If you have any questions, please raise your hand, and I will come to you. Closure: 2 or 3 minutes before dismissal, ask students to use a scratch piece of paper to write down the Pythagorean Theorem and Distance Formula without looking at their notes or book. This is not for a grade; it is to help them memorize the formulas because they will use them often in the future. Adaptations/Enrichments: Student with a learning disability in mathematical calculation: The visuals for the opening problem should make the student understand how the triangle we used to solve the

5 problem came from a real-world situation. Providing the students with a step by step instruction should make it easier for them to solve the problems. Student with ADHD: Giving the students an activity and a chance to get up to write on the board will keep them from getting bored as easily and therefore, less likely distracted during the lesson. Student with Gift/Talents in Math: Instead of solving # s 2 and 4, ask student to use what he or she has learned today to explain how they would find whether any triangle was a right triangle or not, given all of the lengths of the sides. (Answer: See if the Pythagorean Theorem still applies.) Ask them to come up with their own problem using the Pythagorean Theorem if they get done early. Self-Reflection: Did the students seem to understand how to do the problems, or did they depend on my help? Was there something I could have explained better? Did the activity seem to help the students understand the concept?