1. PYTHAGORAS WORD PROBLEMS

2. Introduction/Course Description  Introduction  Introductory notes

3. Agenda  Math Minute  Discuss Agenda  Applying the Pythagorean Theorem to Word Problems

4. Word Problems  Today’s focus is going to be on the dreaded word problems. More often problems that need to be solved using the Pythagorean Theorem appear as word problems. This is where everything we have been doing in this unit comes together.

5. Word Problem  We actually see the Pythagorean Theorem used in our daily lives. For example a television is measured by the diagonal dimension of its screen. For example, a 24-in television has a diagonal measure of 24 in.

6. Word Problems  The first thing you always want to do when solving a word problem, is draw a picture. The pictures aids in making sure that you label the correct sides.

7. Word Problems – Example 1  A television screen is 16 in high and 22 in wide. What is its diagonal dimension to the nearest whole number?

8. Word Problems – Example 1  A television screen is 16 in high and 22 in wide. What is its diagonal dimension to the nearest whole number?  Remember to first draw a picture

9. Word Problems – Example 1  A television screen is 16 in high and 22 in wide. What is its diagonal dimension to the nearest whole number?  Remember to first draw a picture 16 in

10. Word Problems – Example 1  A television screen is 16 in high and 22 in wide. What is its diagonal dimension to the nearest whole number?  Remember to first draw a picture 16 in 22 in

11. Word Problems – Example 1  A television screen is 16 in high and 22 in wide. What is its diagonal dimension to the nearest whole number?  Remember to first draw a picture  a 2 + b 2 = c 2 16 in 22 in

12. Word Problems – Example 1  A television screen is 16 in high and 22 in wide. What is its diagonal dimension to the nearest whole number?  Remember to first draw a picture  a 2 + b 2 = c 2  16 2 + 22 2 = c 2 16 in 22 in

13. Word Problems – Example 1  A television screen is 16 in high and 22 in wide. What is its diagonal dimension to the nearest whole number?  Remember to first draw a picture  a 2 + b 2 = c 2  16 2 + 22 2 = c 2  256 + 484 = c 2  740 = c 2  27.2 = c  Therefore the television has a diagonal measurement of 27.2 inches. 16 in 22 in

14. Word Problems – Example 2  Two hikers start a trip from a camp walking 1.5 km due east. They turn due north and walk 1.7 km to a waterfall. To the nearest tenth of kilometer, how far is the waterfall from the camp?  Remember to first draw a picture

15. Word Problems – Example 2  Two hikers start a trip from a camp walking 1.5 km due east. They turn due north and walk 1.7 km to a waterfall. To the nearest tenth of kilometer, how far is the waterfall from the camp?  Remember to first draw a picture 1.5 km

16. Word Problems – Example 2  Two hikers start a trip from a camp walking 1.5 km due east. They turn due north and walk 1.7 km to a waterfall. To the nearest tenth of kilometer, how far is the waterfall from the camp?  Remember to first draw a picture 1.5 km 1.7 km

17. Word Problems – Example 2  Two hikers start a trip from a camp walking 1.5 km due east. They turn due north and walk 1.7 km to a waterfall. To the nearest tenth of kilometer, how far is the waterfall from the camp?  a 2 + b 2 = c 2 1.7 km 1.5 km

18. Word Problems – Example 2  Two hikers start a trip from a camp walking 1.5 km due east. They turn due north and walk 1.7 km to a waterfall. To the nearest tenth of kilometer, how far is the waterfall from the camp?  a 2 + b 2 = c 2  1.7 2 + 1.5 2 = c 2 1.7 km 1.5 km

19. Word Problems – Example 2  Two hikers start a trip from a camp walking 1.5 km due east. They turn due north and walk 1.7 km to a waterfall. To the nearest tenth of kilometer, how far is the waterfall from the camp?  a 2 + b 2 = c 2  1.7 2 + 1.5 2 = c 2  2.89 + 2.25 = c 2 1.7 km 1.5 km

20. Word Problems – Example 2  Two hikers start a trip from a camp walking 1.5 km due east. They turn due north and walk 1.7 km to a waterfall. To the nearest tenth of kilometer, how far is the waterfall from the camp?  a 2 + b 2 = c 2  1.7 2 + 1.5 2 = c 2  2.89 + 2.25 = c 2  5.14 = c 2 1.7 km 1.5 km

21. Word Problems – Example 2  Two hikers start a trip from a camp walking 1.5 km due east. They turn due north and walk 1.7 km to a waterfall. To the nearest tenth of kilometer, how far is the waterfall from the camp?  a 2 + b 2 = c 2  1.7 2 + 1.5 2 = c 2  2.89 + 2.25 = c 2  5.14 = c 2  2.3 = c  Therefore the waterfall is 2.3 km from camp. 1.7 km 1.5 km

22. Word Problems – Example 3  A carpenter is attaching a brace to the back of a frame that measures 30 in. by 40 in. What the is length, in inches, of the brace?  Remember to first draw a picture

23. Word Problems – Example 3  A carpenter is attaching a brace to the back of a frame that measures 30 in. by 40 in. What the is length, in inches, of the brace?  Remember to first draw a picture 40 in 30 in

24. Word Problems – Example 3  A carpenter is attaching a brace to the back of a frame that measures 30 in. by 40 in. What the is length, in inches, of the brace?  a 2 + b 2 = c 2 30 in 40 in

25. Word Problems – Example 3  A carpenter is attaching a brace to the back of a frame that measures 30 in. by 40 in. What the is length, in inches, of the brace?  a 2 + b 2 = c 2  30 2 + 40 2 = c 2 30 in 40 in

26. Word Problems – Example 3  A carpenter is attaching a brace to the back of a frame that measures 30 in. by 40 in. What the is length, in inches, of the brace?  a 2 + b 2 = c 2  30 2 + 40 2 = c 2  900 + 1600 = c 2 30 in 40 in

27. Word Problems – Example 3  A carpenter is attaching a brace to the back of a frame that measures 30 in. by 40 in. What the is length, in inches, of the brace?  a 2 + b 2 = c 2  30 2 + 40 2 = c 2  900 + 1600 = c 2  2500 = c 2 30 in 40 in

28. Word Problems – Example 3  A carpenter is attaching a brace to the back of a frame that measures 30 in. by 40 in. What the is length, in inches, of the brace?  a 2 + b 2 = c 2  30 2 + 40 2 = c 2  900 + 1600 = c 2  2500 = c 2  50 = c  Therefore the brace is 50 inches long. 30 in 40 in

29. Word Problems – Example 4  A diver swims 20 m under water to the anchor of a bout that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter.  Remember to first draw a picture

30. Word Problems – Example 4  A diver swims 20 m under water to the anchor of a bout that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter.  Remember to first draw a picture 20 m

31. Word Problems – Example 4  A diver swims 20 m under water to the anchor of a bout that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter.  Remember to first draw a picture 20 m 10 m

32. Word Problems – Example 4  A diver swims 20 m under water to the anchor of a bout that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter.  a 2 + b 2 = c 2 20 m 10 m

33. Word Problems – Example 4  A diver swims 20 m under water to the anchor of a bout that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter.  a 2 + b 2 = c 2  10 2 + b 2 = 20 2 20 m 10 m

34. Word Problems – Example 4  A diver swims 20 m under water to the anchor of a bout that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter.  a 2 + b 2 = c 2  10 2 + b 2 = 20 2  100 + b 2 = 400 20 m 10 m

35. Word Problems – Example 4  A diver swims 20 m under water to the anchor of a bout that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter.  a 2 + b 2 = c 2  10 2 + b 2 = 20 2  100 + b 2 = 400  b 2 = 400 – 100 20 m 10 m

36. Word Problems – Example 4  A diver swims 20 m under water to the anchor of a bout that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter.  a 2 + b 2 = c 2  10 2 + b 2 = 20 2  100 + b 2 = 400  b 2 = 400 – 100  b 2 = 300  Therefore the brace is 50 inches long. 20 m 10 m

37. Word Problems – Example 4  A diver swims 20 m under water to the anchor of a bout that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter.  a 2 + b 2 = c 2  10 2 + b 2 = 20 2  100 + b 2 = 400  b 2 = 400 – 100  b 2 = 300  b = 17.3 m  Therefore the buoy is 17.3 meters from where the diver started. 20 m 10 m

38. Word Problems  As a team, work on the worksheet. Your exit slip will be one of the problems from the worksheet.

39. Exit Slip  Your exit slip is problem number  Group 6: Problem #1  Group 7: Problem #2  Group 8: Problem #3  Group 9 and 10: Problem #4