1. 3.3 How Can I Find the Height? Pg. 9 Heights and Areas of Triangles

2. 3.3 – How Can I Find the Height?__________ Heights and Area of Triangles What if you want the area of only half a rectangle? Today you will discover how to find the area of a triangle. Then you will find a shortcut to find either the hypotenuse or leg of a right triangle.

3. 3.10 –AREA OF SHADED PART Find the area of the shaded figure. Be ready to share any observations.

4. 3.11 –AREA OF A TRIANGLE Isaiah claimed that he didn't need to calculate the area for (b) and (c) because it must be the same as the area for the triangle in part (a). a. Explain why the area of any triangle is A triangle is half of a rectangle

5. http://hotmath.com/util/hm_flash_movie_full. html?movie=/hotmath_help/gizmos/triangleA rea.swf Area of Triangle

6. b. Sketch triangles that meet the three conditions below. Height InsideHeight on SideHeight Outside

7. They all have the same area

8. 3.12 –AREA OF A TRIANGLE, CONT How do you know which dimensions to use when finding the area of a triangle? a. Find the area of each shaded figure. Draw any lines on the paper that will help. Turning the triangles may help you discover a way to find their areas.

9. A = 126un 2

10. A = 15un 2

11. 7 8 A = 28un 2

12. 3.13 –SIDE OF A SQUARE What do you notice about the square at right? a. Eunice does not know how to solve for x. Explain to her how to find the missing dimension. = 10

13. b. What if the area of the shape above is instead 66ft 2 ? What would x be in that case? 66

14. Between 5 and 6

15. Between 4 and 5 Find what integers the square root is between.

16. Between 11 and 12

17. Find what integers the square root is between. Between 10 and 11

18. d. You can simplify square roots by using a factor tree. Pairs come out of the square root and "loners" are stuck inside. If possible, simplify the square roots.

19. 3.14 –LENGTH OF THE LONG SIDE OF A TRIANGLE While Alexandria was doodling on graph paper, she made the design at right. She started with the shaded right triangle. She then rotated it 90° clockwise and translated the result so that the right angle of the image was at B. She continued this pattern until she completed the square.

21. a. Draw Alexandria's design below. Be sure it is to scale. It is already started for you. What is the shape of quadrilateral ABCD? How can you tell?

22. 3 73 7 7 3

23. square

24. b. What is the shape of the inner quadrilateral? How do you know? square

25. c. What is the area of the inner quadrilateral? Area = Area of Big square – Area of 4 triangles

26. 3 7 3 7 3 7 3 7 10.5 10

27. c. What is the area of the inner quadrilateral? Area = Area of Big square – Area of 4 triangles 100 – 42 58

28. 3 7 3 7 3 7 3 7 10.5 10 58

29. d. What's the length of the longest side of the shaded triangle?

30. 3.16 – ANOTHER WAY Robert complained that while the method from the previous problem works, it seems like too much work! He decides to use the area of the triangles, the inner square, and the large outer square to look for a short cut.

31. a. Find the area of the large square around the entire shape. (a+b)(a+b) a 2 + 2ab + b 2

32. b. Find the area of the inner square. Then find the TOTAL area of the four triangles. inner = c 2 4(½bh) = 2ab

33. c. Since the large square has the same area as the 4 triangles and inner square, set them equal to each other and reduce. a 2 + 2ab + b 2 - 2ab = c 2 a 2 + b 2 = c 2

34. 3.17 – A SHORTER WAY Build a square off of each side of the following triangles (the first one is done for you.) Then find the area of each square. What is the relationship between the areas of the squares?

35. 9 16 25 9 + 16 = 25

36. 144 169 25 + 144 =169

37. 64 225 289 64 + 225 =289

38. http://www.cpm.org/flash/technology/pyt hagoreanv1.2.swf

39. PYTHAGOREAN THEOREM: (Peh-Tha-Gore-Ian) leg hypotenuse a b c c 2 = a 2 + b 2 If a triangle is a right triangle, then (hypotenuse) 2 = (leg) 2 + (leg) 2

40. 3.18 – PYTHAGOREAN THEOREM For each triangle below, find the value of the variable. Write answers in simplified square root form.

41. 121= y 2 +49 72 = y 2 11 2 = y 2 + 7 2

42. x 2 = 25 + 16 x 2 = 41 x 2 = 5 2 + 4 2

43. 196 = x 2 +64 132 = x 2 14 2 = x 2 + 8 2

44. 49 = x 2 +25 24 = x 2 7 2 = x 2 + 5 2

45. 45 = x 2 +9 36 = x 2

46. x 2 = 12 + 36 x 2 = 48