1. The
Pythagorean
Theorem c a b

2. This is a right triangle:

3. We call it a right triangle because it contains a right angle.

4. The measure of a right angle is 90o

5. The little square
in the
angle tells you it is a
right angle.
90o

6. About 2,500 years ago, a Greek mathematician named Pythagorus discovered a special relationship between the sides of right triangles.

7. Pythagorus realized that if you have a right triangle,3 4 5

8. and you square the lengths of the two sides that make up the right angle,3 4 5

9. and add them together, 3 4 5

10. you get the same number you would get by squaring the other side.3 4 5

11. Is that correct? ? ?

12. It is. And it is true for any right triangle.8 6 10

13. The two sides which come together in a right angle are called

14. The two sides which come together in a right angle are called

15. The two sides which come together in a right angle are called
legs.

16. The lengths of the legs are usually called a and b.

17. The side across from the right angle is called the
hypotenuse. a b

18. And the length of the hypotenuse is usually labeled c.

19. The relationship Pythagorus discovered is now called The Pythagorean Theorem:b

20. The Pythagorean Theorem says, given the right triangle with legs a and b and hypotenuse c,

21. then c a b

22. You can use The Pythagorean Theorem to solve many kinds of problems.
Suppose you drive directly west for 48 miles, 48

23. Then turn south and drive for 36 miles.48 36

24. How far are you from where you started?48 36 ?

25. Using The Pythagorean Theorem,48
482 +
362 = c2 36 c

26. Why?
Can you see that we have a right triangle? 48 36 c
482
362 + = c2

27. Which side is the hypotenuse?
Which sides are the legs? 48 36 c
482
362 + = c2

28. Then all we need to do is calculate:

29. And you end up 60 miles from where you started.
So, since c2 is 3600, c is
60.
So, since c2 is 3600, c is 48 36 60

30. Find the length of a diagonal of the rectangle:
15"
8" ?

31. Find the length of a diagonal of the rectangle:
15"
8" ?
b = 8 c
a = 15

32. b = 8
a = 15 c

33. Find the length of a diagonal of the rectangle:
15"
8" 17

34. Practice using The Pythagorean Theorem to solve these right triangles:

35. 5 12 c
= 13

36. 10 b 26

37. = 24 10 b 26
(a)
(c)

38. 12 b 15
= 9

39. 7.2 The Converse of the Pythagorean Theorem
If c2 = a2 + b2, then ∆ABC is a right triangle. B c a C A b

40. Verify a Right Triangle
Is ∆ABC a right triangle?
Yes, it is a right triangle. C 16 12 A B 20

41. Classifying Triangles
In ABC with longest side c:
If c2 < a2 + b2, then ABC is acute.
If c2 = a2 + b2, then ABC is right.
If c2 > a2 + b2, then ABC is obtuse. C b a B A c B c a A C b B c a A C b

42. Acute Triangles Show that the triangle is an acute triangle.
Because c2 < a2 + b2, the triangle is acute. 5 4

43. Obtuse Triangles Show that the triangle is an obtuse triangle.
Because c2 > a2 + b2, the triangle is obtuse. 12 8

44. Classify Triangles Classify the triangle as acute, right, or obtuse.
Because c2 > a2 + b2, the triangle is obtuse. 6 5

45. Classify Triangles
Classify the triangle with the given side lengths as acute, right, or obtuse.
A. 4, 6, 7
Because c2 < a2 + b2, the triangle is acute.

46. Classify Triangles
Classify the triangle with the given side lengths as acute, right, or obtuse.
B. 12, 35, 37
Because c2 = a2 + b2, the triangle is right.

47. 7.3 Similar Right Triangles
Geometry
Mr. Lopiccolo

48. Objectives/Assignment
Solve problems involving similar right triangles formed by the altitude drawn to the hypotenuse of a right triangle.
Use a geometric mean to solve problems such as estimating a climbing distance.
Assignment: pp (4-26) even

49. Proportions in right triangles
In Lesson 6.4, you learned that two triangles are similar if two of their corresponding angles are congruent. For example ∆PQR ~ ∆STU. Recall that the corresponding side lengths of similar triangles are in proportion.

50. Activity: Investigating similar right triangles. Do in pairs or threes
Cut an index card along one of its diagonals.
On one of the right triangles, draw an altitude from the right angle to the hypotenuse. Cut along the altitude to form two right triangles.
You should now have three right triangles. Compare the triangles. What special property do they share? Explain.
Tape your group’s triangles to a piece of paper and place in labwork.

51. Theorem 7.5
If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other.
∆CBD ~ ∆ABC, ∆ACD ~ ∆ABC, ∆CBD ~ ∆ACD

52. A plan for proving thm. 7.5 is shown below:
Given: ∆ABC is a right triangle; altitude CD is drawn to hypotenuse AB.
Prove: ∆CBD ~ ∆ABC, ∆ACD ~ ∆ABC, ∆CBD ~ ∆ACD
Plan for proof: First prove that ∆CBD ~ ∆ABC. Each triangle has a right triangle and each includes B. The triangles are similar by the AA Similarity Postulate. You can use similar reasoning to show that ∆ACD ~ ∆ABC. To show that ∆CBD ~ ∆ACD, begin by showing that ACD  B because they are both complementary to DCB. Then you can use the AA Similarity Postulate.

53. Ex. 1: Finding the Height of a Roof
Roof Height. A roof has a cross section that is a right angle. The diagram shows the approximate dimensions of this cross section.
A. Identify the similar triangles.
B. Find the height h of the roof.

54. Solution:
You may find it helpful to sketch the three similar triangles so that the corresponding angles and sides have the same orientation. Mark the congruent angles. Notice that some sides appear in more than one triangle. For instance XY is the hypotenuse in ∆XYW and the shorter leg in ∆XZY.
∆XYW ~ ∆YZW ~ ∆XZY.

55. Solution for b. Use the fact that ∆XYW ~ ∆XZY to write a proportion.
Corresponding side lengths are in proportion. = ZY XZ h
3.1 =
Substitute values.
5.5
6.3
6.3h = 5.5(3.1)
Cross Product property
Solve for unknown h.
h ≈ 2.7
The height of the roof is about 2.7 meters.

56. Using a geometric mean to solve problems
In right ∆ABC, altitude CD is drawn to the hypotenuse, forming two smaller right triangles that are similar to ∆ABC From Theorem 9.1, you know that ∆CBD ~ ∆ACD ~ ∆ABC.

57. Write this down!
Notice that CD is the longer leg of ∆CBD and the shorter leg of ∆ACD. When you write a proportion comparing the legs lengths of ∆CBD and ∆ACD, you can see that CD is the geometric mean of BD and AD.
Longer leg of ∆CBD.
Shorter leg of ∆CBD. BD CD = CD AD
Shorter leg of ∆ACD
Longer leg of ∆ACD.

58. Copy this down!
Sides CB and AC also appear in more than one triangle. Their side lengths are also geometric means, as shown by the proportions below:
Shorter leg of ∆ABC.
Hypotenuse of ∆ABC. AB CB = CB DB
Hypotenuse of ∆CBD
Shorter leg of ∆CBD.

59. Copy this down!
Sides CB and AC also appear in more than one triangle. Their side lengths are also geometric means, as shown by the proportions below:
Longer leg of ∆ABC.
Hypotenuse of ∆ABC. AB AC = AC AD
Hypotenuse of ∆ACD
Longer leg of ∆ACD.

60. Geometric Mean Theorems
Theorem 7.6: In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of the altitude is the geometric mean of the lengths of the two segments
Theorem 7.7: In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of each leg of the right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg. BD CD = CD AD AB CB = CB DB AB AC = AC AD

61. What does that mean? 6 x 5 + 2 y = = x 3 y 2 18 = x2 7 y = √18 = x y 2

62. Ex. 3: Using Indirect Measurement.
MONORAIL TRACK. To estimate the height of a monorail track, your friend holds a cardboard square at eye level. Your friend lines up the top edge of the square with the track and the bottom edge with the ground. You measure the distance from the ground to your friend’s eye and the distance from your friend to the track.

63. In the diagram, XY = h – 5.75 is the difference between the track height h and your friend’s eye level. Use Theorem 9.2 to write a proportion involving XY. Then you can solve for h.

64. 7.4 Special Right Triangles
*You will be able to find the lengths of sides of special right triangles
And

65. Special Right Triangles
Leg:Leg:Hypotenuse
Short Leg:Long Leg:Hypotenuse

66. LEGS ARE THE SAME LENGTH
In a triangle…
LEGS ARE THE SAME LENGTH
We will use a reference triangle to set up a proportion then solve.

67. This is our reference triangle for the 45-45-90.
Right Triangle 1 1
This is our reference triangle for the

68. Right Triangle x x

69. EX: 1 Solve for x
Let’s set up a proportion by using our reference triangle. x 3 1 3 x 3 1 1

70. EX: 2 Solve for x x 5 1 5 x 5 1 1

71. EX: 3 Solve for x x 3 45 3 1 1 1 x

73. We will use a reference triangle to set up a proportion then solve.
Right Triangle 60 2 1 30
This is our reference triangle for the triangle.
We will use a reference triangle to set up a proportion then solve.

74. Right Triangle
60 2x x
30

75. Ex: 1 60 8 60 2 x 1 30 30 y y 8 x 8 2 1 2

76. Solve for x
Ex: 2 30 60 2 x 1 24 60 30 24 x 2 1
2x = 24
x = 12

77. Ex: 3 30 60 2 14 1 y 30 60 x 14 y 14 x 2 2 1
2x = 14
x = 7

78. Ex: 4 x 60 2 1 60 30 y 30 x y 1 2
y = 10
x = 5

79. Ratios in Right Triangles
WHAT YOU WILL LEARN
To find trigonometric ratios using right triangles, and
To solve problems using trigonometric ratios.

80. Great Chief Soh Cah Toa Chief Soh Cah Toa.
A young brave, frustrated by his inability to understand the geometric constructions of his tribe's battle dress, kicked out in anger against a stone and crushed his big toe. Fortunately, he learned from this experience, and began to use study and concentration to solve his problems rather than violence. This was especially effective in his study of math, and he went on to become the wisest man of his tribe. He studied many aspects of trigonometry; and even today we remember many of the functions by his name. When he became an adult, the tribal priest gave him a name that reflected his special nature -- one that reminded them of his great discoveries and of the event which changed his life. Because he was troubled throughout his life by the problematic foot, he was constantly at the edge of the river, soaking his aches in the cooling waters. For that behavior, he was named
Chief Soh Cah Toa.

81. DEFINITIONS Sine: Opposite side over hypotenuse.
Cosine: Adjacent side over hypotenuse.
Tangent: Opposite side over adjacent.

82. Example
Find the sin S, cos S, tan S, sin E, cos E and tan E. Express each ratio as a fraction and as a decimal. M
Sin S = ME/SE = 3/5 or 0.6
Cos S = SM/SE = 4/5 or 0.8 4 3
Tan S = ME/SM = ¾ or 0.75
Sin E = SM/SE = 4/5 or 0.8 S E
Cos E = ME/SE = 3/5 or 0.6 5
Tan E = SM/ME = 4/3 or 1.3

83. Example
Find each value using a calculator. Round to the nearest ten thousandths.
Cos 41
Sin 78

84. Example
A plane is one mile about sea level when it begins to climb at a constant angle of 2 for the next 70 ground miles. How far above sea level is the plane after its climb? 2 h
1 mi
Sea level
70 mi
Tan 2 = h/70
70 tan 2 = h
h = 2.44

85. Example Show that the sin and cos give you the same hypotenuse. 4 3 3753 3

86. The End!

88. Pronounced “sign inverse”
SINE-1
Pronounced “sign inverse”

89. Pronounced “co-sign inverse”
COSINE-1
Pronounced “co-sign inverse”

90. Pronounced “tan-gent inverse”

91. Represents an unknown angle
Greek Letter q
Prounounced “theta”
Represents an unknown angle

92. hypotenuse
hypotenuse
opposite
opposite
adjacent
adjacent

93. We need a way to remember all of these ratios…

94. Some
Old
Hippie
Came A
Hoppin’
Through
Our
Old Hippie
Apartment

95. Sin
SOHCAHTOA
Opp
Hyp
Cos
Adj
Hyp
Tan
Opp
Adj
Old Hippie

96. Finding an angle. (Figuring out which ratio to use and getting to use the 2nd button and one of the trig buttons.)

97. Ex. 1: Find . Round to four decimal places.
tan
17.2 9 )
17.2 9
Shrink yourself down and stand where the angle is.
Now, figure out which trig ratio you have and set up the problem.
Make sure you are in degree mode (not radians).

98. Ex. 2: Find . Round to three decimal places.7
2nd
cos 7 23 ) 23
Make sure you are in degree mode (not radians).

99. Ex. 3: Find . Round to three decimal places.
200
400
2nd
sin
200
400 )
Make sure you are in degree mode (not radians).

100. When we are trying to find a side we use sin, cos, or tan.
When we are trying to find an angle
we use sin-1, cos-1, or tan-1.

101. Class Work
WS Packet