1. 9.1 Similar Right Triangles

2. Objectives
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.

3. Proportions in right triangles
In Lesson 8.4, 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.

4. Theorem 9.1
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

5. 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.

6. 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.

7. 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.

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

9. ASSIGNMENT
pp #1-34

10. 9.2 The Pythagorean Theorem

11. Theorem 9.4: Pythagorean Theorem
In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the legs.
c2 = a2 + b2

12. Using the Pythagorean Theorem
A Pythagorean triple is a set of three positive integers a, b, and c that satisfy the equation a2 + b2 = c2 For example, the integers 3, 4 and 5 form a Pythagorean Triple because 52 =

13. Ex. 1: Finding the length of the hypotenuse.
Find the length of the hypotenuse of the right triangle. Tell whether the sides lengths form a Pythagorean Triple.

14. Ex. 2: Finding the Length of a Leg
Find the length of the leg of the right triangle.

15. Assignment :
pp #1-31 all

16. 9.3 The Converse of the Pythagorean Theorem

17. Using the Converse
In Lesson 9.2, you learned that if a triangle is a right triangle, then the square of the length of the hypotenuse is equal to the sum of the squares of the length of the legs. The Converse of the Pythagorean Theorem is also true, as stated on the following slide.

18. Theorem 9.5: Converse of the Pythagorean Theorem
If the square of the length of the longest side of the triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right triangle.
If c2 = a2 + b2, then ∆ABC is a right triangle.

19. Ex. 1: Verifying Right Triangles
The triangles on the slides that follow appear to be right triangles. Tell whether they are right triangles or not.
√113
4√95

20. Classifying Triangles
Sometimes it is hard to tell from looking at a triangle whether it is obtuse or acute. The theorems on the following slides can help you tell.

21. Theorem 9.6—Triangle Inequality
If the square of the length of the longest side of a triangle is less than the sum of the squares of the lengths of the other two sides, then the triangle is acute.
If c2 < a2 + b2, then ∆ABC is acute
c2 < a2 + b2

22. Theorem 9.7—Triangle Inequality
If the square of the length of the longest side of a triangle is greater than the sum of the squares of the lengths of the other two sides, then the triangle is obtuse.
If c2 > a2 + b2, then ∆ABC is obtuse
c2 > a2 + b2

23. Ex. 2: Classifying Triangles
Decide whether the set of numbers can represent the side lengths of a triangle. If they can, classify the triangle as right, acute or obtuse.
8, 7, 6 b. 10, 36, 40
You can use the Triangle Inequality to confirm that each set of numbers can represent the side lengths of a triangle. Compare the square o the length of the longest side with the sum of the squares of the two shorter sides.

24. Assignment :
pp #1-35

25. 9.4 Special Right Triangles

26. Side lengths of Special Right Triangles
Right triangles whose angle measures are 45°-45°-90° or 30°-60°-90° are called special right triangles. The theorems that describe these relationships of side lengths of each of these special right triangles follow.

27. Theorem 9.8: 45°-45°-90° Triangle Theorem
In a 45°-45°-90° triangle, the hypotenuse is √2 times as long as each leg.
45°
√2x
45°
Hypotenuse = √2 ∙ leg

28. Theorem 9.8: 30°-60°-90° Triangle Theorem
In a 30°-60°-90° triangle, the hypotenuse is twice as long as the shorter leg, and the longer leg is √3 times as long as the shorter leg.
60°
30°
√3x
Hypotenuse = 2 ∙ shorter leg
Longer leg = √3 ∙ shorter leg

29. Ex. 1: Finding the hypotenuse in a 45°-45°-90° Triangle
Find the value of x
By the Triangle Sum Theorem, the measure of the third angle is 45°. The triangle is a 45°-45°-90° right triangle, so the length x of the hypotenuse is √2 times the length of a leg. 3 3
45° x

30. Ex. 1: Finding the hypotenuse in a 45°-45°-90° Triangle3 3
45° x
45°-45°-90° Triangle Theorem
Substitute values
Simplify
Hypotenuse = √2 ∙ leg
x = √2 ∙ 3
x = 3√2

31. Ex. 2: Finding a leg in a 45°-45°-90° Triangle
Find the value of x.
Because the triangle is an isosceles right triangle, its base angles are congruent. The triangle is a 45°-45°-90° right triangle, so the length of the hypotenuse is √2 times the length x of a leg. 5 x x

32. Ex. 3: Finding side lengths in a 30°-60°-90° Triangle
Find the values of s and t.
Because the triangle is a 30°-60°-90° triangle, the longer leg is √3 times the length s of the shorter leg.
60°
30°

33. Ex. 4: Finding the area of a sign
Road sign. The road sign is shaped like an equilateral triangle. Estimate the area of the sign by finding the area of the equilateral triangle.
18 in. h
36 in.

34. Ex. 4: Solution
18 in.
First, find the height h of the triangle by dividing it into two 30°-60°-90° triangles. The length of the longer leg of one of these triangles is h. The length of the shorter leg is 18 inches.
h = √3 ∙ 18 = 18√3
30°-60°-90° Triangle Theorem h
36 in.
Use h = 18√3 to find the area of the equilateral triangle.

35. Ex. 5: Solution Area = ½ bh = ½ (36)(18√3) ≈ 561.18
18 in.
Area = ½ bh
= ½ (36)(18√3) ≈ h
36 in.
The area of the sign is a bout 561 square inches.

36. Assignment
pp #1-30

37. 9.5 Trigonometric Ratios

38. Trigonometric Ratios
Let ∆ABC be a right triangle. The since, the cosine, and the tangent of the acute angle A are defined as follows.
Side adjacent to A b
cos A = =
hypotenuse c
Side opposite A a
sin A = =
hypotenuse c
Side opposite A a
tan A = =
Side adjacent to A b

39. Ex. 1: Finding Trig Ratios
Compare the sine, the cosine, and the tangent ratios for A in each triangle beside.

40. Ex. 1: Finding Trig Ratios
Large
Small
opposite 8
sin A = ≈
0.4706 4 ≈
0.4706
hypotenuse 17
8.5
adjacent
7.5
cosA = 15 ≈
0.8824 ≈
0.8824
hypotenuse
8.5 17
opposite
tanA = 8 4 ≈
0.5333 ≈
0.5333
adjacent 15
7.5
Trig ratios are often expressed as decimal approximations.

41. Ex. 2: Finding Trig Ratios
opposite 5
sin S = ≈
0.3846
hypotenuse 13
adjacent
cosS = 12 ≈
0.9231
hypotenuse 13
opposite
tanS = 5 ≈
0.4167
adjacent 12

42. Ex. 2: Finding Trig Ratios—Find the sine, the cosine, and the tangent of the indicated angle.
opposite 12
sin S = ≈
0.9231
hypotenuse 13
adjacent
cosS = 5 ≈
0.3846
hypotenuse 13
opposite
tanS = 12 ≈
2.4
adjacent 5

43. Ex. 3: Finding Trig Ratios—Find the sine, the cosine, and the tangent of 45
opposite 1 √2
sin 45= = ≈
0.7071
hypotenuse √2 2
adjacent 1 √2
cos 45= = ≈
0.7071
hypotenuse √2 2
opposite 1
tan 45=
adjacent = 1 1
Begin by sketching a 45-45-90 triangle. Because all such triangles are similar, you can make calculations simple by choosing 1 as the length of each leg. From Theorem 9.8 on page 551, it follows that the length of the hypotenuse is √2. √2
45

44. Ex. 4: Finding Trig Ratios—Find the sine, the cosine, and the tangent of 30
opposite 1
sin 30= =
0.5
hypotenuse 2
adjacent √3
cos 30= ≈
0.8660
hypotenuse 2
opposite 1 √3
tan 30= =
adjacent ≈
0.5774 √3 3
Begin by sketching a 30-60-90 triangle. To make the calculations simple, you can choose 1 as the length of the shorter leg. From Theorem 9.9, on page 551, it follows that the length of the longer leg is √3 and the length of the hypotenuse is 2.
30 √3

45. Assignment :
pp #3-38 all

46. 9.6 Solving Right Triangles

47. Solving a right triangle
Every right triangle has one right angle, two acute angles, one hypotenuse and two legs. To solve a right triangle, means to determine the measures of all six (6) parts. You can solve a right triangle if the following one of the two situations exist:
Two side lengths
One side length and one acute angle measure

48. Example 1:
Solve the right triangle. Round the decimals to the nearest tenth.
HINT: Start by using the Pythagorean Theorem. You have side a and side b. You don’t have the hypotenuse which is side c—directly across from the right angle.

49. Example 1: (hypotenuse)2 = (leg)2 + (leg)2 Pythagorean Theorem
Substitute values
c2 =
Simplify
c2 = 9 + 4
Simplify
c2 = 13
Find the positive square root
c = √13
Use a calculator to approximate
c ≈ 3.6

50. Example 1 continued Then use a calculator to find the measure of B:
2nd function
Tangent button 2
Divided by symbol
3 ≈ 33.7°

51. Finally Because A and B are complements, you can write
mA = 90° - mB ≈ 90° ° = 56.3°
The side lengths of the triangle are 2, 3 and √13, or about The triangle has one right angle and two acute angles whose measure are about 33.7° and 56.3°.

52. Ex. 2: Solving a Right Triangle (h)
Solve the right triangle. Round decimals to the nearest tenth.
25°
You are looking for opposite and hypotenuse which is the sin ratio.
sin H =
opp.
hyp.
Set up the correct ratio
13 sin 25° = h 13
Substitute values/multiply by reciprocal
Substitute value from table or calculator
13(0.4226) ≈ h
Use your calculator to approximate.
5.5 ≈ h

53. Ex. 2: Solving a Right Triangle (g)
Solve the right triangle. Round decimals to the nearest tenth.
25°
You are looking for adjacent and hypotenuse which is the cosine ratio.
cos G =
adj.
hyp.
Set up the correct ratio
13 cos 25° = g 13
Substitute values/multiply by reciprocal
Substitute value from table or calculator
13(0.9063) ≈ g
11.8 ≈ h
Use your calculator to approximate.

54. Assignment
pp #4-38