7.1 Ratio and Proportion

1. 7.1 Ratio and Proportion Students will be able to write and simplify ratios and to use proportions to solve problems

2. Ratios • A ratiocompares two numbers by division. • The ratio of two numbers a and b can be written as a to b, a:b, or where b ≠ 0 • A ratio can involve more than two numbers. For the rectangle here, the ratio of the side lengths is 3:7:3:7. 7 3 3 7

3. Ratios l • Write a ratio expressing the slope of line l. Slope = = run rise B (4,3) A (-2,-1) Given that two points on m are C(-2,3) and D(6,5), write a ratio expressing the slope of m. The slope of m is

4. Using Ratios • The ratio of the side lengths of a quadrilateral is 2:3:5:7, and its perimeter is 85 feet. What is the length of the longest side? • We want to let the lengths of the sides be 2x, 3x, 5x and 7x. Then we can add them together to get the perimeter. 2x + 3x + 5x + 7x = 85 17x = 85 x = 5 The longest side is 7x or35 ft.

5. Proportions • A proportion is an equation stating that two ratios are equal. • In the proportion ,the values a and d are the extremes. The values b and c are the means. • When a proportion is written as a:b = c:d, the extremes are in the first and last position. The means are in the two middle positions. • The product of the extremesadand the product of the meansbc are called the cross products.

6. Cross Products Property • In a proportion, if and b and d ≠ 0, then ad = bc. ad = bc • Example: solve this proportion. 5·63 = y·45 → 315 = 45y y = 7

7. Geometric Mean • The geometric meanof two numbers is the positive square root of their product. We use the following proportion to find the geometric mean: • Notice that the means both have x. That is the geometric mean. How do you solve for x?

8. Examples of Geometric Means • Find the geometric mean between 4 and 9. → x2 = 36 → → x = 6 • Now, find the geometric mean between 2 and 8. → x2 = 16 → → x = 4

9. More examples • Find the arithmetic and geometric mean between 32 and 2.

10. 7.2 Ratios in Similar Polygons Students will be able to apply properties of similar polygons to solve problems

11. Similarity • Figures that are similar (∼) have the same shape but not necessarily the same size. • Two polygons are similar polygons if and only if their corresponding angles are congruent and their corresponding side lengths are proportional. • A similarity ratio is the ratio of the lengths of the corresponding sides of two similar polygons.

12. Similar Figures • When we write a similarity statement, ΔABC ∼ΔLMN for two triangles, we always write each corresponding part in the same order. For example: If, then ∠A corresponds to ∠L, ∠B corresponds to ∠M, and ∠C corresponds to ∠N. Also, corresponds to ___ corresponds to ___ and corresponds to ___. B M N L A C

13. Example 1: Determining Similarity Determine if ∆MLJ ~ ∆NPS. If so, write the similarity ratio and a similarity statement. Step 1: Identify pairs of congruent angles. M  N, L  P, J  S Step 2: Compare corresponding sides. Thus, ∆MLJ ~ ∆NPS

15. 2 More examples

16. Find the ratio of the perimeter of ABCD and EFGH

17. Challenge

18. Applications • Find the length of the model to the nearest tenth of a centimeter. • Let x be the length of the model in centimeters. • The rectangular model of the racing car is similar to the rectangular racing car, so the corresponding lengths are proportional.

19. Example 2 (Continued) 5(6.3) = x(1.8) 31.5 = 1.8x 17.5 = x The length of the model is 17.5 centimeters.

20. 7.3 Triangle Similarity Students will be able to use triangle similarity to solve problems

21. Similar Triangles • If we know that all three pairs of corresponding angles are congruent to each other and that all three pairs of corresponding sides of the triangle are in proportion, then we know that we have similar triangles. • But do we need to know all of this information? NO ! There are shortcuts to find similarity. We will be learning these shortcuts.

22. AA Similarity Postulate • If two angles of a triangle are congruent to two angles of another triangle, then the triangles are similar. If ∠A ≅ ∠D and ∠B ≅∠E, then ∆ABC∼∆DEF B E C F A D

23. SSS Similarity Theorem • If all three pairs of corresponding sides of two triangles are in proportion, then the triangles are similar. = = E B D F A C

24. SAS Similarity Theorem • If an angle of one triangle is congruent to an angle of another triangle, and the two pairs of sides including these angles are in proportion, then the triangles are similar. E B Since ∠A≅∠D And = D F A C

25. Example 2: Using the Similarity Postulate and Theorems Explain why ∆RSV ~ ∆RTU and then find RT. Step 1: Are the triangles similar? Since RSV T and R R, ∆RSV ~ ∆RTU by AA ~. Step 2: Find RT. → RT(8) = 10(12) →RT = 15 → 8RT = 120

26. 7.4-7.5 Applying Properties of Similar Triangles Students will be able to use properties of similar triangles to solve problems

27. 1. Triangle Proportionality Theorem (Side Splitter) • If a line parallel to a side of a triangle intersects the other two sides, then it divides those sides proportionally. If then = Q S T R P

28. Example 1: Using Triangle Proportionality Theorem J 18 Find the length of JN Use the Triangle Proportionality Theorem: K 12 L N 8 M 12 · JN = 8 · 18 JN = JN = 12

29. 2. Converse of the Triangle Proportionality Theorem • If a line divides two sides of a triangle proportionally, then it is parallel to the third line. If = then Q S T R P

30. Example 2: Using the Converse of Triangle Proportionality AC = 36 cm, and BC = 27 cm. Verify that . Since , by the Converse of the Triangle Proportionality Theorem.

31. 3. Two-Transversal Proportionality (Parallel Proportion Thm) • If three or more parallel lines intersect two transversals, then they divide the transversals proportionally. A B • If are parallel lines, then C D E F

32. Example

33. Example 3: Using the Two-Transversal • An artist used perspective to draw guidelines to sketch a row of parallel trees. What length would LN be? 8 in. 4 in. A BD = BC + CD = 4 + 5 = 9 in. 8 · LN = 90 LN = 11 in. 5 in. B C D N M L K 10 in. Unit F

34. 4. Triangle Angle Bisector Theorem • An angle bisector of a triangle divides the opposite side into two segments whose lengths are proportional to the lengths of the other two sides. If bisects ∠BAC, then

35. Example 4: Using Triangle Angle Bisector Theorem • Find AC and DC. (8) = 4.5(y– 2) 4y = 4.5y - 9 → -.5y = -9 → y = 18 AC = y – 2 = 16 and

36. You try!

37. 7.4-7.5 Part 2 Similarity in Right Triangles Students will be able to find segment lengths in right triangles, and to apply similarity relationships in right triangles to solve problems.

38. Geometric Mean • Remember that in a proportion such as , aand b are called the extremes and r and q are called the means. • The geometric meanof two numbers is the positive square root of their product. We use the following proportion to find the geometric mean: • Notice that the means both have x. That is the geometric mean. How do you solve for x? → →

39. Examples of Geometric Means • Find the geometric mean between 4 and 9. → x2 = 36 → → x = 6 • Now, find the geometric mean between 2 and 8. → x2 = 16 → → x = 4

40. Similar Right Triangles Theorem • The altitude to the hypotenuse of a right triangle forms two triangles that are similar to each other and to the original triangle. C ∆CBD ∆ACD∼ ∆ABC∼ A B D

41. Geometric Means Corollary • The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the two lengths of the segments of the hypotenuse. • This means: The altitude is the geometric mean of the two segments of the hypotenuse • and . • Or we could say: • CD2 = AD ∙ BD C B A D

42. The Heartbeat • This is the memory device to help you remember how to set up the proportion you need to use the theorem. If AD = 9 and BD = 4, what is CD? C CD AD BD CD A B D

43. Example of Corollary To estimate the height of a Douglas fir, Jan positions herself so that her lines of sight to the top and bottom of the tree form a 90º angle. Her eyes are about 1.6 m above the ground, and she is standing 7.8 m from the tree. What is the height of the tree to the nearest meter? Use the heartbeat to set up the proportion: → 1.6x = 7.8 ∙ 7.8 x 1.6x = 60.84 → x ≈ 38 So the tree is about 40 meters tall.

44. Geometric Means Corollary • If the altitude is drawn to the hypotenuse of a right triangle, then the length of each leg is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse adjacent to that leg. C and We can also say this as: AC2 = AB ∙ AD and BC2 = AB ∙ BD A B D

45. More examples

46. Challenge: Find DB and AD

47. Using Proportional Relationships Students will be able to use ratios to make measurements, and be able to use scale drawings to solve problems

48. Indirect Measurements • Indirect measurement is any method that uses formulas, similar figures, and/or proportions to measure an object. • In this example we can use similar figures to find the height of the tree. x → 4 ft. ft. 3 ft. 18 ft.

49. Example 1: Using Indirect Measurement A student who is 66 in. tall measured shadows to find the height LM of a flagpole. What is LM? h 66 in. 60 in. 170 in. 60(h) = 66 • 170 →h = 187 The height of the flagpole is 187 in., or 15 ft. 7 in.

50. Scale Drawing • A scale drawingrepresents an object as smaller than or larger than its actual size. • The drawing’s scaleis the ratio of any length in the drawing to the corresponding actual length. • For example, on a map with a scale of 1 cm: 1500 m, one centimeter on the map represents 1500 meters in actual distance To convert a distance on the map to actual distance, use a proportion: