1. Lesson 6-R Chapter 6 Review

2. Objectives Review Chapter 6

3. Vocabulary None new

4. Ratios and Proportions Proportion is two ratios set equal to each other Scaling Factor is a Ratio –Used in recipes to increase (sf > 1) or decrease (sf < 1) amount it makes 4 x + 2 --- = -------- 4  14 = 7  (x + 2) 7 14 56 = 7x + 14 42 = 7x 6 = x cross-multiply to solve (remember distributive property with + or -) Top Bottom

5. Similar Polygons Similar Polygons have the same shape and are different sizes based on the scaling factor (called in chapter 9, dilations – a transformation not  ) Like with congruent triangles – Order Rules! H K J L 16 t + 3 r - 3 AB C D 12 10 8 Trapezoid HJLJ ~ Trapezoid ACDB so side HJ goes with AC Remember to separate the variables with the key (constant ratio) 8 10 8 12 --- = ------- and ---- = ------------ 16 t + 3 16 r - 3 T B

6. Triangle Similarity Theorems TheoremPictureCheck SSS Only 3 sides given All three sides have the same ratios SAS Two sides have the same ratios and the included angle is congruent AA Two angle are congruent A B C D F E 5 4 3 9 12 15 A B C D F E A B C D F E 5 4 12 15

7. Parallel Lines and Proportions Parallel lines cut the sides of a triangle (or sides of two lines) into the same ratio AB AC so ------ = ------- in ∆AED to the right BE CD B C A N P M 3x - 7 17 – x y + 6 3y J N M K L 9 7 20 A B C E D KL is a midsegment, which means K and L are midpoints and ∆JKL is half of ∆JMN If we have more than one variable, then look for congruent marks; otherwise, we can’t solve it. TB

8. Parts of Similar Triangles If two triangles are similar then all parts of the triangles (perimeter, medians, altitudes, etc) have to be in the same ratio as the two triangles A B C D P Q R S 6 y 18 24 4 ∆ABC ~ ∆PRQ Side AB matches to side PR and gives a 1 : 3 ratio. So y is 3 times 4 or 12. All parts of ∆PRQ are 3 times larger than corresponding parts of ∆ABC Reminder PS and AD are altitudes T B

9. Angle Bisector Theorem An angle bisector cuts the side opposite in the same ratio as the sides that form the original angle. J K L M 6 x 4 x + 3 The Ratio of JL to JK must be the same as ML to MK x x + 3 --- = -------- 4  (x + 3) = 6  x 4 6 4x + 12 = 6x 12 = 2x 6 = x T B

10. Summary & Homework Summary: –Cross-multiply to solve proportions –Remember the distributive property –Order rules for matching corresponding sides of similar figures like congruent triangles –Parallel lines cut into equal ratios –All parts of similar triangles have the same ratio –Angle bisector cuts divided side into same ratio as the sides forming the angle Homework: Study for Ch 6 Test