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Ch 11 Note sheet Shortened L1 Key Algebra Review: Proportion and Reasoning Name ___________________________ If you don’t understand the notes, read page 578 in the book, including Example A. To solve a proportion, use “cross multiplication”. ratio An expression that compares two quantities by division. a or a to b or a : b b proportion A statement of equality between two ratios. Proportions are useful for solving problems involving comparisons. Ways to write a ratio: Example A In a photograph, Robbie is 2.5 inches tall and his sister Paige is 1.5 inches tall. Robbie’s actual height is 70 inches. What is Paige’s actual height? Photo Paige then solve. Example A: Additional Notes You can set up these proportions any way you want as long as you match the measures and the units! For example: Just One Alternate Set Up: The Set Up: Robbie a c = b d aid = cib Actual 2.5 in. 1.5 in. 70 in. ? in. Actual Photo Robbie Paige 70 in. 2.5 in. ? in. 1.5 in. Now solve the proportion Now solve the proportion 2.5 70 = Use cross multiplication … 1.5 x 2.5 x = 105 now divide both sides by 2.5 x = 42 70 x = Use cross multiplication … 2.5 1.5 105 = 2.5x See, algebraically it is still the same. x = 42 Example B (alternate) Solve the proportion: 3x − 13 23 = x + 40 10 First realize that the fraction bar will group the terms in the numerator and the denominator. ( 3x − 13) = 23 Now, cross multiply. ( x + 40 ) 10 10 ( 3x − 13) = 23 ( x + 40 ) You’ll have to distribute! 30 x − 130 = 23x + 920 Add and subtract to solve. −23x + 130 − 23x + 130 Remember, BOTH sides! 7 x = 1050 Divide both sides by 7. x = 150 S. Stirling Additional Example Since ratios are fractions, you can solve a proportion simply by making equal fractions. Remember, to make equivalent fraction, you must multiply the numerator and the denominator by the same number. (You are simply multiplying by 1.) Example: Solve 2 x = 7 21 If you multiply the top and bottom by 3, you get an equal fraction (ratio). 2 3 6 i = 7 3 21 So x = 6 Page 1 of 6 Ch 11 Note sheet Shortened L1 Key Name ___________________________ Lesson 11.1 Similar Polygons Introduction similar Identical in shape but proportional in size. similar polygons Polygons whose corresponding angles are congruent and whose corresponding side lengths are proportional. scale factor The ratio of corresponding lengths in similar figures. The Statue of Liberty’s Nose Consider this problem: The Statue of Liberty in New York City has a nose that is 4 feet 6 inches long. What is the approximate length of one of her arms? 1. Solve the problem. (Hint: Think about your own nose and arms.) Explain how you got your answer and tell why you think your method works. Give your answer in feet. 4 ft 12 in i = 48 in , 48 + 6 = 54 in. 1 1 ft Nose Liberty You Arm 54 in. ? in. 2.125in. 27.5 in. 54 x = 2.125 27.5 1485 = 2.125x x = 698.82 inches 698.82 in 1 ft i = 58.24 ft 1 12 in Scale factor: (Liberty:You) 54 ≈ 25.4118 2.125 Means that she is 25.41 times larger than you. In length, that is. 2. Pick two other body parts and find the approximate length that these parts should be on the Statue of Liberty. Again, show how you got your answer and explain fully. S. Stirling Page 2 of 6 Ch 11 Note sheet Shortened L1 Key Lesson 11.1 Similar Polygons Name ___________________________ Read page 581 - 584 in the book, including Investigation 1. Congruence VS Similar Notes Definition of Congruence: Two figures are congruent if they have 1) the same shape: the corresponding angles are congruent and 2) the same size: the corresponding sides are congruent Definition of Similarity: Two figures are similar if they have 1) the same shape: the corresponding angles are congruent and 2) proportional size: the corresponding sides are in proportion (When you compare the corresponding sides from one figure to the other, they are in the same ratio, scale factor.) When writing congruent figures, you must match the corresponding vertices. Use the ≅ symbol. When writing similar figures, you must match the corresponding vertices. Use the ~ symbol. EXAMPLE (alternate) Given: QUAD ∼ SIML Find the missing parts: The quadrilaterals are similar, so corresponding angles are congruent QUAD ∼ SIML m∠Q = m∠S = 75° , m∠D = m∠L = 120° , m∠U = m∠I = 85° , m∠A = m∠M = 360 − (85 + 75 + 120) = 80° Corresponding sides are in proportion QU:SI UA:IM AD:ML QUAD 20 25 AD SIML 8 IM ML DQ:LS Scale 13 LS 5 2 Set up (include scale factor) 20 5 25 13 = = = 8 2 IM LS Solve: SL = 5.2 MI = 10 Scale factor = 5:2 5 5 25 i = so IM = 10 2 5 IM Solve: 5 13 26 = ≈ 5.2 then 5i LS = 26 , LS = 2 LS 5 m∠D = 120 m∠U = 85 m∠A = 80 S. Stirling Can you find AD or ML? Why or Why not? You would need one of the measures. You only know that they would be in a 20:8 or 5:2 ratio. Page 3 of 6 Ch 11 Note sheet Shortened L1 Key Name ___________________________ EXAMPLE (alternate) Are the triangles similar? The corresponding angles are congruent m∠A = m∠A same angle m∠ADE = m∠DBC DE BC so corresponding angles equal. m∠AED = m∠ECB same reason The corresponding sides are in proportion AB:AD BC:DE CA:EA ABC 4 8 6 ADE 2 4 3 See if they all equal the scale factor (or that the cross products are equal). 2 4 8 6 = = = 1 2 4 3 They all equal the scale factor 2:1! So ΔABC ∼ ΔADE Lesson 11.2 Similar Triangles (the short cuts) Read pages 589 – 591 in the book if necessary. Remember that to show that two triangles are congruent, by the definition of congruence, you would need to show that all six corresponding parts are congruent. So, we investigated and found 4 congruence short cuts. There are 3 shortcuts for showing that two triangles are similar. Congruence Short Cuts: ASA Congruence If two angles of one triangle and the included side are congruent to two angles and the included side of another triangle, then the two triangles are congruent. AAS Congruence If two angles of one triangle and an non-included side are congruent to two angles and the non-included side of another triangle, then the two triangles are congruent. SAS Congruence If two sides of one triangle and the included angle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent. SSS Congruence If three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent. Similarity Short Cuts: AA Similarity Postulate If two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar. SAS Similarity Theorem If two sides of one triangle are proportional to two sides of another triangle and the included angles are congruent, then the triangles are similar. SSS Similarity Theorem If the three sides of one triangle are proportional to the three sides of another triangle, then the two triangles are similar. Remember that the “included” part is important! The only difference between congruence and similarity properties is that the corresponding sides are proportional (not congruent). S. Stirling Page 4 of 6 Ch 11 Note sheet Shortened L1 Key Name ___________________________ EXAMPLE 1 EXAMPLE 2 Is ΔXYO ∼ ΔMKO ? Are the triangles similar? If so, explain why they are similar. Shoot for SAS Similarity since m∠O = m∠O . See if the corresponding sides are in proportion. OX:OM OY:OK XYO 30 70 48 108 Since the lines are parallel, the alternate interior angles are congruent. m∠A = m∠E m∠B = m∠D MKO So ΔABC ∼ ΔEDC by AA Similarity. ratios are not equal, the triangles are not similar. Simplify each ratio: OR Remember to match corresponding vertices! Cross multiply EXAMPLE 3 30 3 48 4 = and = , since the 70 7 108 9 30 48 = , 3240 ≠ 3360 . 70 108 Since you are not given any sides, shoot for SSS Similarity Match corresponding sides BA:JO AK:OL BAK 60 38 JOL 50 28 KB:LJ 36 26 See if they all equal the scale factor (or that the cross products are equal). 60 6 38 19 = = 50 5 28 14 So the triangles are not similar. OR 60 38 ≠ , 1680 ≠ 1900 50 28 S. Stirling Page 5 of 6 Ch 11 Note sheet Shortened L1 Key Name ___________________________ Lesson 11.3 Indirect Measurement with Similar Triangles Read pages 598 – 599 in the book. indirect measurement Finding a distance or length by using properties of similar triangles or trigonometry. Typical problems include shadows and mirrors. EXAMPLE 1 At a certain time of day, a 6 ft man casts a 4 ft shadow. At the same time of day, how tall is a tree that casts an 18 ft shadow? Make a drawing: Since the triangles will be similar. Height Shadow Man Tree 6 ft 4 ft ? ft 18 ft Solve the proportion: 6 x = 4 18 4 x = 108 x = 27 So the tree is 27 feet tall. EXAMPLE 2 Samantha has her friend Sophie place a mirror between her and a tall building so that she can see the very top of the building in the mirror. Sam is 5 feet tall. She is standing 79 feet from the building and the mirror is 4 feet in front of her. Find the height of the building. Make a drawing: Since the triangles will be similar. Height Distance S. Stirling Sam Building 5 ft 4 ft ? ft 75 ft Solve the proportion: 5 x = 4 75 4 x = 375 x = 93.75 So the building is 93.75 feet tall. Page 6 of 6