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# Module 6

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• Remember: congruent polygons are always similar but not all similar polygons are congruent
• Taken to the properties of similar polygons that all corresponding angles are congruent, then AA similarity Postulate is taken from AAA Similarity Postulate. Obviously, if two pairs corresponding angles of two triangles are congruent, it follows that the last pair of corresponding angles are also congruent for the sum of interior angles in a triangle is 1800.
• To have more emphasis in this postulate, we have
• ### Module 6

1. 1. Module 6 - SIMILARITY Regional Mass Training of Grade 9 Mathematics Teachers May 17 – 21, 2014
2. 2. 1. Solve problems involving similar polygons. 2. Prove certain triangles are similar by using AA, SSS, and SAS. 3. Solve problems involving Basic Triangle Proportionality Theorem and Triangle Angle Bisector Theorem. 4. Solve problems involving Right Triangle Similarities and Special Right Triangles. DEPARTMENT OF EDUCATION
3. 3. Warm Up Solve each proportion. 1. 2. 3. 4. If ∆QRS ~ ∆XYZ, identify the pairs of congruent angles and write 3 proportions using pairs of corresponding sides. Review: Proportion DEPARTMENT OF EDUCATION z = ±10 x = 8 Q  X; R  Y; S  Z;
4. 4. Similar Polygons DEPARTMENT OF EDUCATION Two polygons are said to be similar if and only if: i. corresponding angles are congruent. ii. corresponding sides are proportional. A B C D E M N O P Q ∠A↔∠M ∠E↔∠Q ∠C↔∠O ∠D↔∠P ∠B↔∠N AB↔ MN BC↔ NO CD↔ OP DE↔ PQ AE↔ MQ ∠A≅ ∠M ∠E≅ ∠Q ∠C≅ ∠O ∠D≅ ∠P ∠B≅ ∠N AB MN = BC NO = CD OP = DE PQ = AE MQ ABCDE∼MNOPQ
5. 5. Similar Polygons DEPARTMENT OF EDUCATION Example: If ABCDE∼MNOPQ, determine the value of x and y. Figure is drawn not to scale. A B C D E Answer: y=3 x=2 9 M N O P Q 6 y+3 4 x+2 5 2x+2 6 7.5 4y-3
6. 6. Triangle Similarities DEPARTMENT OF EDUCATION
7. 7. Example 1: Using the AA Similarity Postulate Explain why the triangles are similar and write a similarity statement. Since 𝐴𝐵 ∥ 𝐷𝐸, B  E by the Alternate Interior Angles Theorem. Also, A  D by the Right Angle Congruence Theorem. Therefore ∆ABC ~ ∆DEC by AA~. Similar Triangles DEPARTMENT OF EDUCATION
8. 8. Similar Triangles DEPARTMENT OF EDUCATION In ∆ABC and ∆DEF, AB DE = BC EF = AC DF . Prove that ∆ABC∼∆DEF. B A C D F E
9. 9. Similar Triangles DEPARTMENT OF EDUCATION Proof: Construct GH in ∆DEF such that G and H are in DE and EF respectively, GH ∥ DF, GE ≅ AB and BC ≅ EH. B A C D F E G H
10. 10. Similar Triangles DEPARTMENT OF EDUCATION B A C D F E G H By AA Sim. Postulate, ∆EGH∼∆EDF which implies EG ED = GH DF = EH EF . Since EG ≅ AB, BC ≅ EH, then EG ED = AB ED and EH EF = BC EF by substitution.
11. 11. Similar Triangles DEPARTMENT OF EDUCATION B A C D F E G H And since AB ED = AC DF from the given and EG ED = GH DF , then by transitivity PE, AC DF = GH DF which implies AC=GH by multiplication PE.
12. 12. Similar Triangles DEPARTMENT OF EDUCATION B A C D F E G H Meaning ∆ABC ≅∆GEH by SSS congruence postulate. Since ∆EGH∼∆EGF, then ∆ABC∼∆DEF by substitution.
13. 13. Similar Triangles DEPARTMENT OF EDUCATION Given ∆ABC and ∆DEF such that AB DE = AC DF and ∠𝐴 ≅ ∠𝐷 B A C D F E
14. 14. Similar Triangles DEPARTMENT OF EDUCATION B A C D F E Locate P on DE so that PE=AB. Draw PQ so that PQ║ DF. By the AA Sim. Theo., we have ∆PEQ∼∆DEF, and thus DE PE = EF EQ = FD QP . P Q
15. 15. Similar Triangles DEPARTMENT OF EDUCATION B A C D F E P Q Since PE=AB, we can substitute this in the given proportion and find EQ=BC and QP=CA. By SSS congruence Theo., it follows that ∆PEQ≅∆ABC.
16. 16. Similar Triangles DEPARTMENT OF EDUCATION B A C D F E P Q And since ∆PEQ∼∆DEF and ∆PEQ≅∆ABC, by substitution, then ∆ABC∼∆DEF
17. 17. Example : Verifying Triangle Similarity Verify that the triangles are similar. ∆PQR and ∆STU Therefore ∆PQR ~ ∆STU by SSS ~. Similar Triangles DEPARTMENT OF EDUCATION
18. 18. Example : Verifying Triangle Similarity ∆DEF and ∆HJK Verify that the triangles are similar. D  H by the Definition of Congruent Angles. Therefore ∆DEF ~ ∆HJK by SAS ~. Similar Triangles DEPARTMENT OF EDUCATION
19. 19. A  A by Reflexive Property of , and B  C since they are both right angles. Example : Finding Lengths in Similar Triangles Explain why ∆ABE ~ ∆ACD, and then find CD. Step 1 Prove triangles are similar. Therefore ∆ABE ~ ∆ACD by AA ~. Similar Triangles DEPARTMENT OF EDUCATION
20. 20. Example Continued Step 2 Find CD. Corr. sides are proportional. Seg. Add. Postulate. Substitute x for CD, 5 for BE, 3 for CB, and 9 for BA. Multiplication PEx(9) = 5(3 + 9) Simplify.9x = 60 Similar Triangles DEPARTMENT OF EDUCATION
21. 21. Example : Writing Proofs with Similar Triangles Given: ∆BAC with medians 𝑫𝑪 and 𝑨𝑬. Prove: ∆BAC ~ ∆BDE Similar Triangles DEPARTMENT OF EDUCATION A B C D E
22. 22. Example Continued Similar Triangles DEPARTMENT OF EDUCATION Statement Reason 1. 𝐷𝐶 and 𝐴𝐸 are medians of ∆BAC 1. Given 2. 2BD=BA ; 2BE = BC 2. Definition of Medians 3. 𝐵𝐷 𝐵𝐴 = 1 2 ; 𝐵𝐸 𝐵𝐶 = 1 2 3. Properties of Ratio 4. 𝐵𝐷 𝐵𝐴 = 𝐵𝐸 𝐵𝐶 4. Transitive PE 5. ∠𝐵 ≅ ∠𝐵 5. Reflective PE 6. ∆BAC ~ ∆BDE 6. SAS ~ A B C D E
23. 23. Check It Out! Given: M is the midpoint of JK. N is the midpoint of KL, and P is the midpoint of JL. Similar Triangles DEPARTMENT OF EDUCATION
24. 24. Statements Reasons Check It Out! Example 4 Continued 1. Given 1. M is the mdpt. of 𝐽𝐾, N is the mdpt. of 𝐾𝐿, and P is the mdpt. of 𝐽𝐿. 2. Midline Theorem2. 3. Mult. PE. 3. 4. SSS ~ Step 34. ∆JKL ~ ∆NPM Similar Triangles DEPARTMENT OF EDUCATION
25. 25. Basic Triangle Proportionality DEPARTMENT OF EDUCATION Proof: In ∆ABC, let D and E be points on 𝐴𝐵 and 𝐵𝐶 respectively, such that 𝐷𝐸║ 𝐴𝐶. We have to prove that 𝐵𝐷 𝐷𝐴 = 𝐵𝐸 𝐸𝐶 . Since parallel lines form congruent corresponding angles, we have ∠𝐷 ≅ ∠𝐴 and ∠𝐸 ≅ ∠𝐶. By the AA Sim. Theo., we say ∆ABC∼∆DBE.
26. 26. Basic Triangle Proportionality DEPARTMENT OF EDUCATION Since corresponding sides of similar triangles are proportional, 𝐵𝐷 𝐵𝐴 = 𝐵𝐸 𝐵𝐶 . By Reciprocal Property, we have 𝐵𝐴 𝐵𝐷 = 𝐵𝐶 𝐵𝐸 . By Subtraction Property of Proportionality, we have 𝐵𝐴−𝐵𝐷 𝐵𝐷 = 𝐵𝐶−𝐵𝐸 𝐵𝐸 and hence 𝐷𝐴 𝐵𝐷 = 𝐸𝐶 𝐵𝐸 or 𝐵𝐷 𝐷𝐴 = 𝐵𝐸 𝐸𝐶 .
27. 27. Check It Out! Basic Triangle Proportionality DEPARTMENT OF EDUCATION Find the value of x in the figure below. The figure is not drawn to scale. 4 6 3 x 6 4 = 𝑥 − 3 3 4(x – 3)=18 4x – 12 =18 4x =30 x= 15 2
28. 28. Triangle Angle Bisector Theorem DEPARTMENT OF EDUCATION If a segment bisects an angle of a triangle, then it divides the opposite side into segments proportional to the other two sides. Proof: 1 2 3 4 Let 𝐴𝑋 bisect ∠A of ∆ABC. We must prove that 𝐵𝑋 𝑋𝐶 = 𝐴𝐵 𝐴𝐶 . B X C A Y Draw 𝐵𝑌 parallel to 𝐴𝑋. Extend 𝐶𝐴 so that it intersects 𝐵𝑌 at Y. Since 𝐵𝑌 ∥ 𝐴𝑋, we have 𝐵𝑋 𝑋𝐶 = 𝐴𝑌 𝐴𝐶 . By theorems involving parallel lines cut by a transversal, we also have ∠1 ≅ ∠3 and ∠4 ≅ ∠2.
29. 29. DEPARTMENT OF EDUCATION If a segment bisects an angle of a triangle, then it divides the opposite side into segments proportional to the other two sides. Proof: 1 2 3 4 B X C A Y Since 𝐴𝑋 bisect ∠A, then ∠1 ≅ ∠2 and ∠4 ≅ ∠3. Thus, by the Isosceles Triangle Theorem, AY = AB. By substitution, we conclude that 𝐵𝑋 𝑋𝐶 = 𝐴𝐵 𝐴𝐶 . Triangle Angle Bisector Theorem
30. 30. Check It Out! DEPARTMENT OF EDUCATION In the figure below, ∠𝐵𝐴𝑋 ≅ ∠𝐶𝐴𝑋. Use the lengths to find the value of y. 𝐴𝐵 𝐴𝐶 = 𝐵𝑋 𝑋𝐶 14 y X C 9 A 15 15 9 = 𝑦 14 − 𝑦 15(14 – y) = 9y y = 8.75 Triangle Angle Bisector Theorem B
31. 31. Consider ∆ABC such that ∠𝐶 is right and 𝐶𝐷 is the altitude to the hypotenuse. We have to prove that ∆ADC∼∆ACB∼∆CDB. Since 𝐶𝐷 ⊥ 𝐴𝐵, then ∠𝐴𝐷𝐶 = ∠𝐶𝐷𝐵 = ∠𝐴𝐶𝐵 = 900. Since ∠𝐴=∠𝐴, then by AA Sim. Theo., we have ∆ADC∼∆ACB. Using the same theorem and that ∠𝐵=∠𝐵, we have ∆ACB∼∆CDB. Hence, ∆ADC∼∆CDB by transitivity. Right Triangle Similarity DEPARTMENT OF EDUCATION
32. 32. Consider the proportion . In this case, the means of the proportion are the same number, and that number is the geometric mean of the extremes. The geometric mean of two positive numbers is the positive square root of their product. So the geometric mean of a and b is the positive number x such that , or x2 = ab. Right Triangle Similarity DEPARTMENT OF EDUCATION
33. 33. You can use Theorem on Right Triangle Similarity to write proportions comparing the side lengths of the triangles formed by the altitude to the hypotenuse of a right triangle. All the relationships in red involve geometric means. Right Triangle Similarity DEPARTMENT OF EDUCATION
34. 34. Right Triangle Similarity DEPARTMENT OF EDUCATION
35. 35. Example 3: Finding Side Lengths in Right Triangles Find x, y, and z. 62 = (9)(x) 6 is the geometric mean of 9 and x. x = 4 Divide both sides by 9. y2 = (4)(13) = 52 y is the geometric mean of 4 and 13. Find the positive square root. z2 = (9)(13) = 117 z is the geometric mean of 9 and 13. Find the positive square root. Right Triangle Similarity DEPARTMENT OF EDUCATION
36. 36. Once you’ve found the unknown side lengths, you can use the Pythagorean Theorem to check your answers. Helpful Hint Right Triangle Similarity DEPARTMENT OF EDUCATION
37. 37. Special Right Triangles DEPARTMENT OF EDUCATION
38. 38. Example : Craft Application Jana is cutting a square of material for a tablecloth. The table’s diagonal is 36 inches. She wants the diagonal of the tablecloth to be an extra 10 inches so it will hang over the edges of the table. What size square should Jana cut to make the tablecloth? Round to the nearest inch. Jana needs a 45°-45°-90° triangle with a hypotenuse of 36 + 10 = 46 inches. Special Right Triangles DEPARTMENT OF EDUCATION
39. 39. Check It Out! Example What if...? Tessa’s other dog is wearing a square bandana with a side length of 42 cm. What would you expect the circumference of the other dog’s neck to be? Round to the nearest centimeter. Tessa needs a 45°-45°-90° triangle with a hypotenuse of 42 cm. Special Right Triangles DEPARTMENT OF EDUCATION
40. 40. Example: Using the 30º-60º-90º Triangle Theorem An ornamental pin is in the shape of an equilateral triangle. The length of each side is 6 centimeters. Josh will attach the fastener to the back along AB. Will the fastener fit if it is 4 centimeters long? Step 1 The equilateral triangle is divided into two 30°-60°-90° triangles. The height of the triangle is the length of the longer leg. Special Right Triangles DEPARTMENT OF EDUCATION
41. 41. Example Continued Step 2 Find the length x of the shorter leg. Step 3 Find the length h of the longer leg. The pin is approximately 5.2 centimeters high. So the fastener will fit. Hypotenuse = 2(shorter leg)6 = 2x 3 = x Divide both sides by 2. Special Right Triangles DEPARTMENT OF EDUCATION
42. 42. Check It Out! Example What if…? A manufacturer wants to make a larger clock with a height of 30 centimeters. What is the length of each side of the frame? Round to the nearest tenth. Step 1 The equilateral triangle is divided into two 30º-60º-90º triangles. The height of the triangle is the length of the longer leg. Special Right Triangles DEPARTMENT OF EDUCATION
43. 43. Check It Out! Example Continued Step 2 Find the length x of the shorter leg. Each side is approximately 34.6 cm. Step 3 Find the length y of the longer leg. Rationalize the denominator. Hypotenuse = 2(shorter leg)y = 2x Simplify. Special Right Triangles DEPARTMENT OF EDUCATION