MATH STUDENT BOOK 10th Grde Unit 5
Unit 5 Similr Polygons MATH 1005 Similr Polygons INTRODUCTION 3 1. PRINCIPLES OF ALGEBRA 5 RATIOS AND PROPORTIONS 5 PROPERTIES OF PROPORTIONS 11 SELF TEST 1 16 2. SIMILARITY 18 MEANING OF SIMILARITY 18 THEOREMS ABOUT SIMILAR POLYGONS 31 THEOREMS ABOUT SIMILAR TRIANGLES 36 SELF TEST 2 61 3. RIGHT TRIANGLES 64 GEOMETRY 64 TRIGONOMETRY 69 INDIRECT MEASURE 79 SELF TEST 3 85 GLOSSARY 89 LIFEPAC Test is locted in the center of the booklet. Plese remove before strting the unit. Section 1 1
Similr Polygons Unit 5 Author: Milton R. Christen, M.A. Editor-in-Chief: Richrd W. Wheeler, M.A.Ed. Editor: Robin Hintze Kreutzberg, M.B.A. Consulting Editor: Robert L. Zenor, M.A., M.S. Revision Editor: Aln Christopherson, M.S. 804 N. 2nd Ave. E. Rock Rpids, IA 51246-1759 MCMXCVII by Alph Omeg Publictions division of Glynlyon, Inc. All rights reserved. LIFEPAC is registered trdemrk of Alph Omeg Publictions, Inc. All trdemrks nd/or service mrks referenced in this mteril re the property of their respective owners. Alph Omeg Publictions, Inc. mkes no clim of ownership to ny trdemrks nd/ or service mrks other thn their own nd their ffilites, nd mkes no clim of ffilition to ny compnies whose trdemrks my be listed in this mteril, other thn their own. 2 Section 1
Unit 5 Similr Polygons Similr Polygons Introduction In our study of congruent tringles, we lerned tht congruent tringles hve the sme size nd the sme shpe. We re now going to study objects tht hve the sme shpe, but not necessrily the sme size. We hve ll looked t photogrphs. The photogrph shows smller or lrger version of the object tht ws photogrphed. The picture is the sme shpe but different size. We hve ll tken trips nd used rod mp to help get us from here to there. A mp is smller version of the rel thing. Some of you my hve built model irplnes or my hve model trin. These models re scled-down versions of rel item. The model is the sme shpe s the rel thing but the size is different. These exmples re ll prcticl exmples of similrly shped objects. In mthemtics we lso study similr shpes. In this LIFEPAC we shll lern nd use some principles of lgebr in our study of similrity nd its ppliction to the right tringle. Objectives Red these objectives. The objectives tell you wht you will be ble to do when you hve successfully completed this LIFEPAC. When you hve finished this LIFEPAC, you should be ble to: 1. Write rtios in simplest form. 2. Nme the properties of proportions. 3. Solve proportion problems. 4. Identify similr polygons by using definitions, postultes, nd theorems. 5. Use theorems bout specil segments in similr tringles. 6. Solve problems bout similr right tringles. 7. Use the Pythgoren Theorem. 8. Solve tringle problems using trigonometry. 9. Find mesurements indirectly. Section 1 3
Unit 5 Similr Polygons 1. PRINCIPLES OF ALGEBRA Before we lern nd use some of the properties of similr polygons, we should review some bsic principles from lgebr. Section Objectives Review these objectives. When you hve completed this section, you should be ble to: 1. Express rtios in simplest form. 2. Nme the properties of proportion. 3. Solve proportion problems. RATIOS AND PROPORTIONS Two ides from lgebr tht we need to review t this time re rtios nd proportions. DEFINITION Rtio: the comprison of two numbers by division. The quotient is the rtio of the two numbers. The rtio of 3 to 15 is 5. The rtio of 8 to 2 is 1. The rtio of to b is b. Notice the quotients, 1 5, 4 1, b, re written s frctions. We cn rrive t the rtio by dividing the to number into the of number. Keep in mind tht you re not finding the rtio of one object to nother, but rther the rtio of two numbers tht re mesures of the object in the sme unit. Rther thn go through division process to find rtio, we cn set up the two numbers s frction nd reduce the frction. The of number will be the numertor (top) nd the to number will be the denomintor (bottom). Rtio is simply number. No units re connected to rtio. 1 4 Model 1: Find the rtio of 6 to 8. 6 8 = 3 4 Model 2: If AB = 6 inches nd CD = 18 inches, find the rtio of AB to CD. Model 3: Model 4: AB 6 1 CD = 18 = 3 If +A = 35, +B = 50, find the rtio of +A to +B. A 35 7 B = 50 = 10 If the side of one tringle is 2 feet nd the side of nother tringle is 18 inches, find the rtio of the smll side to the lrge side. 18 24 = 3 4 3 1 1 2 2 3 2 = 2 = 4 (2 feet ws chnged to 24 inches) (18 inches ws chnged 1 to 1 2 feet) Section 1 5
Similr Polygons Unit 5 The rtio of two numbers cn lso be written in the form :b. This form is useful when we re compring three or more numbers. Model 5: The rtio of 3 to 4 to 5 cn be written s 3:4:5. This rtio mens tht the rtio of 3 the first to the second is 4, the rtio 4 of the second to the third is 5, nd 3 the rtio of the first to the third is 5. The following sets of numbers ll hve the rtio 3:4:5. 12 3 16 4 12 3 {12, 16, 20} 16 = 4, 20 = 5, 20 = 5 3x 3 4x 4 3x 3 {3x, 4x, 5x} 4x = 4, 5x = 5, 5x = 5 2 2 2 15x 3 20x 4 15x 3 {15x 2, 20x 2, 25x 2 } 20x 2 = 4, 25x 2 = 5, 25x 2 = 5 Express ech rtio in simplest form. 6 6 1.1 12 = 18 36 1.3 29 = 1.2 9 = 1.4 72 = 1.5 21:28 = 1.7 10:20 = 1.6 25:45 = 1.8 15:25:35 = If = 3 nd b = 5, find ech rtio. 1.9 b = 1.11 ( + b) = b 1.10 = b 1.12 ( + b) = If c = 4 nd d = 5, find ech rtio. 1.13 c to d = 1.14 d to c = d (d c) 1.15 (c + d) = 1.16 (d + c) = 6 Section 1
Unit 5 Similr Polygons Use the following figures to find ech rtio in simplest form. A 12 14 D E 6 7 AD 1.17 DB = DB 1.19 AB = AC 1.21 AE = B P 6 AD C 1.18 AB = AE 1.20 AC = EC 1.22 AC = R 15 O 9 10 PO 1.23 OS = OS 1.25 PS = OR 1.27 QR = Q S 1.24 PO PS = OR 1.26 QO = QR 1.28 PS = Work the following problems. 1.29 The mesures of the ngles of tringle re in the rtio of 1:2:3. Find the mesure of ech ngle. (Hint: The sum of the ngles of tringle = 180.) 1.30 The distnce from A to B is 60 feet. The distnce from B to C is 10 yrds. The distnce from C to D is 20 inches. Find the rtio of AB:BC:CD. Section 1 7
Similr Polygons Unit 5 DEFINITION Proportion: n eqution tht sttes tht two rtios re equl. c The proportion b = d tells us tht the rtio to b nd the rtio c to d re equl rtios. The proportion cn be red, is to b s c is to d; or, the quotient of nd b equls the quotient of c nd d. The proportion cn lso be written :b = c:d. Ech of the four numbers, b, c, nd d is clled term of the proportion: is the first term, b is the second term, c is the third term, nd d is the fourth term. The first nd the fourth terms re known s the extremes of the proportion. The second nd third terms re clled the mens. 3 4 mens = extremes 15 20 3:4 = 15:20 mens extremes The fct tht more thn two rtios re equl is often expressed in the form of n extended proportion. c e g b = d = f = h is n extended proportion stting tht ll four rtios re the sme number. Another extended proportion is 1 2 3 4 5 6 7 8 2 = 4 = 6 = 8 = 10 = 12 = 14 = 16. We cn pick ny two of the rtios nd form 3 6 7 2 regulr proportion: 6 = 12, 14 = 4. Since proportion is n eqution, we cn use the properties of equlity to trnsform proportion to nother form. For exmple, b c = d cn be written s d = bc by using the multipliction property of equlity (multiply ech side of the eqution by bd). b = c d (bd) b = (bd) d = bc c d Nme the mens nd the extremes in these proportions. 3 15 1.31 4 = 20 5 20 1.32 7 = 28 mens: extremes: mens: extremes: 6 x 1.33 11 = y mens: extremes: 1.34 3:9 = 2:6 mens: extremes: 1.35 1:2 = 4:8 mens: extremes: 1.36 x:y = 3:7 mens: extremes: 8 Section 1
Unit 5 Similr Polygons Find the vlue of x in ech of these proportions. x 2 x 3 1.37 25 = 5 x = 9 3 10 x 1.39 x = 12 x = 1.38 6 = 2 x = 1.40 7 = 5 x = 1.41 9:x = x:4 x = 1 2 3 1.42 2 : x = 3 : 4 x = (x + 3) 5 (x + 1) 2 1.43 6 = 4 x = 3 x 1.45 2 = 4 x = 1.44 (x + 2) = 3 x = Find the rtio of x to y. 1.46 2x = 3y x y 1.48 3 = 2 1.47 5x = 7y 1.49 2x 3y = 0 1.50 x 5y = 0 Along with our knowledge of complementry nd supplementry ngles, rtios cn help us solve geometry problems. Model: Two complementry ngles hve mesures in the rtio of 4 to 5. Find the mesure of ech ngle. Solution: Let 4x = mesure of the smller ngle. 5x = mesure of the lrger ngle. Then 4x + 5x = 90 (ngles re complementry) 9x = 90 x = 10 4x = 40 5x = 50 Section 1 9
Similr Polygons Unit 5 Solve the following problems. Show your work nd circle your nswer. 1.51 Two complementry ngles hve mesures in the rtio of 1 to 5. Find the mesure of ech ngle. 1.52 The rtio of the mesures of two supplementry ngles is 3:7. Find the mesure of ech ngle. 1.53 A 30-inch segment is cut into two prts whose lengths hve the rtio 3 to 5. Find the length of ech prt. 1.54 The perimeter of tringle is 48 inches nd the sides re in the rtio of 3:4:5. Find the length of ech side. 1.55 A tringle hs perimeter of 18 inches. If one side hs length of 8 inches, find the other two sides if their lengths re in the rtio of 2 3. 10 Section 1
Unit 5 Similr Polygons PROPERTIES OF PROPORTIONS You will often wish to chnge proportion into some equivlent eqution. Although we cn mke this chnge by using bsic properties of lgebr, we will sve time nd steps by using some specil properties of proportions. When we use these properties for reson on proof, we shll simply write POP (Property of Proportion). In these sttements the vribles used represent nonzero numbers. EQUIVALENT FORMS PROPERTY c b d c b d b = d, c = d, b =, = c re equivlent proportions. The cross product of ech proportion gives the sme eqution, d = bc. 2 5 Model 1: 4 = 10 20 = 20 CROSS PRODUCT PROPERTY c If b = d, then d = bc. 3 15 Model 1: 5 = 25 10 5 Model 2: 4 = 2 20 = 20 3 25 = 5 15 75 = 75 2 4 Model 3: 5 = 10 20 = 20 6 12 Model 2: 11 = 22 6 22 = 11 12 132 = 132 4 10 Model 4: 2 = 5 20 = 20 x 3 Model 3: 10 = 5 5x = 30 x = 6 Section 1 11
Similr Polygons Unit 5 DENOMINATOR SUM PROPERTY c ( + b) (c + d) If b = d, then b = d. DENOMINATOR DIFFERENCE PROPERTY c ( b) (c d) If b = d, then b = d. We cn dd 1 to both sides of the eqution: b = c d b + 1 = d + 1 b c d b + b = d + d ( + b) (c + d) b = d c We cn subtrct 1 from both sides of the eqution: b = c d b 1 = d 1 b c d b b = d d ( b) b = (c d) d c 2 5 Model 1: 4 = 10 (2 + 4) 4 = (5 + 10) 10 6 4 = 15 10 8 16 Model 1: 3 = 6 (8 3) 3 = (16 6) 6 5 3 = 10 6 3 1 Model 2: 9 = 3 (3 + 9) 9 = (1 + 3) 3 12 9 = 4 3 12 4 Model 2: 3 = 1 (12 3) 3 = (4 1) 1 9 3 = 3 1 2 3 Model 3: 8 = 12 (2 + 8) 8 = (3 + 12) 12 10 8 = 15 12 6 3 Model 3: 8 = 4 (6 8) 8 = (3 4) 4 2 1 8 = 4 12 Section 1
Unit 5 Similr Polygons NUMERATOR-DENOMINATOR SUM PROPERTY c e g ( + c + e + g +...) c e If b = d = f = h =..., then (b + d + f + h +...) = b = d = f... c e Let b = x, d = x, f = x,... then = bx, c = dx, e = fx,... ( + c + e +...) (bx + dx + fx +...) so (b + d + f +...) = (b + d + f +...) = x (b + d + f +...) = x (b + d + f +...) c e = b = d = f... 1 2 3 4 Model 1: 2 = 4 = 6 = 8 (1 + 2 + 3 + 4) (2 + 4 + 6 + 8) = 1 2 10 20 = 1 2 3 6 9 Model 2: 5 = 10 = 15 (3 + 6 + 9) (5 + 10 + 15) = 6 10 18 30 = 6 10 Nme the POP illustrted. b ( + 2) (b + 5) 1.56 If 2 = 5, then 2 = 5. ( + 1) b 1.57 If 3 = 5, then 5( + 1) = 3b. x r t (x + r + t) x 1.58 If y = s = u, then (y + s + u) = y. x y x 4 1.59 If 4 = 3, then y = 3. (x + 3) (y + 6) (x + 1) (y + 3) 1.60 If 2 = 3, then 2 = 3. r s 1.61 If x = t, then sx = rt. Section 1 13
Similr Polygons Unit 5 SELF TEST 1 Complete the following sttements (ech nswer, 3 points). 1.01 A rtio compres two numbers by. 1.02 The rtio of 3 to 4 is written s. 1.03 Another wy of writing the rtio of 3 to 4 is. 1.04 In writing the rtio of mesurement numbers, the units must be the. 1.05 The rtio of 3 to 4 nd the rtio of 4 to 3 (re/re not) the sme number. 1.06 A proportion is n eqution tht sttes two re equl. 2 x 1.07 If 3 = y, then 2y = 3x is n exmple of the POP. x 2 (x + 2 + ) 2 1.08 If y = 3 = b, then (y + 3 + b) = 3 is n exmple of the POP. p 1.09 In the proportion b = q, the third term is. 5 10 1.010 In the proportion 6 = 12, the mens re. nd b.. Express ech rtio in simplest form (ech nswer, 2 points). 1.011 Rtio of 6 feet to 3 feet. 1.012 Rtio of 7 yrds to 6 feet. 1.013 Rtio of 12 to 100. A B C } 3 } 9 1.014 Rtio of AB to BC. 1.015 Rtio of BC to AC. 16 Section 1
Unit 5 Similr Polygons Find x in the following proportions (ech nswer, 2 points). x 3 1.016 7 = 5 x = 5 25 1.018 2x = 4 x = 16 x 1.020 x = 4 x = 1.017 3:8 = x:32 x = x x + 2 1.019 3 = 5 x = x + 2 6 + 2 1.021 2 = 2 x = Find the required numbers (ech nswer, 2 points). TU UW WT Given: TR = RS = ST TU = 3 RS = 15 TW = 4 TR = 9 U T W 1.022 UW = 1.023 ST = R S 1.024 WS = 1.025 UR = 54 67 SCORE TEACHER initils dte Section 1 17
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