1. Geometry Connections ExtraPracticeManaging EditorLeslie DietikerPhillip and Sala Burton Academic High SchoolSan Francisco, CAContributing EditorsElizabeth Ellen CoynerLew DouglasChristian Brothers High SchoolThe College Preparatory SchoolSacramento, CAOakland, CADavid GulickPhillips Exeter AcademyExeter, NHLara LomacPhillip and Sala Burton AcademicHigh School, San Francisco, CADamian MolinariPhillip and Sala Burton AcademicHigh School, San Francisco, CAJason Murphy-ThomasGeorge Washington HighSchool, San Francisco, CALeslie NielsenIsaaquah High SchoolIssaquah, WAKaren OConnellSan Lorenzo High SchoolSan Lorenzo, CAWard QuinceyGideon Hausner Jewish DaySchool, Palo Alto, CABarbara ShreveSan Lorenzo High SchoolSan Lorenzo, CAMichael TitelbaumUniversity of CaliforniaBerkeley, CAIllustratorKevin CoffeySan Francisco, CATechnical AssistantsErica Andrews Elizabeth Fong Marcos RojasElizabeth Burke Rebecca Harlow Susan RyanCarrie Cai Michael LeongDaniel Cohen Thomas LeongProgram DirectorsLeslie Dietiker Judy Kysh, Ph.D.Phillip and Sala Burton Academic High School Departments of Mathematics and EducationSan Francisco, CA San Francisco State UniversityBrian Hoey Tom Sallee, Ph.D.Christian Brothers High School Department of MathematicsSacramento, CA University of California, Davis

2. Contributing Editor of theParent Guide:Karen Wootten, Managing EditorOdenton, MDTechnical Manager ofParent Guide and ExtraPractice:Rebecca HarlowStanford UniversityStanford, CAEditor of Extra Practice:Bob Petersen, Managing EditorRosemont High SchoolSacramento, CAAssessment Contributors:John Cooper, Managing EditorDel Oro High SchoolLoomis, CALeslie DietikerPhillip and Sala Burton Academic High SchoolSan Francisco, CADamian MolinariPhillip and Sala Burton Academic High SchoolSan Francisco, CABarbara ShreveSan Lorenzo High SchoolSan Lorenzo, CACopyright 2007 by CPM Educational Program. All rights reserved. No part of thispublication may be reproduced or transmitted in any form or by any means, electronic ormechanical, including photocopy, recording, or any information storage and retrievalsystem, without permission in writing from the publisher. Requests for permission should bemade in writing to: CPM Educational Program, 1233 Noonan Drive, Sacramento, CA95822. Email: cpm@cpm.org.1 2 3 4 5 6 7 8 9 10 09 08 07 06 ISBN-10: 1-931287-63-5Printed in the United States of America ISBN-13: 978-1-931287-63-0 3. GEOMETRY CONNECTIONSEXTRA PRACTICEIntroduction to Students and Their TeachersLearning is an individual endeavor. Some ideas come easily; others take time--sometimes lots of time--to grasp. In addition, individual students learn the same ideain different ways and at different rates. The authors of this textbook designed theclassroom lessons and homework to give students time--often weeks and months--topractice an idea and to use it in various settings. This section of the textbook offersstudents a brief review of 33 geometry and algebra topics followed by additionalpractice with answers. Not all students will need extra practice. Some will need todo a few topics, while others will need to do many of the sections to help developtheir understanding of the ideas. This resource of the text may also be useful toprepare for tests, especially final examinations.How these problems are used will be up to your teacher, your parents, andyourself. In classes where a topic needs additional work by most students, yourteacher may assign work from one of the extra practice sections that follow. In mostcases, though, the authors expect that these resources will be used by individualstudents who need to do more than the textbook offers to learn an idea. This willmean that you are going to need to do some extra work outside of class. In the casewhere additional practice is necessary for you individually or for a few students inyour class, you should not expect your teacher to spend time in class going over thesolutions to the extra practice problems. After reading the examples and trying theproblems, if you still are not successful, talk to your teacher about getting a tutor orextra help outside of class time.Warning! Looking is not the same as doing. You will never become good atany sport just by watching it. In the same way, reading through the worked outexamples and understanding the steps are not the same as being able to do theproblems yourself. An athlete only gets good with practice. The same is true ofdeveloping your algebra and geometry skills. How many of the extra practiceproblems do you need to try? That is really up to you. Remember that your goal is tobe able to do problems of the type you are practicing on your own, confidently andaccurately.Two other sources for help with the geometry and algebra topics in thiscourse are the Geometry Connections Parent Guide and the Algebra ConnectionsParent Guide. These resources are available free from the internet at www.cpm.org.There is also an order form there in the event you wish to purchase them. 4. Geometry Extra Practice Topics1. Transformations 12. Properties of angles, lines, & triangles 33. Area 64. Areas by dissection 95. The Pythagorean Theorem 126. Triangle similarity 147. Ratio of similarity 168. Geometric probability 189. Right triangle trigonometry 2110. Laws of sines and cosines 2511. Triangle congruence expressions 2812. Proof 3113. Quadrilaterals 3514. Interior-exterior angles of polygons 3915. Area of sectors 4116. Similarity of length, area, and volume 4417. Volume and surface area of polyhedra 4718. Central and inscribed angles 5119. Tangents, secants, and chords 5320. Cones and spheres 58Algebra Extra Practice Topics21. Writing and graphing linear equations 6222. Solving linear systems 6723. Linear inequalities 7024. Multiplying polynomials 7525. Factoring polynomials 7726. Zero product property and quadratics 8027. The quadratic formula 8228. Laws of Exponents 8529. Radicals 8730. Simplifying rational expressions 8931. Multiplication and division of rational expressions 9132. Addition and subtraction of rational expressions 9333. Solving mixed equations and inequalities 95 5. RIGID TRANSFORMATIONS #1A RIGID TRANSFOMATION is a transformation of a figure that preservessize and shape. The three kinds of rigid transformations studied are translation(slide), reflection (flip), and rotation (turn). These transformations may also becombined together. The original figure is called the pre-image and the newfigure is called the image. For !ABC it is common to say that the original pointsA, B, and C are mapped to the new points A' , B' , and C' respectively. For moreinformation about rigid transformations see the Math Notes boxes on pages 33and 38.Example 1yTranslate (slide) !ABC right six units and up three units. Givethe coordinates of the image triangle.The original vertices are A(-5, -2), B(-3, 1), and C(0, -5). The newvertices are A' (1, 1), B' (3, 4), and C' (6, -2). Notice that eachpoint (x, y) is mapped to (x + 6, y + 3).Example 2Reflect (flip) !ABC with coordinates A(5, 2), B(2, 4), and C(4, 6)across the y-axis to get !A'B'C' . The key is that the reflection isthe same distance from the axis as the original figure. The newpoints are A' (-5, 2), B' (-2, 4), and C' (-4, 6). Notice that inreflecting across the y-axis, each point (x, y) is mapped to the point(-x, y).If you reflect !ABC across the x-axis to get !PQR , then the new points are P(5, -2), Q(2, -4), andR(4, -6). In this case, reflecting across the x-axis, each point (x, y) is mapped to the point (x, -y).Example 3Rotate (turn) !ABC with coordinates A(2, 0), B(6, 0), and C(3, 4)90 counterclockwise about the origin (0, 0) to get !A'B'C' withcoordinates A' (0, 2), B' (0, 6), and C' (-4, 3). Notice that this 90counterclockwise rotation about the origin maps each point(x, y) to the point (-y, x).Rotating another 90 (180 from the starting location) yields!A"B"C"with coordinates A"(-2, 0), B" (-6, 0), and C" (-3, -4).This 180 counterclockwise rotation about the origin maps each point (x, y) to the point (-x, -y).Similarly a 270 counterclockwise or 90 clockwise rotation about the origin maps eachpoint (x, y) to the point (y, -x).xABA'B'C'CxyB'A'CyA BC'A"B"C"GEOMETRY Connections 1xC'A'B'CBAPQR 6. For the following problems, refer to the figures below:Figure ACxyABFigure BxyA BCFigure CxyA BCTell the new coordinates after each transformation.1. Slide figure A left 2 units and down 3 units.2. Slide figure B right 3 units and down 5 units.3. Slide figure C left 1 unit and up 2 units.4. Flip figure A across the x-axis.5. Flip figure B across the x-axis.6. Flip figure C across the x-axis.7. Flip figure A across the y-axis.8. Flip figure B across the y-axis.9. Flip figure C across the y-axis.10. Rotate figure A 90 counterclockwise about the origin.11. Rotate figure B 90 counterclockwise about the origin.12. Rotate figure C 90 counterclockwise about the origin.13. Rotate figure A 180 counterclockwise about the origin.14. Rotate figure C 180 counterclockwise about the origin.15. Rotate figure B 270 counterclockwise about the origin.16. Rotate figure C 90 clockwise about the origin.Answers (given in order A' , B' , C' )1. (-1, -3) (1, 2) (3, -1) 2. (-2, -3) (2, -3) (3, 0)3. (-5, 4) (3, 4) (-3, -1) 4. (1, 0) (3, -4) (5, -2)5. (-5, -2) (-1, -2) (0, -5) 6. (-4, -2) (4, -2) (-2, 3)7. (-1, 0) (-3, 4) (-5, 2) 8. (5, 2) (1, 2) (0, 5)9. (4, 2) (-4, 2) (2, -3) 10. (0, 1) (-4, 3) (-2, 5)11. (-2, -5) (-5, 0) (-2, -1) 12. (-2, -4) (-2, 4) (3, -2)13. (-1, 0) (-3, -4) (-5, -2) 14. (4, -2) (-4, -2) (2, 3)15. (2, 5) (2, 1) (5, 0) 16. (2, 4) (2, -4) (-3, 2)2 Extra Practice 7. PROPERTIES OF ANGLES, LINES, AND TRIANGLES #2Parallel lines Triangles123 498 7 61011 corresponding angles are equal:m!1 = m!3 alternate interior angles are equal:m!2 = m!3 m!2 + m!4 = 180 m!7 + m!8 + m!9 = 180 m!6 = m!8 + m!9(exterior angle = sum remote interior angles) m!10 + m!11 = 90(complementary angles)Also shown in the above figures: vertical angles are equal: m!1 = m!2 linear pairs are supplementary: m!3 + m!4 = 180and m!6 + m!7 = 180In addition, an isosceles triangle, ABC, hasBA=BC and m!A = m!C. An equilateraltriangle, GFH, has GF = FH = HG andm!G = m!F = m!H = 60 . ABCFG HExample 1Solve for x.Use the Exterior Angle Theorem: 6x + 8 = 49 + 676x= 108 ! x =1086! x = 186x+8 49Example 2Solve for x.There are a number of relationships in this diagram. First, 1and the 127 angle are supplementary, so we know thatm!1+ 127= 180 so m1 = 53. Using the same idea,m2 = 47. Next, m!3 + 53+ 47= 180 , so m3 = 80.Because angle 3 forms a vertical pair with the angle marked7x + 3, 80 = 7x + 3, so x = 11.Example 3Find the measure of the acute alternate interior angles.Parallel lines mean that alternate interior angles are equal, so5x + 28= 2x + 46 ! 3x = 18 ! x = 6 . Use either algebraic anglemeasure: 2(6) + 46= 58 for the measure of the acute angle.6717x+323127 1332x + 465x+28GEOMETRY Connections 3 8. Use the geometric properties and theorems you have learned to solve for x in eachdiagram and write the property or theorem you use in each case.1.7560x2.6580x3.100x x4.112x x5.6060 4x + 106.6060 8x 607.45 3x8.1255x9.685x + 1210.12810x+211.3019x+312.583x13.3814220x + 214.3814220x 215.128527x + 316.525x + 3 12817.x235818.117578x19.189x5x + 3620.8x + 12 in.12x 18 in.21.8x18 cm5x + 3 cm4 Extra Practice 9. 22.5x 1823.50705x 1024.13x + 215x 225.403x + 2026.452x + 527.5x + 87x 428.5x + 87x 46x 4Answers1. 45 2. 35 3. 40 4. 34 5. 12.5 6. 157. 15 8. 25 9. 20 10. 5 11. 3 12. 103 213. 7 14. 2 15. 7 16. 25 17. 81 18. 7.519. 9 20. 7.5 21. 7 22. 15.6 23. 26 24. 225. 40 26. 65 27. 71 28. 106 GEOMETRY Connections 5 10. AREA #3AREA is the number of square units in a flat region. The formulas to calculatethe area of several kinds of polygons are:RECTANGLE PARALLELOGRAM TRAPEZOID TRIANGLEhbhbhbb12hbhbA = bh A = bh A =12( b+ b)h A =1 2 12bhNote that the legs of any right triangle form a base and a height for the triangle.The area of a more complicated figure may be found by breaking it into smallerregions of the types shown above, calculating each area, and finding the sum.Example 1Find the area of each figure. All lengths are centimeters.a)2381A = bh = (81)(23) = 1863cm2b)412A =12(12)(4) = 24 cm2c)42108A =12(108)(42) = 2268cm2d)10 821A = (21)8 = 168 cm2Note that 10 is a side of the parallelogram, not the height.e)121434A =12(14 + 34)12 =12(48)(12) = 288cm26 Extra Practice 11. Example 2Find the area of the shaded region.The area of the shaded region is the area of thetriangle minus the area of the rectangle.triangle: A 1=(7)(10) = 35cm22rectangle: A = 5(4) = 20 cm2shaded region: A = 35 ! 20 = 15cm24 cm.7 cm.5 cm.10 cm.Find the area of the following triangles, parallelograms and trapezoids. Pictures are not drawnto scale. Round answers to the nearest tenth.1. 2. 3. 4.102036510865465. 6. 7. 8.1326267 151512828201649. 10. 11. 12.39 41.56741082!10613. 14. 15. 16.1513592355!2713964 4GEOMETRY Connections 7 12. Find the area of the shaded region.17.4810243663218.182419. 18341620.10612316Answers (in square units)1. 100 2. 90 3. 24 4. 15 5. 338 6. 1057. 93 8. 309.8 9. 126 10. 19.5 11. 36.3 12. 5413. 84 14. 115 15. 110.8 16. 7.9 17. 1020 18. 21619. 272 20. 1388 Extra Practice 13. AREAS BY DISSECTION #4DISSECTION PRINCIPLE: Every polygon can be dissected (or broken up)into triangles which have no interior points in common. This principle is anexample of the problem solving strategy of subproblems. Finding simplerproblems that you know how to solve will help you solve the larger problem.Example 1Find the area of the figure below.4245226233First fill in the missing lengths. The rightvertical side of the figure is a total of8 units tall. Since the left side is also8 units tall, 8 4 2 = 2 units. Themissing vertical length on the top middlecutout is also 2.Horizontally, the bottom length is 8 units(6 + 2). Therefore, the top is 2 units(8 4 2). The horizontal length at theleft middle is 3 units.Enclose the entire figure in an 8 by 8square and then remove the area of thethree rectangles that are not part of thefigure. The area is:Enclosing square: 8 ! 8 = 64Top cut out: 2 ! 2 = 4Left cut out: 3! 2 = 6Lower right cut out: 2 ! 3 = 6Area of figure: 64 ! 4 ! 6 ! 6 = 48Example 2Find the area of the figure below.355 422 2This figure consists of five familiarfigures: a central square, 5 units by 5units; three triangles, one on top withb = 5 and h = 3, one on the right withb = 4 and h = 3, and one on the bottomwith b = 5 and h = 2; and a trapezoidwith an upper base of 4, a lower base of 2and a height of 2.The area is:square 5 ! 5 = 25top triangle 12! 5 ! 3 = 7.5bottom triangle 12! 5 ! 2 = 5right triangle 12! 3! 4 = 6trapezoid (4 + 2) !22 = 6total area = 49.5 u2GEOMETRY Connections 9 14. Find the area of each of the following figures. Assume that anything that looks like aright angle is a right angle.1.64232322.3332 23 33. 75594.4321 185.22211334456.21 210322227.44588.3213 2439.44225610. 11 and 12: Find the surface area of each polyhedra.3314 441015232343610Find the area and perimeter of each of the following figures.13.3 542314.3425615.455645516.3135 1353 26042210 Extra Practice 15. Answers1. 42 u2 2. 33 u2 3. 85 u2 4. 31 u2 5. 36 u26. 36 u2 7. 36 u2 8. 29.5 u2 9. 46 u2 10. 28 u211. SA = 196 sq.un. 12. 312 sq.un.13. A = 32u2P = 34 + 3 2 + 12 ! 38.3162u14. A = 41.5u2P = 18 + 4 2 + 10 ! 26.8191u15. A = 21u2P = 16 + 6 2 ! 24.4853u16. A = 14 + 6 3 ! 24.3923u2P = 24 + 2 6 ! 28.899uGEOMETRY Connections 11 16. PYTHAGOREAN THEOREM #5acbAny triangle that has a right angle is called a RIGHTTRIANGLE. The two sides that form the right angle, aand b, are called LEGS, and the side opposite (that is,across the triangle from) the right angle, c, is called theHYPOTENUSE.For any right triangle, the sum of the squares of the legs of the triangle is equalto the square of the hypotenuse, that is, a2 + b2 = c2. This relationship is knownas the PYTHAGOREAN THEOREM. In words, the theorem states that:(leg)2 + (leg)2 = (hypotenuse)2.ExampleDraw a diagram, then use the Pythagorean Theorem to write an equation to solve each problem.a) Solve for the missing side.13c17c2 + 132 = 172c2 + 169 = 289c2 = 120c = 120c = 2 30c ! 10.95b) Find x to the nearest tenth:5xx20(5x)2 + x2 = 20225x2 + x2 = 40026x2 = 400x2 ! 15.4x ! 15.4x ! 3.9c) One end of a ten foot ladder is four feetfrom the base of a wall. How high onthe wall does the top of the ladder touch?x1044x2 + 42 = 102x2 + 16 = 100x2 = 84x 9.2The ladder touches the wall about 9.2feet above the ground.d) Could 3, 6 and 8 represent the lengthsof the sides of a right triangle?Explain.32 + 62?82=?649 + 36 =45! 64Since the Pythagorean Theoremrelationship is not true for theselengths, they cannot be the sidelengths of a right triangle.12 Extra Practice 17. Use the Pythagorean Theorem to find the value of x. Round answers to the nearest tenth.1. 2. 3. 4. 5.3722xx9620x42x16 4683x72 656. 7. 8. 9. 10.x16 221532xx381675105x12530xSolve the following problems.11. A 12 foot ladder is six feet from a wall. How high on the wall does the ladder touch?12. A 15 foot ladder is five feet from a wall. How high on the wall does the ladder touch?13. A 9 foot ladder is three feet from a wall. How high on the wall does the ladder touch?14. A 12 foot ladder is three and a half feet from a wall. How high on the wall does the ladder touch?15. A 6 foot ladder is one and a half feet from a wall. How high on the wall does the ladder touch?16. Could 2, 3, and 6 represent the lengths of sides of a right angle triangle? Justify your answer.17. Could 8, 12, and 13 represent the lengths of sides of a right triangle? Justify your answer.18. Could 5, 12, and 13 represent the lengths of sides of a right triangle? Justify your answer.19. Could 9, 12, and 15 represent the lengths of sides of a right triangle? Justify your answer.20. Could 10, 15, and 20 represent the lengths of sides of a right triangle? Justify your answer.Answers1. 29.7 2. 93.9 3. 44.9 4. 69.1 5. 31.06. 15.1 7. 35.3 8. 34.5 9. 73.5 10. 121.311. 10.4 ft 12. 14.1 ft 13. 8.5 ft 14. 11.5 ft 15. 5.8 ft16. no 17. no 18. yes 19. yes 20. noGEOMETRY Connections 13 18. TRIANGLE SIMILARITY #6Two triangles are SIMILAR if they have the same shape but not necessarily thesame size. There are three similarity statements using only sides and angles thatguarantee similar triangles. They are side-side-side similarity (SSS ! ), angle-anglesimilarity (AA ! ), and side-angle-side similarity (SAS ! .) In side-angle-sidesimilarity the angle must be between the two sides.Example 1The two triangles shown at right illustrate the SSS !theorem. The ratios of corresponding sides are equal:93=186=124. Therefore !ABC "!DEF by SSS ! .Example 2AA ! is demonstrated by the two triangles at right..Since the sum of the angles of any triangle equal 180degrees, mF = 40. The measures of two angles in thefirst triangle are equal to the measures of two angles inthe second triangle. Therefore !ABC "!DEF by AA ! .Example 3The two triangles at right illustrate the SAS ! theorem.The ratios of two sets of corresponding sides are equal:510=816. Also the measures of the included angles areequal: mC = mF. Therefore !ABC "!DEF bySAS ! .18B C129AD3 6E F4A85 40 85BC D55EAB8CDE16F2020510Determine if each pair of triangles is similar. If they are similar, justify your answer.1.2.3.F363647 4792432125316 48 124.55555 7214 Extra Practice 19. 4.5. 6.45 155454398 1010207.8. 9.10121579122137391410.11. 12.123246504050 402025 2010Answers1. AA ! 2. SSS ! 3. AA ! 4. SAS !5. not ! 6. not ! 7. SAS ! or SSS ! 8. not !9. AA ! 10. SSS ! 11. AA ! 12. AA !74614GEOMETRY Connections 15 20. 35RATIO OF SIMILARITY #7The RATIO OF SIMILARITY between any two similar figures is the ratio of anypair of corresponding sides. Simply stated, once it is determined that two figuresare similar, all of their pairs of corresponding sides have the same ratio.The ratio of similarity offigure P to figure Q, writtenP : Q, is . 4.5and 127.5 20 havethe same ratio. a34.5P12~ Qb57.520Note that the ratio of similarity is always expressed in lowest possible terms. Also,the order of the statement, P : Q or Q : P, determines which order to state and usethe ratios between pairs of corresponding sides.An Equation that sets one ratio equal to another ratio is called a312=proportion. An example is at right.520ExampleFind x in the figure. Be consistent in matchingcorresponding parts of similar figures.ABC ~ DEC by AA (BAC EDC and C C).Therefore the corresponding sides are proportional.102022xB12 24ACA ==> 221116 Extra PracticeCDEDEAB= CDx = 2436 ==> 24x = 22(36) ==> 24x = 792 ==> x = 33Note: This problem also could have been solved with the proportion: DEAB=CECB .For each pair of similar figures below, find the ratio of similarity, for large:small.1.121592.553573.128644.8 7245.66116.415 1068 21. For each pair of similar figures, state the ratio of similarity, then use it to find x.7.141628x8.243612x9.6x184510.x20182411.1645x12.1011121518xFor problems 13 to 18, use the given information and the figure to find each length.13. JM = 14, MK = 7, JN = 10 Find NL.J14. MN = 5, JN = 4, JL = 10 Find KL.15. KL = 10, MK = 2, JM = 6 Find MN.16. MN = 5, KL = 10, JN = 7 Find JL.M N17. JN = 3, NL = 7, JM = 5 Find JK.18. JK = 37, NL = 7, JM = 5 Find JN.K L19. Standing 4 feet from a mirror laying on the flat ground, Palmer, whose eye height is 5feet, 9 inches, can see the reflection of the top of a tree. He measures the mirror to be24 feet from the tree. How tall is the tree?20. The shadow of a statue is 20 feet long, while the shadow of a student is 4 ft long. Ifthe student is 6 ft tall, how tall is the statue?Answers1. 43 2. 51 3. 21 4. 2475. 61 6. 158 7. 78 ; x = 32 8. 21 ; x = 729. 13 ; x = 15 10. 56 ; x = 15 11. 45 ; x = 20 12. 32 ; x = 16.513. 5 14. 12.5 15. 7.5 16. 1417. 163 18. 11 2532 19. 34.5 ft. 20. 30 ft.2GEOMETRY Connections 17 22. GEOMETRIC PROBABILITY #8The PROBABILITY of an event involving outcomes with different probabilitiesis represented using an area model and a tree diagram. For one example and acomplete explanation of the two methods see the Math Notes box on page 212 ofthe textbook. Two additional examples are shown below.Example 1A popular game at a county fair is Flip-to-Spin-or-Roll. You start by flipping a coin. If headscomes up, you get to spin the big wheel, whichhas ten equal sectors: three red, three blue, andfour yellow. If the coin shows tails, you roll acube with three sides red, two yellow sides, andone blue side. If your spin lands on blue or theblue side of the cube comes up you win a prize.What is the probability of winning a prize?Using the two blue boxes from the area model or the two blue branches from the treediagram, the probability of winning a prize is 3area model tree diagramR310 ( )B310 ( )Y25 ( )15 320 32014 12 ( )H 12 ( )T 112 ( )R20 + 112 = 1460 = 730 .2 ( )H 12 ( )T 110 ( )R 310 ( )B 35 ( )Y 22 ( )R 13 ( )Y 16 ( )B 16 11213 ( )Y16 ( )BExample 2There are five children in Sarahs family.Each night one child has to help at thefamily business. A spinner is used todetermine who has to work each night.Sarah has a big math test tomorrow andknows there is an 80% chance of getting anA if she can study but only a 10% chanceof getting an A if she can not study. Whatis the probability of Sarah getting an A?Using the two A boxes from the area model or the two A branches from the treediagram, the probability of getting an A is 0.64 + 0.02 = 0.66.320320151416112.64.16.02.18A(.8)not A(.2)A(.1)not A(.9)study(.8)work(.2)A(.8)not A(.2).64 .16.02 .18study(.8)work(.2)(.1)A(.9)not A18 Extra Practice 23. For each question use an area model or atree diagram to compute the desiredprobability.For problems 1-3 use the spinners atrightBRY RBRB YR1. If each spinner is spun once, what is the probability that both spinners show blue?2. If each spinner is spun once, what is the probability that both spinners show thesame color?3. If each spinner is spun once, what is the probability of getting a red-bluecombination?4. A pencil box has three yellow pencils, one blue, and two red pencils. There arealso two red erasers and one blue. If you randomly choose one pencil and oneeraser, what is the probability of getting the red-red combination?5. Sallys mother has two bags of candy but she says that Sally can only have onepiece. Bag #1 has 70% orange candies and 30% red ones. Bag #2 has 10%orange, 50% white, and 40% green. Sallys eyes are covered and she chooses onebag and pulls out one candy. What is the probability that she choose her favoritecolor-orange?6. You throw a die and flip a coin. What is the probability of getting tails on thecoin and a number less than five on the die?7. A spinner is evenly divided into eight sectionsthree are red, three are white andtwo are blue. If the spinner is spun twice, what is the probability of getting thesame color twice?8. Your friend and you have just won achance to collect a million dollars. Youplace the money in one room at right andyou friend has to walk through the maze.In which room should you place the moneyso that your friend will have the bestchance of finding the million dollars?A BGEOMETRY Connections 19 24. 9. Find the probability of randomlyentering each room.a) P(A)b) P(B)c) P(C)ABC10. The weather forecast is a 60% chance of rain. If there is no rain then there is an80% chance of going to the beach. What is the probability of going to the beach?11. A baseball player gets a hit 40% of the time if the weather is good but only 20%of the time if it is cold or windy. The weather forecast is a 70% chance of nice,20% chance of cold, and 10% chance of windy. What is the probability of gettinga hit?12. If you have your assignment completed before the next day there is an 80%chance of a good grade. If the assignment is finished during class or late thenthere is only a 30% chance of a good grade. If assignments are not done at allthen there is only a 5% chance of a good grade. In a certain class, 50% of thestudents have the assignment completed before class, 40% finish during class, and10% do not do their assignments. If a student is selected at random, what is theprobability that student has a good grade?Answers1. 1122. 924=383. 7244. 295. 256. 137. 11328. P(B) = 599. 1118, 518, 21810. 0.32 11. 0.34 12. 0.52520 Extra Practice 25. RIGHT TRIANGLE TRIGONOMETRY #9The three basic trigonometric ratios for right trianglesare the sine (pronounced "sign"), cosine, and tangent.BEach one is used in separate situations, and the easiestway to remember which to use when is the mnemonicSOH-CAH-TOA. With reference to one of the acuteACangles in a right triangle, Sine uses the Opposite andtan A = opposite legthe Hypotenuse - SOH. The Cosine uses the Adjacentside and the Hypotenuse - CAH, and the Tangent usesthe Opposite side and the Adjacent side -TOA. Ineach case, the position of the angle determines whichleg (side) is opposite or adjacent. Remember thatopposite means across from and adjacent meansnext to.adjacent leg = BCACsin A = opposite leghypotenuse = BCABcos A = adjacent leghypotenuse = ACABExample 1Use trigonometric ratios to find the lengthsof each of the missing sides of the trianglebelow.y h4217 ftThe length of the adjacent side with respectto the 42 angle is 17 ft. To find the lengthy, use the tangent because y is the oppositeside and we know the adjacent side.tan 42 = y1717 tan 42 = y15.307 ft ! yThe length of y is approximately 15.31 feet.To find the length h, use the cosine ratio(adjacent and hypotenuse).cos 42 = 17hhcos 42 = 17h = 17cos 42 ! 22.876 ftThe hypotenuse is approximately 22.9 feet long.Example 2Use trigonometric ratios to find the size of eachangle and the missing length in the triangle below.v18 ft hu21 ftTo find mu, use the tangent ratio because youknow the opposite (18 ft) and the adjacent (21 ft)sides.tan u = 1821m!u = tan"1 1821 # 40.601The measure of angle u is approximately40.6. By subtraction we know thatmv 49.4.Use the sine ratio for mu and the oppositeside and hypotenuse.sin 40.6 = 18hhsin 40.6 = 18h = 18sin 40.6 ! 27.659 ftThe hypotenuse is approximately 27.7 feet long.GEOMETRY Connections 21 26. Use trigonometric ratios to solve for the variable in each figure below.1.3815h2.268 h3.4923x4.4137x5.1538y6.5543y7.38z158.52z 189.3823w10.38w1511.15x3812.2991x13.x7514.u9715.y121816.v788822 Extra Practice 27. Draw a diagram and use trigonometric ratios to solve each of the following problems.17. Juanito is flying a kite at the park and realizes that all 500 feet of string areout. Margie measures the angle of the string with the ground with herclinometer and finds it to be 42.5. How high is Juanitos kite above theground?18. Nells kite has a 350 foot string. When it is completely out, Ian measures theangle to be 47.5. How far would Ian need to walk to be directly under thekite?19. Mayfield High Schools flagpole is 15 feet high. Using a clinometer, Tamarameasured an angle of 11.3 to the top of the pole. Tamara is 62 inches tall.How far from the flagpole is Tamara standing?20. Tamara took another sighting of the top of the flagpole from a differentposition. This time the angle is 58.4. If everything else is the same, how farfrom the flagpole is Tamara standing?GEOMETRY Connections 23 28. Answers1.h = 15sin 38 ! 9.2352. h = 8sin26 ! 3.507 3. x = 23cos 49 ! 15.0894.x = 37cos 41 ! 27.924 5. y = 38 tan15 ! 10.182 6. y = 43tan 55 ! 61.41047.sin 38 ! 24.364 8. z = 18sin52 ! 22.8423 9. w = 23z = 15cos 38 ! 29.187410.cos 38 ! 19.0353 11. x = 38tan15 ! 141.818 12. x = 91tan29 ! 164.168w = 1513.7 " 35.5377 14. u = tan!1 7x = tan!1 59 " 37.87515. y = tan!1 1218 " 33.69016. y = tan!1 7888 " 41.552617.500 ft42.5h ftsin 42.5 = h500h = 500 sin 42.5 337.795 ft18.350 ft47.5d ftcos 47.5= d350d = 350 cos 47.5 236.46 ft19.20.15 ft11.3x ft62 inh15 feet = 180 inches, 180" 62" =118" = hx 590.5 inches or 49.2 ft.58.4x ft62 inh 15 fth = 118", tan 58.4= 118!!x ,x tan 58.4 = 118!! , x = 118!!tan 58.4x 72.59 inches or 6.05 ft.24 Extra Practice 29. LAW OF SINES AND LAW OF COSINES #10LAW OF SINES is used to solve for the missing parts of any triangledetermined by ASA or AAS. For any ABC, it is always true that:sin Aa =sin Bb =sin CcaCbA c BExample 1 ASA or AASIf mA = 53, a = 18', and mB = 39, you can solve for b by substituting the valuesand solving the equation. Remember that the sine values represent numbers in theequation.sin 5318 =sin 39b Substitute.b(sin 53) = 18(sin 39) Fraction Busters.b =18(sin 39)sin 53 Solving for b.b 14.18' Calculations.Note that you can use your calculator to convert the sine values into decimalapproximations at any time during the solution, but it is most efficient to wait until theend of the problem.LAW OF COSINES is used to solve for the missing parts of any triangledetermined by SSS or SAS. For any ABC, it is always true that:a2 = b2 + c2 - 2bc cos Aorb2 = a2 + c2 2ac cos Borc2 = b2 + a2 - 2ba cos CaCbA c BGEOMETRY Connections 25 30. Example 2 SASIf mB = 38, c = 25', and = 18', you can solve for b by substituting the values andsolving the equation.b2 = 252 + 182 2(25)(18) cos 38 Substitute.b2 = 625 + 324 900(0.7880) Evaluate.b2 239.79 b 15.49' SimplifyExample 3 SSSIf c = 18', a = 5', and b = 16' then you can solve for A by substituting the values andsolving the equation.52 = 162 + 182 2(16)(18) cos A Substitute.25 = 256 + 324 576 cos A Simplify25 = 580 576 cos A Simplify0.9635 = cos A Solve for cos A15.52 A Use cos-1 on calculator.Note: With SSA triangles it is possible to have zero, one, or two solutionsUse the law of sines or the law of cosines to find the required part of the triangle.1.1023414x'2.x25 12015'ABC3.4. 5.x'1872656.16811x67120x107 8x26 Extra Practice 31. Draw and label a triangle similar to the one in the examples. Use the giveninformation to find the required part(s).7. mA = 40, mB = 88, a = 15.Find b.8. mB = 75, a = 13, c = 14.Find b.9. mB = 50, mC = 60, b = 9.Find a.10. mA = 62, mC = 28,c = 24. Find a.11. mA = 51, c = 8, b = 12.Find a.12. mB = 34, a = 4, b = 3.Find c.13. a = 9, b = 12, c = 15.Find mB .14. mB = 96, mA = 32, a = 6.Find c.15. mC = 18, mB = 54, b = 18.Find c.16. a = 15, b = 12, c = 14.Find mC.17. mC = 76, a = 39. B = 19.Find c.18. mA = 30, mC = 60, a = 8.Find b.19. a = 34, b = 38, c = 31.Find mB.20. a = 8, b = 16, c = 7.Find mC.21. mC = 84, mB = 23, c = 11.Find b.22. mA = 36, mB = 68,and b = 8. Find a and c.23. mB = 40, b = 4, and c = 6.Find a, mA , and mC.24. a = 2, b = 3, c = 4.Find mA , mB , and mC.Answers1. 24.49 2. 22.6 3. 4.0 4. 83.35. 17.15 6. 11.3 7. 23.32 8. 16.469. 11.04 10. 45.14 11. 9.34 12. 5.32 or 1.3213. 53.13 14. 8.92 15. 6.88 16. 61.2817. 39.03 18. 16 19. 71.38 20. no triangle21. 4.32 22. 5.07, 8.37 23. 5.66, 65.4, 74.6 24. 28.96, 46.57, 104.45GEOMETRY Connections 27 32. TRIANGLE CONGRUENCEIf two triangles are congruent, then allsix of their corresponding pairs of partsare also congruent. In other words, ifABC XYZ, then all thecongruence statements at right are#11AB CXY Zcorrect. Note that the matching parts follow the order ofthe letters as written in the triangle congruence statement.For example, if A is first in one statement and X is firstin the other, then A always corresponds to X incongruence statements for that relationship or figure.AB ! XYBC ! YZAC ! XZ!A " !X!B " !Y!C " !ZYou only need to know that three pairs of parts are congruent (sometimes in acertain order) to prove that the two triangles are congruent. The TriangleCongruence Properties are: SSS (Side-Side-Side), SAS (Side-Angle-Side; mustbe in this order), ASA (Angle-Side-Angle), and HL (Hypotenuse-Leg). AAS isan acceptable variation of ASA.The four Triangle Congruence Properties are the only correspondences that maybe used to prove that two triangles are congruent. Once two triangles are knowto be congruent, then the other pairs of their corresponding parts are congruent.In this course, you may justify this conclusion with the statement "congruenttriangles give us congruent parts." Remember: only use this statement after youhave shown the two triangles are congruent.CAT DOGExample 1In the figure at left, CA ! DO, CT ! DG, andAT ! OG. Thus, !CAT " !DOG because of SSS.Now that the triangles are congruent, it is also truethat !C " !D , !A " !O , and !T " !G .Example 2RDAIn the figure at right, RD ! CP , D P, andED ! AP . Thus RED CAP because of SAS.Now RE ! CA, R C, and E A. EPCX BCOARExample 3In the figure at left, O A, OX ! AR, andX R. Thus, BOX CAR because ofASA. Now B C, BO ! CA, and BX ! CR .28 Extra Practice 33. Example 4In the figure at right, there are two right angles(mN = mY = 90) and MO ! KE , andON ! EY. Thus, MON KEY because ofHL, so MN ! KY, M K, and O E.Example 5In the figure S Q, SR ! QR , and !SRT " !QRUbecause of vertical angles. Thus SRT QRU by ASA.EN KTQOSRYUMBriefly explain if each of the following pairs of triangles are congruent or not. If so, statethe Triangle Congruence Property that supports your conclusion.1.AF EB CD2.GJ KH IL3.MPN O4.RQS T5.V WYT UX6.A EDC BF7.IGHJ8.NKLM9.P QOS R10.T UXW V11.ABCD1312512.EG3HI54GEOMETRY Connections 29 34. 13.RJQ11239K11239PL14.B1284ACD12E4 F815.WXYZAnswers1. ABC DEFby ASA.2. GIH LJK by SAS. 3. PNM PNOby SSS.4. QS ! QS , soQRS QTS by HL.5. The triangles are notnecessarily congruent.6. ABC DFE byASA or AAS.7. GI ! GI , soGHI IJG by SSS.8. Alternate interiorangles = used twice, soKLN NMK byASA.9. Vertical angles at 0,so POQ ROSby SAS.10. Vertical angles and/oralternate interior angles=, soTUX VWXby ASA.11. No, the length of eachhypotenuse is different.12. Pythagorean Theorem,so EGH IHGby SSS.13. Sum of angles oftriangle = 180, butsince the equal anglesdo not correspond, thetriangles are notconguent.14. AF + FC = FC + CD,so ABC DEF bySSS.15. XZ ! XZ , soWXZ YXZby AAS.30 Extra Practice 35. PROOF #12A proof convinces an audience that a conjecture is true for ALL cases (situations)that fit the conditions of the conjecture. For example, "If a polygon is a triangleon a flat surface, then the sum of the measures of the angles is 180." Because weproved this conjecture in chapter two, it is always true. There are many formatsthat may be used to write a proof. This course explores three of them, namely,paragraph, flow chart, and two-column.ExampleIf BD is a perpendicular bisector of AC, prove that ABC isosceles.Paragraph proofBTo prove that ABC is isosceles, show that BA ! BC . We cando this by showing that the two segments are correspondingparts of congruent triangles.Since BDis perpendicular to AC , mBDA = mBDC = 90. AD CSince BDbisects AC , AD ! CD . With BD ! BD (reflexive property), ADB CDBby SAS.Finally, BA ! BC because corresponding parts of congruent triangles are congruent.Therefore, ABC must be isosceles since two of the three sides are congruent.Flow chart proofGiven: BD is the perpendicular bisector of AC!AD CDADB BDC!BD BD" !"bisector reflexive!!CBD !ABDSAS! !BA BC!!ABC is isosceles!s have partsDefinition of isoscelesTwo-Column ProofGiven: BD is a bisector of AC.BD is perpendicular to AC.Prove: ABC is isoscelesStatement ReasonBD bisects AC . GivenBD ! AC GivenAD ! CD Def. of bisectorADB and BDC Def. of perpendicularare right anglesADB BDC All right angles are .BD ! BD Reflexive propertyABD CBD S.A.S.AB ! CB 's have partsABC is isosceles Def. of isoscelesGEOMETRY Connections 31 36. In each diagram below, are any triangles congruent? If so, prove it. (Note: It is goodpractice to try different methods for writing your proofs.)1.ABD C2.BACDE3.ABC D4.DACB5.C DB A6.BACEDFComplete a proof for each problem below in the style of your choice.7. Given: TR and MN bisect each other.Prove: !NTP " !MRPNTRPM8. Given: CD bisects !ACB; !1 " !2 .Prove: !CDA " !CDBCADB129. Given: AB||CD, !B " !D,AB ! CDProve: !ABF " !CEDAFBECD10. Given: PG ! SG , TP ! TSProve: !TPG " !TSGPTSG11. Given: OE!MP , OE bisects !MOPProve: !MOE " !POEMO EP12. Given: AD||BC , DC||BAProve: !ADB " !CBDD CA B32 Extra Practice 37. 13. Given: AC bisects DE , !A " !CProve: !ADB " !CEBABCDE14. Given: PQ!RS , !R " !SProve: !PQR " !PQSRP QS15. Given: !S " !R , PQ bisects!SQRProve: !SPQ " !RPQSPRQ16. Given: TU ! GY, KY||HU ,KT! TG , HG! TGProve: !K " !HKTYUGH17. Given: MQ||WL , MQ ! WLProve: ML||WQQ WM LConsider the diagram below.ABCDE18. Is !BCD " !EDC ? Prove it!19. Is AB ! DC ? Prove it!20. Is AB ! ED? Prove it!Answers1. Yes !BAD"!BCD !BDC !BDA"BD BDReflexive"! ABD ! CBDGivenAAS"right ! 's are "2. Yes !B "!E !BCA !BCD"BC CE"! ABC ! DECGivenASA"vertical !s are "GivenGEOMETRY Connections 33 38. 3. Yes ! BCD ! BCABC BCReflexive"! ABC ! DBCAC " CDGivenright ! 's are =SAS""4. Yes!BA CDCA !CA GivenReflexive!! ABC ! CDA!AD BCGivenSSS5. Not necessarily. Counterexample:6. Yes! !AC FDGivenBC EFGiven! ABC ! DEFHL!7. NP ! MP and TP ! RP by definition of bisector. !NPT " !MPR because vertical angles are equal.So, !NTP " !MRP by SAS.8. !ACD"!BCD by definition of angle bisector. CD ! CD by reflexive so !CDA " !CDB by ASA.9. !A " !C since alternate interior angles of parallel lines congruent so !ABF " !CED by ASA.10. TG ! TG by reflexive so !TPG " !TSG by SSS.11. !MEO " !PEO because perpendicular lines form ! right angles !MOE " !POE by anglebisector and OE ! OE by reflexive. So, !MOE " !POE by ASA.12. !CDB " !ABD and !ADB " !CBD since parallel lines give congruent alternate interior angles.DB ! DB by reflexive so !ADB " !CBD by ASA.13. DB ! EB by definition of bisector. !DBA " !EBC since vertical angles are congruent. So!ADB " !CEB by AAS.14. !RQP " !SQP since perpendicular lines form congruent right angles. PQ ! PQ by reflexive so!PQR " !PQS by AAS.15. !SQP " !RQP by angle bisector and PQ ! PQ by reflexive, so !SPQ " !RPQ by AAS.16. !KYT " !HUG because parallel lines form congruent alternate exterior angles.TY+YU=YU+GU so TY ! GU by subtraction. !T " !G since perpendicular lines formcongruent right angles. So !KTY " !HGU by ASA. Therefore, !K " !H since ! triangleshave congruent parts.17. !MQL " !WLQ since parallel lines form congruent alternate interior angles. QL ! QL byreflexive so !MQL " !WLQ by SAS so !WQL " !MLQ since congruent triangles havecongruent parts. So ML||WQ since congruent alternate interior angles are formed byparallel lines.18. Yes !DB " CE ""DC DCReflexiveBDC !ECD! BCD ! EDCSAS"19. Not necessarily.20. Not necessarily.34 Extra Practice 39. PROPERTIES OF QUADRILATERALS #13Polygons with four sides are called QUADRILATERALS. There are variousproperties associated with specific kinds of quadrilaterals. The names ofquadrilaterals studied in the book are listed below along with their properties.Quadrilateral ABCD is a parallelogram. Oppositesides are parallel and congruent. Consecutive anglesare supplementary. The diagonals bisect.XB C!BAD " !DCB!ADC " !ABCAX " CXBX " DXQuadrilateral KYSH is a rhombus. It is aparallelogram with four congruent sides. In additionto all the properties of a parallelogram, the diagonalsare perpendicular and bisect the opposite angles.K YX2345 678H SHY ! KS"1 # "2 # "3 # "4"5 # "6 # "7 # "8Quadrilateral THOM is a rectangle. It is aparallelogram with four right angles. In addition toall the properties of a parallelogram, the diagonalsare congruent.If a rectangle is also a rhombus, then it is called asquare. A square would have all the propertieslisted previously.T HM OTO ! MHAD1GEOMETRY Connections 35 40. Two additional quadrilaterals are studied.Quadrilateral HOEY is a trapezoid. It has exactlyone set of parallel sides called bases. The segmentconnecting the midpoints of the non-parallel sides isparallel to the bases and has length equal to theaverage of the bases.Quadrilateral WXYZ is a kite. It has one pair ofopposite angles congruent and the diagonals areperpendicular.BR = 12(HO + EY)XZ ! WY"WXY # "WZYYExample 1Use the information in the parallelogram at right tofind m!BAD, m!ADB, AT, and BD.Solution: m!BAD = 100 since consecutiveangles are supplementary.m!ADB = 80 - 32 = 48 = 80 ! 32 = 48 sinceopposite angles are congruent. AT = 7 and BD = 20since diagonals bisect.Example 2Use the information in the rhombus at right to findRQ, m!SPQ ,m!STR , and RT.Solution: RQ = 7 since all sides congruent.m!SPQ = 70 since diagonals bisect oppositeangles. m!STR = 90 since diagonals areperpendicular. RT = 6 since diagonals bisect.HBORA BD C36 Extra PracticeEXYZWPQRST67 3510732T 80 41. Example 3Given that Q and R are midpoints in the trapezoid atright to find m!QMN , m!QRN, and QR.Solution: m!QMN = 120 since MN is parallel to PO.m!QRN= 70 since Q and R are midpoints, then QRis parallel to the bases and with parallel lines,corresponding angles are congruent.QR = 12(5 + 8) = 6.5 .5Q R60 70P OFind the required information and justify your answers.For problems 1-4 use the parallelogram at right.1. Find the perimeter.2. If CT = 9, find AT.3. If m!CDA = 60 , find m!CBA andm!BAD.TD C4. If AT = 4x 7 and CT = x + 13 , solve for x.For problems 5-8 use the rhombus at right.5. If PS = 6 , what is the perimeter of PQRS?6. If PQ = 3x + 7 and QR = x + 17, solve for x.7. If m!PSM= 22 , find m!RSM andm!SPQ .8. If m!PMQ = 4x 5, solve for x.For problems 9-12 use the quadrilateral at right.9. If WX = YZ and WZ = XY, must WXYZ bea rectangle?10. If m!WZY = 90 , must WXYZ be arectangle?W X11. If the information in problems 9-10 are bothtrue, must WXYZ be a rectangle?Z Y12. If WY = 15 and WZ = 9, what is YZ and XZ?M N8A B1012P QMS RGEOMETRY Connections 37 42. For problems 13-16 use the trapezoid at right withmidpoints E and F.13. If m!EDC = 60 , find m!AEF .14. Ifm!DCB= 5x + 20 and m!ABC = 3x + 10,solve for x.15. If AB = 6 and DC = 10, find EF.16. If EF = 9 and DC = 15, find AB.For problems 17-20 use the kite at right.17. If m!XWZ = 95 , find m!XYZ.18. If m!WZT = 110 and m!WXY = 40 ,find m!ZWX.19. If WZ = 5 and WT = 4, find ZT.20. If WT = 4, TZ = 3, and TX = 10, find theperimeter of WXYZ.A BE FD CWTYZAnswers1. 44 units 2. 9 units 3. 60 , 120 4. 45. 4 6 6. 2.5 7. 22,!136 8. 23.759. no 10. no 11. yes 12. 12, 1513. 60 14. 18.75 15. 8 16. 317. 95 18. 105 19. 3 20. 10 + 2 116X38 Extra Practice 43. INTERIOR AND EXTERIOR ANGLES OF POLYGONS #14The sum of the measures of the interior angles of an n-gon is sum = (n ! 2)180 .The measure of each angle in a regular n-gon is m! = (n"2)180n .The sum of the exterior angles of any n-gon is 360.Example 1Find the sum of the interior angles of a 22-gon.Since the polygon has 22 sides, we can substitute this number for n:(n ! 2)180= (22 ! 2)180= 20 "180= 3600 .Example 2If the 22-gon is regular, what is the measure of each angle? Use the sum from theprevious example and divide by 22: 360022! 163.64Example 3Each angle of a regular polygon measures 157.5. How many sides does this n-gon have?a) Solving algebraically: 157.5= (n!2)180n" 157.5n = (n ! 2)180! 157.5n = 180n " 360 ! " 22.5n = "360 ! n = 16b) If each interior angle is 157.5, then each exterior angle is 180! 157.5= 22.5 .Since the sum of the exterior angles of any n-gon is 360, 360 22.5 ! 16sides .Example 4Find the area of a regular 7-gon with sides of length 5 ft.Because the regular 7-gon is made up of 7 identical, isoscelestriangles we need to find the area of one, and then multiply it by7. In order to find the area of each triangle we need to start withthe angles of each triangle.h!12.5 2.5Each interior angle of the regular 7-gon measures(7 ! 2)1807=(5)1807= 9007" 128.57 . Theangle in the triangle is half the size of the interior angle, so m!1 "128.572" 64.29 . Findthe height of the triangle by using the tangent ratio:tan1 h!=2.5" h = 2.5 # tan!1 $ 5.19 ft. The area of the triangle is: 5 ! 5.192" 12.98 ft2 .Thus the area of the 7-gon is 7 !12.98 " 90.86 ft2 .GEOMETRY Connections 39 44. Find the measures of the angles in each problem below.1. Find the sum of the interior anglesin a 7-gon.2. Find the sum of the interior anglesin an 8-gon.3. Find the size of each of the interiorangles in a regular 12-gon.4. Find the size of each of the interiorangles in a regular 15-gon.5. Find the size of each of the exteriorangles of a regular 17-gon.6. Find the size of each of the exteriorangles of a regular 21-gon.Solve for x in each of the figures below.7.5x 3x3x4x8.4x x2x 5x9.x1.5xx0.5x1.5x10.3x5x4x5x2x4xAnswer each of the following questions.11. Each exterior angle of a regular n-gon measures 16 411. How many sides does this n-gon have?12. Each exterior angle of a regular n-gon measures 13 13. How many sides does this n-gon have?13. Each angle of a regular n-gon measures 156. How many sides does this n-gon have?14. Each angle of a regular n-gon measures 165.6. How many sides does this n-gon have?15. Find the area of a regular pentagon with side length 8 cm.16. Find the area of a regular hexagon with side length 10 ft.17. Find the area of a regular octagon with side length 12 m.18. Find the area of a regular decagon with side length 14 in.Answers1. 900 2. 1080 3. 150 4. 1565. 21.1765 6. 17.1429 7. x = 24 8. x = 309. x = 98.18 10. x = 31.30 11. 22 sides 12. 27 sides13. 15 sides 14. 25 sides 15. 110.1106 cm2 16. 259.8076 ft217. 695.2935 m2 18. 1508.0649 in240 Extra Practice 45. AREA OF SECTORSSECTORS of a circle are formed by the two radii of a centralangle and the arc between their endpoints on the circle. Forexample, a 60 sector looks like a slice of pizza. The area of asector is found by multiplying the fraction of the circle by thearea of the whole circle.#1560CExample 1CAB454'Find the area of the 45 sector. First find the fractional part ofthe circle involved.!mZY= 45=36036018 of the circle's area. Next findthe area of the circle: A = !42 = 16!sq.ft. Finally, multiply thetwo results to find the area of the sector.1!16" = 2" # 6.28sq.ft.8Example 2CAB8'150Find the area of the 150 sector.Fractional part of circle is!mAB360= 150360=512Area of circle is A = !r2 = !82 = 64!sq.ft.Area of the sector: 512! 64" = 320"12= 80"3# 83.76sq.ft.1. Find the area of the shaded sector in each circle below. Points A, B and C are thecenters.a)454Ab)120 7Bc)C 11Calculate the area of the following shaded sectors. Point O is the center of each circle.2.452O3.120O64.O71705.O1018GEOMETRY Connections 41 46. 6. The shaded region in the figure is called a segment of thecircle. It can be found by subtracting the area of MILfrom the sector MIL. Find the area of the segment of thecircle.MIL2Find the area of the shaded regions.7. IH5N C8. 29. YARD is a square; A andD are the centers of thearcs.Y AD R410. Find the area of a circular garden if the diameter of the garden is 30 feet.11. Find the area of a circle inscribed in a square whose diagonal is 8 feet long.12. The area of a 60 sector of a circle is 10 m2. Find the radius of the circle.13. The area of a sector of a circle with a radius of 5 mm is 10 mm2. Find the measureof its central angle.Find the area of each shaded region.14.10 m10 m60 6015.14 ft14 ft16.4 in17.r = 818.60 rsm= 5rlg= 819.121220.Find the radius. The shaded areais 12 cm2.120 r42 Extra Practice 47. Answers1. a) 2 u2 b) 493! u2 c) 363!4u22. !2 3. 12 4. 931!36 5. 56. 2 u2 7. 1003! " 25 3 u2 8. 10 20 u2 9. 8 16 u210. 225 ft2 11. 8 ft2 12. 2 15 m 13. 14414. 100 25!3"m3 15. 196 49 ft2 16. 10 in2 17. 48 + 32 u218. 652! u2 19. 61.8 u2 20. 6 uGEOMETRY Connections 43 48. SIMILARITY OF LENGTH, AREA, AND VOLUMEThe relationships for similarity of length, area, and volume aredeveloped in Chapter 8 and Chapter 9 and is summarized in theMath Notes box on page 435. The basic relationships of ther:r2:r3 Theorem are stated below.Once you know two figures are similar with a ratio of similarity ab ,the following proportions for the SMALL (sm) and LARGE (lg)figures (which are enlargements or reductions of each other) are true:sidesm= aPsm= aAsma2Vsma3= = sidelgb Plgb Algb2 Vlgb3In each proportion above, the data from the smaller figure iswritten on top (in the numerator) to help be consistent withcorrespondences. When working with area, the basic ratio ofsimilarity is squared; for volume, it is cubed. NEVER square orcube the actual areas or volumes themselves.#16abExample 1The two rectangular prisms above are similar. Suppose the ratio of their vertical edges is3:8. Use the r:r2:r3 Theorem to find the following without knowing the dimensions of theprisms.a) Find the ratio of their surface areas.b) Find the ratio of their volumes.c) The perimeter of the front face of the large prism is 18 units. Find the perimeter ofthe front face of the small prism.d) The area of the front face of the large prism is 15 square units. Find the area of thefront face of the small prism.e) The volume of the small prism is 21 cubic units. Find the volume of the large prism.Use the ratio of similarity for parts (a) and (b): the ratio of the surface areas is 239=; the2864 2ratio of the volumes is 382=27. Use proportions to solve for the rest of the parts.512 (c) 38=P188P= 54P= 6.75 units8 ( )2d) 3=A15964=A1564A= 135A 2.11 sq. units8 ( )3e) 3=21V27512=21V27V= 10752V 398.22 cu. units44 Extra Practice 49. Solve each of the following problems.1. Two rectangular prisms are similar. The smaller, A,has a height of four units while the larger, B, has aheight of six units.A4 Bxy6a) What is the magnification factor from prism A to prism B?b) What would be the ratio of the lengths of the edges labeled x and y?c) What is the ratio, small to large, of their surface areas? their volumes?d) A third prism, C is similar to prisms A and B. Prism C's height is ten units. Ifthe volume of prism A is 24 cubic units, what is the volume of prism C?2. If rectangle A and rectangle B have a ratio of similarity of 5:4, what is the area of rectangleB if the area of rectangle A is 24 square units?3. If rectangle A and rectangle B have a ratio of similarity of 2:3, what is the area of rectangleB if the area of rectangle A is 46 square units?4. If rectangle A and rectangle B have a ratio of similarity of 3:4, what is the area of rectangleB if the area of rectangle A is 82 square units?5. If rectangle A and rectangle B have a ratio of similarity of 1:5, what is the area of rectangleB if the area of rectangle A is 24 square units?6. Rectangle A is similar to rectangle B. The area of rectangle A is 81 square units while thearea of rectangle B is 49 square units. What is the ratio of similarity between the tworectangles?7. Rectangle A is similar to rectangle B. The area of rectangle B is 18 square units while thearea of rectangle A is 12.5 square units. What is the ratio of similarity between the tworectangles?8. Rectangle A is similar to rectangle B. The area of rectangle A is 16 square units while thearea of rectangle B is 100 square units. If the perimeter of rectangle A is 12 units, what isthe perimeter of rectangle B?9. If prism A and prism B have a ratio of similarity of 2:3, what is the volume of prism B ifthe volume of prism A is 36 cubic units?10. If prism A and prism B have a ratio of similarity of 1:4, what is the volume of prism B ifthe volume of prism A is 83 cubic units?11. If prism A and prism B have a ratio of similarity of 6:11, what is the volume of prism B ifthe volume of prism A is 96 cubic units?GEOMETRY Connections 45 50. 12. Prism A and prism B are similar. The volume of prism A is 72 cubic units while thevolume of prism B is 1125 cubic units. What is the ratio of similarity between these twoprisms?13. Prism A and prism B are similar. The volume of prism A is 27 cubic units while thevolume of prism B is approximately 512 cubic units. If the surface area of prism B is 128square units, what is the surface area of prism A?14. The corresponding diagonals of two similar trapezoids are in the ratio of 1:7. What is theratio of their areas?15. The ratio of the perimeters of two similar parallelograms is 3:7. What is the ratio of theirareas?16. The ratio of the areas of two similar trapezoids is 1:9. What is the ratio of their altitudes?17. The areas of two circles are in the ratio of 25:16. What is the ratio of their radii?18. The ratio of the volumes of two similar circular cylinders is 27:64. What is the ratio of thediameters of their similar bases?19. The surface areas of two cubes are in the ratio of 49:81. What is the ratio of their volumes?20. The ratio of the weights of two spherical steel balls is 8:27. What is the ratio of thediameters of the two steel balls?Answers1. a) 64=3b) 42 6=3 c) 16236or49,64216or1627 d) 375 cu unit2. 15.36 u2 3. 103.5 u2 4. 145.8 u2 5. 600 u26. 97 7. 65 8. 30 u 9. 121.510. 5312 11. 591.6 12. 25 13. 18 u214. 149 15. 949 16. 13 17. 5418. 34 19. 343729 20. 2346 Extra Practice 51. VOLUME AND SURFACE AREA OF POLYHEDRA #17The VOLUME of various polyhedra, that is, the number of cubic units needed tofill each one, is found by using the formulas below.for prisms and cylindersV = base area ! height , V = Bhbase area (B)hhBhBfor pyramids and conesV =13Bhbase area (B)h hIn prisms and cylinders, you may use either base, since they are congruent. Sincethe bases of cylinders and cones are circles, their area formulas may be expressedas: cylinder V r2h and 1= !cone !V =3!r2hThe SURFACE AREA of a polyhedron is the sum of the areas of its base(s) and faces.Example 1Use the appropriate formula(s) to find the volume of each figure below:a)5'8'22'b) 5'8'c)18'8'd)1581214a) This is a triangular pyramid. The base is a right triangle so the area of the base isB = 12 8 5 = 20 square units, so V = 13(20)(22) 146.7 cubic feet.b) This is a cylinder. The base is a circle, so B= !52 , V = (25)(8) = 200 628.32cubic feet.c) This is a cone. The base is a circle, so B = 82. V = 13(64!)(18) = 13(64)(18)! =13(1152)! = 384! " 1206.37 feet3d) This prism has a trapezoidal base, so B = 12(12 + 8)(15) = 150 .Thus, V = (150)(14) = 2100 cubic feet.GEOMETRY Connections 47 52. Example 2Find the surface area of the triangularprism shown at right. The figure ismade up of two triangles (the top andbottom) and three rectangles as shownat right. Find the area of each of theseshapes.To find the area of the triangle and thelast rectangle, use the PythagoreanTheorem to find the length of thesecond leg of the right triangular base.Since 32 + leg2 = 52 , leg = 4.53823?3 + 8 +858 +?5Calculate all of the areas, and find theirsum.SA = 2 12( (3)(4)) + 3(8) + 5(8) + 4(8)= 12 + 24 + 40 + 32 = 108square unitsExample 3Find the total surface area of aregular square pyramid with a slantheight of 10 inches and a base withsides 8 inches long.The figure is made up of 4 identicaltriangles and a square base.10 in8 inSA = 410 ! 82"# $%& '+ 8(8)= 160 + 64= 224 square inchesFind the volume of each figure.1.3 m4 m4 m2.15 cm10 cm12 cm3.36 ft36 ft40 ft25 ft4.123 in2 in5.12.1 m6.3 m6.2 m5 m48 Extra Practice 53. 7.2 ft4 ft9 ft8.11 cm11 cm9.8 cm4 cm 2 3 cm10.8 in12 in10 in11.13 in5 in12.8 cm11 cm13. 1.0 m2.6 m14. 10 in16 in16 in15.11 m18 m16. 6.1 cm8.3 cm6.2 cm17.5 cm2 cm18. 7 in8 in8 in19. Find the volume of the solid shown.18 ft9 ft35 ft14 ft20. Find the volume of the remainingsolid after a hole with a diameter of 4mm is drilled through it.8 mm15 mm10 mmFind the total surface area of the figures in the previous volume problems.21. Problem 1 22. Problem 2 23. Problem 3 24. Problem 525. Problem 6 26. Problem 7 27. Problem 8 28. Problem 929. Problem 10 30. Problem 14 31. Problem 18 32. Problem 19GEOMETRY Connections 49 54. Answers1. 48 m3 2. 540 cm3 3. 14966.6 ft3 4. 76.9 m35. 1508.75 m3 6. 157 m3 7. 72 ft3 8. 1045.4 cm39. 332.6 cm3 10. 320 in3 11. 314.2 in3 12. 609.7 cm313. 2.5 m3 14. 512 m3 15. 514.4 m3 16. 2.3 cm317. 20.9 cm3 18. 149.3 in3 19. 7245 ft3 20. 1011.6 mm321. 80 m2 22. 468 cm2 23. 3997.33 ft2 24. 727.98 m225. 50 + 20 219.8 m2 26. 124 ft227. 121 + 189.97 569.91 cm2 28. 2192 + 48 3 ! 275.14 cm29. 213.21 in2 30. 576 in2 31. 193.0 in2 32. 2394.69 ft250 Extra Practice 55. CENTRAL AND INSCRIBED ANGLESA central angle is an angle whose vertex is the center of a circleand whose sides intersect the circle. The degree measure of acentral angle is equal to the degree measure of its intercepted arc.For the circle at right with center C, ACB is a central angle.An INSCRIBED ANGLE is an angle with its vertex on the circleand whose sides intersect the circle. The arc formed by theintersection of the two sides of the angle and the circle is called anINTERCEPTED ARC. ADB is an inscribed angle, !ABis anintercepted arc.The INSCRIBED ANGLE THEOREM says that the measure ofany inscribed angle is half the measure of its intercepted arc.Likewise, any intercepted arc is twice the measure of anyinscribed angle whose sides pass through the endpoints of the arc.mADB =12!= 2mADB!and ABAB#18CABABDExample 1In P, m!CPQ = 70 . Find!andmCQ!.mCRQR PCQ!= m!CPQ = 70mCQ!mCRQ"= 360 ! mCQ= 360 = 70 = 290Example 2In the circle shown below, the vertex ofthe angles is at the center. Find x.BAx2xCSince AB is the diameter of the circle,x + 2x = 180 3x = 180 x = 60Example 3Solve for x.A100xBC120!= 220, mBCSince mBAC!= 360 220 = 140. The!or 70.m!BAC equals half the mBCGEOMETRY Connections 51 56. Find each measure in P if m!WPX = 28 , m!ZPY = 38 , and WZ and XV arediameters.1. !YZ!2. WX3. !VPZ4. !VWX5. !XPY6. !XY!7. XWY!8. WZXVZPYWXIn each of the following figures, O is the center of the circle. Calculate the values of x andjustify your answer.9.x136O10.x146O11.xO4912.xO6213.xO10014.xO15010015.18 Ox16. O27x17.55x O18. O 9112x19. O35 x20.x29 O101Answers1. 38 2. 28 3. 28 4. 180 5. 1146. 114 7. 246 8. 332 9. 68 10. 7311. 98 12. 124 13. 50 14. 55 15. 1816. 27 17. 55 18. 77 19. 35 20. 5052 Extra Practice 57. TANGENTS, SECANTS, AND CHORDS #19The figure at right shows a circle with three lines lying ona flat surface. Line a does not intersect the circle at all.Line b intersects the circle in two points and is called aSECANT. Line c intersects the circle in only one pointand is called a TANGENT to the circle.abcTANGENT/RADIUS THEOREMS:1. Any tangent of a circle is perpendicular to a radius of thecircle at their point of intersection.2. Any pair of tangents drawn at the endpoints of a diameterare parallel to each other.A CHORD of a circle is a line segment with its endpoints on the circle.chordDIAMETER/CHORD THEOREMS:1. If a diameter bisects a chord, then it is perpendicular to the chord.2. If a diameter is perpendicular to a chord, then it bisects the chord.ANGLE-CHORD-SECANT THEOREMS:m1 =!12 (mAD!+ mBC)AE EC = DE EBmP =!12 (mRT!! mQS)PQ PR = PS PTDA2EBC1RQTPSExample 1If the radius of the circle is 5 units andAC = 13 units, find AD and AB.ABCDAD!CD and AB!CD by Tangent/RadiusTheorem, so (AD)2+ (CD)2= (AC)2 or(AD)2+ (5)2= (13)2 . So AD = 12 andAB ! AD so AB = 12.Example 2In B, EC = 8 and AB = 5. Find BF.Show all subproblems.AEBDCFThe diameter is perpendicular to the chord,therefore it bisects the chord, so EF = 4. ABis a radius and AB = 5. EB is a radius, soEB = 5. Use the Pythagorean Theorem tofind BF: BF2 + 42 = 52, BF = 3.GEOMETRY Connections 53 58. In each circle, C is the center and AB is tangent to the circle at point B. Find the area ofeach circle.1.BAC = 30CA252.CBA12AC = 453.BCA306! 34. BCA18455. BCAAC = 16126.BAC = 86CA566 37. BCA28AC = 908 .BCA24DAD =189. In the figure at right, point E is the center andmCED = 55. What is the area of the circle?AC55BDEIn the following problems, B is the center of the circle.Find the length of BF given the lengths below.10. EC = 14, AB = 16 11. EC = 35, AB = 2112. FD = 5, EF = 10 13. EF = 9, FD = 6 AEBCDF14. In R, ifAB = 2x ! 7 andCD = 5x ! 22 ,find x.B CxxRA D15. In O, MN ! PQ,MN = 7x + 13 , andPQ = 10x ! 8 . Find PS.N POTSM Q16. In D, if AD = 5and TB = 2, find AT.DAT B5254 Extra Practice 59. 17. In J, radius JL and chord MNhave lengths of 10 cm. Find thedistance from J to MN.JL1010MN18. In O, OC = 13 and OT = 5.Find AB.C13OATB519. If AC is tangent to circle E and EH ! GI , isGEH ~ AEB? Prove your answer.20. If EH bisects GI and AC is tangent to circle E atpoint B, are AC and GI parallel? Prove youranswer.ABCGEDHIFind the value of x.21.8040x22.x25148P23.43x4124.1520x!= 84, mBCIn F, mAB!= 38, mCD!= 64, mDE!= 60. Find the measure of eachangle and arc.25. !mEA27. m!129. m!3!26. mAEB28. m!230. m!4A B4C23DFE1GEOMETRY Connections 55 60. For each circle, tangent segments are shown. Use the measurements given find the value of x.31. AE61FC141x32.P Q70xR S T11033. D GB58H Ix34.x3835.160 35x36. (4x + 5) 50 xFind each value of x. Tangent segments are shown in problems 40, 43, 46, and 48.37.x94338.1142x39.x54340. x102041.3xx6242.1xx843.1.00.8x44.x55x45.6x102x46.54x47.6123x48.3x51056 Extra Practice 61. Answers1. 275 sq. un.2. 1881 sq. un. 3. 36 sq. un.4. 324 sq. un.5. 112 sq. un. 6. 4260 sq. un.7. 7316 sq. un.8. 49 sq. un. 9. 117.047 sq. un.10. 14.411. 11.6 12. 7.513. 3.7514. 5 15. 3116. 417. 5 3 cm. 18. 5 319. Yes, GEH AEB (reflexive). EB is perpendicular to AC since it is tangentso GHE ABE because all right angles are congruent. So the triangles aresimilar by AA~.20. Yes. Since EH bisects GI it is also perpendicular to it (SSS). Since AC is atangent, ABE is a right angle. So the lines are parallel since the correspondingangles are right angles and all right angles are equal.21. 16022. 9 23. 42 24. 70 25. 11426. 27627. 87 28. 49 29. 131 30. 3831. 4032. 55 33. 64 34. 38 35. 4536. 22.537. 12 38. 5 12 39. 2 40. 3041. 242. 2 2 43. 1.2 44. 5 45. 3046. 6 47. 7.5 48. 5GEOMETRY Connections 57 62. CONES AND SPHERES #20For a CONE, the formula for volume is the same as thatfor a pyramid except now the base is a circle.V = 13! r2hThe lateral surface area is the same as the area of thesector found by unrolling the cone.LA = ! r lIn these formulas, r represents the radius of the base, h isthe height and l is the slant height.For a SPHERE with radius r, the volume and surface areais found using: V = 4! r3 and SA = 4! r23Example 1Find the volume and lateral surface area of the cone atright. Note that r = 3 cm.V = 13! 32 ! 4 = 12! " 37.70#cm3LA = ! ! 3! 5 = 15! " 47.12 cm2radius2rcenter6 cm5 cm4 cmlhExample 2A cone with volume 63.62 cubic units has a height of 3 units. What is the radius of thecone and the lateral surface area?First use the formula for volume to find the radius.13! r2h = 13! r2 ! 3 = 63.62 Substituting! r2 = 63.62 Simplifyingr2 =63.62!r =63.62!= 4.5 units Solve for r>>Problem continues next page.>>58 Extra Practice 63. To find the lateral area we need the slant height. Notice that the radius, height, and slantheight form a right triangle so we use r2+ h2 = l2 (the Pythagorean theorem.)4.52 + 32 = 12 Substituting29.25 = 12 Simplifyingl ! 5.41 units SolvingLA = "rl = " # 4.5 # 5.41 SubstitutingLA ! 76.48 square unitsExample 3Find the volume and surface area of the sphere at right.V = 43! r3 = 43! 23 = 32!3! 33.51"ft3SA = 4! ! 22 = 16! " 50.27#ft22 feetExample 4A sphere has a volume of 972 ! . Find the surface area.First use the formula for volume and solve for the radius. Then substitute the value forthe radius into the formula for surface area.V = 43! r3 = 972! Substituting4! r3 = 2916! Remove fractionr3 =2916!4!= 729 Solvingr = 729 3 = 9 SolvingSA = 4! r2 = 4! ! 92 SubstitutingSA = 324! SimplifyingGEOMETRY Connections 59 64. Use the given information to find the volume of the cone.1. radius = 1.5 in,height = 4 in2. diameter = 6 cm,height = 5 cm3. base area = 25 ! ,height = 34. base circum. = 12 ! ,height = 105. diameter =12,slant height = 106. lateral area = 12 ! ,radius = 1.5Use the given information to find the lateral area of the cone.7. radius = 8 in,slant height = 1.75 in8. slant height = 10 cmheight = 8 cm9. base area = 25 ! ,slant height = 610. radius = 8 cm,height = 15 cm11. volume = 100 ! ,height = 512. volume = 36 ! ,radius = 3Use the given information to find the volume of the sphere.13. radius = 10 cm 14. diameter = 10 cm 15. circumference ofgreat circle = 12 !16. surface area = 256! 17. circumference ofgreat circle = 20 cm18. surface area = 100Use the given information to find the surface area of the sphere.19. radius = 5 in 20. diameter = 12 in 21. circumference ofgreat circle = 1422. volume = 250 23. circumference ofgreat circle = 24. volume = 9!! 260 Extra Practice 65. Answers1. 3! " 9.42 in32. 15! " 47.12 cm33. 25! " 78.54 u3 4. 120! " 376.99 u35. 96! " 301.59 u3 6. ! 18.51 u37. 14! " 43.98 in28. 60! " 188.50 cm29. 30! " 94.25 u2 10. 136! " 427.26 cm211. ! 224.35 u2 12. 116.58 u213. 4000!! 4188.79 cm314. 500!33! 523.60 cm315. 288! " 904.79 u3 16. 2048!3! 2144.66 u317. ! 135.09 u3 18. ! 94.03 u319. 100! " 314.16 u220. 144! " 452.39 u221. ! 62.39 u222. ! 191.91 u223. ! " 3.14 u2 24. 9! " 28.27 u2GEOMETRY Connections 61 66. WRITING AND GRAPHING LINEAR EQUATIONS ON A FLAT SURFACE #21SLOPE is a number that indicates the steepness (or flatness) of a line, as well asits direction (up or down) left to right.SLOPE is determined by the ratio:vertical changehorizontal change between any two pointson a line.For lines that go up (from left to right), the sign of the slope is positive. For linesthat go down (left to right), the sign of the slope is negative.Any linear equation written as y = mx + b, where m and b are any real numbers,is said to be in SLOPE-INTERCEPT FORM. m is the SLOPE of the line. b isthe Y-INTERCEPT, that is, the point (0, b) where the line intersects (crosses)the y-axis.If two lines have the same slope, then they are parallel. Likewise, PARALLELLINES have the same slope.Two lines are PERPENDICULAR if the slope of one line is the negativereciprocal of the slope of the other line, that is, m and !1.m m ( ) = "1.Note that m! "1Examples: 3 and !3 , !13 and 32 , 524 and !45Two distinct lines on a flat surface that are not parallel intersect in a single point.See "Solving Linear Systems" to review how to find the point of intersection.Example 1Graph the linear equation y = 47 x + 2Using y = mx + b , the slope in y = 47x + 2 is 47and they-intercept is the point (0, 2). To graph, begin at they-intercept (0, 2). Remember that slope isvertical changehorizontal change so go up 4 units (since 4 is positive) from(0, 2) and then move right 7 units. This gives a secondpoint on the graph. To create the graph, draw a straightline through the two points.yx554-57-562 Extra Practice 67. Example 2A line has a slope of 34and passes through (3, 2). What is the equation of the line?Using y = mx + b , write y = 34 x + b . Since (3, 2) represents a point (x, y) on the line,substitute 3 for x and 2 for y, 2 =34(3) + b , and solve for b.2 =94+ b ! 2 "94= b ! "14= b . The equation is y = 34 x ! 14 .Example 3Decide whether the two lines at right are parallel,perpendicular, or neither (i.e., intersecting).5x 4y = -6 and -4x + 5y = 3.First find the slope of each equation. Then compare the slopes.5x 4y = -6-4y = -5x 6y = -5x 6-4y = 54x + 32The slope of this line is54.-4x + 5y = 35y = 4x + 3y = 4x + 35y = 45x + 35The slope of this line is45.These two slopes are notequal, so they are not parallel.The product of the two slopesis 1, not -1, so they are notperpendicular. These twolines are neither parallel norperpendicular, but dointersect.Example 4Find two equations of the line through the given point, one parallel and one perpendicularto the given line: y = ! 52 x + 5 and (-4, 5).For the parallel line, use y = mx + bwith the same slope to writey = !5x + b .2 Substitute the point (4, 5) for x andy and solve for b.5 = ! 52(!4) + b " 5 = 202+ b " ! 5 = bTherefore the parallel line through(-4, 5) is y = ! 52 x ! 5For the perpendicular line, usey = mx + b where m is the negativereciprocal of the slope of the originalequation to write y = 25 x + b .Substitute the point (4, 5) and solvefor b.5 = 25(!4) + b " 335= bTherefore the perpendicular linethrough (-4, 5) is y = 25 x + 335 .GEOMETRY Connections 63 68. Identify the y-intercept in each equation.1. y = 12x 2 2. y = ! 35 x ! 53 3. 3x + 2y = 124. x y = -13 5. 2x 4y = 12 6. 4y 2x = 12Write the equation of the line with:7. slope =12and passing through (4, 3). 8. slope =23and passing through (-3, -2).9. slope = -13and passing through (4, -1). 10. slope = -4 and passing through (-3, 5).Determine the slope of each line using the highlighted points.11. 12. 13.xyxyxyUsing the slope and y-intercept, determine the equation of the line.14. 15. 16. 17.xyxyxyxy3512Graph the following linear equations on graph paper.18. y = x + 3 19. y = - x 1 20. y = 4x21. y = -6x + 1222. 3x + 2y = 1264 Extra Practice 69. State whether each pair of lines is parallel, perpendicular, or intersecting.23. y = 2x 2 and y = 2x + 4 24. y = 12x + 3 and y = -2x 425. x y = 2 and x + y = 3 26. y x = -1 and y + x = 327. x + 3y = 6 and y = -13x 3 28. 3x + 2y = 6 and 2x + 3y = 61229. 4x = 5y 3 and 4y = 5x + 3 30. 3x 4y = 12 and 4y = 3x + 7Find an equation of the line through the given point and parallel to the given line.31. y = 2x 2 and (-3, 5) 32. y = x + 3 and (-4, 2)1233. x y = 2 and (-2, 3) 34. y x = -1 and (-2, 1)35. x + 3y = 6 and (-1, 1) 36. 3x + 2y = 6 and (2, -1)37. 4x = 5y 3 and (1, -1) 38. 3x 4y = 12 and (4, -2)Find an equation of the line through the given point and perpendicular to the given line.39. y = 2x 2 and (-3, 5) 40. y = x + 3 and (-4, 2)41. x y = 2 and (-2, 3) 42. y x = -1 and (-2, 1)43. x + 3y = 6 and (-1, 1) 44. 3x + 2y = 6 and (2, -1)45. 4x = 5y 3 and (1, -1) 46. 3x 4y = 12 and (4, -2)Write an equation of the line parallel to each line below through the given point.47.xy5(3,8)5-5(-3,2)-5(-2,-7)48.(7,4)xy55(-2,3)-5(-8,6)-549. Write the equation of the line through(7, -8) which is parallel to the linethrough (2, 5) and (8, -3).50. Write the equation of the linethrough (1, -4) which is parallel tothe line through (-3, -7) and (4, 3).GEOMETRY Connections 65 70. Answers1. (0, 2) 2. 0,! 53 ( ) 3. (0, 6) 4. (0, 13)5. (0, 3) 6. (0, 3) 7. y = 12 x + 1 8. y = 23 x9. y = ! 13 x + 13 10. y = !4x ! 7 11. !112. 32 413. 2 14. y = 2x ! 2 15. y = !x + 2 16. y = 13 x + 217. y = !2x + 4 18. line with slope 12 and y-intercept (0, 3)19. line with slope !3and y-intercept (0, 1) 20. line with slope 4 and y-intercept (0, 0)5 !21. line with slope 6 and y-intercept 0, 12" #$% &22. line with slope !32 and y-intercept (0, 6)1223. parallel 24. perpendicular 25. perpendicular 26. perpendicular27. parallel 28. intersecting 29. intersecting 30. parallel31. y = 2x + 11 32. y = x + 4 33. y = x + 5 34. y = x + 335. y = - 13x + 2336. y = - 32x + 2 37. y = 45x 9538. y = 34x 539. y = - 12x + 7240. y = -2x 6 41. y = -x + 1 42. y = -x 143. y = 3x + 4 44. y = 23x 7345. y = - 54x + 1446. y = - 43x + 10347. y = 3x + 11 48. y = - 12x + 152 49. y = 43x + 4350. y = 107 x 38766 Extra Practice 71. SOLVING LINEAR SYSTEMS #22You can find where two lines intersect (cross) by using algebraic methods. The two mostcommon methods are SUBSTITUTION and ELIMINATION (also known as the additionmethod).Example 1Solve the following system of equations at right by substitution.y = 5x + 1Check your solution.y = -3x - 15When solving a system of equations, you are solving to find the x- and y-values thatresult in true statements when you substitute them into both equations. You generallyfind each value one at a time. Since both equations are in y-form (that is, solved for y),and we know y = y, we can substitute the right side of each equation for y in the simpleequation y = y and write5x + 1 = !3x ! 15Now solve for x. 5x + 1 = !3x ! 15 " 8x + 1 = !15 " 8x = !16 " x = !2Remember you must find x and y. To find y, use either one of the two originalequations. Substitute the value of x into the equation and find the value of y. Using thefirst equation,y = 5x + 1 ! y = 5("2) + 1 ! y = "10 + 1 = "9 .The solution appears to be (-2, -9). In order for this to be a solution, it must make bothequations true when you replace x with -2 and y with -9. Substitute the values in bothequations to check.y = 5x + 1 y = -3x - 15-9 = ?5(-2) + 1 -9 = ?-3(-2) - 15-9 = ?-10 + 1 -9 = ?6 - 15-9 = -9 Check! -9 = -9 Check!Therefore, (-2, -9) is the solution.Example 2Substitution can also be used when the equations are notin y-form.Use substitution to rewrite the two equations as4(-3y + 1) - 3y = -11 by replacing x with (-3y + 1), thensolve for y as shown at right.x = !3y + 14x ! 3y = !11x = !3y + 14( ) ! 3y = !114(!3y + 1) ! 3y = !11y = 1Substitute y = 1 into x = -3y + 1. Solve for x, and write the answer for x and y as anordered pair, (1, 2). Substitute y = 1 into 4x - 3y = -11 to verify that either original equationmay be used to find the second coordinate. Check your answer as shown in example 1.GEOMETRY Connections 67 72. Example 3When you have a pair of two-variable equations, sometimes it is easier to ELIMINATEone of the variables to obtain one single variable equation. You can do this by adding thetwo equations together as shown in the example below.Solve the system at right:To eliminate the y terms, add the two equations together.then solve for x.Once we know the x-value we can substitute it into either of theoriginal equations to find the corresponding value of y.Using the first equation:2x + y = 11x ! y = 42x + y = 11x ! y = 43x = 153x = 15x = 52x + y = 112(5) + y = 1110 + y = 11y = 1Check the solution by substituting both the x-value and y-value into the other originalequation, x - y = 4: 5 - 1 = 4, checks.Example 4You can solve the system of equations at right by elimination,3x + 2y = 11but before you can eliminate one of the variables, you must4x + 3y = 14adjust the coefficients of one of the variables so that they areadditive opposites.9x + 6y = 33To eliminate y, multiply the first equation by 3, then multiply-8x - 6y = -28the second equation by 2 to get the equations at right.9x + 6y = 33Next eliminate the y terms by adding the two adjusted equations.-8x - 6y = -28x = 5Since x = 5, substitute in either original equation to find that y = -2. Thesolution to the system of equations is (5, -2). You could also solve the system byfirst multiplying the first equation by 4 and the second equation by -3 toeliminate x, then proceeding as shown above to find y.68 Extra Practice 73. Solve the following systems of equations to find the point of intersection (x, y) for eachpair of lines.1. y = x ! 6y = 12 ! x2. y = 3x ! 5y = x + 33. x = 7 + 3yx = 4y + 54. x = !3y + 10x = !6y ! 25. y = x + 7y = 4x ! 56. y = 7 ! 3xy = 2x ! 87. y = 3x - 12x - 3y = 108. x = - 12 y + 48x + 3y = 319. 2y = 4x + 106x + 2y = 1010. y = 35 x - 2y = x10 + 111. y = -4x + 5y = x12. 4x - 3y = -10x = 14 y - 113. x + y = 12x y = 414. 2x y = 64x y = 1215. x + 2y = 75x 4y = 1416. 5x 2y = 64x + y = 1017. x + y = 10x 2y = 518. 3y 2x = 16y = 2x + 419. x + y = 11x = y 320. x + 2y = 15y = x 321. y + 5x = 10y 3x = 1422. y = 7x - 34x + 2y = 823. y = 12 xy = x 424. y = 6 2xy = 4x 12Answers1. (9, 3) 2. (4, 7) 3. (13, 2) 4. (22, -4)5. (4, 11) 6. (3, -2) 7. (-1, -4) 8. 72$, 1!" #% &9. (0, 5) 10. (6, 1.6) 11. (1, 1) 12. (-0.25, 3)13. (8, 4) 14. (3, 0) 15. (4, 1.5) 16. (2, 2)17. 253, 53!" #$% & 18. (1, 6) 19. (4, 7) 20. (7, 4)!21. (-0.5, 12.5) 22. 79, 229" #$% &23. (8, 4) 24. (3, 0)GEOMETRY Connections 69 74. LINEAR INEQUALITIES #23To graph a linear inequality, first graph the line of the corresponding equality.This line is known as the dividing line, since all the points that make theinequality true lie on one side or the other of the line. Before you draw the line,decide whether the dividing line is part of the solution or not, that is, whether theline is solid or dashed. If the inequality symbol is either or , then the dividingline is part of the inequality and it must be solid. If the symbol is either < or >,then the dividing line is not part of the inequality and it must be dashed.Next, decide which side of the dividing line must be shaded to show the part ofthe graph that represents all values that make the inequality true. Choose a pointnot on the dividing line. Substitute this point into the original inequality. If theinequality is true for this test point, then shade the graph on this side of thedividing line. If the inequality is false for the test point, then shade the oppositeside of the dividing line.CAUTION: If the inequality is not in slope-intercept form and you have to solveit for y, always use the original inequality to test a point, NOT the solved form.Example 1Graph the inequalityy > 3x ! 2 . First, graph theline y = 3x ! 2 ,but draw it dashed since >means the dividing line is notpart of the solution.Next, test the point (-2, 4) tothe left of the dividing line.?3(!2) ! 2, so 4 > !84 >xy55-5-5xy55-5(-2,4)-5Since the inequality is true for this point, shade the left side of the dividing line.70 Extra Practice 75. Example 2Graph the system of inequalitiesy ! 12 x + 2 and y > ! 23 x ! 1 .Graph the lines y = 12 x + 2 and3 x ! 1. The first line isy = ! 2solid, the second is dashed. Testthe point (-4, 5) in the firstinequality.xy55-5-5xy55(-4, 5)-5-5?5!12("4) + 2, so 5 ! 0 This inequality is false, so shade on the oppositeside of the dividing line, from (-4, 5). Test thesame point in the second inequality.?! 25 >33 This inequality is true, so shade on the same side(!4) ! 1, so 5 > 5of the dividing line as (-4, 5).The solution is the overlap of the two shaded regions shown by the darkest shading in thesecond graph above right.23Graph each of the following inequalities on separate sets of axes.1. y 3x + 1 2. y 2x 13. y 2x 3 4. y 3x + 45. y > 4x + 2 6. y < 2x + 17. y < -3x 5 8. y > 5x 49. y 3 10. y -211. x > 1 12. x 813. y > x + 8 14. y ! " 2x + 33 15. y < 35x 7 16. y 14x 217. 3x + 2y 7 18. 2x 3y 519. -4x + 2y < 3 20. -3x 4y > 4GEOMETRY Connections 71 76. Graph each of the following pairs of inequalities on the same set of axes.21. y > 3x 4 and y -2x + 5 22. y -3x 6 and y > 4x 423. y ! " 35 x + 4 and y ! 13 x + 3 24. y < ! 37 x ! 1 and y > 45 x + 12 x + 2 26. x 3 and y < 3425. y < 3 and y ! " 1x 4Write an inequality for each of the following graphs.27.xy55-5-528.xy55-5-529.xy55-5-530.xy55-5-531.xy55-5-532.xy55-5-5Answers1.xy55-5-52.xy55-5-53.xy55-5-572 Extra Practice 77. 4.xy55-5-55.xy55-5-56.xy55-5-57.xy55-5-58.xy55-5-59.xy55-5-510.xy55-5-511.xy55-5-512.xy55-5-513.xy55-5-514.xy55-5-515.xy55-5-5GEOMETRY Connections 73 78. 16.xy55-5-517.xy55-5-518.xy55-5-519.xy55-5-520.xy55-5-521.xy55-5-522.xy55-5-523.xy55-5-524.xy55-5-525.xy55-5-526.xy55-5-527. y x + 528. y - 52x + 529. y 15x + 430. y > !4x + 131. x -432. y 374 Extra Practice 79. MULTIPLYING POLYNOMIALS #24We can use generic rectangles as area models to find the products of polynomials. Ageneric rectangle helps us organize the problem. It does not have to be drawnaccurately or to scale.Example 1Multiply (2x + 5)(x + 3)2x + 52x 2x+35x6x 152(2x + 5)(x + 3) = 2x+ 11x + 15area as a product area as a sum2x + 5x+3Example 2Multiply (x + 9) x2 ( ! 3x + 5)x - 3x + 5 2x+9x - 3x + 5 2x+93x-3x25x9x2-27x 45Therefore (x + 9) x2 ( ! 3x + 5) = x3 + 9x2 ! 3x2 ! 27x + 5x + 45 =x3 + 6x2 ! 22x + 45Another approach to multiplying binomials is to use the mnemonic "F.O.I.L." F.O.I.L. isan acronym for First, Outside, Inside, Last in reference to the positions of the terms in thetwo binomials.Example 3Multiply (3x ! 2)(4x + 5) using the F.O.I.L. method.F. multiply the FIRST terms of each binomial (3x)(4x) = 12x2O. multiply the OUTSIDE terms (3x)(5) = 15xI. multiply the INSIDE terms (-2)(4x) = -8xL. multiply the LAST terms of each binomial (-2)(5) = -10Finally, we combine like terms: 12x2 + 15x - 8x - 10 = 12x2 + 7x - 10.GEOMETRY Connections 75 80. Find each of the following products.1. (3x + 2)(2x + 7) 2. (4x + 5)(5x + 3) 3. (2x ! 1)(3x + 1)4. (2a ! 1)(4a + 7) 5. (m ! 5)(m + 5) 6. (y ! 4)(y + 4)7. (3x ! 1)(x + 2) 8. (3a ! 2)(a ! 1) 9. (2y ! 5)(y + 4)10. (3t ! 1)(3t + 1) 11. (3y ! 5)2 12. (4x ! 1)213. (2x + 3)2 14. (5n + 1)2 15. (3x ! 1) 2x2 ( + 4x + 3)16. (2x + 7) 4x2 ( ! 3x + 2) 17. (x + 7) 3x2 ( ! x + 5) 18. (x ! 5) x2 ( ! 7x + 1)19.(3x + 2) x3 ! 7x2 ( + 3x) 20. (2x + 3) 3x2 ( + 2x ! 5)Answers1. 6x2 + 25x + 14 2. 20x2 + 37x + 15 3. 6x2 ! x ! 14. 8a2 + 10a ! 7 5. m2 ! 25 6. y2 ! 167. 3x2 + 5x ! 2 8. 3a2 ! 5a + 2 9. 2y2 + 3y ! 2010. 9t2 ! 1 11. 9y2 ! 30y + 25 12. 16x2 ! 8x + 113. 4x2 + 12x + 9 14. 25n2 + 10n + 1 15. 6x3 + 10x2 + 5x ! 316. 8x3 + 22x2 ! 17x + 14 17. 3x3 + 20x2 ! 2x + 35 18. x3 ! 12x2 + 36x ! 519. 3x4 ! 19x3 ! 5x2 + 6x 20. 6x3 + 13x2 ! 4x ! 1576 Extra Practice 81. FACTORING POLYNOMIALS #25Often we want to un-multiply or FACTOR a polynomial P(x). This process involvesfinding a constant and/or another polynomial that evenly divides the given polynomial.In formal mathematical terms, this means P(x) = q(x) ! r (x) , where q and r are alsopolynomials. For elementary algebra there are three general types of factoring.1) Common term (finding the largest common factor):6x + 18 = 6(x + 3) where 6 is a common factor of both terms.2x3 ! 8x2 ! 10x = 2x x2 ( ! 4x ! 5) where 2x is the common factor.2x2 (x ! 1) + 7(x ! 1) = (x ! 1) 2x2 ( + 7) where x 1 is the common factor.2) Special productsa2 ! b2 = (a + b)(a ! b) x2 ! 25 = (x + 5)(x ! 5)9x2 ! 4y2 = (3x + 2y)(3x ! 2y)x2 + 2xy + y2 = (x + y)2 x2 + 8x + 16 = (x + 4)2x2 ! 2xy + y2 = (x ! y)2 x2 ! 8x + 16 = (x ! 4)23a) Trinomials in the form x2 + bx + c where the coefficient of x2 is 1.Consider x2 + (d + e)x + d ! e = (x + d)(x + e) , where the coefficient of x isthe sum of two numbers d and e AND the constant is the product of thesame two numbers, d and e. A quick way to determine all of the possiblepairs of integers d and e is to factor the constant in the original trinomial.For example, 12 is 1!12 , 2 ! 6 , and 3! 4 . The signs of the two numbers aredetermined by the combination you need to get the sum. The "sum andproduct" approach to factoring trinomials is the same as solving a "DiamondProblem" in CPM's Algebra 1 course (see below).x2 + 8x + 15 = (x + 3)(x + 5); 3 + 5 = 8, 3! 5 = 15x2 ! 2x ! 15 = (x ! 5)(x + 3); ! 5 + 3 = !2, ! 5 " 3 = !15x2 ! 7x + 12 = (x ! 3)(x ! 4); ! 3 + (!4) = !7, (!3)(!4) = 12The sum and product approach can be shown visually using rectangles for anarea model. The figure at far left below shows the "Diamond Problem"format for finding a sum and product. Here is how to use this method tofactor x2 + 6x + 8 .>> Explanation and examples continue on the next page. >>GEOMETRY Connections 77 82. 8product# # 2 4(x + 4)(x + 2)sum6x2 4x2x 8x + 4x2 4x2x 8x+23b) Trinomials in the form ax2 + bx + c where a 1.Note that the upper value in the diamond is no longer the constant. Rather, itis the product of a and c, that is, the coefficient of x2 and the constant.22x + 7x + 3multiply6? ?766 17x + 32 2x+12 2x 6x2x 6x1x 31x 3(2x + 1)(x + 3)Below is the process to factor 5x2 ! 13x + 6 .30? ?2 ?5x-13 ?6? ?5x-3(5x - 3)(x-2)25x-10x-3x 6Polynomials with four or more terms are generally factored by grouping the termsand using one or more of the three procedures shown above. Note thatpolynomials are usually factored completely. In the second example in part (1)above, the trinomial also needs to be factored. Thus, the complete factorization of2x3 ! 8x2 ! 10x = 2x x2 ( ! 4x ! 5) = 2x(x ! 5)(x + 1) .Factor each polynomial completely.1. x2 ! x ! 42 2. 4x2 ! 18 3. 2x2 + 9x + 9 4. 2x2 + 3xy + y25. 6x2 ! x ! 15 6. 4x2 ! 25 7. x2 ! 28x + 196 8. 7x2 ! 8479. x2 + 18x + 81 10. x2 + 4x ! 21 11. 3x2 + 21x 12. 3x2 ! 20x ! 3213. 9x2 ! 16 14. 4x2 + 20x + 25 15. x2 ! 5x + 6 16. 5x3 + 15x2 ! 20x17. 4x2 + 18 18. x2 ! 12x + 36 19. x2 ! 3x ! 54 20. 6x2 ! 2121. 2x2 + 15x + 18 22. 16x2 ! 1 23. x2 ! 14x + 49 24. x2 + 8x + 1525. 3x3 !12x2 ! 45x 26. 3x2 + 24 27. x2 + 16x + 6478 Extra Practice 83. Factor completely.28. 75x3 ! 27x 29. 3x3 ! 12x2 ! 36x 30. 4x3 ! 44x2 + 112x31. 5y2 ! 125 32. 3x2y2 ! xy2 ! 4y2 33. x3 + 10x2 ! 24x34. 3x3 ! 6x2 ! 45x 35. 3x2 ! 27 36. x4 ! 16Factor each of the following completely. Use the modified diamond approach.37. 2x2 + 5x ! 7 38. 3x2 ! 13x + 4 39. 2x2 + 9x + 1040. 4x2 ! 13x + 3 41. 4x2 + 12x + 5 42. 6x3 + 31x2 + 5x43. 64x2 + 16x + 1 44. 7x2 ! 33x ! 10 45. 5x2 + 12x ! 9Answers1. (x + 6)(x ! 7) 2. 2 2x2 ( ! 9) 3. (2x + 3)(x + 3)4. (2x + y)(x + y) 5. (2x + 3)(3x ! 5) 6. (2x ! 5)(2x + 5)7. (x ! 14)2 8. 7(x ! 11)(x + 11) 9. (x + 9)210. (x + 7)(x ! 3) 11. 3x(x + 7) 12. (x ! 8)(3x + 4)13. (3x ! 4)(3x + 4) 14. (2x + 5)2 15. (x ! 3)(x ! 2)16. 5x(x + 4)(x ! 1) 17. 2 2x2 ( + 9) 18. (x ! 6)219. (x ! 9)(x + 6) 20. 3 2x2 ( ! 7) 21. (2x + 3)(x + 6)22. (4x + 1)(4x ! 1) 23. (x ! 7)2 24. (x + 3)(x + 5)25. 3x x2 ( ! 4x ! 15) 26. 3 x2 ( + 8) 27. (x + 8)228. 3x(5x ! 3)(5x + 3) 29. 3x(x ! 6)(x + 2) 30. 4x(x ! 7)(x ! 4)31. 5(y + 5)(y ! 5) 32. y2 (3x ! 4)(x + 1) 33. x(x + 12)(x ! 2)34. 3x(x ! 5)(x + 3) 35. 3(x ! 3)(x + 3) 36. (x ! 2)(x + 2) x2 ( + 4)37. (2x + 7)(x ! 1) 38. (3x ! 1)(x ! 4) 39. (x + 2)(2x + 5)40. (4x ! 1)(x ! 3) 41. (2x + 5)(2x + 1) 42. x(6x + 1)(x + 5)43. (8x + 1)2 44. (7x + 2)(x ! 5) 45. (5x ! 3)(x + 3)GEOMETRY Connections 79 84. ZERO PRODUCT PROPERTY AND QUADRATICS #26If a b = 0, then either a = 0 or b = 0.Note that this property states that at least one of the factors MUST be zero. It isalso possible that all of the factors are zero. This simple statement gives us apowerful result which is most often used with equations involving the products ofbinomials. For example, solve (x + 5)(x ! 2) = 0 .By the Zero Product Property, since (x + 5)(x ! 2) = 0 , either x + 5 = 0 orx ! 2 = 0 . Thus, x = !5 or x = 2 .The Zero Product Property can be used to find where a quadratic crosses the x-axis.These points are the x-intercepts. In the example above, they would be (-5, 0) and (2, 0).Example 1Where does y = (x + 3)(x ! 7) cross thex-axis? Since y = 0 at the x-axis, then(x + 3)(x ! 7) = 0 and the Zero ProductProperty tells you that x = !3 and x = 7so y = (x + 3)(x ! 7) crosses the x-axisat (!3,0) and (7,0) .Example 2Where does y = x2 ! x ! 6 cross thex-axis? First factor x2 ! x ! 6 into(x + 2)(x ! 3) to get y = (x + 2)(x ! 3) .By the Zero Product Property, thex-intercepts are (-2, 0) and (3, 0).Example 3Graph y = x2 ! x ! 6Since you know the x-intercepts fromexample 2, you already have two points tograph. You need a table of values to getadditional points.x! 2! 1! 0! 1! 2! 3! 4y! 0! 4! 6! 6! 4! 0! 6Example 4Graph y > x2 ! x ! 6First graph y = x2 ! x ! 6 as you did at left.Use a dashed curve. Second, pick a point noton the parabola and substitute it into theinequality. For example, testing point (0, 0) iny > x2 ! x ! 6 gives 0 > !6 which is a truestatement. This means that (0, 0) is a solutionto the inequality as wellas all points inside thecurve. Shade the interiorof the parabola.1y4 3 2 1 1 2 3 41234567x1y4 3 2 1 1 2 3 4123456780 Extra Practicex 85. Solve the following problems using the Zero Product Property.1. (x ! 2)(x + 3) = 0 2. 2x(x + 5)(x + 6) = 03. (x ! 18)(x ! 3) = 0 4. 4x2 ! 5x ! 6 = 05. (2x ! 1)(x + 2) = 0 6. 2x(x ! 3)(x + 4) = 07. 3x2 ! 13x ! 10 = 0 8. 2x2 ! x = 15Use factoring and the Zero Product Property to find the x-intercepts of each parabolabelow. Express your answer as ordered pair(s).9. y = x2 3x + 2 10. y = x2 10x + 25 11. y = x2 x - 1212. y = x2 4x 5 13. y = x2 + 2x - 8 14. y = x2 + 6x + 915. y = x2 8x + 16 16. y = x2 9Graph the following inequalities. Be sure to use a test point to determine which region toshade. Your solutions to the previous problems might be helpful.17. y < x2 ! 3x + 2 18. y > x2 ! 10x + 25 19. y ! x2 " x " 1220. y ! x2 " 4x " 5 21. y > x2 + 2x ! 8 22. y ! x2 + 6x + 923. y < x2 ! 8x + 16 24. y ! x2 " 9Answers1. x = 2, x = !3 2. x = 0, x = !5, x = !6 3. x = 18, x = 34. x = !0.75, x = 2 5. x = 0.5, x = !2 6. x = 0, x = 3, x = !47. x = ! 23, x = 5 8. x = !2.5, x = 3 9. (1, 0) and (2, 0)10. (5, 0) 11. (3, 0) and (4, 0) 12. (5, 0) and (1, 0)13. (4, 0) and (2, 0) 14. (3, 0) 15. (4, 0)16. (3, 0) and (3, 0)17. 18. 19. 20.44 444 44844 4421. 22. 23. 24.xyxy4444 44GEOMETRY Connections 81 86. THE QUADRATIC FORMULA #27You have used factoring and the Zero Product Property to solve quadraticequations. You can solve any quadratic equation by using the QUADRATICFORMULA.If ax2 + bx + c = 0, then x = -b b2 - 4ac2a .For example, suppose 3x2 + 7x - 6 = 0. Here a = 3,