geometry skill builders

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Geometry 463 GEOMETRY SKILL BUILDERS (Extra Practice) Introduction to Students and Their Teachers Learning is an individual endeavor. Some ideas come easily; others take time--sometimes lots of time--to grasp. In addition, individual students learn the same idea in different ways and at different rates. The authors of this textbook designed the classroom lessons and homework to give students time--often weeks and months--to practice an idea and to use it in various settings. This section of the textbook offers students a brief review of 27 topics followed by additional practice with answers. Not all students will need extra practice. Some will need to do a few topics, while others will need to do many of the sections to help develop their understanding of the ideas. This section of the text may also be useful to prepare for tests, especially final examinations. How these problems are used will be up to your teacher, your parents, and yourself. In classes where a topic needs additional work by most students, your teacher may assign work from one of the skill builders that follow. In most cases, though, the authors expect that these resources will be used by individual students who need to do more than the textbook offers to learn an idea. This will mean that you are going to need to do some extra work outside of class. In the case where additional practice is necessary for you individually or for a few students in your class, you should not expect your teacher to spend time in class going over the solutions to the skill builder problems. After reading the examples and trying the problems, if you still are not successful, talk to your teacher about getting a tutor or extra help outside of class time. Warning! Looking is not the same as doing. You will never become good at any sport just by watching it. In the same way, reading through the worked out examples and understanding the steps are not the same as being able to do the problems yourself. An athlete only gets good with practice. The same is true of developing your algebra and geometry skills. How many of the extra practice problems do you need to try? That is really up to you. Remember that your goal is to be able to do problems of the type you are practicing on your own, confidently and accurately. Two other sources for help with the algebra and geometry topics in this course are the Parent's Guide with Review to Math 2 (Geometry) and the Parent's Guide with Review to Math 1 (Algebra 1). Information about ordering these supplements can be found inside the front page of the student text. These resources are also available free from the internet at www.cpm.org. Online homework help is underwritten by CPM at www.hotmath.com. Skill Builder Topics 1. Writing and graphing linear equations 15. Right triangle trigonometry 2. The Pythagorean Theorem 16. Ratio of similarity 3. Area 17. Similarity of length, area, and volume 4. Solving linear systems 18. Interior and exterior angles of polygons 5. Properties of angles, lines, and triangles 19. Areas by dissection 6. Linear inequalities 20. Central and inscribed angles 7. Multiplying Polynomials 21. Area of sectors 8. Factoring Polynomials 22. Tangents, secants, and chords 9. Zero Product Property and quadratics 23. Volume and surface area of polyhedra 10. The Quadratic Formula 24. Simplifying rational expressions 11. Triangle Congruence 25. Multiplication and division of rational expressions 12. Laws of Exponents 26. Addition and subtraction of rational expressions 13. Proof 27. Solving mixed equations and inequalities 14. Radicals

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Geometry Skill Builders

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  • 1. GEOMETRY SKILL BUILDERS(Extra Practice)Introduction to Students and Their TeachersLearning is an individual endeavor. Some ideas come easily; others take time--sometimeslots of time--to grasp. In addition, individual students learn the same idea in different ways and atdifferent rates. The authors of this textbook designed the classroom lessons and homework to givestudents time--often weeks and months--to practice an idea and to use it in various settings. Thissection of the textbook offers students a brief review of 27 topics followed by additional practicewith answers. Not all students will need extra practice. Some will need to do a few topics, whileothers will need to do many of the sections to help develop their understanding of the ideas. Thissection of the text may also be useful to prepare for tests, especially final examinations.How these problems are used will be up to your teacher, your parents, and yourself. In classeswhere a topic needs additional work by most students, your teacher may assign work from one ofthe skill builders that follow. In most cases, though, the authors expect that these resources will beused by individual students who need to do more than the textbook offers to learn an idea. Thiswill mean that you are going to need to do some extra work outside of class. In the case whereadditional practice is necessary for you individually or for a few students in your class, you shouldnot expect your teacher to spend time in class going over the solutions to the skill builder problems.After reading the examples and trying the problems, if you still are not successful, talk to yourteacher about getting a tutor or extra help outside of class time.Warning! Looking is not the same as doing. You will never become good at any sport justby watching it. In the same way, reading through the worked out examples and understanding thesteps are not the same as being able to do the problems yourself. An athlete only gets good withpractice. The same is true of developing your algebra and geometry skills. How many of the extrapractice problems do you need to try? That is really up to you. Remember that your goal is to beable to do problems of the type you are practicing on your own, confidently and accurately.Two other sources for help with the algebra and geometry topics in this course are theParent's Guide with Review to Math 2 (Geometry) and the Parent's Guide with Review to Math 1(Algebra 1). Information about ordering these supplements can be found inside the front page ofthe student text. These resources are also available free from the internet at www.cpm.org.Online homework help is underwritten by CPM at www.hotmath.com.Skill Builder Topics1. Writing and graphing linear equations 15. Right triangle trigonometry2. The Pythagorean Theorem 16. Ratio of similarity3. Area 17. Similarity of length, area, and volume4. Solving linear systems 18. Interior and exterior angles of polygons5. Properties of angles, lines, and triangles 19. Areas by dissection6. Linear inequalities 20. Central and inscribed angles7. Multiplying Polynomials 21. Area of sectors8. Factoring Polynomials 22. Tangents, secants, and chords9. Zero Product Property and quadratics 23. Volume and surface area of polyhedra10. The Quadratic Formula 24. Simplifying rational expressions11. Triangle Congruence 25. Multiplication and division of rational expressions12. Laws of Exponents 26. Addition and subtraction of rational expressions13. Proof 27. Solving mixed equations and inequalities14. RadicalsGeometry 463

2. WRITING AND GRAPHING LINEAR EQUATIONS ON A FLAT SURFACE#1SLOPE is a number that indicates the steepness (or flatness) of a line, as well as its direction (upor down) left to right.SLOPE is determined by the ratio:vertical changehorizontal change between any two points on a line.For lines that go up (from left to right), the sign of the slope is positive. For lines that 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 inSLOPE-INTERCEPT FORM. m is the SLOPE of the line. b is the 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, PARALLEL LINES have thesame slope.Two lines are PERPENDICULAR if the slope of one line is the negative reciprocal of the slopeof the other line, that is, m and 1mm ( )= 1.. Note that m 1Examples: 3 and 13, 23and 32, 54and 45Two distinct lines on a flat surface that are not parallel intersect in a single point. See "SolvingLinear Systems" to review how to find the point of intersection.Example 1Graph the linear equation y =47x + 2Using y = mx + b, the slope in y = 47x + 2 is 47 and they-intercept is the point (0, 2). To graph, begin at they-intercept (0, 2). Remember that slope is vertical changehorizontal change so goup 4 units (since 4 is positive) from (0, 2) and then move right 7units. This gives a second point on the graph. To create the graph,draw a straight line through the two points.yx554-57-5464 SKILL BUILDERS 3. Example 2A line has a slope of 34 and passes through (3, 2). What is the equation of the line?Using y = mx + b, write y =34x + 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 =34x 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 = 54 x + 32The slope of this line is 5.4-4x + 5y = 35y = 4x + 3y = 4x + 35y = 45 x + 35The slope of this line is 45.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 perpendicular to the givenline: y = 52x + 5 and (-4, 5).For the parallel line, use y = mx + b with thesame slope to write y = 52x + b .Substitute the point (4, 5) for x and y andsolve for b.5 = 52(4) + b 5 =202+ b 5 = bTherefore the parallel line through (-4, 5) isy = 52x 5.For the perpendicular line, usey = mx + b where m is the negativereciprocal of the slope of the originalequation to write y =25x + b.Substitute the point (4, 5) and solve for b.5 =25(4) + b 335= bTherefore the perpendicular line through(-4, 5) is y =25x +335.Geometry 465 4. Identify the y-intercept in each equation.1. y = 12 x 2 2. y = 35x 533. 3x + 2y = 124. x y = -13 5. 2x 4y = 12 6. 4y 2x = 12Write the equation of the line with:7. slope = 12 and passing through (4, 3). 8. slope = 23 and passing through (-3, -2).9. slope = - 13 and 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.xyxyxyxyGraph the following linear equations on graph paper.18. y = 12 x + 3 19. y = - 35 x 1 20. y = 4x21. y = -6x + 12 22. 3x + 2y = 12466 SKILL BUILDERS 5. State whether each pair of lines is parallel, perpendicular, or intersecting.23. y = 2x 2 and y = 2x + 4 24. y = 12 x + 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 = -13 x 3 28. 3x + 2y = 6 and 2x + 3y = 629. 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 = 12 x + 3 and (-4, 2)33. 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 = 12 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 point47.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 line through(2, 5) and (8, -3)50. Write the equation of the line through(1, -4) which is parallel to the line through(-3, -7) and (4, 3)Geometry 467 6. Answers1. (0, 2) 2. 0, 53 ( ) 3. (0, 6) 4. (0, 13)125. (0, 3) 6. (0, 3) 7. y =x +1 8. y=23x9. y = 13x +1310. y = 4x 7 11. 1212. 3413. 2 14. y = 2x 2 15. y = x + 2 16. y =13x + 217. y = 2x + 4 18. line with slope 12and y-intercept (0, 3)19. line with slope 35and y-intercept (0, 1) 20. line with slope 4 and y-intercept (0, 0)21. line with slope 6 and y-intercept 0, 1222. line with slope 32and y-intercept (0, 6)23. parallel 24. perpendicular 25. perpendicular 26. perpendicular27. parallel 28. intersecting 29. intersecting 30. parallel31. y = 2x + 11 32. y = 12 x + 4 33. y = x + 5 34. y = x + 335. y = - 13 x + 23 36. y = - 32 x + 2 37. y = 45 x 95 38. y = 34 x 539. y = - 12 x + 72 40. y = -2x 6 41. y = -x + 1 42. y = -x 143. y = 3x + 4 44. y = 23 x 73 45. y = - 54 x + 14 46. y = - 43 x + 10347. y = 3x + 11 48. y = - 12 x + 152 49. y = 43 x + 43 50. y = 107 x 387468 SKILL BUILDERS 7. PYTHAGOREAN THEOREM #2ac Any triangle that has a right angle is called a RIGHTbTRIANGLE. The two sides that form the right angle, a and b,are called LEGS, and the side opposite (that is, across thetriangle from) the right angle, c, is called the HYPOTENUSE.For any right triangle, the sum of the squares of the legs of the triangle is equal to the squareof the hypotenuse, that is, a2 + b2 = c2. This relationship is known as thePYTHAGOREAN 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 or use area pictures(as shown on page 22, problem RC-1) on each side of the triangle 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 on thewall does the top of the ladder touch?x1044x2 + 42 = 102x2 + 16 = 100x2 = 84x 9.2The ladder touches the wall about 9.2 feetabove the ground.d) Could 3, 6 and 8 represent the lengths ofthe sides of a right triangle? Explain.32 + 62 ?=82?649 + 36 =45 64Since the Pythagorean Theoremrelationship is not true for these lengths,they cannot be the side lengths of a righttriangle.Geometry 469 8. Use the Pythagorean Theorem to find the value of x. Round answers to the nearest tenth.1. 2. 3. 4. 5.3722x20 96xx42x 8316 46x72 656. 7. 8. 9. 10.x16 221532xx3816x 1057512530xSolve 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 laddertouch?15. A 6 foot ladder is one and a half feet from a wall. How high on the wall does the laddertouch?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. no470 SKILL BUILDERS 9. AREA #3AREA is the number of square units in a flat region. The formulas to calculate the area ofseveral kinds of polygons are:RECTANGLE PARALLELOGRAM TRAPEZOID TRIANGLEhbhbhb1b2hbhbA = bh A = bh A =12(b1 + b2 )h A =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 smaller regions of thetypes shown above, calculating each area, and finding the sum of the areas.Example 1Find the area of each figure. All lengths are centimeters.a)2381A = bh = (81)(23) = 1863 cm2b)412A =12(12)(4) = 24 cm2c)42108A =12(108)(42) = 2268 cm2d)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) = 288 cm2Geometry 471 10. Example 2Find the area of the shaded region.The area of the shaded region is the area of the triangleminus the area of the rectangle.triangle: A =12(7)(10) = 35 cm2rectangle: A = 5(4) = 20 cm2shaded region: A = 35 20 = 15 cm24 cm.7 cm.5 cm.10 cm.Find the area of the following triangles, parallelograms and trapezoids. Pictures are not drawn toscale. Round answers to the nearest tenth.1. 2. 3. 4.102036 510865465. 6. 7. 8.1326267 151512828201649. 10. 11. 12.3181239 41.5674108210613. 14. 15. 16.15135923135 752964 4472 SKILL BUILDERS 11. Find the area of the shaded region.17.4810243663218.182419. 18341620.106 31216Answers (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. 138Geometry 473 12. SOLVING LINEAR SYSTEMS #4You 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.Check your solution.y = 5x + 1y = -3x - 15When solving a system of equations, you are solving to find the x- and y-values that result in truestatements when you substitute them into both equations. You generally find 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 theright side of each equation for y in the simple equation y = y and write5x + 1 = 3x 15.Now 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 original equations.Substitute the value of x into the equation and find the value of y. Using the first 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 both equationstrue when you replace x with -2 and y with -9. Substitute the values in both equations 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 equationsare not in y-form.Use substitution to rewrite the two equations as4(-3y + 1) - 3y = -11 by replacing x with (-3y + 1),then solve 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 asan ordered pair, (1, 2). Substitute y = 1 into 4x - 3y = -11 to verify that either originalequation may be used to find the second coordinate. Check your answer as shown inexample 1.474 SKILL BUILDERS 13. Example 3When you have a pair of two-variable equations, sometimes it is easier to ELIMINATE one ofthe variables to obtain one single variable equation. You can do this by adding the two equationstogether 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 the originalequations 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 original equation,x - y = 4: 5 - 1 = 4, checks.Example 4You can solve the system of equations at right by elimination, but before youcan eliminate one of the variables, you must adjust the coefficients of one of thevariables so that they are additive opposites.To eliminate y, multiply the first equation by 3, then multiply the secondequation by 2 to get the equations at right.Next eliminate the y terms by adding the two adjusted equations.3x + 2y = 114x + 3y = 149x + 6y = 33-8x - 6y = -289x + 6y = 33-8x - 6y = -28x = 5Since x = 5, substitute in either original equation to find that y = -2. Therefore, the solution tothe system of equations is (5, -2).You could also solve the system by first multiplying the first equation by 4 and the secondequation by -3 to eliminate x, then proceeding as shown above to find y.Geometry 475 14. Solve the following systems of equations to find the point of intersection (x, y) for each pair 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 = 10 8. 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. 7, 1 29. (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, 5318. (1, 6) 19. (4, 7) 20. (7, 4)21. (-0.5, 12.5) 22. 79, 22923. (8, 4) 24. (3, 0)476 SKILL BUILDERS 15. PROPERTIES OF ANGLES, LINES, AND TRIANGLES #5Parallel lines Triangles123 498 7 61011 corresponding angles are equal:m1 = m3 alternate interior angles are equal:m2 = m3 m2 + m4 = 180 m7 + m8 + m9 = 180 m6 = m8 + m9(exterior angle = sum remote interior angles) m10 + m11 = 90(complementary angles)Also shown in the above figures: vertical angles are equal: m1 = m2 linear pairs are supplementary: m3 + m4 = 180and m6 + m7 = 180In addition, an isosceles triangle, ABC, hasBA = BC and mA = mC. An equilateraltriangle, GFH, has GF = FH = HG andmG = mF = mH = 60.ABCFG HExample 1Solve for x.Use the Exterior Angle Theorem: 6x + 8 = 49 + 676x= 108 x = 10866x+8 49 x = 18 67Example 2Solve for x.There are a number of relationships in this diagram. First, 1 and the127 angle are supplementary, so we know that m1 + 127= 180 som1 = 53. Using the same idea, m2 = 47. Next,m3 + 53+ 47= 180, so m3 = 80. Because angle 3 forms avertical pair with the angle marked 7x + 3, 80 = 7x + 3, so x = 11.17x+323127 133Example 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 algebraicangle measure: 2(6) + 46= 58 for the measure of the acute angle.2x + 465x+28Geometry 477 16. Use the geometric properties and theorems you have learned to solve for x in each diagram and writethe 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+312.583x13.142 3820x + 214.3814220x 215.128527x + 316.5x + 3 1285217.x235818.8x117 5719.189x5x + 3620.8x + 12 in.12x 18 in.21.8x18 cm5x + 3 cm478 SKILL BUILDERS 17. 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. 10 2313. 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. 7 16 28. 10Geometry 479 18. LINEAR INEQUALITIES #6To graph a linear inequality, first graph the line of the corresponding equality. This line is knownas the dividing line, since all the points that make the inequality true lie on one side or the other ofthe line. Before you draw the line, decide whether the dividing line is part of the solution or not,that is, whether the line is solid or dashed. If the inequality symbol is either or , then thedividing line is part of the inequality and it must be solid. If the symbol is either < or >, then thedividing 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 of the graph thatrepresents all values that make the inequality true. Choose a point not on the dividing line.Substitute this point into the original inequality. If the inequality is true for this test point, thenshade the graph on this side of the dividing line. If the inequality is false for the test point, thenshade the opposite side of the dividing line.CAUTION: If the inequality is not in slope-intercept form and you have to solve it for y, alwaysuse the original inequality to test a point, NOT the solved form.Example 1Graph the inequality y > 3x 2.First, graph the line y = 3x 2 ,but draw it dashed since > meansthe dividing line is not part of thesolution.Next, test the point (-2, 4) to the leftof the dividing line.4> ?3(2) 2, so 4 > 8xy55-5-5xy55-5(-2,4)-5Since the inequality is true for this point, shade the left side of the dividing line.480 SKILL BUILDERS 19. Example 2Graph the system of inequalitiesy 12x + 2 and y > 23x 1.Graph the lines y =12x + 2 andy = 23x 1. The first line is solid,the second is dashed. Test the point(-4, 5) in the first inequality.xy55-5-5xy55(-4, 5)-5-55 ?12(4) + 2, so 5 0This inequality is false, so shade on the opposite side ofthe dividing line, from (-4, 5). Test the same point inthe second inequality.5 > ?23(4) 1, so 5 >53This inequality is true, so shade on the same side ofthe 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.Graph 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 > 23 x + 8 14. y 23x + 315. y < -35x 7 16. y 14x 217. 3x + 2y 7 18. 2x 3y 519. -4x + 2y < 3 20. -3x 4y > 4Graph 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 35x + 4 and y 13x + 3 24. y < 37x 1 and y >45x + 125. y < 3 and y 12x + 2 26. x 3 and y < 34x 4Geometry 481 20. Write 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.xy55-5-54.xy55-5-55.xy55-5-56.xy55-5-5482 SKILL BUILDERS 21. 7.xy55-5-58.xy55-5-59.xy55-5-510.xy55-5-511.xy55-5-512.xy55-5-513.xy55-5-514.xy55-5-515.xy55-5-516.xy55-5-517.xy55-5-518.xy55-5-5Geometry 483 22. 19.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 - 52 x + 529. y 15 x + 430. y > 4x + 131. x -432. y 3484 SKILL BUILDERS 23. MULTIPLYING POLYNOMIALS #7We can use generic rectangles as area models to find the products of polynomials. A genericrectangle helps us organize the problem. It does not have to be drawn accurately or to scale.Example 1Multiply (2x + 5)(x + 3)2x + 52x 2x+35x6x 15(2x + 5)(x + 3) = 2 2x+ 11x + 15area as a product area as a sum2x + 5x+3Example 2Multiply (x + 9)(x2 3x + 5)x 2 - 3x + 5x+9x 2 - 3x + 5x+9x3 -3x2 5x9x2 -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. is anacronym for First, Outside, Inside, Last in reference to the positions of the terms in the two 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 485 24. 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) 2x( 2 + 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 15486 SKILL BUILDERS 25. FACTORING POLYNOMIALS #8Often 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 is thesum of two numbers d and e AND the constant is the product of the same twonumbers, d and e. A quick way to determine all of the possible pairs of integersd 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 are determined by the combinationyou need to get the sum. The "sum and product" approach to factoring trinomials isthe same as solving a "Diamond Problem" 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 an areamodel. The figure at far left below shows the "Diamond Problem" format for findinga sum and product. Here is how to use this method to factor x2 + 6x + 8.8product# # 2 4(x + 4)(x + 2)sum6x2 4x2x 8x + 4x2 4x2x 8x+2>> Explanation and examples continue on the next page. >>Geometry 487 26. 3b) Trinomials in the form ax2 + bx + c where a 1.Note that the upper value in the diamond is no longer the constant. Rather, it is the productof a and c, that is, the coefficient of x2 and the constant.2x + 7x + 3 2multiply6? ?766 17x + 32x 2+12 2x 6x2x 6x1x 31x 3(2x + 1)(x + 3)Below is the process to factor 5x2 13x + 6 .30? ?2 ??? ?5x-13 65x-3(5x - 3)(x-2)5x2-3x 6-10xPolynomials with four or more terms are generally factored by grouping the terms and using oneor more of the three procedures shown above. Note that polynomials are usually factoredcompletely. In the second example in part (1) above, the trinomial also needs to be factored.Thus, the complete factorization of 2x3 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 + 64Factor 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 16488 SKILL BUILDERS 27. Factor 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 489 28. ZERO PRODUCT PROPERTY AND QUADRATICS #9If 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 is also possible thatall of the factors are zero. This simple statement gives us a powerful result which is most oftenused with equations involving the products of binomials. For example, solve (x + 5)(x 2) = 0.By the Zero Product Property, since (x + 5)(x 2) = 0, either x + 5 = 0 or x 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 pointsare 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 = 7 soy = (x + 3)(x 7) crosses the x-axis at(-3, 0) and (7, 0).Example 2Where does y = x2 x 6 cross the x-axis?First factor x2 x 6 into (x + 2)(x 3) toget y = (x + 2)(x 3). By the Zero ProductProperty, the x-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 40 4 6 6 4 0 61y4 3 2 1 1 2 3 41234567xExample 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 well as all points inside thecurve. Shade the interior of the parabola.1y4 3 2 1 1 2 3 41234567x490 SKILL BUILDERS 29. Solve the following problems using the Zero Product Property.1. (x 2)(x + 3) = 0 2. 2x(x + 5)(x + 6) = 0 3. (x 18)(x 3) = 04. 4x2 5x 6 = 0 5. (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 parabola below. Expressyour answer as ordered pair(s).9. y = x2 3x + 2 10. y = x2 10x + 25 11. y = x2 x - 12 12. y = x2 4x - 513. y = x2 + 2x - 8 14. y = x2 + 6x + 9 15. y = x2 8x + 16 16. y = x2 9Graph the following inequalities. Be sure to use a test point to determine which region to shade.Your solutions to the previous problems might be helpful.17. y < x2 3x + 2 18. y > x2 10x + 25 19. y x2 x 12 20. y x2 4x - 521. y > x2 + 2x 8 22. y x2 + 6x + 9 23. 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 491 30. THE QUADRATIC FORMULA #10You have used factoring and the Zero Product Property to solve quadratic equations. You cansolve any quadratic equation by using the QUADRATIC FORMULA.If ax2 + bx + c = 0, then x =-b b2 4ac2a .For example, suppose 3x2 + 7x - 6 = 0. Here a = 3, b = 7, and c = -6.Substituting these values into the formula results in:x =-(7) 72 - 4(3)(- 6)2(3) x =-7 1216 x =-7 116Remember that non-negative numbers have both a positive and negative square root.The sign represents this fact for the square root in the formula and allows us to write theequation once (representing two possible solutions) until later in the solution process.-7 + 11Split the numerator into the two values: x =6 or x =-7 - 116Thus the solution for the quadratic equation is: x = 23or -3.Example 1Solve x2 + 3x 2 = 0 using the quadratic formula.First, identify the values for a, b, and c. In this case they are 1, 3, and 2, respectively. Next,substitute these values into the quadratic formula.x =(3) 32 4(1)(2)2(1) x =3 172Then split the numerator into the two values: x =3 + 172or x =3 172Using a calculator, the solution for the quadratic equation is: x = 0.56 or 3.56.Example 2Solve 4x2 + 4x = 3 using the quadratic formula.To solve any quadratic equation it must first be equal to zero. Rewrite the equation as4x2 + 4x 3 = 0. Identify the values for a, b, and c: 4, 4, and -3, respectively. Substitute these valuesinto the quadratic formula.x =-(4) 42 - 4(4)(-3)2(4) x =-4 648 x =-4 88Split the numerator into the two values: x =-4 + 88 or x =-4 88 , so x = 12or 32.492 SKILL BUILDERS 31. Use the quadratic formula to solve each of the following equations.1. x2 x 6 = 0 2. x2 + 8x + 15 = 03. x2 + 13x + 42 = 0 4. x2 10x + 16 = 05. x2 + 5x + 4 = 0 6. x2 9x + 18 = 07. 5x2 x 4 = 0 8. 4x2 11x 3 = 09. 6x2 x 15 = 0 10. 6x2 + 19x + 15 = 011. 3x2 + 5x 28 = 0 12. 2x2 x 14 = 013. 4x2 9x + 4 = 0 14. 2x2 5x + 2 = 015. 20x2 + 20x = 1 16. 13x2 16x = 417. 7x2 + 28x = 0 18. 5x2 = -125x19. 8x2 50 = 0 20. 15x2 = 3Answers1. x = -2, 3 2. x = -5, -3 3. x = -7, -6 4. x = 2, 85. x = -4, -16. x = 3, 6 7. x = - 45, 1 8. x = - 14, 39. x = - 32, 53 10. x = - 32, - 53 11. x = - 4, 73 12. x = 1 113413. x = 9 178 14. x = 2, 12 15. x = - 20 48040=-5 301016. x = 16 46426=8 2 2913 17. x = -4, 0 18. x = -25, 019. x = - 52, 52 20. x = 55Geometry 493 32. TRIANGLE CONGRUENCEIf two triangles are congruent, then all six oftheir corresponding pairs of parts are alsocongruent. In other words, if ABC XYZ,then all the congruence statements at right are#11AB CXY Zcorrect. Note that the matching parts follow the order of the lettersas written in the triangle congruence statement. For example, if Ais first in one statement and X is first in the other, then A alwayscorresponds to X in congruence statements for that relationship orfigure.AB XYBC YZAC XZA XB YC ZYou only need to know that three pairs of parts are congruent (sometimes in a certain order) toprove that the two triangles are congruent. The Triangle Congruence Properties are:SSS (Side-Side-Side), SAS (Side-Angle-Side; must be in this order), ASA (Angle-Side-Angle),and HL (Hypotenuse-Leg). AAS is an acceptable variation of ASA.The four Triangle Congruence Properties are the only correspondences that may be used toprove that two triangles are congruent. Once two triangles are know to be congruent, then theother pairs of their corresponding parts are congruent. In this course, you may justify thisconclusion with the statement "congruent triangles give us congruent parts." Remember: onlyuse this statement after you have shown the two triangles are congruent.CAG Example 1T DOIn 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 2In the figure at right, RD CP, D P, andED AP. Thus RED CAP because of SAS.Now RE CA, R C, and E A.CAPREDX BCOARExample 3In the figure at left, O A, OX AR,and X R. Thus, BOX CAR because ofASA. Now B C, BO CA, andBX CR.494 SKILL BUILDERS 33. Example 4In the figure at right, there are two right angles(mN = mY = 90) and MO KE, and ON EY.Thus, MON KEY because of HL, so MN KY,M K, and O E.MOEN YKExample 5Refer to page 196 in the textbook (problem CG-41) and do each part, (a) through (i). Briefexplanations and the answers appear below.a) PQR ZXY by SSS b) BAE DEC (vertical angles),BAE DEC by SAS .c) PR PR , QPR SPR by SSS. d) These triangles are not necessarilycongruent because we dont know therelative sizes of the sides. Rememberthat AAA is not a congruence property.e) E H because the three angles in anytriangle must add up to 180.DEF GHJ by SAS.f) YXW LKJ by SAS.g) Not enough information; pairs ofcorresponding parts.h) QS QS, RSQ PQS by ASA.i) Not necessarily congruent since the one side we know is not between the two angles ofthe second triangle.Briefly explain if each of the following pairs of triangles are congruent or not. If so, state the TriangleCongruence 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 RGeometry 495 34. 10.T UXW V11.ABCD1312512.EG3HI5413.RP JQ11239K11239L14.BC12A 4128DE4 F815.WXYZAnswers1. ABC DEF by ASA 2. GIH LJK by SAS 3. PNM PNO by SSS4. QS QS, soQRS QTS by HL5. The triangles are notnecessarily congruent.6. ABC DFE by ASAor AAS.7. GI GI, soGHI IJG by SSS.8. Alternate interior angles =used twice, soKLN NMK by ASA.9. Vertical angles = at 0, soPOQ ROS by SAS.10. Vertical angles and/oralternate interior angles =,so TUX VWX byASA.11. No, the length of eachhypotenuse is different.12. Pythagorean Theorem, soEGH IHG by SSS.13. Sum of angles of triangle= 180, but since the equalangles do not correspond,the triangles are notconguent.14. AF + FC = FC + CD, soABC DEF by SSS.15. XZ XZ, soWXZ YXZ by AAS.496 SKILL BUILDERS 35. LAWS OF EXPONENTS #12BASE, EXPONENT, AND VALUEIn the expression 25 , 2 is the base, 5 is the exponent, and the value is 32.25 means 2 2 2 2 2 = 32 x3 means x x xLAWS OF EXPONENTSHere are the basic patterns with examples:1) xa . xb = xa + b examples: x3 . x4 = x3 + 4 = x7; 27 24 = 2112)xaxb = xa - b examples: x10 x4 = x10 - 4 = x6;2427= 2-33) (xa)b = xab examples: (x4)3 = x4 3 = x12; (2x3)4 = 24x12 = 16x124) xa =1xa and 1xb = xb examples: 3x3y2 =3y2x3 ; 2x5y2 = 2x5y2Example 1Simplify: (2xy3 )(5x2y4 )Multiply the coefficients: 2 5 xy3 x2y4 = 10xy3 x2y4Add the exponents of x, then y: 10x1+2y3+4 = 10x3y7Example 2Simplify: 14x2y127x5y7Divide the coefficients: (147)x2y12x5y7 =2x2y12x5y7Subtract the exponents: 2x25y127 = 2x3y5 OR 2y5x3Example 3Simplify: (3x2y4 )3Cube each factor: 33 (x2 )3 (y4 )3= 27(x2 )3(y4 )3Multiply the exponents: 27x6y12Geometry 497 36. Simplify each expression:1. y5 y7 2. b4 b3 b2 3. 86 82 4. (y5 )2m3 7. 12x94x4 8. (x3y2 )35. (3a)4 6. m89.(y4 )2(y3 )2 10. 15x2y73x4y5 11. (4c4 )(ac3 )(3a5c) 12. (7x3y5 )2x2 ( )313. (4xy2 )(2y)3 14. 415.(2a7 )(3a2 )16. 5m3n6a3 m5 317. (3a2x3 )2(2ax4 )318. x3yy4419. 6y2x812x3y7 220.(2x5y3 )3(4xy4 )28x7y12Answers1. y12 2. b9 3. 88 4. y105. 81a4 6. m5 7. 3x5 8. x9y69. y2 10. 5y2x2 11. 12a6c8 12. 49x6y10x6 15. a6 16. 125n313. 32xy5 14. 64m617. 72a7x18 18. x12y12 19. x104y10 20. 16x10y5498 SKILL BUILDERS 37. PROOF #13A proof convinces an audience that a conjecture is true for ALL cases (situations) that fit theconditions of the conjecture. For example, "If a polygon is a triangle on a flat surface, thenthe sum of the measures of the angles is 180." Because we proved this conjecture near theend of Unit 3, it is always true. There are many formats that 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 proof__To prove that ABC is isosceles, show that BA__ BC. We can do thisby showing that the two segments are corresponding parts of congruenttriangles.____Since BDis perpendicular to AC, mBDA = mBDC = 90. ABD C__Since BD__bisects AC__, AD__ CD__. With BD__ BD(reflexive property), ADB CDB by SAS.__Finally, BA__ BCbecause 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 ACAD CDADB BDCBD BD bisector reflexiveCBD ABDSAS BA BCABC is isosceless have partsDefinition of isoscelesTwo-Column ProofGiven: BD is a bisector of AC.BD is perpendicular to AC.Prove: ABC is isoscelesStatement R easonBD 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 499 38. In each diagram below, are any triangles congruent? If so, prove it. (Note: It is good practice to trydifferent methods for writing your proofs.)1.ABD C2.BACDE3.ABC D4.D 5.ACBC D 6.B ABACEDFComplete 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 B500 SKILL BUILDERS 39. 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 HKTYU GH17. 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 BADBCD BDC BDABD BDReflexive ABD CBDGivenAASright 's are 2. Yes B E BCA BCDBC CE ABC DECGivenASAvertical s are GivenGeometry 501 40. 3. Yes BCD BCABC BCReflexive ABC DBCAC CDGiven 4. Yesright 's are =SASBA CDCA CA GivenReflexive ABC CDAAD BCGivenSSS5. Not necessarily. Counterexample: 6. Yes AC FDBC EFGiven Given ABC DEFHL7. NPMP and TPRP by definition of bisector. NPTMPR because vertical angles are equal.So, NTP MRP by SAS.8. ACD BCD by definition of angle bisector. CDCD by reflexive so CDA CDB by ASA.9. AC since alternate interior angles of parallel lines congruent so ABF CED by ASA.10. TGTG by reflexive so TPG TSG by SSS.11. MEOPEO because perpendicular lines form right angles MOEPOE by angle bisectorand OEOE by reflexive. So, MOE POE by ASA.12. CDBABD and ADBCBD since parallel lines give congruent alternate interior angles.DBDB by reflexive so ADB CBD by ASA.13. DBEB by definition of bisector. DBAEBC since vertical angles are congruent. SoADB CEB by AAS.14. RQPSQP since perpendicular lines form congruent right angles. PQPQ by reflexive soPQR PQS by AAS.15. SQPRQP by angle bisector and PQPQ by reflexive, so SPQ RPQ by AAS.16. KYTHUG because parallel lines form congruent alternate exterior angles. TY+YU=YU+GUso TYGU by subtraction. TG since perpendicular lines form congruent right angles. SoKTY HGU by ASA. Therefore, KH since triangles have congruent parts.17. MQLWLQ since parallel lines form congruent alternate interior angles. QLQL by reflexiveso MQL WLQ by SAS so WQLMLQ since congruent triangles have congruent parts. SoML||WQ since congruent alternate interior angles are formed by parallel lines.18. YesDB CE BDC ECDDC DCReflexive BCD EDCSAS19. Not necessarily.20. Not necessarily.502 SKILL BUILDERS 41. RADICALS #14Sometimes it is convenient to leave square roots in radical form instead of using a calculator tofind approximations (decimal values). Look for perfect squares (i.e., 4, 9, 16, 25, 36,49, ...) as factors of the number that is inside the radical sign (radicand) and take the squareroot of any perfect square factor. Multiply the root of the perfect square times the reducedradical. When there is an existing value that multiplies the radical, multiply any root(s) timesthat value.For example:9 =3 59 = 5 3 =1518 = 9 2 = 9 2 = 3 2 3 98 = 3 49 2 = 3 7 2 = 21 280 = 16 5 = 16 5 = 4 5 45 +4 20= 9 5 +4 4 5 =3 5+ 4 2 5=11 5When there are no more perfect square factors inside the radical sign, the product of the wholenumber (or fraction) and the remaining radical is said to be in SIMPLE RADICAL FORM.Simple radical form does not allow radicals in the denominator of a fraction. If there is aradical in the denominator, RATIONALIZE THE DENOMINATOR by multiplying thenumerator and denominator of the fraction by the radical in the original denominator. Thensimplify the remaining fraction. Examples:22=2222=2 22= 24 56=4 5666=4 306=2 303In the first example, 2 2 = 4 = 2 and 22= 1. In the second example,6 6 = 36 = 6 and 46=23.Example 1Add 27 + 12 48 . Factor each radical and simplify.9 3 + 4 3 16 3 = 3 3 +2 3 4 3 = 1 3 or 3Example 2Simplify36. Multiply by66and simplify:3666=3 66=62.Geometry 503 42. Write each of the following radicals in simple radical form.1. 24 2. 48 3. 174. 31 5. 75 6. 507. 96 8. 243 9. 8 + 1810. 18 + 32 11. 27 12 12. 50 3213. 6 + 63 14. 44 + 99 15. 50 + 32 2716. 75 8 32 17.3318.52719.3520.5721. 4 5 105Answers1. 2 6 2. 4 3 3. 17 4. 31 5. 5 36. 5 2 7. 4 6 8. 9 3 9. 5 2 10. 7 211. 3 12. 2 13. 3 7 + 6 14. 5 11 15. 9 2 - 3 316. 5 3 6 2 17. 3 18.5 39 19.155 20.35721. 2 5504 SKILL BUILDERS 43. RIGHT TRIANGLE TRIGONOMETRYThe three basic trigonometric ratios for right triangles arethe sine (pronounced "sign"), cosine, and tangent. Eachone is used in separate situations, and the easiest way toremember which to use when is the mnemonic SOH-CAH-TOA. With reference to one of the acute angles ina right triangle, Sine uses the Opposite and theHypotenuse - SOH. The Cosine uses the Adjacent sideand the Hypotenuse - CAH, and the Tangent uses theOpposite side and the Adjacent side -TOA. In each case,the position of the angle determines which leg (side) isopposite or adjacent. Remember that opposite meansacross from and adjacent means next to.#15ABCtan A =opposite legadjacent leg =BCACsin A =opposite leghypotenuse =BCABcos A =adjacent leghypotenuse =ACABExample 1Use trigonometric ratios to find the lengths ofeach of the missing sides of the triangle below.y h4217 ftThe length of the adjacent side with respect tothe 42 angle is 17 ft. To find the length y, usethe tangent because y is the opposite side andwe 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 =17hh cos 42 = 17h =17cos 42 22.876 ftThe hypotenuse is approximately 22.9 feetlong.Example 2Use trigonometric ratios to find the size ofeach angle and the missing length in thetriangle below.v18 ft hu21 ftTo find mu, use the tangent ratio becauseyou know the opposite (18 ft) and theadjacent(21 ft) sides.tan u =1821mu = tan-11821 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 =18hh sin 40.6 = 18h =18sin 40.6 27.659 ftThe hypotenuse is approximately 27.7 feetlong.Geometry 505 44. 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.v7888Draw 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 are out. Margiemeasures the angle of the string with the ground with her clinometer and finds it to be 42.5.How high is Juanitos kite above the ground?18. Nells kite has a 350 foot string. When it is completely out, Ian measures the angle to be47.5. How far would Ian need to walk to be directly under the kite?19. Mayfield High Schools flagpole is 15 feet high. Using a clinometer, Tamara measured anangle of 11.3 to the top of the pole. Tamara is 62 inches tall. How far from the flagpole isTamara standing?506 SKILL BUILDERS 45. 20. Tamara took another sighting of the top of the flagpole from a different position. This timethe angle is 58.4. If everything else is the same, how far from the flagpole is Tamarastanding?Answers1. h = 15 sin 38 9.235 2. h = 8 sin 26 3.507 3. x = 23 cos 49 15.0894. x = 37 cos 41 27.924 5. y = 38 tan 15 10.182 6. y = 43 tan 55 61.41047.z =15sin 38 24.3648.z =18sin 52 22.84239.w =23cos 38 29.187410.w =15cos 38 19.035311.x =38tan 15 141.81812.x =91tan 29 164.16813.x = tan-157 35.537714.u = tan-179 37.87515.y = tan-11218 33.69016.v = tan-17888 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 ft19.15 ft11.3x ft62 inh15 feet = 180 inches, 180" 62" = 118" = hx 590.5 inches or 49.2 ft.20.58.4x ft62 inh 15 fth = 118", tan 58. 4 =118x,x tan 58. 4 = 118 , x =118tan 58.4x 72.59 inches or 6.05 ft.Geometry 507 46. RATIO OF SIMILARITY #16The RATIO OF SIMILARITY between any two similar figures is the ratio of any pair ofcorresponding sides. Simply stated, once it is determined that two figures are similar, all of theirpairs of corresponding sides have the same ratio.The ratio of similarity of figure Pto figure Q, written P : Q, is 35.4.57.5 and 1220 have the same ratio.a34.5P12~ Qb57.520Note that the ratio of similarity is always expressed in lowest possible terms. Also, the order ofthe statement, P : Q or Q : P, determines which order to state and use the ratios between pairs ofcorresponding sides.An Equation that sets one ratio equal to another ratio is called a proportion.An example is at right.35=1220ExampleFind 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.102022xBED C12 24ADEAB = CDCA==>22x = 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.1.121592.55351173.128644.8 7245.66116.415 1068508 SKILL BUILDERS 47. For each pair of similar figures below, state the ratio of similarity, then use it to find x.7.141628x8.243612x9.6x184510.x20182411.1645x12.1011121518xFor problems 13 through 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 5 feet, 9inches, can see the reflection of the top of a tree. He measures the mirror to be 24 feet fromthe 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. If thestudent is 6 ft tall, how tall is the statue?Answers1. 432. 513. 214. 2475. 616. 1587. 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. 16 2318. 11 253219. 34.5 ft. 20. 30 ft.Geometry 509 48. SIMILARITY OF LENGTH, AREA, AND VOLUMEThe relationships for similarity of length, area, and volume are developed inUnit 8 and summarized in problem S-68 on page 322. The basic relationshipsof the r:r2:r3 Theorem are stated below.Once you know two figures are similar with a ratio of similarityab, the followingproportions for the SMALL (sm) and LARGE (lg) figures (which areenlargements or reductions of each other) are true:sidesmaPsmaAsma2Vsma3====sidelgbPlgbAlgb2Vlgb3In each proportion above, the data from the smaller figure is written on top (inthe numerator) to help be consistent with correspondences. When workingwith area, the basic ratio of similarity is squared; for volume, it is cubed.NEVER square or cube the actual areas or volumes themselves.#17abExample 1The two rectangular prisms above are similar. Suppose the ratio of their vertical edges is 3:8. Use ther:r2:r3 Theorem to find the following without knowing the dimensions of the prisms.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 of the front faceof the small prism.d) The area of the front face of the large prism is 15 square units. Find the area of the front face ofthe 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 3282 =964; the ratio ofthe volumes is 3282 =27512. Use proportions to solve for the rest of the parts.(c) 38=P188P = 54P = 6.75 units8 ( )2d) 3=A15964=A1564A = 135A 2.11 sq. units8 ( )3e) 3=21V27512=21V27V = 10752V 398.22 cu. units510 SKILL BUILDERS 49. Solve each of the following problems.1. Two rectangular prisms are similar. The smaller, A, has a height of4 Bfour units while the larger, B, has a height of six units.a) 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?Axy6c) 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. If the volume ofprism 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 rectangle Bif 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 rectangle Bif 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 rectangle Bif 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 rectangle Bif 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 is theperimeter 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 if thevolume 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 if thevolume 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 if thevolume of prism A is 96 cubic units?12. Prism A and prism B are similar. The volume of prism A is 72 cubic units while the volumeof prism B is 1125 cubic units. What is the ratio of similarity between these two prisms?13. Prism A and prism B are similar. The volume of prism A is 27 cubic units while the volumeof prism B is approximately 512 cubic units. If the surface area of prism B is 128 squareunits, what is the surface area of prism A?Geometry 511 50. 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 the diametersof the two steel balls?Answers1. a) 64=32b) 46=23c) 1636or49,64216or1627d) 375 cu unit2. 15.36 u2 3. 103.5 u2 4. 145.8 u2 5. 600 u26. 977. 658. 30 u 9. 121.510. 5312 11. 591.6 12. 2513. 18 u214. 14915. 94916. 1317. 5418. 3419. 34372920. 23512 SKILL BUILDERS 51. INTERIOR AND EXTERIOR ANGLES OF POLYGONS #18The 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 =(n2)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 the previousexample 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 (=n2)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 16 sides .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, isosceles triangles weneed to find the area of one, and then multiply it by 7. (See page 349 foraddition figures and details that parallel this solution.) In order to find thearea of each triangle we need to start with the angles of each triangle.Each interior angle of the regular 7-gon measuresh12.5 2.5(7 2)1807=(5)1807=9007 128. 57. The angle in the triangle is half the size of theinterior angle, so m1 128.572 64. 29. Find the height of the triangle by using thetangent ratio: tan 1 =h2.5 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 513 52. Find the measures of the angles in each problem below.1. Find the sum of the interior angles in a7-gon.2. Find the sum of the interior angles in an8-gon.3. Find the size of each of the interior anglesin a regular 12-gon.4. Find the size of each of the interior anglesin a regular 15-gon.5. Find the size of each of the exterior anglesof a regular 17-gon.6. Find the size of each of the exterior anglesof 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-gonhave?12. Each exterior angle of a regular n-gon measures 13 13. How many sides does this n-gonhave?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. 156 5. 21.17656. 17.1429 7. x = 24 8. x = 30 9. x = 98.18 10. x = 31.3011. 22 sides 12. 27 sides 13. 15 sides 14. 25 sides 15. 110.1106 cm216. 259.8076 ft2 17. 695.2935 m2 18. 1508.0649 in2514 SKILL BUILDERS 53. AREAS BY DISSECTION #19DISSECTION PRINCIPLE: Every polygon can be dissected (or broken up) intotriangles which have no interior points in common. This principle is an example of theproblem solving strategy of subproblems. Finding simpler problems that you know howto solve will help you solve the larger problem.ExampleFind the area of the figure below.4245226233First fill in the missing lengths. The rightvertical side of the figure is a total of 8 units tall.Since the left side is also 8 units tall,8 4 2 = 2 units. The missing vertical lengthon the top middle cutout 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 the leftmiddle is 3 units.Enclose the entire figure in an 8 by 8 square andthen remove the area of the three rectangles thatare not part of the figure. 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 = 48ExampleFind the area of the figure below.355 422 2This figure consists of five familiar figures:a central square, 5 units by 5 units;three triangles, one on top with b = 5 andh = 3, one on the right with b = 4 and h = 3,and one on the bottom with b = 5 and h = 2;and a trapezoid with an upper base of 4, a lowerbase of 2 and a height of 2.The area is:square 55 = 25top triangle 12 5 3 = 7.5bottom triangle 12 5 2 = 5right triangle 12 3 4 = 6trapezoid (4 + 2) 22= 6total area = 49.5 u2Find the area of each of the following figures. Assume that anything that looks like a right angleis a right angle.1.64232322.3332 23 33. 7559Geometry 515 54. 4.4321 185.224 32113456.21 210322227.44588.3213 2439.44225610. 11 and 12: Find the surface area of each polyhedra.3314 44101523234 3610Find the area and perimeter of each of the following figures.13.3 523414.34 25615.455645516.3135 1353 260422Answers1. 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. 899u516 SKILL BUILDERS 55. CENTRAL AND INSCRIBED ANGLESA central angle is an angle whose vertex is the center of a circle andwhose sides intersect the circle. The degree measure of a central angleis equal to the degree measure of its intercepted arc. For the circle atright with center C, ACB is a central angle.An INSCRIBED ANGLE is an angle with its vertex on the circle andwhose sides intersect the circle. The arc formed by the intersection ofthe two sides of the angle and the circle is called an INTERCEPTEDARC. ADB is an inscribed angle, AB is an intercepted arc.The INSCRIBED ANGLE THEOREM says that the measure of anyinscribed angle is half the measure of its intercepted arc. Likewise, anyintercepted arc is twice the measure of any inscribed angle whose sidespass through the endpoints of the arc.mADB =12 AB and AB = 2mADB#20CABABDExample 1In P, CPQ = 70. Find CQ and mCRQ.R PCQmCQ = m CPQ = 70mCRQ = 360 CQ = 360 70 = 290Example 2In the circle shown below, the vertex of theangles is at the center. Find x.BAx2xCSince AB is the diameter of the circle,x + 2x = 180 3x = 180 x = 60Example 3Solve for x.A100xBC120Since mBAC = 220, mBC = 360 220 = 140. ThemBAC equals half the mBC or 70.Geometry 517 56. Find each measure in P if mWPX = 28, mZPY = 38, and WZ and XV are diameters.1. YZ2. WX3. VPZ4. VWX5. XPY6. XY7. XWY8. WZXVZPYWXIn each of the following figures, O is the center of the circle. Calculate the values of x and justifyyour answer.9.xO13610.xO14611.xO4912.xO6213.xO10014.xO15010015.18 Ox16. Ox 2717.55x O18.12 Ox9119. O 20.35 xx29 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. 50518 SKILL BUILDERS 57. AREA OF SECTORSSECTORS of a circle are formed by the two radii of a central angleand the arc between their endpoints on the circle. For example, a 60sector looks like a slice of pizza. See page 376, problem CS-16 formore details and examples.#2160CExample 1CAB454'Find the area of the 45 sector. First find the fractional part of the circleinvolved. mAB360 =45360=18of the circle's area. Next find the area ofthe circle: A = 42 = 16 sq. ft. Finally, multiply the two results tofind the area of the sector.1 16 = 2 6. 28 sq. 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 =32012=803 83. 76 sq. ft.1. Find the area of the shaded sector in each circle below. Points A, B and C are the centers.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.O10186. The shaded region in the figure is called a segment of the circle. It canbe found by subtracting the area of MIL from the sector MIL. Findthe area of the segment of the circle.MIL2Geometry 519 58. Find the area of the shaded regions.7. IH5N C8.29. YARD is a square; A and Dare the centers of the arcs.Y AD 4 R10. 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 measure of itscentral 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 area is 12 cm2.120 rAnswers1. a) 2 un2 b) 493 un2 c) 3634un22. 23. 12 4. 931365. 56. 2 sq. un. 7. 1003 25 3 un2 8. 10 20 sq. un. 9. 8 16 sq. un.10. 225 ft2 11. 8 ft2 12. 2 15 m 13. 14414. 100 253 m3 15. 196 49 ft2 16. 10 in2 17. 48 + 32 un218. 652 un2 19. 61.8 un2 20. 6 un.520 SKILL BUILDERS 59. TANGENTS, SECANTS, AND CHORDS #22The figure at right shows a circle with three lines lyingaon a flat surface. Line a does not intersect the circle atall. Line b intersects the circle in two points and isbcalled a SECANT. Line c intersects the circle in onlyone point and is called a TANGENT to the circle.cTANGENT/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 diameter areparallel to each other.A CHORD of a circle is a line segment with its endpoints on thecircle.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.AFDEB CThe diameter is perpendicular to the chord,therefore it bisects the chord, so EF = 4.AB is a radius and AB = 5. EB is a radius,so EB = 5. Use the Pythagorean Theorem tofind BF: BF2 + 42 = 52, BF = 3.Geometry 521 60. In each circle, C is the center and AB is tangent to the circle at point B. Find the area of each circle.1.AC = 30 ABC252.CBA12AC = 453.BCA306 34. BCA18455.AC = 16BCA126.AC = 86 ABC567. B28 8. BCAAC = 90CA24DAD =189. In the figure at right, point E is the center and mCED = 55.What is the area of the circle?AC5E D5BIn 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 = 6AEBCDF14. In R, if AB = 2x 7and CD = 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 = 5 andTB = 2, find AT.DAT B52522 SKILL BUILDERS 61. 17. In J, radius JL andchord MN have lengthsof 10 cm. Find thedistance from J to MN.JL1010 NM18. In Q, QC = 13 andQT = 5. Find AB.C13OA5 TB19. 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 at point B,are AC and GI parallel? Prove your answer.ABCGEDHIFind the value of x.21.x 22.8040x25148P23.43x4124.1520xIn F, mAB = 84, mBC = 38, mCD = 64, and mDE = 60. Find the measure of each angleand arc.A B25. mEA26. mAEB142C27. m128. m2F329. m330. m4E DFor each circle, tangent segments are shown. Use the measurements given find the value of x.31. AE61xF C14132.P Q70xR S T 11033. D G B58H Ix34.x3835.160 35x36.(4x + 5) 50 xGeometry 523 62. Find each value of x. Tangent segments are shown in problems 40, 43, 46, and 48.37.x94338.1142x39.x54340. x102041.3xx6242.1 xx843.1.00.8x44.x55x45.6x102x46.54x47.6123x48.3x510Answers1. 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. 14.4 11. 11.6 12. 7.513. 3.75 14. 5 15. 3116. 4 17. 5 3 cm. 18. 5 319. Yes, GEH AEB (reflexive). EB is perpendicular to AC since it is tangent soGHE ABE because all right angles are congruent. So the triangles are similar by AA~.20. Yes. Since EH bisects GI it is also perpendicular to it (SSS). Since AC is a tangent, ABE is aright angle. So the lines are parallel since the corresponding angles are right angles and all rightangles are equal.21. 160 22. 9 23. 42 24. 70 25. 11426. 276 27. 87 28. 49 29. 131 30. 3831. 40 32. 55 33. 64 34. 38 35. 4536. 22.5 37. 12 38. 5 1239. 2 40. 3041. 2 42. 2 2 43. 1.2 44. 5 45. 3046. 6 47. 7.5 48. 5524 SKILL BUILDERS 63. VOLUME AND SURFACE AREA OF POLYHEDRA #23The VOLUME of various polyhedra, that is, the number of cubic units needed to filleach 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. Since thebases of cylinders and cones are circles, their area formulas may be expressed as:cylinder V = r2h and cone V =13r2h .The 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)1215814a) 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) = 200 628.32 cubic 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.3This figure is for Example 2, on next page: 58Geometry 525 64. Example 2Find the surface area of the triangular prismshown on the bottom of the previous page.The figure is made up of two triangles (thetop and bottom) and three rectangles asshown at right. Find the area of each ofthese shapes. To find the area of the triangleand the last rectangle, use the PythagoreanTheorem to find the length of the second legof the right triangular base. Since32 + leg2 = 52, leg = 4.23?3 5 + 8 +858 +?Calculate all of the areas, and find their sum.SA = 2 12( (3)(4)) + 3(8) + 5(8) + 4(8)= 12 + 24 + 40 + 32 = 108 square 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 = 4 108+ 88= 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 ft4.3 in 122 in5.12.1 m6.3 m6.2 m5 m7.2 ft4 ft9 ft8.11 cm11 cm9.8 cm4 cm 2 3cm526 SKILL BUILDERS 65. 10.8 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 remaining solidafter a hole with a diameter of 4 mm isdrilled 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 19Answers1. 48 m3 2. 540 cm3 3. 14966.6 ft3 4. 76.9 m3 5. 1508.75 m36. 157 m3 7. 72 ft3 8. 1045.4 cm3 9. 332.6 cm3 10. 320 in311. 314.2 in3 12. 609.7 cm3 13. 2.5 m3 14. 512 m3 15. 514.4 m316. 52.3 cm3 17. 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 ft2 27. 121 + 189.97 569.91 cm228. 192 + 48 3 275.14 cm2 29. 213.21 in2 30. 576 in2 31. 193.0 in232. 2394.69 ft2Geometry 527 66. SIMPLIFYING RATIONAL EXPRESSIONS #24RATIONAL EXPRESSIONS are fractions that have algebraic expressions in their numeratorsand/or denominators. To simplify rational expressions find factors in the numerator anddenominator that are the same and then write them as fractions equal to 1. For example,66= 1x2x2 = 1(x + 2)(x + 2) = 1(3x - 2)(3x - 2) = 1Notice that the last two examples involved binomial sums and differences. Only when sums ordifferences are exactly the same does the fraction equal 1. Rational expressions such as theexamples below CANNOT be simplified:(6 + 5)6x3 + yx3xx + 23x - 22Most problems that involve rational expressions will require that you factor the numerator anddenominator. For example:1254 =2232333 =29 Notice that22 and33 each equal 1.6x3y215x2y4 =23x2xy253x2 y2y2 =2x5y2 Notice that33,x2x2, andy2y2 = 1.x2 x 6x2 5x + 6=(x + 2)(x 3)(x 2)(x 3)=x + 2x 2wherex 3x 3= 1.All three examples demonstrate that all parts of the numerator and denominator--whetherconstants, monomials, binomials, or factorable trinomials--must be written as products beforeyou can look for factors that equal 1.One special situation is shown in the following examples:22= 1 xx= 1 x 2x + 2= (x + 2)x + 2= 1 5 xx 5= (x 5)x 5= 1Note that in all cases we assume the denominator does not equal zero.Example 1Simplify 12(x 1)3 (x + 2)3(x 1)2 (x + 2)2 completely.Factor: 43(x1)2 (x1)(x+2)3(x1)2 (x+2)(x+2)Note: 33, (x1)2, and x+2(x1)2 x+2= 1, sothe simplified form is 4(x1)(x+2), x 1 or -2.Example 2Simplify x2 6x + 8x2 + 4x 12completely.Factor: x2 6x + 8x2 + 4x 12=(x 4)(x 2)(x + 6)(x 2)Since x2x2= 1, x4x+6is the simplified form,x -6 or 2.528 SKILL BUILDERS 67. Simplify each of the following rational expression completely. Assume the denominator is notequal to zero.1. 2(x + 3)4(x 2)2. 2(x 3)6(x + 2)3. 2(x + 3)(x 2)6(x 2)(x + 2)4. 4(x 3)(x 5)6(x 3)(x + 2)5. 3(x 3)(4 x)15(x + 3)(x 4)6. 15(x 1)(7 x)25(x + 1)(x 7)7. 24(y 4)(y 6)16(y + 6)(6 y)8. 36(y + 4)(y 16)32(y + 16)(16 y)9. (x + 3)2 (x 2)4(x + 3)4 (x 2)310. (x + 3)4 (x 2)5(x + 3)7 (x 2)11. (x + 5)4 (x 3)2(x + 5)5(x 3)312. (2x 5)2 (x + 3)2(2x 5)5(x + 3)313. (5 x)2 (x 2)2(x + 5)4 (x 2)314. (5 x)4 (3x 1)2(x 5)4 (3x 2)315. 12(x 7)(x + 2)420(x 7)2 (x + 2)516. 24(3x 7)(x + 1)620(3x 7)3(x + 1)517. x2 1(x + 1)(x 2)18. x2 4(x + 1)2(x 2)19. x2 4x2 + x 620. x2 16x3 + 9x2 + 20xAnswers1. x + 32(x 2)2. x 33(x + 2)3. x + 33(x + 2)4. 2(x 5)3(x + 2)5.x 35(x + 3)6.3(x 1)5(x + 1)7.3(y 4)2(y + 6)8.9(y + 4)8(y + 16)9. x 2(x + 3)210. (x 2)4(x + 3)311. 1(x + 5)(x 3)12. 1(2x 5)3(x + 3)13. (5 x)2(x + 5)4(x 2)14. (3x 1)2(3x 2)315. 35(x 7)(x + 2)16. 6(x + 1)5(3x 7)217. x 1x 218. x + 2(x + 1)219. x + 2x + 320. x 4x(x + 5)Geometry 529 68. MULTIPLICATION AND DIVISION OF RATIONAL EXPRESSIONS #25To multiply or divide rational expressions, follow the same procedures used with numericalfractions. However, it is often necessary to factor the polynomials in order to simplify.Example 1Multiply x 2x( x+ )2x xx2+ +2667 61+and simplify the result.( + )After factoring, the expression becomes: x x6( x + )( x+ )( x + )( x+ )( x + )( x )6 66 11 1After multiplying, reorder the factors: (x + 6)(x + 6).(x + 6)(x + 6).x(x - 1).(x + 1)(x + 1)( + )( + )Since xx66( + )( + )= 1 and xx11= 1, simplify: 1 1 xx - 1 1 => xx - 1 .Note: x -6, -1, or 1.Example 222Divide x x4 54 4 +x x22x xx x2 154 12 + and simplify the result.First, change to a multiplication expression byinverting (flipping) the second fraction:22x xx x22x xx x4 54 44 122 15 ++ ( )( + )( )( )After factoring, the expression is: x x5 12 2x x( x + )( x )( x )( x+ )6 25 3( )( )Reorder the factors (if you need to): xx( x )( x )( x+ )( x )( x+ )( x+ )55221263( )( )Since xx55( )( )= 1 and xx22( + )( + )( )( + )= 1, simplify: x x1 62 3x xThus,22x xx x22x xx x4 54 42 154 12 + + ( + )( + )( )( + )= x x1 62 3x x22or x x7 6+ ++ x x6. Note: x -3, 2, or 5.Multiply or divide each pair of rational expressions. Simplify the result. Assume the denominator is notequal to zero.1.x2 + 5x + 6x2 4x4xx + 22.x2 2xx2 4x + 44x2x 23.x2 16(x 4)2 x2 3x 18x2 2x 244.x2 x 6x2 + 3x 10x2 + 2x 15x2 6x + 95.x2 x 6x2 x 20x2 + 6x + 8x2 x 66.x2 x 30x2 + 13x + 40x2 + 11x + 24x2 9x + 18530 SKILL BUILDERS 69. 7.15 5xx2 x 65xx2 + 6x + 88.17x + 119x2 + 5x 149x 1x2 3x + 29.2x2 5x 33x2 10x + 39x2 14x2 + 4x + 110.x2 1x2 6x 7x3+ x2 2xx 711.3x 21x2 493xx2 + 7x12.x2 y2x + y1x y13.y2 yw2 y2 y2 2y + 11 y14.y2 y 12y + 2y 4y2 4y 1215.x2 + 7x + 10x + 2x2 + 2x 15x + 2Answers1. 4(x+3)(x4)2. 14x3. (x+3)x44. (x + 2)(x 2)5. x+2x56. x+3x37. x4x8. 17(x1)9x19. 3x+12x+110. 1x(x + 2)11. 1 12. 1 13. yw2 y2 14. (y + 3)(y 6) 15. x+2x3Geometry 531 70. ADDITION AND SUBTRACTION OF RATIONAL EXPRESSIONS #26Addition and subtraction of rational expressions is done the same way as addition and subtractionof numerical fractions. Change to a common denominator (if necessary), combine thenumerators, and then simplify.ExampleThe Least Common Multiple (lowest commondenominator) of (x + 3)(x + 2) and (x + 2) is(x + 3)(x + 2).4(x+2)(x+3)+2xx+2The denominator of the first fraction already is theLeast Common Multiple. To get a commondenominator in the second fraction, multiply the fractionbyx + 3x + 3, a form of one (1).=4(x+2)(x+3)+2xx+2(x+3)(x+3)Multiply the numerator and denominator of the secondterm: =4(x+2)(x+3)+2x(x+3)(x+2)(x+3)Distribute in the second numerator.=4(x+2)(x+3)+2x2 +6x(x+2)(x+3)Add, factor, and simplify. Note: x -2 or -3.=2x2 +6x+4(x+2)(x+3)=2(x+1)(x+2)(x+2)(x+3)=2(x+1)(x+3)Simplify each of the following sums and differences. Assume the denominator does not equal zero.1. 7x3y2 +4y6x2 2. 189xy+72x23x2 3. 7y868y4. y 1 +1y15. xx+36xx2 96. 6x2 +4x+4+5x+27. 2a3a15+ 16a+203a2 12a 158. w+124w16w+42w89. 3x+1x2 163x+5x2 +8x+1610. 7x1x2 2x36xx2 x211. 3x1+41x+1x12. 3y9y2 4x2 13y+2x13. 2x+4x4x2 1614. 5x+9x2 2x3+6x2 7x+1215. x+4x2 3x28+x5x2 +2x35Answers1. 14x3 + 4y312x 2. + 21xy 4y6x2y2 6x2y3. 13y84. y2 2y+2y15. x(x9)(x+3)(x3)(x+2)2 7. 2(a2)6. 5x+163(a+1)8. 149. 4(5x+6)(x4)(x+4)2 10. x+2(x3)(x2)11. 1x(x1)12. 2x(3y+2x)(3y2x)13. 1x+414. 5(x+2)(x4)(x+1)15. 2x(x+7)(x7)532 SKILL BUILDERS 71. SOLVING MIXED EQUATIONS AND INEQUALITIES #27Solve these various types of equations.1. 2(x - 3) + 2 = -4 2. 6 - 12x = 108 3. 3x - 11 = 04. 0 = 2x - 5 5. y = 2x - 3x + y = 156. ax - b = 0(solve for x)7. 0 = (2x - 5)(x + 3) 8. 2(2x - 1) = -x + 5 9. x2 + 52 = 13210. 2x + 1 = 7x - 15 11. 5 - 2x3=x512. 2x - 3y + 9 = 0(solve for y)13. x2 + 5x + 6 = 0 14. x2 = y100 = y15. x - y = 7y = 2x - 116. x2 - 4x = 0 17. x2 - 6 = -2 18. x2+x3= 219. x2 + 7x + 9 = 3 20. y = x + 3x + 2y = 321. 3x2 + 7x + 2 = 022. xx + 1=5723. x2 + 2x - 4 = 0 24. 1x+13x= 225. 3x + y = 5x - y = 1126. y = -34x + 414x - y = 827. 3x2 = 8x28. x = 4 29. 23x + 1 =12x - 3 30. x2 - 4x = 531. 3x + 5y = 15(solve for y)32. (3x)2 + x2 = 152 33. y = 11y = 2x2 + 3x - 934. (x + 2)(x + 3 )(x - 4) = 0 35. x + 6 = 8 36. 2(x + 3) = y + 2y + 2 = 8x37. 2x + 3y = 13x - 2y = -1138. 2x2 = -x + 7 39. 1 - 56x=x640. x - 15=3x + 141. 2x + 1 = 5 42. 2 2x 1 + 3 = 743. 3x 1 + 1 = 7 44. (x + 3)2 = 49 45. 4x - 1x - 1= x + 1Geometry 533 72. Solve these various types of inequalities.46. 4x - 2 6 47. 4 - 3(x + 2) 19 48. x2> 3749. 3(x + 2) -9 50. - 23x < 6 51. y < 2x - 352. x > 4 53. x2 - 6x + 8 0 54. x + 3 > 555. 2x2 - 4x 0 56. y -23x + 2 57. y -x + 2y 3x - 658. 2x 1 9 59. 5 - 3(x - 1) -x + 2 60. y 4x + 16y > - 43 x - 4Answers1. 0 2. -8.5 3. 113 4. 525. (6, 9) 6. x = ba 7. 52, -3 8. 759. 12 10. 165 11. 2513 12. y = 23x + 313. -2, -3 14. (10, 100) 15. (-6, -13) 16. 0, 417. 2 18. 125 19. -1. -6 20. (-1, 2)21. -13, -2 22. 52 23. 2 202 24. 2325. (4, -7) 26. (12, -5) 27. 0, 83 28. 429. -24 30. 5, -1 31. y = -35x + 3 32. 4.742( ,11) 34. -2, -3, 4 35. 2, -14 36. (1, 6)33. (4,11) and 537. (-1, 5) 38. 1 57 39. 1, 5 40. 4441. 12 42. 32, - 12 43. 373 44. 4, -1045. 0, 4 46. x 2 47. x -7 48. x > 6749. x -5 50. x > -9 51. below 52. x>4, x2 or x< -8 55. x0 or x2 56. below57. below 58. -4 x 5 59. x 3 60. below534 SKILL BUILDERS 73. 51.y 56.xxy57.y = 3x - 6xyy = -x + 260.xyy = 4x + 16y = 3x - 6Geometry 535 74. 536 SKILL BUILDERS