Chapter 6. Worked-Out Solutions. Chapter 6 Maintaining Mathematical Proficiency (p. 299)

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1 hapter 6 hapter 6 Maintaining Mathematical Proficiency (p. 99) 1. Slope perpendicular to y = 1 x 5 is. y = x + b 1 = + b 1 = 9 + b 10 = b n equation of the line is y = x Slope perpendicular to y = x 5 is 1. y = x + b = b = 4 + b 7 = b n equation of the line is y = x 7.. Slope perpendicular to y = 4x + 1 is 1 4. y = 1 4 x + b = 4 1 ( 1) + b = b 4( ) = 4 ( 4) 1 + 4b 8 = 1 + 4b 7 = 4b 7 4 = b n equation of the line is y = 1 4 x w and w 8, or w 8 5. m > 0 and m < 11, or 0 < m < s 5 or s > 7. d < 1 or d 7 8. yes; s with Exercises 6 and 7, if the graphs of the two inequalities overlap going in opposite directions and the variable only has to make one or the other true, then every number on the number line makes the compound inequality true. hapter 6 Mathematical Practices (p. 00) 1. perpendicular bisector is perpendicular to a side of the triangle at its midpoint.. n angle bisector divides an angle of the triangle into two congruent adjacent angles.. median of a triangle is a segment from a vertex to the midpoint of the opposite side. 4. n altitude of a triangle is the perpendicular segment from a vertex to the opposite side or to the line that contains the opposite side. 5. midsegment of a triangle is a segment that connects the midpoints of two sides of the triangle. 6.1 Explorations (p. 01) 1. a. heck students work. b. heck students work. c. heck students work (for sample in text, 1.97, 1.97); For all locations of, and have the same measure. d. Every point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.. a. heck students work. b. heck students work. c. heck students work (for sample in text, E 1.4, F 1.4); For all locations of on the angle bisector, E and F have the same measure. d. Every point on an angle bisector is equidistant from both sides of the angle.. ny point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment. ny point on the angle bisector is equidistant from the sides of the angle. 4. The distance point is from is 5 units, which is the same as the distance is from. Point is in the angle bisector, so it is equidistant from either side of the angle. Therefore, E F. 6.1 Monitoring Progress (pp. 0 05) 1. WZ = YZ WZ = 1.75 y the Perpendicular isector Theorem, WZ = WZ = YZ 4n 1 = n + 17 n = 0 n = 10 YZ = n + 17 YZ = YZ = 7 y the Perpendicular isector Theorem, YZ = 7.. WX = 1 WY WX = 1 (14.8) WX = = 6.9 by the ngle isector Theorem. opyright ig Ideas Learning, LL Geometry 185 ll rights reserved.

2 hapter 6 5. = z + 7 = z + 11 z + 7 = 11 z = 4 = z + 11 = = = ecause =, is the angle bisector of and m = m. Therefore, m = (9) = no; In order to use the onverse of the ngle isector Theorem (Thm. 6.4), PS would have to be perpendicular to QP, and RS would have to be perpendicular to QR. 1 ( 5) 8. Slope: m = = ( 1) + 1 = 4 4 = 1 The slope of the perpendicular line is m = 1. midpoint = ( ( 1), ) ( =, 6 ) = (1, ) y = 1 x + b = 1 (1) + b = 1 + b = b So, an equation of the perpendicular bisector is y = x. 6.1 Exercises (pp ) Vocabulary and ore oncept heck 1. Point is in the interior of EF. If E and EF are congruent, then E is the bisector of EF.. The question that is different is: Is point collinear with X and Z? is not collinear with X and Z. ecause the two segments containing points X and Z are congruent, is the same distance from both X and Z, point is equidistant from X and Z, and point is in the perpendicular bisector of XZ. Monitoring Progress and Modeling with Mathematics. GH = 4.6; ecause GK = KJ and HK GJ, point H is on the perpendicular bisector of GJ. So, by the Perpendicular isector Theorem (Thm. 6.1), GH = HJ = QR = 1.; ecause point T is equidistant from Q and S, point T is on the perpendicular bisector of QS by the onverse of the Perpendicular isector Theorem (Thm. 6.). So, by definition of segment bisector, QR = RS = = 15; ecause and point is equidistant from and, point is on the perpendicular bisector of by the onverse of the Perpendicular isector Theorem (Thm. 6.). y definition of segment bisector, =. = 5x = 4x + x = = 5 = UW = 55; ecause V W and UX VW, point U is on the perpendicular bisector of VW. So, by the Perpendicular isector Theorem (Thm. 6.1), VU = WU. VU = UW 9x + 1 = 7x + 1 x = 1 x = 6 UW = = yes; ecause point N is equidistant from L and M, point N is on the perpendicular bisector of LM by the onverse of the Perpendicular isector Theorem (Thm. 6.). ecause only one line can be perpendicular to LM at point K, NK must be the perpendicular bisector of LM, and P is on NK. 8. no; You would need to know that either LN = MN or LP = MP. 9. no; You would need to know that PN ML. 10. yes; ecause point P is equidistant from L and M, point P is on the perpendicular bisector of LM by the onverse of the Perpendicular isector Theorem (Thm. 6.). lso, LN MN, so PN is a bisector of LM. ecause P can only be on one of the bisectors, PN is the perpendicular bisector of LM. 11. ecause is equidistant from and, bisects by the onverse of the ngle isector Theorem (Thm. 6.4). So, m = m = QS is an angle bisector of PQR, PS QP, and SR QR. So, by the ngle isector Theorem (Thm. 6.), PS = RS = JL bisects KJM. ngle isector Theorem (Thm. 6.) m KJL = m MJK efinition of angle bisector 7x = x x = 16 x = 4 m KJL = 7x = 7 4 = EG bisects FEH. ngle isector Theorem (Thm. 6.) FG = GH onverse of the ngle isector Theorem (Thm. 6.4) x + 11 = x + 1 x = 10 x = 5 FG = = yes; ecause H is equidistant from EF and EG, EH bisects FEG by the ngle isector Theorem (Thm. 6.). 16. no; ongruent segments connect H to both EF and EG, but unless those segments are also perpendicular to EF and EG, you cannot conclude that H is equidistant from EF and EG. 186 Geometry opyright ig Ideas Learning, LL ll rights reserved.

3 hapter no; ecause neither nor are marked as perpendicular to or respectively, you cannot conclude that =. 18. yes; is on the angle bisector of, and. So, = by the ngle isector Theorem (Thm. 6.). 19. Slope of MN : m = = 6 6 = 1 The slope of the perpendicular line is m = 1. midpoint = ( 1 + 7, 5 + ( 1) ) ( = 8, 4 = (4, ) ) y = 1 x + b = 1 (4) + b = 4 + b = b n equation of the perpendicular bisector of MN is y = x. 0. Slope of QR 1 0 : m = 6 ( ) = 1 8 = The slope of the perpendicular line is m =. midpoint of QR = ( , ) ( = 4, 1 ) = (, 6) y = x + b 6 = () + b 6 = 4 + b 18 = 4 + b = b = b b = n equation of the perpendicular bisector of QR is y = x Slope of UV 8 4 : m = 9 ( ) = 4 1 = 1 The slope of the perpendicular line is m = 1 =. midpoint of UV = ( , ) ( = 6, 1 ) = (, 6) y = x + b 6 = + b 6 = 9 + b 15 = b n equation of the perpendicular bisector of UV is y = x Slope of YZ : m = 1 ( 7) 4 10 = 8 14 = 4 7 The slope of the perpendicular line is m = 7 4. midpoint of YZ 10 + ( 4) = (, ) ( = 6, 6 ) = (, ) y = 7 4 x + b = b = b 1 = 1 + 4b = 4b 4 = b n equation of the perpendicular bisector of YZ is y = 7 4 x 4.. ecause is not necessarily congruent to E, will not necessarily pass through point. The reasoning should be: ecause = E, and E, is the perpendicular bisector of E. 4. ecause P is not necessarily perpendicular to, you do not have sufficient evidence to say that P = P. The reasoning should be: y the ngle isector Theorem (Thm. 6.), point P is equidistant from and. 5. The Perpendicular isector Theorem (Thm. 6.1) will allow the conclusion. 6. a. The relationship between PG and P is that PG is the angle bisector of P. b. m P gets larger. overing the goal becomes more difficult if the goalie remains at the same distance from the puck on the perpendicular bisector. s the angle increases, the goalie is farther away from each side of the angle. 7. raw XY, using a radius that is greater than 1 the distance of XY. raw two arcs of equal radii, using X and Y as centers, so that the arcs intersect. raw a line through both intersections of the arcs. Set a compass at centimeters Z by its own scale or with a ruler. cm Retaining this setting, place the.9 cm.9 cm compass point on the midpoint of XY and mark the point on the perpendicular bisector as X.5 cm.5 cm Y point Z. The distance between point Z and XY is centimeters because of the compass setting and in this example, XZ and YZ are both equal to.9 centimeters. This construction demonstrates the Perpendicular isector Theorem (Thm. 6.1). opyright ig Ideas Learning, LL Geometry 187 ll rights reserved.

4 hapter 6 8. ecause every point on a compass arc is the same distance from one endpoint, and every point on the other compass arc with the same setting is the same distance from the other endpoint, the line connecting the points where these arcs intersect contains the points that are equidistant from both endpoints. You know from the onverse of the Perpendicular isector Theorem (Thm. 6.) that the set of points that are equidistant from both endpoints make up the perpendicular bisector of the given segment. 9. ; (x 9) = 45 x = 54 x = ; Slope of MN : m = = 0 8 = 0 The slope of the perpendicular line is undefined. midpoint of MN = ( 7 + ( 1) 5 + 5, ) ( = 6, 10 ) = (, 5) The equation of a line that has a slope that is undefined through the point (, 5) is x =. So, the point that lies on the perpendicular bisector is (, 9). 1. no; In isosceles triangles, for example, the ray that has an endpoint of the vertex and passes through the base (the opposite side of the vertex) is not only an angle bisector of the vertex, but also a perpendicular bisector of the base.. Given = Prove Point lies on the perpendicular bisector of. P Given isosceles, construct P such that point P is on and P. So, P and P are right angles by the definition of perpendicular lines, and P and P are right triangles. lso, because and P P by the Reflexive Property of ongruence (Thm..1), P P by the HL ongruence Theorem (Thm. 5.9). So, P P because corresponding parts of congruent triangles are congruent, which means that point P is the midpoint of, and P is the perpendicular bisector of.. a. Given bisects,, Prove = STTEMENTS RESONS 1. bisects. 1. Given.. efinition of angle bisector., 4. and are right angles.. Given 4. efinition of perpendicular lines Right ngles ongruence Theorem (Thm..) Reflexive Property of ongruence (Thm..1) S ongruence Theorem (Thm. 5.11) orresponding parts of congruent triangles are congruent. 9. = 9. efinition of congruent segments b. Given =,, Prove bisects. STTEMENTS 1. =,. and are right angles.. and are right triangles , RESONS 1. Given. efinition of perpendicular lines. efinition of a right triangle 4. efinition of congruent segments 5. Reflexive Property of ongruence (Thm..1) HL ongruence Theorem (Thm. 5.9) orresponding parts of congruent triangles are congruent. 8. bisects. 8. efinition of angle bisector 188 Geometry opyright ig Ideas Learning, LL ll rights reserved.

5 hapter 6 4. a. Roosevelt School; ecause the corner of Main and rd Street is exactly blocks of the same length from each hospital, and the two streets are perpendicular, rd Street is the perpendicular bisector of the segment that connects the two hospitals. ecause Roosevelt school is on rd Street, it is the same distance from both hospitals by the Perpendicular isector Theorem (Thm. 6.1). b. no; ecause the corner of Maple and nd Street is approximately the midpoint of the segment that connects Wilson School to Roosevelt School, and nd Street is perpendicular to Maple, nd Street is the perpendicular bisector of the segment connecting Wilson and Roosevelt Schools. y the contrapositive of the onverse of the Perpendicular isector Theorem (Thm. 6.), the Museum is not equidistant from the two schools because it is not on nd Street. 5. a. y = x b. y = x c. y = x 6. no; In spherical geometry, all intersecting lines meet in two points which are equidistant from each other because they are the two endpoints of a diameter of the circle. 7. Given and E E Prove E ecause and E E, by the onverse of the Perpendicular isector Theorem (Thm. 6.), both points and E are on the perpendicular bisector of. So, E is the perpendicular bisector of. So, if, then by the onverse of the Perpendicular isector Theorem (Thm. 6.), point is also on E. So, points, E, and are collinear. onversely, if points, E, and are collinear, then by the Perpendicular isector Theorem (Thm. 6.), point is also on the perpendicular bisector of. So,. 8. Given Plane P is a perpendicular bisector of XZ at point Y. Prove a. XW ZW b. XV ZV c. VZW VZW X Y V W P Z a. ecause YW is on plane P, and plane P is a perpendicular bisector of XZ at point Y, YW is a perpendicular bisector of XZ by definition of a plane perpendicular to a line. So, by the Perpendicular isector Theorem (Thm. 6.1), XW ZW. b. ecause YV is on plane P, and plane P is a perpendicular bisector of XZ at point Y, YV is a perpendicular bisector of XZ by definition of a plane perpendicular to a line. So, by the Perpendicular isector Theorem (Thm. 6.1), XV ZV. c. First, WV WV by the Reflexive Property of ongruence (Thm..1). Then, because XW ZW and XV ZV, WVX WVZ by the SSS ongruence Theorem (Thm. 5.8). So, VXW VZW because corresponding parts of congruent triangles are congruent. Maintaining Mathematical Proficiency 9. The triangle is isosceles because it has two congruent sides. 40. The triangle is scalene because no sides are congruent. 41. The triangle is equilateral because all sides are congruent. 4. The triangle is an acute triangle because all angles measure less than The triangle is a right triangle because one angle measures The triangle is obtuse because one angle measure is greater than Explorations (p. 09) 1. a c. Sample answer: a. The perpendicular bisectors of the sides of all intersect at one point. c. The circle passes through all three vertices of.. a c. Sample answer: E a. The angle bisectors all intersect at one point. c. distance.06; The circle passes through exactly one point of each side of.. The perpendicular bisectors of the sides of a triangle meet at a point that is the same distance from each vertex of the triangle. The angle bisectors of a triangle meet at a point that is the same distance from each side of the triangle. opyright ig Ideas Learning, LL Geometry 189 ll rights reserved.