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Chapter 6 ⋅
Chapter 6 Chapter 6 Maintaining Mathematical Proficiency (p. 303) 1. Slope perpendicular to y = 13 x — − 5 is −3. y = mx + b y = −3x + b 1 = −3 1 = −9 + b 10 = b An equation of the line is y = −3x + 10. ⋅ 3+b y = mx + b y=x+b −3 = 1 −3 = 4 + b −7 = b An equation of the line is y = x − 7. ⋅4 + b 1.a.Check students’ work. b. Check students’ work. c. Check students’ work (for sample in text, CA ≈ 1.97, d. Every point on the perpendicular bisector of a segment is — — CB ≈ 1.97); For all locations of C, CA and CB have the same measure. equidistant from the endpoints of the segment. 2.a.Check students’ work. 3. Slope perpendicular to y = −4x + 13 is — 14 . 5. A midsegment of a triangle is a segment that connects the midpoints of two sides of the triangle. 6.1 Explorations (p. 305) 2. Slope perpendicular to y = −x − 5 is 1. 4. An altitude of a triangle is the perpendicular segment from a vertex to the opposite side or to the line that contains the opposite side. y = mx + b 1 y = — x + b 4 1 −2 = — (−1) + b 4 1 −2 = − — + b 4 1 4(−2) = 4 −— + 4b 4 −8 = −1 + 4b ⋅ ( ) −7 = 4b −7 = b — 4 An equation of the line is y = — 14 x − — 74 . b. Check students’ work. c. Check students’ work (for sample in text, DE ≈ 1.24, DF ≈ 1.24); For all locations of D on the angle bisector, — and FD — ED have the same measure. d. Every point on an angle bisector is equidistant from both sides of the angle. 3. Any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment. Any point on the angle bisector is equidistant from the sides of the angle. is 5 units, which is the same 4. The distance point D is from AB . Point D is in the angle bisector, as the distance D is from AC so it is equidistant from either side of the angle. Therefore, — ≅ DF — DE . 6.1 Monitoring Progress (pp. 307–309) 4. w ≥ −3 and w ≤ 8, or −3 ≤ w ≤ 8 1. WZ = YZ WZ = 13.75 5. m > 0 and m < 11, or 0 < m < 11 By the Perpendicular Bisector Theorem, WZ = 13.75. 6. s ≤ 5 or s > 2 2. 7. d < 12 or d ≥ −7 8. yes; As 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. Chapter 6 Mathematical Thinking (p. 304) 1. A perpendicular bisector is perpendicular to a side of the triangle at its midpoint. WZ = YZ 4n − 13 = n + 17 3n = 30 n = 10 YZ = n + 17 YZ = 10 + 17 YZ = 27 By the Perpendicular Bisector Theorem, YZ = 27. 3. WX = — 12 WY 1 2. An angle bisector divides an angle of the triangle into two congruent adjacent angles. WX = — 2 (14.8) WX = 7.4 3. A median of a triangle is a segment from a vertex to the midpoint of the opposite side. 4. DA = 6.9 by the Angle Bisector Theorem. Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 189 Chapter 6 5. AD = CD 3z + 7 = 2z + 11 z + 7 = 11 z=4 CD = 2z + 11 = 2 4 + 11 = 8 + 11 = 19 ⋅ is the angle bisector of ∠ ABC and 6. Because AD = CD, BD m∠ ABC = 2m∠ CBD. Therefore, m∠ ABC = 2(39) = 78°. 7. no; In order to use the Converse of the Angle Bisector — Theorem (Thm. 6.4), PS would have to be perpendicular — to QP , and RS would have to be perpendicular to QR . −1 − (−5) −1 + 5 4 = — = — = 1 8. Slope: m = — 3 − (−1) 3+1 4 The slope of the perpendicular line is m = −1. ( −12+ 3 −5 +2 (−1) ) ( 22 −62 ) , = (1, −3) midpoint = — = — , — — y = mx + b ⋅ ⋅ (1) + b y = −1 x + b −3 = −1 −3 = −1 + b −2 = b So, an equation of the perpendicular bisector is y = −x − 2. 6.1 Exercises (pp. 310–312) Vocabulary and Core Concept Check — — — ⊥ VW ≅ WD and UX , point U 6. UW = 55; Because VD — is on the perpendicular bisector of VW . So, by the Perpendicular Bisector Theorem (Thm. 6.1), VU = WU. VU = UW 9x + 1 = 7x + 13 2x = 12 x=6 UW = 7 ⋅ 6 + 13 = 55 7. yes; Because point N is equidistant from L and M, point N is — on the perpendicular bisector of LM by the Converse of the Perpendicular Bisector Theorem (Thm. 6.2). Because only — must be one line can be perpendicular to LM at point K, NK — . the perpendicular bisector of LM , and P is on NK 8. no; You would need to know that either LN = MN or LP = MP. ⊥ ML . 9. no; You would need to know that PN 10. yes; Because point P is equidistant from L and M, point P is — on the perpendicular bisector of LM by the Converse of the — — Perpendicular Bisector Theorem (Thm. 6.2). Also, LN ≅ MN , — is a bisector of LM so PN . Because P can only be on one of — is the perpendicular bisector of LM the bisectors, PN . , BD 11. Because D is equidistant from BC and BA bisects ∠ ABC by the Converse of the Angle Bisector Theorem (Thm. 6.4). So, m∠ ABD = m∠ CBD = 20°. 1. Point C is in the interior of ∠DEF. If ∠DEC and ∠CEF are congruent, then EC is the bisector of ∠DEF. — — is an angle bisector of ∠ PQR, PS , and SR . 12. QS ⊥ QP ⊥ QR So, by the Angle Bisector Theorem (Thm. 6.3), PS = RS = 12. 2. The question that is different is: Is point B collinear with X and Z? B is not collinear with X and Z. Because the two segments containing points X and Z are congruent, B is the same distance from both X and Z, point B is equidistant from — X and Z, and point B is in the perpendicular bisector of XZ . m∠ KJL = m∠ MJK 7x = 3x + 16 4x = 16 Monitoring Progress and Modeling with Mathematics x = 4 m∠ KJL = 7x = 7 , point H is on ⊥ GJ 3. GH = 4.6; Because GK = KJ and HK — the perpendicular bisector of GJ . So, by the Perpendicular Bisector Theorem (Thm. 6.1), GH = HJ = 4.6. 4. QR = 1.3; Because point T is equidistant from Q and S, — point T is on the perpendicular bisector of QS by the Converse of the Perpendicular Bisector Theorem (Thm. 6.2). So, by definition of segment bisector, QR = RS = 1.3. and point D is equidistant from ⊥ AC 5. AB = 15; Because DB A and C, point D is on the perpendicular bisector of AC — by the Converse of the Perpendicular Bisector Theorem (Thm. 6.2). By definition of segment bisector, AB = BC. AB = BC 5x = 4x + 3 x=3 AB = 5 3 = 15 ⋅ 190 Geometry Worked-Out Solutions bisects ∠ KJM. 13.JL Angle Bisector Theorem (Thm. 6.3) Definition of angle bisector ⋅ 4 = 28° 14. EG bisects ∠ FEH. Angle Bisector Theorem (Thm. 6.3) FG = GHConverse of the Angle Bisector Theorem (Thm. 6.4) x + 11 = 3x + 1 −2x = −10 x=5 FG = 5 + 11 = 16 , EH and EG bisects 15. yes; Because H is equidistant from EF ∠ FEG by the Angle Bisector Theorem (Thm. 6.3). , but and EG 16. no; Congruent segments connect H to both EF , and EG unless those segments are also perpendicular to EF . and EG you cannot conclude that H is equidistant from EF Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 — — 17. no; Because neither BD nor DC are marked as perpendicular to AB or AC respectively, you cannot conclude that DB = DC. — and 18. yes; D is on the angle bisector of ∠ BAC, DB ⊥ AB — . So, DB = DC by the Angle Bisector Theorem ⊥ AC CD (Thm. 6.3). 4 8 1 − (−7) — = — = − — 22. Slope of YZ : m = — 7 −4 − 10 −14 The slope of the perpendicular line is m = — 74 . — ( 10 +2(−4) −72+ 1) ( 62 −62 ) , = (3, −3) midpoint of YZ = — = — , — — y = mx + b 7 y = — x + b 4 7 −3 = — 3 + b 4 21 −3 = — + b 4 −12 = 21 + 4b y=1 x+b −33 = 4b 2 = 1 (4) + b 2=4+b −2 = b 33 = b − — 4 — An equation of the perpendicular bisector of YZ is 7 33 y = — 4 x − — 4 . −1 − 5 −6 — 19. Slope of MN = — = −1 : m = — 7−1 6 The slope of the perpendicular line is m = 1. 1 + 7 5 + (−1) 8 4 = (4, 2) midpoint = — , — = — , — 2 2 2 2 y = mx + b ( ) ( ) ⋅ ⋅ — An equation of the perpendicular bisector of MN is y = x − 2. 12 3 12 − 0 — = — = — 20. Slope of QR : m = — 6 − (−2) 8 2 2 The slope of the perpendicular line is m = − — 3 . — ( −22+ 6 0 +2 12 ) ( 24 122 ) , = (2, 6) midpoint of QR = — = — , — — 22 = 3b — = b y = mx + b 2 y = − — x + b 3 2 6 = − — (2) + b 3 4 6 = − — + b 3 18 = −4 + 3b ⋅ — , you 24. Because BP is not necessarily perpendicular to CB do not have sufficient evidence to say that BP = AP. The reasoning should be: By the Angle Bisector Theorem and CA (Thm. 6.3), point P is equidistant from CB . and ∠ APB is that PG is the 26.a.The relationship between PG angle bisector of ∠ APB. 22 b = — 3 — An equation of the perpendicular bisector of QR is 2 22 — y = − 3 x + — 3 . 4 1 8−4 — = — 21. Slope of UV : m = — = — 9 − (−3) 12 3 3 The slope of the perpendicular line is m = − — 1 = −3. −3 + 9 4 + 8 6 12 — , = — = (3, 6) midpoint of UV = — , — — y = mx + b y = −3x + b 6 = −3 6 = −9 + b 15 = b — — will is not necessarily congruent to EC , AB 23. Because DC not necessarily pass through point C. The reasoning ⊥ DE is the should be: Because AD = AE, and AB —, AB — perpendicular bisector of DE . 25. The Perpendicular Bisector Theorem (Thm. 6.1) will allow — ≅ CD — the conclusion AD . 22 3 ( ⋅ 2 2 ) ( 2 2 ) ⋅3 + b — An equation of the perpendicular bisector of UV is y = −3x + 15. Copyright © Big Ideas Learning, LLC All rights reserved. b.m∠ APB gets larger. Covering the goal becomes more difficult if the goalie remains at the same distance from the puck on the perpendicular bisector. As the angle increases, the goalie is farther away from each side of the angle. — 27. Draw XY , using a radius that is greater than — 12 the distance — of XY . Draw two arcs of equal radii, using X and Y as centers, so that the arcs intersect. Draw a line through both intersections of the arcs. Set a compass at 3 centimeters Z by its own scale or with a ruler. 3 cm Retaining this setting, place the 3.9 cm 3.9 cm compass point on the midpoint — of XY and mark the point on the perpendicular bisector as X 2.5 cm 2.5 cm Y point Z. The distance between — point Z and XY is 3 centimeters because of the compass setting and in this example, XZ and YZ are both equal to 3.9 centimeters. This construction demonstrates the Perpendicular Bisector Theorem (Thm. 6.1). Geometry Worked-Out Solutions 191 Chapter 6 28. Because 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 Converse of the Perpendicular Bisector Theorem (Thm. 6.2) that the set of points that are equidistant from both endpoints make up the perpendicular bisector of the given segment. 3x = 54 x = 18 bisects ∠ BAC. 1. AD 1. Given 2. Definition of angle bisector — — , DC 3. Given 3. DB ⊥ AB ⊥ AC 4. ∠ ABD and ∠ACD are 4. Definition of right angles. perpendicular lines 5. ∠ ABD ≅ ∠ACD −1 − 7 5. Right Angles Congruence Theorem (Thm. 2.3) 6. — AD ≅ — AD 0 5−5 — = — = 0 : m = — 30. B; Slope of MN −8 6. Reflexive Property of Congruence (Thm. 2.1) 7. △ADB ≅ △ADC The slope of the perpendicular line is undefined. , = — = (3, 5) midpoint of MN = — , — — The equation of a line that has a slope that is undefined — REASONS 2. ∠ BAD ≅ ∠ CAD 29. B; (3x − 9)° = 45° STATEMENTS ( 7 + 2(−1) 5 +2 5 ) ( 62 102 ) 7. AAS Congruence Theorem (Thm. 5.11) 8. — DB ≅ — DC 8. Corresponding parts of congruent triangles are congruent. through the point (3, 5) is x = 3. So, the point that lies on the perpendicular bisector is (3, 9). 31. 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. 32. Given CA = CB — Prove Point C lies on the perpendicular bisector of AB . 9. DB = DC 9. Definition of congruent segments B b.Given BD = CD, — , ⊥ AB DB — ⊥ AC DC bisects ∠ BAC. Prove AD D A C C STATEMENTS A B P such that point P is on Given isosceles △ACB, construct CP — and CP — ⊥ AB AB . So, ∠ CPB and ∠ CPA are right angles by the definition of perpendicular lines, and △CPB and △CPA — — — — are right triangles. Also, because AC ≅ BC and CP ≅ CP by the Reflexive Property of Congruence (Thm. 2.1), △CPB ≅ △CPA by the HL Congruence Theorem (Thm. 5.9). — — So, AP ≅ BP because corresponding parts of congruent triangles are congruent, which means that point P is the — midpoint of AB —, and CP is the perpendicular bisector of AB . — bisects ∠ BAC, DB , DC 33.a.Given AD ⊥ AB — ⊥ AC Prove DB = DC 1. Given 2. ∠ ABD and ∠ ACD are right angles. 2. Definition of perpendicular lines 3. △ ABD and △ACD are right triangles. 3. Definition of a right triangle 4. — BD ≅ — CD 4. Definition of congruent segments 5. — AD ≅ — AD 5. Reflexive Property of Congruence (Thm. 2.1) 6. △ ABD ≅ △ACD 6. HL Congruence Theorem (Thm. 5.9) 7. ∠ BAD ≅ ∠ CAD 7. Corresponding parts of congruent triangles are congruent. B D A bisects ∠ BAC. 8. AD C 192 Geometry Worked-Out Solutions REASONS , DB ⊥ AB 1. BD = CD, — — DC ⊥ AC 8. Definition of angle bisector Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 — — c.First, WV ≅ WV by the Reflexive Property of Congruence 34.a.Roosevelt School; Because the corner of Main and 3rd Street is exactly 2 blocks of the same length from each hospital, and the two streets are perpendicular, 3rd Street is the perpendicular bisector of the segment that connects the two hospitals. Because Roosevelt school is on 3rd Street, it is the same distance from both hospitals by the Perpendicular Bisector Theorem (Thm. 6.1). 39. The triangle is isosceles because it has two congruent sides. b. no; Because the corner of Maple and 2nd Street is approximately the midpoint of the segment that connects Wilson School to Roosevelt School, and 2nd Street is perpendicular to Maple, 2nd Street is the perpendicular bisector of the segment connecting Wilson and Roosevelt Schools. By the contrapositive of the Converse of the Perpendicular Bisector Theorem (Thm. 6.2), the Museum is not equidistant from the two schools because it is not on 2nd Street. 35.a.y = x b. y = −x c. y = x 36. 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. — ≅ CD — — ≅ CE — 37. Given AD and AE — — AB ≅ CB Prove — — — — (Thm. 2.1). Then, because XW ≅ ZW and XV ≅ ZV , △WVX ≅ △WVZ by the SSS Congruence Theorem (Thm. 5.8). So, ∠ VXW ≅ ∠ VZW because corresponding parts of congruent triangles are congruent. Maintaining Mathematical Proficiency 40. The triangle is scalene because no sides are congruent. 41. The triangle is equilateral because all sides are congruent. 42. The triangle is an acute triangle because all angles measure less than 90°. 43. The triangle is a right triangle because one angle measures 90°. 44. The triangle is obtuse because one angle measure is greater than 90°. 6.2 Explorations (p. 313) 1. a–c. Sample answer: 5 B 4 3 A 2 D −1 — — — — Perpendicular Bisector Theorem (Thm. 6.2), both points — is D and E are on the perpendicular bisector of AC . So, DE — — — the perpendicular bisector of AC . So, if AB ≅ CB , then by the Converse of the Perpendicular Bisector Theorem (Thm. 6.2), . So, points D, E, and B are collinear. point B is also on DE Conversely, if points D, E, and B are collinear, then by the Perpendicular Bisector Theorem (Thm. 6.2), point B is also — — — on the perpendicular bisector of AC . So, AB ≅ CB . 3 4 5 6 7 a.The perpendicular bisectors of the sides of △ABC all c.The circle passes through all three vertices of △ABC. intersect at one point. 2. a–c. Sample answer: 5 A E 4 B 3 D 2 1 X Y 0 V −2 P −1 0 1 2 3 4 5 6 7 −1 W C −2 −3 Z 2 −3 — 38. Given Plane P is a perpendicular bisector of XZ at point Y. — — — — XW ≅ ZW b. XV ≅ ZV c. ∠ VZW ≅ ∠ VZW Prove a. C 1 −2 Because AD ≅ CD and AE ≅ CE , by the Converse of the 0 −1 C A 0 −2 D 1 B E — — bisector of XZ — at point Y, YW is a perpendicular bisector of — XZ by definition of a plane perpendicular to a line. So, by — ≅ ZW — the Perpendicular Bisector Theorem (Thm. 6.1), XW . a.Because YW is on plane P, and plane P is a perpendicular — b.Because 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 — ≅ ZV — the Perpendicular Bisector Theorem (Thm. 6.1), XV . Copyright © Big Ideas Learning, LLC All rights reserved. a. The angle bisectors all intersect at one point. c.distance ≈ 2.06; The circle passes through exactly one point of each side of △ABC. 3. 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. Geometry Worked-Out Solutions 193 Chapter 6 6.2 Monitoring Progress (pp. 315–318) 4. 1. The pretzel distributor is located at point F, which is the circumcenter of △ABE. 3x + 8 = 7x + 2 QM = 3x + 8 = 3 B F A E S(−6, 5) R(−2, 5) 6 4 y=2 −6 x ( −2 +2 (−6) 5 +2 5 ) ( −82 102 ) −2 + (−2) 5 + (−1) — , midpoint of RT = ( ) = ( −42 , 24 ) = (−2, 2) 2 2 — , — = — = (−4, 5) midpoint of RS = — , — The equation of the perpendicular bisector of RS through — — — — — its midpoint (−4, 5) is x = −4, and the equation of the perpendicular bisector of RT — through its midpoint (−2, 2) is y = 2. The point of intersection of the two perpendicular bisectors is (−4, 2). So, the coordinates of the circumcenter of △RST is (−4, 2). 3. Graph △WXY. 3 2 9 2 3 2 21 2 By the Incenter Theorem, QM = QN = QP and −2 2 (0, −1) E 6.2 Exercises (pp. 319–322) Vocabulary and Core Concept Check 1. When three or more lines, rays, or segments intersect in the same point, they are called concurrent lines, rays, or segments. Monitoring Progress and Modeling with Mathematics 4 x 3. Because G is the circumcenter of △ABC, AG = BG = CG. Therefore, because AG = 9, BG = 9. y = −1 −4 4. Because G is the circumcenter of △ABC, AG = BG = CG. Therefore, because GC = 11, GA = 11. x=0 −6 L 2. The triangle that does not belong is the fourth triangle because it shows the incenter of the triangle. The other three show the circumcenter. y X(1, 4) W(−1, 4) −4 ⋅ — + 8 = — + 8 = 4.5 + 8 = 12.5 QN = 7x + 2 = 7 ⋅ ( — ) + 2 = — + 2 = 10.5 + 2 = 12.5 T(−2, −1) x = −4 x = — 46 = — 23 5. Draw two angle bisectors and label the intersection of the bisectors, the incenter, as L. Draw a perpendicular segment from the incenter L to any one side of the triangle. Label that point E. Draw a circle with center L and radius LE. It should touch all sides of the triangle. The location of the lamppost is at L. y (−4, 2) −4x = −6 QP = 12.5 units. 2. Graph △RST. QM = QN Y(1, −6) ( −12+ 1 4 +2 4) ( 20 28) — = 1 + , midpoint of XY ( 2 1 4 +2(−6) ) = ( 22 , −22 ) = (1, −1) — , = — = (0, 4) midpoint of WX = — , — — 5. Because P is the incenter of △XYZ, PA = PB = PC. Therefore, because PC = 9, PB = 9. — The equation of the perpendicular bisector of WX through its 6. Because P is the incenter of △XYZ, KP = HP = FP. Therefore, because KP = 15, HP = 15. — — — — midpoint (0, 4) is x = 0, and the equation of the perpendicular bisector of XY — through its midpoint (1, −1) is y = −1. The point of intersection of the two perpendicular bisectors is (0, −1). So, the coordinates of the circumcenter of △WXY are (0, −1). 194 Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 7. Graph △ABC. y 9. Graph △HJK. x=5 12 y C(8, 10) H(−10, 7) (5, 8) 4 4 A(2, 6) B(8, 6) 8 12 −12 x ( 2 +2 8 6 +2 6 ) ( 102 122 ) 8 + 8 6 + 10 16 16 — midpoint of BC = ( , = , = (8, 8) 2 2 ) (2 2) — = — = (5, 6) midpoint of AB = — , — , — The equation of the perpendicular bisector of AB through — — — — — its midpoint (5, 6) is x = 5, and the equation of the perpendicular bisector of BC — through its midpoint (8, 8) is y = 8. The point of intersection of the two perpendicular bisectors is (5, 8). So, the coordinates of the circumcenter of △ABC are (5, 8). 8. Graph △DEF. y 4 −4 x E(−1, −1) y = −5 F(−7, −9) −12 (−4, −5) ( 2 ) 2 −8 −2 = ( , ) = (−4, −1) 2 2 −7 + (−7) −1 + (−9) — , midpoint of DF = ( ) 2 2 −14 −10 = ( , = (−7, −5) 2 2 ) −7 + (−1) −1 + (−1) — , midpoint of DE = — — — — — — — — — The equation of the perpendicular bisector of DE through its midpoint (−4, −1) is x = −4, and the equation of the perpendicular bisector of DF — through its midpoint (−7, −5) is y = −5. The point of intersection of the two perpendicular bisectors is (−4, −5). So, the coordinates of the circumcenter of △DEF are (−4, −5). x −4 4 7−3 −10 − (−6) −4 — The slope of the line perpendicular to JH is m = 1. , — midpoint of JH = —— = −1 Slope of JH: m = —— = — — 7+3 ( −10 + (−6) 2 2 ) −16 10 = ( , ) = (−8, 5) 2 2 — — y = mx + b 5=1 5 = −8 + b 13 = b D(−7, −1) −8 4 J(−6, 3) x = −4 K(−2, 3) y=8 4 8 ⋅ (−8) + b — The equation of the line perpendicular to JH is y = x + 13. — 1 4 7−3 2 −10 − (−2) −8 — The slope of the line perpendicular to HK is m = 2. , — midpoint of HK = —— = − — Slope of HK : m = —— = — — — — y = mx + b ⋅ (−6) + b 5=2 5 = −12 + b 17 = b 7+3 ( −10 + (−2) 2 2 ) −12 10 = ( , ) = (−6, 5) 2 2 — The equation of the line perpendicular to HK is y = 2x + 17. — 3+3 ( −6 + (−2) 2 2 ) −8 6 = ( , ) = (−4, 3) 2 2 , — midpoint of JK = — The equation of the perpendicular bisector of JK through its — — — midpoint (−4, 3) is x = −4. The intersection of y = x + 13 and x = −4: The intersection of y = 2x + 17 and x = −4: So, the coordinates of the circumcenter of △HJK are (−4, 9). y = −4 + 13 = 9, the point of intersection is (−4, 9). y = 2(−4) + 17 = 9, the point of intersection is (−4, 9). Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 195 Chapter 6 10. Graph △LMN. y 2 −2 11. 4 6 8 x N(8, −6) M(5, −3) −4 −6 L(3, −6) — −3 − (−6) −3 + 6 3 5−3 2 2 2 — The slope of the line perpendicular to LM is m = − — 3 . = — = — Slope of LM : m = — ( 2 2 ) 8 9 = ( , − ) = (4, −4.5) 2 2 3 + 5 −6 + (−3) — midpoint of LM = — , — — — y = mx + b 2 9 − — = − — 4 + b 2 3 9 8 − — = − — + b 2 3 9 8 6 −— = 6 −— + 6 b 2 3 −27 = −16 + 6b ⋅ ( ) ( ) −11 = 6b 11 = b − — 6 2 — The equation of the line perpendicular to LM is y = − — 11 . 3 x − — 6 −6 − (−3) −6 + 3 −3 — = — = — = −1 Slope of MN : m = — 8−5 3 3 — The slope of the line perpendicular to MN is m = 1. 5 + 8 −3 + (−6) 13 9 — midpoint of MN = — , — = — , − — ( 2 y = mx + b 9 13 + b − — = — 2 2 9 13 2 −— = 2 — + 2b 2 2 −9 = 13 + 2b ) ( 2 2 ) 2 ( ) ( ) −11 = b — The equation of the line perpendicular to MN is y = x − 11. −6 − (−6) −6 + 6 0 — = — = — = 0 Slope of LN : m = — 8−3 5 5 3 + 8 −6 + (−6) — midpoint of LN = — , — 2 2 11 −12 11 = — = — , — , −6 2 2 2 — The equation of the line perpendicular to LN is x = — 11 . 2 2 11 11 The intersection of y = − — x − and x = : — — 3 6 2 ( ( ) ) ( ) — y = − — 2 − — 6 = − — 6 = − — 3 — 6 − — 6 = − 2 , the point of intersection is — 11 , − — 2 . So, the coordinates of the 2 ( ) 11 ( 22 11 ) 33 11 ( 11 11 ) circumcenter of △LMN are — 2 , − — 2 . 196 Geometry Worked-Out Solutions 11 3x = 9 x = — 93 = 3 ⋅ 3 − 2 = 18 − 2 = 16 ⋅ 3 + 7 = 16 ND = 6x − 2 = 6 NE = 3x + 7 = 3 By the Incenter Theorem, ND = NE = NF and NF = 16 units. 12. NG = NH x + 3 = 2x − 3 −x = −6 x=6 NG = x + 3 = 6 + 3 = 9 NH = 2x − 3 = 2 By the Incenter Theorem, NG = NH = NJ and NJ = 9 units. ⋅ 6 − 3 = 12 − 3 = 9 NK = NL 2x − 2 = −x + 10 3x = 12 x=4 NK = 2x − 2 = 2 NL = −x + 10 = −4 + 10 = 6 By the Incenter Theorem, NK = NL = NM and NM = 6 units. ⋅4 − 2 = 8 − 2 = 6 14. NQ = NR 2x = 3x − 2 −1x = −2 x=2 ⋅2 = 4 ⋅2 − 2 = 6 − 2 = 4 NQ = 2x = 2 NR = 3x − 2 = 3 By the Incenter Theorem, NQ = NR = NS and NS = 4 units. 15. −22 = 2b 2 11 6x − 2 = 3x + 7 13. ⋅ ND = NE PX = PY 3x + 2 = 4x − 8 −1x = −10 x = 10 ⋅ 10 + 2 = 30 + 2 = 32 ⋅ 10 − 8 = 40 − 8 = 32 PX = 3x + 2 = 3 PY = 4x − 8 = 4 By the Incenter Theorem, PX = PY = PZ and PZ = 32 units. 16. PX = PZ 4x + 3 = 6x − 11 −2x = −14 x=7 PX = 4x + 3 = 4 ⋅ 7 + 3 = 28 + 3 = 31 PZ = 6x − 11 = 6 ⋅ 7 − 11 = 42 − 11 = 31 By the Circumcenter Theorem, PX = PY = PZ and PY = 31 units. Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 17. Sample answer: 22. Construct an angle bisector of ∠BCA and ∠ABC. Label the intersection of the two angle bisectors as D. Draw a — perpendicular line from D to AB , label that point E. Using D — as the center and DE as the radius, construct a circle. Point D is the incenter of △ABC. 4 A 3 2 D 1 B 0 −1 C 0 1 2 3 5 4 5 6 4 −1 A 3 −2 E 2 18. Sample answer: 5 −1 A 3 D 1 0 0 B 0 1 2 3 5 4 6 −1 2 1 B 2 3 5C 4 6 −1 19. Sample answer: C 0 4 −1 D 1 23. Construct an angle bisector of ∠ABC and ∠CAB. Label the intersection of the two angle bisectors as D. Draw a — perpendicular line from D to AB , label that point E. Using D — as the center and DE as the radius, construct a circle. Point D is the incenter of △ABC. 4 6 5 3 A 4 2 3 1 B 0 1 C 2 3 5 4 1 6 D 0 −1 0 1 2 C3 B 4 5 6 7 8 9 −1 −2 20. Sample answer: E 2 D 0 −1 A 24. Construct an angle bisector of ∠ABC and ∠BCA. Label the intersection of the two angle bisectors as D. Draw a — perpendicular line from D to BC , label that point E. Using D — as the center and DE as the radius, construct a circle. Point D is the incenter of △ABC. 5 4 A 3 2 D 5 1 B 4 C 0 −1 0 1 2 3 4 5 6 A 3 −1 2 21. Construct an angle bisector of ∠ABC and ∠BCA. Label the intersection of the two angle bisectors as D. Draw a — perpendicular line from D to AC , label that point E. Using D — as the center and DE as the radius, construct a circle. Point D is the incenter of △ABC. A 3 E 2 D 1 C 0 −1 B 0 1 2 E 0 −1 B 0 1 2 C 3 4 5 6 −1 25. Because point G is the intersection of the angle bisectors, it — — is the incenter. But, because GD and GF are not necessarily perpendicular to a side of the triangle, there is not sufficient — evidence to conclude that GD — and GF are congruent. Point G is equidistant from the sides of the triangle. 5 4 D 1 3 4 5 −1 Copyright © Big Ideas Learning, LLC All rights reserved. 6 26.Because point T is the intersection of the perpendicular bisectors, it is the circumcenter and is equidistant from the vertices of the triangle, not necessarily the sides. TU = TW = TY Geometry Worked-Out Solutions 197 Chapter 6 27. You could copy the positions of the three houses, and connect the points to draw a triangle. Then draw the three perpendicular bisectors of the triangle. The point where the perpendicular bisectors meet, the circumcenter, should be the location of the meeting place. 33. Graph △ABC. y 12 A(2, 5) 4 28. To find the location of the fountain, use the Incenter Theorem. The incenter of the triangle is equidistance from the sides of the triangle. That point will be the same distance from each edge of the koi pond. So, place the fountain at the incenter of the pond. C(12, 3) −4 D C B 29. The circumcenter of a scalene triangle is sometimes inside the triangle. If the scalene triangle is obtuse or right, then the circumcenter is outside or on the triangle, respectively. However, if the scalene triangle is acute, then the circumcenter is inside the triangle. 30. If the perpendicular bisector of one side of a triangle intersects the opposite vertex, the triangle is always isosceles. If the perpendicular bisector of one side of a triangle intersects the opposite vertex, then it divides the triangle into two congruent triangles. So, two sides of the original triangle are congruent because corresponding parts of congruent triangles are congruent. 31. The perpendicular bisectors of a triangle intersect at a point that is sometimes equidistant from the midpoints of the sides of the triangle. This only happens when the triangle is equilateral. 32. The angle bisectors of a triangle intersect at a point that is always equidistant from the sides of the triangle. This is the Incenter Theorem (Thm. 6.6). 4 8 12 x −4 — 6−5 1 6−2 4 — The slope of the line perpendicular to AB is m = −4. = — = (4, 5.5) midpoint of AB = — , — , — 5.5 = −4 4 + b 5.5 = −16 + b 21.5 = b A B(6, 6) 8 = — Slope of AB : m = — — 5+6 8 11 ( 2 + 6 2 2 ) (2 2 ) y = mx + b ⋅ — The equation of the line perpendicular to AB is y = −4x + 21.5. — 1 −2 3−5 5 12 − 2 10 — The slope of the line perpendicular to AC is m = 5. , — = — = (7, 4) midpoint of AC = — , — = — = − — Slope of AC : m = — — y = mx + b 4=5 7+b 4 = 35 + b 5+3 14 8 ( 2 + 12 2 2 ) ( 2 2) ⋅ −31 = b — The equation of the line perpendicular to AC is y = 5x − 31. = — = − — Slope of BC : m = — — 1 −3 3 − 6 2 12 − 6 6 — The slope of the line perpendicular to BC is m = 2. , — = — midpoint of BC = — , — = (9, 4.5) — 6+3 18 9 ( 6 + 12 2 2 ) ( 2 2) y = mx + b 4.5 = 2 9 + b 4.5 = 18 + b −13.5 = b ⋅ — The equation of the line perpendicular to BC is Find the intersection of y = −4x + 21.5 and y = 5x − 31. −4x + 21.5 = 5x − 31 −9x + 21.5 = −31 y = 2x − 13.5. 198 Geometry Worked-Out Solutions −9x = −52.5 35 x = — 6 11 35 y = 5 — − 31 = − — 6 6 ⋅ ( 11 35 ) The point of intersection is — 6 , − — 6 . So, the coordinates of the circumcenter of △ABC are ( — 35 , 6 11 ) − — 6 . Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 34. Graph △DEF. F(−2, −2) −8 −4 y x −9 + (−5) −5 + (−9) −14 −14 — , midpoint of DE = — = — , — — ( 2 = (−7, −7) 2 ) ( 2 −7 = 1 −7 = −7 + b The equation of the line perpendicular to DE is y = x. — −2 − (−5) −2 + 5 3 — = — = — Slope of DF : m = — −2 − (−9) −2 + 9 7 7 — The slope of the line perpendicular to DF is m = − — . ( 2 ) ( 3 ) 11 7 −9 + (−2) −5 + (−2) — , — midpoint of DF = — = −— , − — ⋅ ( ⋅ ( ) 2 2 2 ) ⋅ ( ) , — = — Slope of EF : m = — — −2 − (−9) −2 + 9 −2 − (−5) −2 + 5 7 10x = −49 35 x = − — = −4.9 10 y = −4.9 The point of intersection is (−4.9, −4.9). So, the coordinates of the circumcenter of △DEF are (−4.9, −4.9). 36. 242 + (14x)2 = 252 576 + 196x2 = 625 x = 6 196x2 = 49 49 x2 = — 196 7 x = — 14 = — 12 The value of x that will The value of x that will make N the incenter is 6.make N the incenter is — 12 . 37. The circumcenter of any right triangle is located at the midpoint of the hypotenuse of the triangle. y A(0, 2b) 7 3 MAB(0, b) B(0, 0) Copyright © Big Ideas Learning, LLC All rights reserved. 49 Find the intersection of y = x and y = − — 3 . 3 x − — 7 49 x = − — x − — 3 3 3x = −7x − 49 −7 = b 3 — The equation of the line perpendicular to EF is y = − — 7 x −7. 35. 352 + (2x)2 = 372 1225 + 4x2 = 1369 4x2 = 144 x2 = 36 −98 = 6b 98 = b − — 6 49 = b − — 3 7 — T he equation of the line perpendicular to DF is y = − — 49 . 3 x − — 3 ) −14 = 2b 0=b ) ( ) ( ) ) ⋅ (−7) + b y = mx + b 7 7 11 − — = − — −— + b 2 3 2 7 77 + b − — = — 2 6 7 77 6 −— = 6 — + 6b 2 6 −21 = 77 + 6b ⋅ ( 2 11 2 y = mx + b 3 11 7 = − — −— − — + b 2 7 2 11 21 = — + b − — 2 14 11 3 — − = — + b 2 2 11 3 2 −— = 2 — + 2b 2 2 −11 = 3 + 2b — 3 ( −5 +2 (−2) −9 +2 (−2) ) ( 72 — −9 − (−5) −9 + 5 −4 −5 − (−9) −5 + 9 4 — The slope of the line perpendicular to DE is m = 1. , — midpoint of EF = — = − — , − — = — = — = −1 Slope of DE : m = — y = mx + b −8 E(−5, −9) The slope of the line perpendicular to EF is m = − — 7 . −4 D(−9, −5) — MAC (a, b) MBC (a, 0) C(2a, 0) x Let A(0, 2b), B(0, 0), and C(2a, 0) represent the vertices of a right triangle where ∠ B is the right angle. The midpoint — — of AB is MAB — ( 0, b). The midpoint of BC is MBC — ( a, 0). The — — midpoint of AC is MAC — ( a, b). Because AB is vertical, its perpendicular bisector is horizontal. So, the equation of the horizontal line passing through M— ( 0, b) is y = b. Because — is horizontal, its perpendicularABbisector is vertical. So, BC the equation of the vertical line passing through MBC — ( a, 0) is x = a. The circumcenter of △ABC is the intersection of perpendicular bisectors, y = b and x = a, which is (a, b). — This point is also the midpoint of AC . Geometry Worked-Out Solutions 199 Chapter 6 — bisects ∠ CAB, BD — 38. Given △ ABC, AD bisects ∠ CBA, — — — — — — ⊥ AB , DF DE ⊥ BC , DG ⊥ CA . Prove The angle bisectors intersect at D, which is — — — equidistant from AB , BC , and CA . C G A 42.a.The archaeologists need to locate the circumcenter of the three stones because that will be the center of the circle that contains all three stones. In order to locate the circumcenter, the archaeologists need to find the point of concurrency of the perpendicular bisectors of the sides of the triangle formed by the three stones. F coordinates of the point at which the archaeologists should look for the fire pit are (7, 7). E B 10 — — — — — — Because DE ⊥ AB , DF ⊥ BC , and DG ⊥ CA , ∠ DFB, ∠ DEB, ∠ DEA, and ∠ DGA are congruent right angles. Also, by definition of angle bisector, ∠ DBF ≅ ∠ DBE and — — — ∠ DAE ≅ ∠ DAG. In addition, DB — ≅ DB and DA ≅ DA by the Reflexive Property of Congruence (Thm. 2.1). So, △DFB ≅ △DEB and △DEA ≅ △DGA by the AAS Congruence Theorem (Thm. 5.11). Next, because corresponding parts of congruent triangles are congruent, — — — — ≅ DE DF and DG ≅ DE . By the Transitive Property of — — Congruence (Thm. 2.1), DF — ≅ DE ≅ DG . So, point D is — — — equidistant from AB , BC , and CA . Because D is equidistant — from CA — and CB , by the Converse of the Angle Bisector Theorem (Thm. 6.4), point D is on the angle bisector of ∠ ACB. So, the angle bisectors intersect at point D. 39. The circumcenter is the point of intersection of the perpendicular bisectors of the sides of a triangle, and it is equidistant from the vertices of the triangle. In contrast, the incenter is the point of intersection of the angle bisectors of a triangle, and it is equidistant from the sides of the triangle. 41.a.To determine the location of the pool so that it touches . the edges, construct two angle bisectors RD and QD Construct a perpendicular bisector through point D to QR —. Label the intersection E. With D as the center, construct a circle with radius DE —. If the incenter point were to move in any direction, the circle contained within the triangle would become smaller and not touch all three sides of the triangle. So, the circle with the center at the incenter is the largest circle that touches all three sides. ( ) E Q R D P y A(2, 10) 8 B(13, 6) D(7.09, 6.87) 6 4 2 C(6, 1) 2 4 6 8 10 12 x 43. B; by the Perpendicular Bisector Theorem 44. no; When you find the circumcenter of three of the points and draw the circle that circumscribes those three points, it does not pass through the fourth point. An example of one circle is shown. A 40. no; Because the incenter is the center of an inscribed circle, it must be inside the triangle. b.This is a circumcenter problem. The approximate D B E C D 45. yes; In an equilateral triangle, each perpendicular bisector passes through the opposite vertex and divides the triangle into two congruent triangles. So, it is also an angle bisector. 46. The incenter is at the center of the hub of the windmill where the blades, acting as angle bisectors, connect to the hub. 47.a.All triangles have exactly three angle bisectors and three perpendicular bisectors. Only three — 20 segments 3 are needed to represent them on an equilateral triangle because each perpendicular bisector also bisects the opposite angle. ( ) b.All six segments are needed to represent the three angle bisectors and three perpendicular bisectors of a scalene triangle because none of the perpendicular bisectors will also bisect an angle. b.Yes, and the radius would be decreased by 1 foot. You would keep the center of the pool as the incenter of the triangle, but you would make the radius of the pool at least 1 foot shorter. 200 Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 48. Sample answer: , — = (13, 11) midpoint of PN = — = — , — y = mx + b 1 11 = — 13 + b 3 1 3 11 = 3 — 13 + 3b 3 33 = 13 + 3b 53.13° 75 ft 45 ft 60 ft 90° ⋅ ⋅ ⋅ ⋅ 36.87° ( 10 +2 16 20 2+ 2 ) ( 262 222 ) — 20 = 3b 49. To determine the radius of the circle, the angle bisectors would be used. 20 3 1 — The equation of the line perpendicular to PN is y = — 3 x + — 20 . 3 Find the intersection of y = − — 3 x + — 3 . 2 x + 15 and y = — E(4, 128) y 1 1 2 y = − — x + 15 D(0, 0) 6 C(8, 0) x The radius of the circle is approximately 3 inches. 50. To determine the coordinates of the center of the circle and the radius of the circle, use perpendicular bisectors. Graph △LMN. y 24 E(13, 11) P(10, 20) F(6, 12) 8 −8 8 T(2, 4) 16 24 x N(16, 2) — 20 − 4 16 10 − 2 8 1 — The slope of the line perpendicular to TP is m = − — 2 . , — = (6, 12) midpoint of TP = — = — , — y = mx + b 12 = − — 6 + b = — = 2 Slope of TP : m = — — 1 2 12 = −3 + b 15 = b ( 10 2+ 2 20 2+ 4 ) ( 122 242 ) ⋅ — The equation of the line perpendicular to TP is y = − — 2 x + 15. = — = −3 Slope of PN : m = — = b — 4 8 1 1 20 x + — − — x + 15 = — 2 3 3 1 1 20 6 −— x + 6 15 = 6 — x + 6 — 2 3 3 −3x + 90 = 2x + 40 10 — 18 20 − 2 10 − 16 −6 — The slope of the line perpendicular to PN is — 13 . Copyright © Big Ideas Learning, LLC All rights reserved. 1 ⋅ ( ) ⋅ ⋅ 20 ⋅ −5x + 90 = 40 −5x = −50 x = 10 1 2 1 — = − 10 + 15 = −5 + 15 = 10 2 The coordinates of the center of the circle are (10, 10). Distance between the center (10, 10) and T(2, 4): ⋅ —— distance = √ (10 − 2)2 + (10 − 4)2 The radius of the circle is 10 units. — — — = √82 + 62 =√ 64 + 36 =√ 100 = 10 ⋅ ⋅ ⋅⋅ ⋅⋅ ⋅⋅ ⋅⋅ ⋅⋅ ⋅ ⋅ ⋅ 1 AB AC 51. Total area of △BAC = — 2 1 Area of △ADC = — x AC 2 1 Area of △BDC = — x BC 2 1 Area of △BDA = — x AB 2 1 1 1 1 AB AC = — x AC + — x BC + — x AB — 2 2 2 2 1 1 AB AC = — x(AC + BC + AB) — 2 2 1 1 2 — AB AC = 2 — x(AC + BC + AB) 2 2 AB AC = x(AC + BC + AB) ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ AB ⋅ AC —— = x (AC + BC + AB) ⋅⋅ — — The expression for x in terms of the lengths of AB , BC , and AC is —— . — ⋅ AB AC (AC + BC + AB) Geometry Worked-Out Solutions 201 Chapter 6 Maintaining Mathematical Proficiency ( ) ( 2 2 ) 58. 2x + 3y = 18 −3 + 3 5 + 5 0 10 — , = — = (0, 5) 52. midpoint of AB = — , — — 2 2 ——— — — AB = √ ( 3 − (−3) )2 + (5 − 5)2 = √ 62 + 0 2 =√ 36 = 6 ) ( ) ( 2 + 10 −1 + 7 12 6 — , — = (6, 3) = — = — , — 53. midpoint of AB 2 2 2 2 ——— AB = √ ( 10 − 2 )2 + ( 7 − (−1) )2 — — — — =√ 82 + 82 = √64 + 64 =√ 128 = 8√ 2 ≈ 11.3 ( ( ) −5 + 4 1 + (−5) — , = — 54. midpoint of AB — 2 2 1 −1 −4 = −— = — , — , −2 2 2 2 ) ( 3y = −2x + 18 y = − — 18 3 x + — 3 y = − — 3 x + 6 The slope of the new line is — 2 . y = mx + b y = — 32 x + b −6 = — 2 (−8) + b −6 = −12 + b AB = √ ( 4 − (−5) ) + (−5 − 1) — — 3 6=b The equation of the line passing through P(−8, −6) and perpendicular to y = 2x + 1 is y = — 32 x + 6. — ) ( 12 ) AB = √ ( 5 − (−7) ) + (9 − 5) — — y = mx + b y = − — 2 x + b 8 = − — 2 2 + b 8 = −1 + b 10 — −12 −8 P(−8, −6) y 8 2x + 3y = 18 4 8 12 x −4 −8 P(2, 8) y= 1 −2 x +9 −12 1 1 6 ⋅ 4 y = 2x + 1 2 9=b The equation of the line passing through P(2, 8) and perpendicular to 1 y = 2x + 1 is y = − — 2 x + 9. 2 4 6 8 10 x 57. The line y = −5 is horizontal. The equation of the line passing through P(6, −3) and perpendicular to y = −5 is x = 6. The slope of the perpendicular line is undefined so the equation is vertical. y 8 3 =√ 122 + 42 =√ 144 + 16 =√ 160 ≈ 12.6 56. The slope of the new 1 line is − — 2 . y y = 2x + 6 —— 2 2 ⋅ −7 + 5 5 + 9 −2 14 — , = — = (−1, 7) = — , — 55. midpoint of AB — 2 2 2 2 3 =√ 92 + (−6)2 =√ 81 + 36 =√ 117 ≈ 10.8 ( 2 ) ——— 2 2 2 2 4 x=6 −2 −4 −6 202 x 59. y + 3 = −4(x + 3) y + 3 = −4x − 12 y = −4x − 15 The slope of the new line is — 14 . y = mx + b y = — 4 x + b 1 = — 4 (−4) + b y 1 1 4 ⋅ 1 y = 4x + 2 1 = −1 + b P(−4, 1) 2=b −8 −6 −4 The equation of the line passing through P(−4, 1) and perpendicular to y + 3 = −4(x + 3) is y = — 14 x + 2. −2 2 x −2 −4 y + 3 = −4(x + 3) P(6, −3) y = −5 Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 6.3 Explorations (p. 323) 2.a.Check students’ work. 1.a.Draw △ABC and plot the midpoints of each side and construct the medians. Sample answer: b.Sample answer: 7 7 6 6 B 5 5 F A 4 E B 4 D 3 G 3 G 2 D 2 0 C 0 1 2 3 4 5 6 7 8 b.The medians of a triangle are concurrent at a point inside 2 23 . — BE ≈ 3, BG ≈ 2; The ratio is The ratio of the length of the longer segment to the length segment is — 2 of the longer segment a = — 2 b . The median equals a + b = — 2 b + b = — 2 b + — 2 b = — 2 b. Because the segment, b, is — 3 of the median. of the whole median is 2 : 3 or — 23 . If the shorter segment of the median is a and the longer segment is b, the shorter 1 1 median is 32 b, — 1 2 ( 1 3 ) by multiplying by the reciprocal, the longer 2 5 C 6 7 8 9 c.The altitudes that connect a vertex and a point on the opposite side are all perpendicular to that side. If the triangle is acute, the altitudes meet inside of the triangle. If the triangle is a right triangle, the legs of the right triangle are the altitudes, and therefore, meet at a point on the triangle. If the triangle is obtuse, the altitudes meet at a point on the outside of the triangle. 3. The medians meet at a point inside the triangle that divides each median into two segments whose lengths have the ratio 1 : 2. The altitudes meet at a point inside, on, or outside the triangle depending on whether the triangle is acute, right, or obtuse. — 4. The two segments of RU have lengths of 1 inch and 2 inches. 6.3 Monitoring Progress (pp. 325–327) ⋅ ⋅ 13 2100 = 700 ft 1. PS = — 2 PC = — 3 2100 = 1400 ft ⋅ 1000 = — ⋅ BC 2 ⋅ 1000 = 2 ⋅ — ⋅ BC 2. Copyright © Big Ideas Learning, LLC All rights reserved. 4 the shorter segment to the length of the longer segment is 1 1 : 2 or — . 2 AD ≈ 6, AG ≈ 4; The ratio is — 3 . 3 The altitudes meet at the same point. c.Segment AD is divided into AG ≈ 4 and GD ≈ 2. So, the ratio is about — 12 . Segment BE is divided into BG ≈ 2 and GE ≈ 1, so the ratio is about — 12 . The ratio of the length of 2 the triangle. 1 −1 9 −1 F 0 1 0 A 1 E 3. PT = — 12 PA 1 2 800 = — 12 PA 1 2 ⋅ ⋅ 2 ⋅ 800 = 2 ⋅ — ⋅ PA BT = — 12 BC 2000 = BC So, BC is 2000 feet. TC = ⋅ BC ⋅ 2000 12 — 12 — 1 2 1600 = PA So, PA is 1600 feet. TA = PT + PA TC = TC = 1000 TA = 800 + 1600 TA = 2400 So, TC is 1000 feet. So, TA is 2400 feet. Geometry Worked-Out Solutions 203 Chapter 6 4. y 6. G(4, 9) A(0, 3) 8 6 4 y 4 2 F(2, 5) M(5, 5) −2 P(4, 5) 2 B(0, −2) H(6, 1) 4 2 6 x ( 4 +2 6 9 +2 1 ) ( 102 102 ) — = — = (5, 5). The midpoint of GH is — , — , — . The centroid is — of FM 2 + (5 − 5)2 FM = √ (5 − 2) The centroid is — 3 3 = 2, so the centroid P is 2 units to — 2 3 —— — — =√ 32 = √ 9 = 3 2 ⋅ the right of F, which are the coordinates (4, 5). So, the coordinates of the centroid of △FGH are (4, 5). 5. L(−1, 4) X(−3, 3) P(−1, 2) −5 5 3 J(0, 1.5) 1 3 x Z(−1, −2) — ( 1 +2(−1) 5 +2(−2)) ( 02 32 ) , = (0, 1.5). The midpoint of YZ is — = — , — — 1 −1.5 1.5 − 3 — = — = − — . The slope of XJ is — 2 0 − (−3) 3 y = mx + b 1 1.5 = − — 0 + b 2 1.5 = b The centroid has the coordinates of the intersection of and x = −1. y = − — x + — ⋅ C(6, −3) — 1 −3 − (−2) 6 6−0 — The slope of the line perpendicular to BC is 6 and the line passes through vertex A(0, 3). = − — slope of BC = — y = mx + b 3=6 3=b ⋅0 + b — The equation of the line perpendicular to BC is y = 6x + 3. = — = −1 slope AC = — — −3 − 3 −6 6−0 6 — The slope of the line perpendicular to AC is 1 and the line passes through vertex B(0, −2). y = mx + b −2 = 1 −2 = b ⋅0 + b — The equation of the line perpendicular to AC is y = x − 2. Find the intersection of y = 6x + 3 and y = x − 2. 6x + 3 = x − 2 5x + 3 = −2 5x = −5 x = −1 y=6 So, the orthocenter is outside the triangle and the coordinates ⋅ (−1) + 3 = −6 + 3 = −3 are (−1, −3). 3 b = — 2 1 3 — though J(0, 1.5) is y = − — x + — . The equation of XJ 2 2 −3 + 1 3 + 5 −2 8 — , = — The midpoint of XY is — , — = (−1, 4). — 2 2 2 2 6 4 − (−2) — = — = undefined. The slope of ZL is — −1 − (−1) 0 — The equation of ZL through J(−1, 4) is x = −1. ( ) ( ) 1 3 2 2 1 3 y = − — x + — 2 2 1 3 y = − — (−1) + — 2 2 1 3 4 = — = 2 y = — + — 2 2 2 So, the centroid has coordinates (−1, 2). 204 y Y(1, 5) −3 6 x 2 Geometry Worked-Out Solutions 7. △ JKL is a right triangle; therefore, the orthocenter is on the triangle at the right angle (−3, 4). K(−3, 4) y L(5, 4) 3 −1 3 5 x −3 J(−3, −4) Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 8. Proving △ABD ≅ △CBD by the SSS Congruence Theorem (Thm. 5.8) at the beginning of the proof would be the same. But then you would state that ∠ ABD ≅ ∠ CBD because corresponding parts of congruent triangles are congruent. — This means that BD is also an angle bisector by definition. Vocabulary and Core Concept Check 1. The four types of triangular concurrencies are circumcenters, incenters, centroids, and orthocenters. The circumcenters are formed by the intersection of the perpendicular bisectors. The incenters are formed by the intersection of the angle bisectors. The centroids are formed by the intersection of the medians. The orthocenters are formed by the intersection of the altitudes. 2. The length of a segment from a vertex to the centroid is two-thirds the length of the median from that vertex. Monitoring Progress and Modeling with Mathematics 3. PN = 23 QN — 23 9 — 18 = 6 — 3 ⋅ PN = PN = PN = 6 units ⋅ QN ⋅ 9 13 — 13 — 4. PN = PN = PN = 23 QN — 23 21 — 42 = 14 — 3 ⋅ PN = 14 units ⋅ QN ⋅ 21 QP = QP = QP = 3 QP = 7 QP = 3 units QP = 7 units ⋅ 23 30 — 60 = 20 — 3 QP = 13 — 13 — 5. PN = — 23 QN QP = 2 6. PN = — 3 QN PN = ⋅ 23 42 — 84 = 28 — 3 PN = PN = PN = 20 units QP = QP = 13 — 13 — QP = 10 QP = 14 QP = 10 units QP = 14 units ⋅ QN ⋅ 30 6.3 Exercises (pp. 328–330) PN = PN = 28 units QP = QP = ⋅ QN ⋅ 42 13 — 13 — ⋅ 5 = — ⋅ CE 3 ⋅ 5 = 3 ⋅ — ⋅ CE ⋅ ⋅ 3 ⋅ 11 = 3 ⋅ — ⋅ CE 7. DE = — 13 CE 1 3 1 3 ⋅ ⋅ CD = CD = CD = 10 units 2 66 CD = — 3 = 22 CD = 22 units ⋅ ⋅ 3 ⋅ 15 = 3 ⋅ — ⋅ CE 1 3 ⋅ ⋅ CD = CD = CD = 18 units 15 = — 13 CE 23 CE — 23 27 — 54 = 18 — 3 DE = — 13 CE CE = 27 units CD = 10. 1 3 27 = CE ⋅ ⋅ CD = — 3 33 1 3 CE = 33 units 2 ⋅ 9 = — ⋅ CE 3 ⋅ 9 = 3 ⋅ — ⋅ CE 33 = CE CD = — 3 CE 9. DE = — 13 CE 11 = — 13 CE 23 CE — 23 15 — 30 = 10 — 3 CD = CE = 15 units DE = — 13 CE 1 3 15 = CE 8. 45 = CE CE = 45 units 2 ⋅ ⋅ CD = — 3 CE 2 CD = — 3 45 90 CD = — 3 = 30 CD = 30 units — 11. Because G is a centroid, BF is a median and F is the — midpoint of AC . Therefore, FC = 12 units. — 12. Because G is a centroid, BF is a median and BG is — 23 BF. ⋅ ⋅ — ⋅ 6 = — ⋅ — ⋅ BF BG = — 3 BF ⋅ 2 6 = — 23 BF 3 2 3 2 2 3 9 = BF Therefore, BF = 9 units. — 13. Because G is a centroid, AE is a median and AG = — 23 AE. 2 ⋅ ⋅ ⋅ AG = — 3 AE AG = — 3 15 AG = 10 Therefore, AG = 10 units. 2 — 14. Because G is a centroid, AE is a median and GE = — 13 AE. Copyright © Big Ideas Learning, LLC All rights reserved. 1 ⋅ ⋅ ⋅ GE = — 3 AE GE = — 3 15 GE = 5 Therefore, GE = 5 units. 1 Geometry Worked-Out Solutions 205 Chapter 6 15. 8 6 4 16. y ( C(5, 7) ( ) 11 5, 3 F(6.5, 4) G(−2, 7) 7 −3 , 5 ) I(−4, 5) A(2, 3) 2 y 8 F(1, 5) 4 E(−2.5, 4) H(−6, 3) D(5, 2) 2 B(8, 1) 4 2 6 −6 8 x — ( 5 +2 8 7 +2 1 ) ( 132 82 ) = — = (6.5, 4). The midpoint of CB is — , — , — = — = — . The slope of AF is — — y = mx + b 4−3 6.5 − 2 ⋅ ⋅ ⋅ ⋅ 1 4.5 2 9 −4 −2 , = — = (−4, 5). — , — — = 0. The slope of FI is — = — ( −6 +2(−2) 3 +2 7 ) ( −82 102 ) — 0 5−5 1 − (−1) 2 — The equation of FI through I(−4, 5) is y = 5. , — = — The midpoint of HF is — , — = (−2.5, 4). = 6. The slope of GE is —— = — = — 2 4 = — 6.5 + b 9 2 9 4 = 9 — 6.5 + 9 b 9 36 = 2 6.5 + 9b 36 = 13 + 9b y = mx + b 23 = 9b 4=6 23 = b — 9 . The equation of AF through F(6.5, 4) is y = — x + — ⋅ ⋅ — 23 2 9 9 2+8 3+1 10 4 — = — = (5, 2). The midpoint of AB is — , — , — 2 2 2 2 2 − 3 −1 — = — = undefined. The slope of CD is — 5−5 0 ( ) ( ) — The equation of CD through D(5, 2) is x = 5. The centroid has the coordinates of the intersection of y = — 29 x + — 23 and x = 5. 9 y = — x + — 23 2 9 9 2 23 y = — 5 + — 9 9 10 23 y = — + — 9 9 33 11 y = — = — 9 3 ( −6 2+ 1 3 +2 5 ) ( −52 82 ) — — 7−4 −2 − (−2.5) 3 −2 + 2.5 3 0.5 ⋅ (−2.5) + b 4 = −15 + b 19 = b — The equation of GE through E(−2.5, 4) is y = 6x + 19. The centroid has the coordinates of the intersection of y = 6x + 19 and y = 5. 5 = 6x + 19 −14 = 6x 7 14 = − — x = − — 6 3 7 The centroid has coordinates −— , 5 . 3 ( ) 17. y S(5, 5) 4 ⋅ 2 (5, 1) E(8, 1) U(−1, 1) −4 8 −2 ( ) 11 The centroid has coordinates 5, — 3 . Geometry Worked-Out Solutions 12 x D(5, −1) T(11, −3) — ( ) ( ) 16 2 5 + 11 5 + (−3) 2 2 2 2 0 1−1 — = 0. The slope of UE is — = — 8 − (−1) 9 — The equation of UE through E(8, 1) is y = 1. , — The midpoint of ST is — = — , — = (8, 1). — The midpoint of UT is , — = (5, −1). — = — , — = — = undefined. The slope of SD is — 206 2 x — The midpoint of HG is ( −1 +2 11 1 + 2(−3) ) ( 102 −22 ) — −1 − 5 −3 5−5 0 — The equation of SD through D(5, −1) is x = 5. The centroid has the coordinates of the intersection of x = 5 and y = 1, which is (5, 1). Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 18. 4 3 19. y X(1, 4) L(0, 5) y 4 D(4, 3) y=1 Z(2, 3) 2 (103, 3) 2 Y(7, 2) E(4.5, 2.5) 1 M(3, 1) −2 4 N(8, 1) 6 8 x −2 2 4 6 8 x — ( y = 2x − 5 ) ( ) 8 6 1+7 4+2 2 2 2 2 3−3 0 — = — = 0. The slope of ZD is — 4−2 2 — The equation of ZD through D(4, 3) is y = 3. = — = (4.5, 2.5). The midpoint of ZY is — , — , — — ( ) ( ) ⋅ ⋅ ⋅ ( ) 62 = 14b 62 31 b = — = — 14 7 . The equation of XE through E(4.5, 2.5) is y = − — x + — — 3 31 7 7 The centroid has the coordinates of the intersection of 3 31 and y = 3. y = − — x + — 7 7 3 31 y = − — x + — 7 7 3 31 3 = − — x + — 7 7 3 31 7 3 = 7 −— x + 7 — 7 7 21 = −3x + 31 ⋅ ⋅ ( −10x =−3x (0, −5) = — = (4, 3). The midpoint of XY is — , — , — 9 5 2 + 7 3 + 2 2 2 2 2 3 15 1.5 4 − 2.5 — = — The slope of XE is — = − — = − — . 35 7 1 − 4.5 −3.5 y = mx + b 3 9 2.5 = − — — + b 7 2 27 5 = − — + b — 14 2 27 5 14 — = 14 −— + 14b 14 2 35 = −27 + 14b −4 ) ⋅ ( — 1−1 0 8−3 5 — The slope of the line perpendicular to MN is undefined and passes through (0, 5). Therefore, the equation of that line is x = 0. 1 4 5−1 — = — = − — . The slope of the line containing LN is — 2 0 − 8 −8 — The slope of the line perpendicular to LN is 2. y = 2x + b 1=2 3+b 1=6+b −5 = b = — = 0. The slope of the line containing MN is — y = mx + b ⋅ — The equation of the line perpendicular to LN containing point (3, 1) is y = 2x − 5. The orthocenter is the intersection of x = 0 and y = 2x − 5. y = 2x − 5 ⋅0 − 5 y=2 y = −5 The orthocenter of △LMN is located on the outside and the coordinates are (0, −5). 20. △ XYZ is a right triangle; therefore, the orthocenter is on the triangle at the intersection of the legs. y Z(−3, 6) ) 6 4 Y(5, 2) X(−3, 2) 10 = x — 3 10 The centroid has coordinates — , 3 . 3 ( ) Copyright © Big Ideas Learning, LLC All rights reserved. −2 2 4 x The coordinates of the orthocenter are (−3, 2). Geometry Worked-Out Solutions 207 Chapter 6 21. C(−1, 3) (−1, 2) y 6 4 (0, 73) 4 V(0, 4) B(1, 0) A(−4, 0) −6 22. y x = −1 T(−2, 1) −2 U(2, 1) 2 x −2 2 8 2 y = 3x + 3 2 4 x 7 y = 3x + 3 — 0 0−0 1 − (−4) 5 — The slope of the line perpendicular to AB is undefined and passes through (−1, 3). Therefore, the equation of that line is x = −1. 3 3 3−0 — = — = − — . The slope of the line containing CB is — 2 −1 − 1 −2 2 — The slope of the line perpendicular to CB is — . 3 y = mx + b = 0. The slope of the line containing AB is — = — = 0. The slope of the line containing TU is — = — 0 = — (−4) + b 2 3 ⋅ 8 0 = − — + b 3 8 = b — 3 — The equation of the line perpendicular to CB containing 8 2 . point (−4, 0) is y = — x + — 3 3 The orthocenter is the intersection of x = −1 and 8 2 . y = — x + — 3 3 8 2 y = — x + — 3 3 2 8 y = — (−1) + — 3 3 2 8 y = − — + — 3 3 6 y = — = 2 3 The orthocenter of △ABC is inside the triangle with coordinates (−1, 2). ⋅ — 0 1−1 2 − (−2) 4 — The slope of the line perpendicular to TU is undefined and passes through (0, 4). Therefore, the equation of that line is x = 0. 3 3 4−1 — = — = − — . The slope of the line containing VU is — 2 0 − 2 −2 2 — The slope of the line perpendicular to VU is — . 3 y = mx + b ⋅ 2 1 = — (−2) + b 3 4 1 = − — + b 3 4 3 1 = 3 −— + 3 b 3 3 = −4 + 3b ⋅ ( ⋅ ) ⋅ 7 = 3b 7 b = — 3 — The equation of the line perpendicular to VU containing 7 2 . point (−2, 1) is y = — x + — 3 3 7 2 . The orthocenter is the intersection of x = 0 and y = — x + — 3 3 7 2 y = — x + — 3 3 2 7 y = — 0 + — 3 3 7 y = — 3 The orthocenter of △TUV is inside the triangle with 7 coordinates 0, — . 3 ⋅ ( ) 23. Construct the medians of an isosceles right triangle by finding the midpoint of each side and connecting the midpoint and the vertex opposite that midpoint. Where the medians intersect is the location of the centroid. The orthocenter is on the triangle at the right angle. centroid orthocenter 208 Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 24. Construct the medians of an obtuse scalene triangle by finding the midpoint of each side and connecting the midpoint and the vertex opposite that midpoint. Where the medians intersect is the location of the centroid. Construct the altitudes by drawing a perpendicular segment from each vertex to the opposite side or to the line that contains the opposite side. The orthocenter is located outside the triangle where the altitudes intersect. — — 27. The length of DE should be — 13 of the length of AE because it is the shorter segment from the centroid to the side. 1 DE = — 3 AE DE = — 3 (18) DE = 6 1 Sample answer: — — 28. The length of DE is — 12 of the length of AD because DE = — 13 AE 2 and AD = — 3 AE. DE = — 2 AD DE = — 2 (24) DE = 12 centroid 1 1 — 29. Given Isosceles △ABC, AD is an angle bisector of ∠ ABC. orthocenter 25. Construct the medians of a right scalene triangle by finding the midpoint of each side and connecting the midpoint and the vertex opposite that midpoint. Where the medians intersect is the location of the centroid. The orthocenter is on the triangle at the right angle. Sample answer: — Prove BD is a median. B A centroid orthocenter 26. Construct the medians of an acute isosceles triangle by finding the midpoint of each side and connecting the midpoint and the vertex opposite that midpoint. Where the medians intersect is the location of the centroid. Construct the altitudes by drawing a perpendicular segment from each vertex to the opposite side. The orthocenter is where the altitudes intersect. Sample answer: centroid D C STATEMENTS REASONS 1. △ ABC is an isosceles triangle. 1. Given 2. — AB ≅ — BC Copyright © Big Ideas Learning, LLC All rights reserved. 2. Definition of isosceles triangle 3. — AD is an angle bisector of ∠ ABC. 3. Given 4. △ABD ≅ ∠ CBD 4. Definition of angle bisector 5. — BD ≅ — BD 5. Reflexive Property of Congruence (Thm. 2.1) 6. SAS Congruence Theorem (Thm. 5.5) 6. △ABD ≅ △CBD 7. — AD ≅ — CD orthocenter 8. D is the midpoint of — AC . BD is a median. 9. — 7. Corresponding parts of congruent triangles are congruent. 8. Definition of midpoint 9. Definition of median Geometry Worked-Out Solutions 209 Chapter 6 30. Given Isosceles △ABC, — — is an altitude to AC BD . 35. The centroid and orthocenter are sometimes the same point. The centroid and the orthocenter are not the same point unless the triangle is equilateral. B — Prove BD is a perpendicular bisector. 36. The centroid is always formed by the intersection of the three medians. This is the definition of a centroid. A STATEMENTS 1. △ ABC is an isosceles triangle. 2. — AB ≅ — BC 3. — BD is an altitude to — AC . 4. — BD ⊥ — AC 5. ∠ ADB and ∠ CDB are right angles. 6. — BD ≅ — BD 7. △ABD ≅ △CBD 8. — AD ≅ — CD 9. D is the midpoint of — AC . — BD is a 10. perpendicular bisector. D C REASONS 1. Given 2. Definition of isosceles triangle 3. Given 4. Definition of altitude 5. Definition of perpendicular 6. Reflexive Property of Congruence (Thm. 2.1) 7. HL Congruence Theorem (Thm. 5.9) 8. Corresponding parts of congruent triangles are congruent. 9. Definition of midpoint 10. Definition of perpendicular bisector 31. The centroid is never on the triangle. Because medians are always inside a triangle, and the centroid is the point of concurrency of the medians, it will always be inside the triangle. 32. The orthocenter is sometimes outside the triangle. An orthocenter can be inside, on, or outside the triangle depending on whether the triangle is acute, right, or obtuse. 33. A median is sometimes the same line segment as a perpendicular bisector. A median is the same line segment as the perpendicular bisector if the triangle is equilateral or if the segment is connecting the vertex angle to the base of an isosceles triangle. Otherwise, the median and the perpendicular bisectors are not the same segment. 34. An altitude is sometimes the same line segment as an angle bisector. An altitude is the same line segment as the angle bisector if the triangle is equilateral or if the segment is connecting the vertex angle to the base of an isosceles triangle. Otherwise, the altitude and the angle bisector are not the same segment. 210 Geometry Worked-Out Solutions 37. Both segments are perpendicular to a side of a triangle, and their point of intersection can fall either inside, on, or outside of the triangle. However, the altitude does not necessarily bisect the side, but the perpendicular bisector does. Also, the perpendicular bisector does not necessarily pass through the opposite vertex, but the altitude does. 38. All are segments that pass through the vertex of a triangle. A median connects a vertex with the midpoint of the opposite side. An altitude is perpendicular to the opposite side. An angle bisector bisects the angle through which it passes. The medians of a triangle intersect at a single point, and the same is true for the altitudes and angle bisectors of a triangle. Medians and angle bisectors always lie inside the triangle, but altitudes may be inside, on, or outside of the triangle. 39. Area = — 12 bh The area of the triangle in solid red is — 2 — 2 3 = — 4 = 6.75 square inches. The special segment of the triangle used was the altitude. 1 ⋅⋅ 9 27 — — — 40. K is the centroid and DH , EJ , and FG are medians. a.EJ = 3KJ b.DK = 2KH c.FG = — 2 FK d.KG = — 3 FG F 3 J 1 K H D G 41. BD = — 23 BF 2 ⋅ 42. E GD = — 13 GC 1 4x + 5 = — 3 9x 2x − 8 = — 3 (3x + 3) 4x + 5 = 6x 2x − 8 = x + 1 5 = 2x 52 = — x=9 x 43. AD = 2DE 44. DF = — 12 BD 1 5x = 2(3x − 2) 4x − 1 = — 2 (6x + 4) 5x = 6x − 4 4x − 1 = 3x + 2 −x = −4 x=3 x=4 Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 45. — 51. Given △ABC is an equilateral triangle. BD is a median. y median 8 3 y2 = 4 x + 5 — Prove BD is angle bisector, 6 4 A median (0, 2) −8 −6 −4 2 median 4 y1 = 3x − 4 −4 y3 = −2 x − 4 x y2 y-intercept (0, 5). STATEMENTS ( 3. Given 4. — AD ≅ — CD 5. — BD ≅ — BD ) 2. Definition of equilateral triangle 3. — BD is a median. 20 3 C 1. Given 2. — AB ≅ — AC ≅ — BC — 43 , D REASONS 1. △ ABC is an equilateral triangle. ( 0 )and the y-intercept (0, −4). = — x + 5: Graph the x-intercept ( −— , 0 )and the 4. Definition of median 5. Reflexive Property of Congruence (Thm. 2.1) — y3 = − — 2 x − 4: Graph the x-intercept − 3 , 0 and the 6. △ABD ≅ △CBD 6. SSS Congruence Theorem (Thm. 5.8) The points of intersection that form the triangle are (4, 8), 7. ∠ADB ≅ ∠CDB, ∠ABD ≅ ∠CBD 7. Corresponding parts of congruent triangles are congruent. 8. ∠ADB and ∠CDB form a linear pair and are supplementary. 8. Linear Pair Postulate 9. ∠ADB and ∠CDB are right angles. 9. If two angles are congruent and supplementary then they are right angles. 3 y-intercept (0, −4). 3 y1 = 3x − 4: Graph the x-intercept 6 −2 3 4 B perpendicular bisector, and altitude. 8 (−4, 2), and (0, −4). The equation of the median from y1 to the opposite vertex is y = 2. The equation of the median from y2 to the opposite vertex is x = 0. The coordinates of the centroid are (0, 2). 46. right triangle; The orthocenter of a right triangle is the vertex of the right angle. 47. PE = — 13 AE, PE = — 12 AP, PE = AE − AP 10. — BD ⊥ — AC — 48.a.KM is a median. It contains the centroid. AD ≅ — CD 11. — — b.KN is an altitude. It contains the orthocenter. c.The area of △JKM is — 2 9 △ KLM is — 2 9 h = — 2 h, which indicates that the two areas are equal. Yes, triangles formed by the median will 1 1 ⋅⋅ 9 10. Definition of perpendicular ⋅ ⋅ h = — h and the area of 11. Corresponding parts of congruent triangles are congruent. 9 2 — AD is a perpendicular 13. Definition of 13. bisector. perpendicular bisector always have the same area because they will have the same base length and height. 49. yes; If the triangle is equilateral, then the perpendicular bisectors, angle bisectors, medians, and altitudes will all be the same three segments. 50. centroid; Because the triangles formed by the median of any triangle will always be congruent, the mass of the triangle on either side of the median is the same. So, the centroid is the point that has an equal distribution of mass on all sides. AD is an altitude. 14. — 14. Definition of altitude 52. The orthocenter, circumcenter, and the centroid are all inside the triangle. They are all three distinct points. The three concurrent points are all collinear. y A 4 orthocenter 3 circumcenter 2 1 B centroid 1 Copyright © Big Ideas Learning, LLC All rights reserved. 12. Definition of midpoint 12. D is the midpoint BC . of — C 2 3 4 5 x Geometry Worked-Out Solutions 211 Chapter 6 53. Sample answer: The circle passes through nine significant points of the triangle. They are the midpoints of the sides, the midpoints between each vertex and the orthocenter, and the points of intersection between the sides and the altitudes. y A 9 8 7 6 K I J 5 D 4 E B L 1 1 2 3 C 4 5 6 x — — 54. Given LP and MQ are medians of scalene △LMN. Point R — — such is on LP such that LP ≅ PR . Point S is on MQ — that MQ — ≅ QS . — — — — — b. NS and NR are both parallel to LM . c. R, N, and S are collinear. Prove a. NS ≅ NR R M REASONS 1. Given 2. Definition of median 3. ∠ LPM ≅ ∠ RPN, ∠ MQL ≅ ∠ SQN 3. Vertical Angles Congruence Theorem (Thm. 2.6) 4. △LPM ≅ △RPN, △MQL ≅ △SQN 4. SAS Congruence Theorem (Thm. 5.5) 6. — NS ≅ — NR H G 1. — LP and — MQ are medians of scalene △LMN; — LP ≅ — PR , — MQ ≅ — QS NP ≅ — MP , — LQ ≅ — NQ 2. — 5. — NR ≅ — LM , — NS ≅ — LM 5. Corresponding parts of congruent triangles are congruent. F 3 2 a. STATEMENTS P N Q L S 6. Transitive Property of Congruence (Thm. 2.1) b.It was shown in part (a) that △LPM ≅ △RPN and c.Because NS and NR are both parallel to the same segment, △MQL ≅ △SQN. So, ∠ LMP ≅ ∠ RNP and ∠ MLQ ≅ ∠ SNQ because corresponding parts of — — congruent triangles are congruent. Then, NS LM and — — LM NR by the Alternate Interior Angles Converse (Thm. 3.6). — — — LM , they would have to be parallel to each other by the Transitive Property of Parallel Lines (Thm. 3.9). However, because they intersect at point N, they cannot be parallel. So, they must be collinear. Maintaining Mathematical Proficiency −3 1 3−6 — 55. Slope of AB = — = — : — −1 − 5 −6 2 −6 −6 1 3−9 — Slope of CD : —— = — = — = — −16 − (−4) −16 + 4 −12 2 — — — — The slopes of AB and CD are equal, so AB CD . 1 −2 −2 4−6 — = − — = — = — : — 56. Slope of AB 4 5 − (−3) 5 + 3 8 3 1 −7 − (−10) −7 + 10 — Slope of CD = — = — = — : —— −2 − (−14) −2 + 14 12 4 — — — The slopes of AB and CD are not equal, so AB is not parallel to CD —. 5 2 − (−3) 2 + 3 — = — = — = −5 : — 57. Slope of AB 5−6 −1 −1 2+4 6 2 − (−4) — Slope of CD = — = — = −6 : — −5 − (−4) −5 + 4 −1 — — — The slopes of AB and CD are not equal, so AB is not parallel to CD —. −4 −4 2−6 — = 2 : — = — = — 58. Slope of AB −7 − (−5) −7 + 5 −2 −5 − 1 −6 — Slope of CD = — = 2 : — 4−7 −3 212 Geometry Worked-Out Solutions — — — — The slopes of AB and CD are equal, so AB is parallel to CD . Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 6.1– 6.3 What Did You Learn? (p. 331) 4. Graph △ABC. 1. Sample answer: yes; You realize that if you construct the —, you have created two isosceles triangles that segment AC share vertices A and C, and a third triangle that shares the same vertices. Then you look back at the Perpendicular Bisector Theorem (Thm. 6.1) and its converse to see that points D and E would have to be on the perpendicular bisector of —. Then, in the same way, in order for B to also be on the same AC — — perpendicular bisector, AB and CB would have to be congruent. 2. Sample answer: These can be constructed with a compass and a straightedge, or they can be constructed with geometry software; If you construct them with geometry software, you create a triangle that fits the description first. Then use the software to draw the perpendicular bisectors of each side. Next, label the point where these three lines meet. Finally, draw a circle with its center at this point of intersection that passes through one vertex of the triangle. It will automatically pass through the other two vertices. 3. Sample answer: Right triangles are the only kind of triangles that have one of the points of intersection on the vertex of the triangle; While all segment types can be inside the triangle, only the perpendicular bisectors and altitudes can be on or outside the triangle. — — , point V is on the ⊥ SU 1. Because SW ≅ UW and VW perpendicular bisector of SU —. SV = UV −6x + 11 = −1 −6x = −12 x=2 UV = 8x − 1 = 8 Perpendicular Bisector Theorem (Thm. 6.1) 2x + 11 = 8x − 1 — is an angle bisector of ∠ PSR, PQ , and RQ . 2. SQ — ⊥ SP ⊥ SR PQ = RQ 6x = 3x + 9 3x = 9 PQ = 6x = 6 Angle Bisector Theorem (Thm. 6.3) x=3 ⋅ 3 = 18 and GK bisects 3. Because J is equidistant from GH , GJ ∠ HGK by the Angle Bisector Theorem (Thm. 6.3). m∠ HGJ = m∠ JGK x−4=3 x=7 m∠ JGK = 4x + 3 = 4 31° + m∠ GJK + 90° = 180° y 2 −6 2 x −2 B(−4, −4) (−2, −4) C(0, −4) ( −4 +2 (−4) 2 + 2(−4) ) −8 −2 = ( , ) = (−4, −1) 2 2 −4 + 0 2 + (−4) — , midpoint of AC = ( ) 2 2 −4 −2 = ( , ) = (−2, −1) 2 2 — , — midpoint of AB = — The equation of the perpendicular bisector of AB through its — — — — — — midpoint (−4, −1) is y = −1. 3 6 2 − (−4) — = — = − — slope of AC = — 2 −4 − 0 −4 — The slope of the perpendicular line to AC is — 23 . y = mx + b 2 −1 = — (−2) + b 3 4 −1 = − — + b 3 4 3 (−1) = 3 −— + 3b 3 −3 = −4 + 3b ⋅ 2 − 1 = 16 − 1 = 15 D(−2, −1) A(−4, 2) (−4, −1) 6.1–6.3 Quiz (p. 332) — ⋅ ⋅ ⋅ ( ) 1 = 3b 1 = b — 3 — The equation of the perpendicular bisector of AC through its 2 1 midpoint (−2, −1) is y = — x + — . 3 3 2 1 Find the point of intersection of y = −1 and y = — 3 x + — 3 . 1 2 y = — x + — 3 3 2 1 −1 = — x + — 3 3 −3 = 2x + 1 −4 = 2x x = −2 The coordinates of the circumcenter of △ABC are (−2, −1). 5x − 4 = 4x + 3 ⋅ 7 + 3 = 28 + 3 = 31° m∠ GJK + 121° = 180° Triangle Sum Theorem (Thm. 5.1) m∠ GJK = 59° Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 213 Chapter 6 9. 5. Graph △DEF. 16 6 8 8 4 −2 16 x — 0 5−5 11 − 3 8 — The slope of the line perpendicular to DF is undefined, so the — equation of the line perpendicular to DF is x = 7. — ( 3 +2 7 5 +2 9 ) ( 102 142 ) , — = (5, 7) midpoint of DE = — , — , — = — = 1 slope of DE = — — 9−5 4 7−3 4 — The slope of the line perpendicular to DE is −1. y = mx + b ⋅5 + b 7 = −1 7 = −5 + b 12 = b — The equation of the line perpendicular to DE through the midpoint (5, 7) is y = −x + 12. The intersection of x = 7 and y = −x + 12 is y = −7 + 12 = 5. The coordinates of the circumcenter of △DFE are (7, 5). NQ = NR 2x + 1 = 4x − 9 −2x + 1 = −9 −2x = −10 x = 5 NU = NV 7. −3x + 6 = −5x 6 = −2x −3 = x NQ = NR = NS = 2x + 1 = −3 (−3) + 6 = 2 5 + 1 =9+6 = 11 = 15 8. 4x − 10 = 3x − 1 x − 10 = −1 NZ = NY = NW = 4x − 10 = 4 x=9 ⋅ 9 − 10 = 36 − 10 = 26 L(5, −2) — , — = — , — = (5, 2). = 0. The slope of JD is — = — ( 5 +2 5 6 + 2(−2) ) ( 102 42 ) — 0 2−2 5 − (−1) 6 — The equation of JD through D(5, 2) is y = 2. — The midpoint F of JL is 5 + (−1) −2 + 2 4 0 , = (2, 0). — = — , — — 2 2 2 2 6−0 6 — = — = 2. The slope of KF is — 5−2 3 y = mx + b 0=2 2+b 0=4+b −4 = b — The equation of KF through F(2, 0) is y = 2x − 4. The centroid has the coordinates of the intersection of y = 2 and y = 2x − 4. 2 = 2x − 4 6 = 2x 3=x The coordinates of the centroid are (3, 2). ( ) ( ) ⋅ 10. y −8 −6 −4 −2 N(−4, −2) H(−4, −4) D(−6, −4) ⋅ NZ = NY x = −2x + 6 6 F(2, 0) — The midpoint D of KL is NU = NV = NT ⋅ 4 −2 = — = 0 slope of DF = — 6. D(5, 2) J(−1, 2) G(9, 7) F(11, 5) (7, 5) 4 D(3, 5) H(3, 2) 4 E(7, 9) K(5, 6) y H(5, 7) y M(−8, −6) 2 x −2 P(0, −4) F(−4, −5) −8 — The midpoint D of MN is , — = (−6, −4). — = — , — ( ) ( ) −4 + (−8) −2 + (−6) −12 −8 2 2 2 2 −4 − (−4) 0 — = — = 0. The slope of DP is — 0 − (−6) 6 — The equation of DP through D(−6, −4) is y = −4. — The midpoint F of MP is = (−4, −5). , — = — , — — ( ) ( ) −8 −10 0 + (−8) −4 + (−6) 2 2 2 2 −2 − (−5) −2 + 5 3 — = — = — = undefined. The slope of NF is — −4 − (−4) −4 + 4 0 — The equation of NF through F(−4, −5) is x = −4. The centroid has the coordinates of the intersection of y = −4 and x = −4. So, the coordinates of the centroid are (−4, −4). 214 Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 11. 13.a.The point of concurrency at the center of the circle is the — — — incenter. BG , CG , and AG are angle bisectors. y T(−2, 5) 6 V(2, 5) b.Sample answer: △BGF ≅ △BGE by HL Congruence Theorem (Thm. 5.9). 1 y = −2 x + 4 c.Because △BGF ≅ △BGE and corresponding parts of congruent triangles are congruent, BE = BF = 3 centimeters. So, AE = 10 − 3 = 7 centimeters. Then you can use the Pythagorean Theorem (Thm. 9.1) for △AEG to find EG, which is the radius of the wheel. U(0, 1) −4 −2 2 4 x — 0 5−5 2 − (−2) 4 — The slope of the line perpendicular to TV is undefined and passes through U(0, 1). Therefore, the equation of that line is x = 0. AG2 = GE2 + EA2 82 = GE2 + 72 64 = GE2 + 49 15 = GE2 5−1 4 — = — = 2. The slope of the line containing UV is — GE = √ 15 ≈ 3.9 The slope of the line perpendicular to UV is − — . The radius of the wheel is about 3.9 centimeters. y = mx + b 14.a.The point of concurrency used was the centroid. = 0. The slope of the line containing TV is — = — 2−0 — 1 y = − — x + b 2 1 5 = − — (−2) + b 2 5=1+b 4=b 2 1 2 6.4 Explorations (p. 333) — The equation of the line perpendicular to UV containing point T(−2, 5) is y = − — x + 4. The orthocenter is the intersection of x = 0 and y = − — x + 4. 1 2 1 2 1 y = − — x + 4 2 1 y = − — 0 + 4 2 y=4 The coordinates of the orthocenter are located on the inside b.The circumcenter should have been used because the Circumcenter Theorem (Thm. 6.5) can be used to find a point equidistant from the three points (the three cities). — ⋅ 1.a.Check students’ work. — — — — — which indicates that DE AC , and the length of DE 1— is — AC . 2 3 − 4.5 −1.5 — = — ≈ −0.43 slope of DE = — b. Sample answer: The slopes of DE and AC are equal c.The segment connecting the midpoints of two sides of a 5 − 1.5 3.5 —— — length of DE = √ (1.5 − 5)2 + (4.5 − 3)2 ≈ 3.8 −3 1−4 — ≈ −0.43 slope of AC = — = — 5 − (−2) 7 —— — 5 − (−2) )2 + (1 − 4)2 ≈ 7.6 length of AC = √ ( of the triangle and are (0, 4). 12. 4 y Z(7, 4) triangle is parallel to the third side and is half as long as that side. 2.a.Check students’ work. b. Sample answer: The triangle formed by the midsegments 2 4 6 of a triangle, △EFD, is similar to the original triangle, △ABC. The side lengths of △EFD are — 12 the side lengths of △ABC. x −2 X(−1, −4) Y(7, −4) △ XYZ is a right triangle. Therefore, the orthocenter is on the triangle at the intersection of the legs, which is (7,−4). 3. The midsegment that is formed by joining the midpoints of two sides is parallel to the third side and is — 12 the length of the third side. 1 4. UV = — RT 2 T 12 = — 12 RT 2 ⋅ 12 = 2 ⋅ — RT V 1 2 12 24 = RT In △RST, if UV = 12, R U S then RT = 24. Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 215 Chapter 6 ( 6.4 Monitoring Progress (pp. 334–336) 2 − (−1) 2 + 1 3 — = — = — 1. Slope of DE : — 2−0 2 2 0 − (−6) 6 3 — = — = — Slope of AC : — 5−1 4 2 — — — — Because the slope of DE equals the slope of AC , DE AC because parallel lines have equal slopes. —— 2 2 — 2 2 — — —— 2 2 — 2 2 — — — — — ) ( ) 0 + 2p 0 + 0 2p , — = — 3. The midpoint F is — , 0 = ( p, 0). 2 2 2 r r−0 — Slope of FE : — = — p+q−p q r 2r − 0 — Slope of OB = — : — 2q − 0 q — — — — Because the slopes of FE and OB are equal, FE OB . —— —— 2 2 — 2 2 — — 2 2 2 2 DE = √ (2 − 0) + ( 2 − (−1) ) = √(2) + (3) OB = √ (2q− 0) + (2r − 0) 13 = √4 + 9 = √ =√ 4q + 4r AC = √ (5 − 1) + ( 0 − (−6) ) =√ 4(q + r ) = 2√ q + r = √(4) + (6) OB = 2FE = √16 + 36 FE = — OB = √52 = √4 13 = 2√ 13 AC = 2√ 13 AC = 2DE DE = — AC V R ( ) ( 26 −62 ) −1 + 5 4 + 0 4 4 — , The midpoint of BC is ( = , = (2, 2). 2 2 ) (2 2) 5 + 1 0 + (−6) 2 2 — — — — — — The midpoint F of AC is (3, −3) and the midpoint E of BC is (2, 2). — 5 2 − (−3) 2−3 −1 −10 −6 − 4 — = — = −5 Slope of AB : — 1 − (−1) 2 — — EF AB because the slopes are equal. AB = √ ( 1 − (−1) )2 + (−6 − 4)2 ⋅ S U = (3, −3). — , — = — , — ⋅ 1 2 — is 2. The midpoint of AC 1 2 — . If UW = 81, then 4. The third midsegment is UW ST = 2 81 = 162; therefore, VS = — 12 162 = 81 inches. ⋅ = — Slope of EF : — = −5 ——— —— 2 2 — — — — —— 2 2 —— 2 2 — — — — FE = √ ( p + q − p)2 + (r − 0)2 = √ q2 + r2 W T — — — 5. In Example 4, DF is a midsegment; therefore, DF AB and 1— — DF = — 2 AB . 6. From Peach Street to Plum Street is 2.25 miles; from Plum Street to Cherry Street is 1.4 miles; from Cherry Street to Pear Street is 1.3 miles; from Pear Street to Peach Street is — 12 1.4 is 0.7 mile; from Pear Street back home is ( ⋅ ) ( ⋅ 2.25 )is 1.125 miles. The total distance is 12 — 2.25 + 1.4 + 1.3 + 0.7 + 1.125 = 6.775 miles. This route was less than that taken in Example 5. 6.4 Exercises (pp. 337–338) = √(2) + (−10) Vocabulary and Core Concept Check = √4 + 100 = √104 1. The midsegment of a triangle is a segment that connects the midpoints of two sides of a triangle. = √4 26 = 2√ 26 ⋅ EF = √ (3 − 2) + (−3 − 2) =√ (1) + (−5) = √1 + 25 = √ 26 AB = 2√ 26 AB = 2EF EF = — AB 216 — — 2. If DE is the midsegment opposite AC in △ABC, then 1 — — — — AC DE and DE = — 2 AC by the Triangle Midsegment Theorem (Thm. 6.8). Monitoring Progress and Modeling with Mathematics 3. The coordinates are D(−4, −2), E(−2, 0), and F(−1, −4). 1 2 Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 2 0 − (−2) — = — = 1 4. Slope of DE : — −2 − (−4) 2 −2 − (−6) −2 + 6 4 — Slope of CB = — = — = 1 : — 1 − (−3) 4 4 — —— — 12 BC 7. DE = — 1 8. DE = — 2 AB x = — 12 (26) 5 = — 12 (AB) x = 13 x = 10 Because the slope of DE equals the slope of CB , DE CB . DE = √ ( −2 − (−4) )2 + ( 0 − (−2) )2 =√ (2)2 + (2)2 =√ 4 + 4 =√ 8 = √ 4 2 = 2√ 2 CB = √ ( 1 − (−3) ) + ( −2 − (−6) ) =√ (4) + (4) =√ 16 + 16 17. =√ 16 2 = 4√ 2 3x + 8 = — 2 (2x + 24) 3y − 5 = — 2 (4y + 2) Because 2√ 2 = — ( 4√ 2 ), DE = — CB. 3x + 8 = x + 12 3y − 5 = 2y + 1 2x + 8 = 12 y−5=1 2x = 4 x = 2 GL = 2 AB = — 2 (28) = 14 ——— — — — ⋅ — — ——— 2 2 — 2 2 — — — ⋅ 1 2 — — 1 2 −4 −4 −4 − 0 — = — = −4 5. Slope of EF : — = — −1 − (−2) −1 + 2 1 −8 −8 −6 − 2 — = — = −4 Slope of AC : — = — −3 − (−5) −3 + 5 2 — — — — Because the slope of EF equals the slope of AC , EF AC . ——— 2 2 —— 2 2 — — ——— 2 2 —— 2 2 — — — EF = √ ( −1 − (−2) ) + (−4 − 0) = √(1) + (−4) = √1 + 16 = √ 17 AC = √ ( −3 − (−5) ) + (−6 − 2) 10. BE = EC 6 = x 12.JL XZ — — — — 13.XY KL 14.JY ≅ JX ≅ KL — — — 15.JL ≅ XK ≅ KZ 16.JK ≅ YL ≅ LZ — — — ⋅ Because √ 17 = — ( 2√ 17 ), EF = — AC. 1 2 1 2 — 2 −4 − (−2) −4 + 2 — 6. Slope of DF = — = − — : — 3 −1 − (−4) −1 + 4 2 −4 −4 −2 − 2 — Slope of AB = — = − — : — = — 3 1 − (−5) 1 + 5 6 — —— — Because the slope of DF equals the slope of AB , DF AB . DF = √ ( −1 − (−4) )2 + ( −4 − (−2) )2 AB = √ ( 1 − (−5) )2 + (−2 − 2)2 2 + (−4)2 =√ (6) =√ 36 + 16 =√ 52 = √ 4 13 = 2√ 13 ——— — —— — 2 + (−2)2 = √ =√ (3) 9 + 4 =√ 13 ——— —— ⋅ — — ⋅ =3 6−5 = 18 − 5 = 13 ⋅ — (7z − 1) 1 2 8z − 6 = 7z − 1 z − 6 = −1 z=5 GA = CB CB = 4z − 3 = 4 GA = 17 ⋅ 5 − 3 = 20 − 3 = 17 — — — 20.DE is not parallel to BC . So, DE is not a midsegment. So, according to the contrapositive of the Triangle Midsegment Theorem (Thm. 6.8), DE — does not connect the midpoints of — — AC and AB . 21. The distance between first base and second base is 90 feet. Because the shortstop is halfway between second and third bases, and the pitcher is halfway between first and third bases, using the Triangle Midsegment Theorem, the distance between the shortstop and the pitcher is — 12 90 = 45. So, the distance between the shortstop and the pitcher is 45 feet. ⋅ — — HB = AC HB = 3y − 5 4z − 3 = — 12 (7z − 1) — y=6 CB = — 12 (GA) ⋅ 1 = √(2) + (−8) = √4 17 = 2√ 17 = 4 + 24 = 28 2(4z − 3) = 2 AC = — 12 (HJ) 1 2 + 24 18. 1 = √4 + 64 — — — AB = — 12 (GL) x=8 — — 11.JK YZ 19. 9. AE = EC Because √ 13 = — 2 ( 2√13 ), DF = — 2 AB. — 1 — 1 Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 217 Chapter 6 — 22. Given F is the midpoint of OC . 1 — — DF BC and DF = — (BC) Prove 2 y 24. Sample answer: ends of the swing set are midsegments. B(2q, 2r) D(q, r) O(0, 0) The crossbars on the E C(2p, 0) x F — ( 0 +2 2p ) ( 2p2 ) , 0 = — midpoint of OC = — , 0 = F(p, 0) 25.a.The perimeter of the shaded triangle is 8 + 8 + 8 = 24 units. slope of DF = — r q−p 2r r 2r − 0 — = — slope of BC = — = — 2q − 2p 2(q − p) q − p — — 8 — — — —— —— 2 2 2 2 —— 2 2 —— 2 2 —— 2 2 —— 2 2 —— 2 2 DF = √ (q − p) + (r − 0) = √ (q − p) + r BC = √ (2q − 2p) + (2r − 0) =√ 4(q − p) + 4r =√ 4(q − p) + r = 2√ (q − p) + r 16 16 8 Because BC = 2DF, DF = b.The perimeter of all the shaded triangles is ⋅ 24 + 3(3 4) = 24 + 36 = 60 units. 12 BC. — 4 — — 23. An eighth segment, FG , would connect the midpoints of DL — and EN . DE = — 2 (XY + LN) 1 1 = — — (LN) + LN 2 2 FG = — 14 LN + — 12 LN 3 = — LN 4 3 = — 4 LN + LN = — 78 LN [ use 8p, 8q, and 8r. So, L(0, 0), M(8q, 8r), and N(8p, 0). Find the coordinates of X, Y, D, E, F, and G. — Because X is the midpoint of LM , the coordinates are X(4q, 4r). — Because Y is the midpoint of MN , the coordinates are Y(4q + 4p, 4r). — Because D is the midpoint of XL , the coordinates are D(2q, 2r). — Because E is the midpoint of YN , the coordinates are E(2q + 6p, 2r). — Because F is the midpoint of DL , the coordinates are F(q, r). — Because G is the midpoint of EN , the coordinates are G(q + 7p, r). — The y-coordinates of D and E are the same, so DE has a slope of 0. The y-coordinates of F and G are also the same, — so FG — has a slope of 0. LN is on the x-axis, so its slope is 0. — — — Because their slopes are the same, DE LN FG . Use the Ruler Postulate (Post. 1.1) to find DE, FG, and LN. DE = 6p, FG = 7p, LN = 8p Because 6p = — 34 (8p), DE = — 34 LN. Because 7p = — 78 (8p), 7 FG = — LN. 8 218 4 8 Geometry Worked-Out Solutions 8 8 16 4 ] Because you are finding quarter segments and eighth segments, 16 4 4 1 8 =√ ( 2(q − p) ) + 4r 16 Because the slope of DF equals the slope of BC , DF BC . 4 4 4 16 4 c.The perimeter of all the shaded triangles is ⋅ ⋅ 24 + 3 12 + 9 6 = 24 + 36 + 54 = 114 units. 16 6 12 6 24 6 16 6 6 12 12 6 6 6 16 6 26. Two sides of the red triangle have a length of 4 tile widths. A yellow segment connects the midpoints, where there are 2 tile lengths on either side. The third red side has a length of 4 tile diagonals, and the other two yellow segments meet at the midpoint, where there are 2 tile diagonals on either side. Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 — 27. Construct a line ( DF )with midpoint P parallel to and twice — the length of QR . Construct a line ( EF — )with midpoint R — parallel to and twice the length of QP . Construct a line ( DE — ) — with midpoint Q parallel to and twice the length of PR . The vertices of the original large triangle are D(−1, 2), E(9, 8), and F(5, 0). 10 y E 8 6 Q 4 R 2.a.Check students’ work. Using the sample in the text: AB ≈ 3.61 AC ≈ 5.10 BC = 5 b.Check students’ work. Using the sample in the text: BC = 5 < 8.71 = AC + AB AC = 5.10 < 8.61 = AB + BC AB = 3.61 < 10.10 = BC + AC c.Sample answer: D P −2 F 2 4 8 10 x −2 A(x, y) B(x, y) C(x, y) AB AC + BC AC A(5, 1) B(7, 4) C(2, 4) 3.61 9.24 4.24 A(2, 4) B(4, −2) C(7, 6) 6.32 13.93 5.39 A(1, 0) B(7, 0) C(1, 7) 6 16.22 7 A(1, 0) B(7, 0) C(5, 1) 6 6.36 4.12 AB + BC BC AB + AC 8.61 5 7.85 14.86 8.54 11.71 Maintaining Mathematical Proficiency 28. Sample answer: A counterexample to show that the difference of two numbers is not always less than the greater of the two numbers is 6 and −2: 6 − (−2) = 8, which is not less than 6, so the original conjecture is false. 29. Sample answer: An isosceles triangle has at least two sides that are congruent. An isoseles triangle whose sides are 5 centimeters, 5 centimeters, and 3 centimeters is not equilateral. 6.5 Explorations (p. 339) 1.a.Check students’ work. Using the sample in the text: AC ≈ 6.08, AB ≈ 4.47, BC ≈ 3.61, m∠ A ≈ 36.03°, m∠ B ≈ 97.13°, m∠ C ≈ 46.85° b.Check students’ work. Using the sample in the text, BC < AB < AC and m∠ A < m∠ C < m∠ B. The shortest side is opposite the smallest angle, and the longest side is opposite the largest angle. c.Sample answer: A(x, y) B(x, y) 15.22 9.22 13 8.24 2.24 10.12 The length of each side is less than the sum of the other two. 3. The largest angle is opposite the longest side, and the smallest angle is opposite the shortest side; The sum of any two side lengths is greater than the third side length. 4. no; The sum 3 + 4 is not greater than 10, and it is not possible to form a triangle when the sum of the lengths of the two sides is less than the length of the third side. 6.5 Monitoring Progress (pp. 340–343) 1. Given △ABC is a scalene triangle. C(x, y) AB AC BC Prove △ABC does not have two congruent angles. 5 Assume temporarily that △ABC is a scalene triangle with A(5, 1) B(7, 4) C(2, 4) 3.61 4.24 A(2, 4) B(4, −2) C(7, 6) 6.32 5.39 8.54 A(1, 0) B(7, 0) C(1, 7) m∠ A m∠ B m∠ C 78.69° 56.31° 45° 93.37° 38.99° 47.64° 90° 49.4° 40.6° 6 7 9.22 If one side of a triangle is longer than another side, then the angle opposite the longer side is larger than the angle opposite the shorter side. Similarly, if one angle of a triangle is larger than another angle, then the side opposite the larger angle is longer than the side opposite the smaller angle. Copyright © Big Ideas Learning, LLC All rights reserved. ∠ A ≅ ∠ B. By the Converse of Base Angles Theorem (Thm. 5.7), if ∠ A ≅ ∠ B, then the opposite sides are — ≅ AC — congruent: BC . A scalene triangle cannot have two congruent sides. So, this contradicts the given information. So, the assumption that △ABC is a scalene triangle with two congruent angles must be false, which proves that a scalene triangle cannot have two congruent angles. — — 2. The sides of △PQR from smallest to largest are PR , RQ , and — PQ . So, by the Triangle Longer Side Theorem, the angles from smallest to largest are ∠Q, ∠P, and ∠R. 3. The angles of △RST from smallest to largest are ∠R, ∠T, and ∠S. So, by the Triangle Larger Angle Theorem, the sides — — — from shortest to longest are ST , RS , and RT . Geometry Worked-Out Solutions 219 Chapter 6 4. Let x represent the length of the third side. By the Triangle Inequality Theorem: x + 12 > 20 and 12 + 20 > x. x + 12 > 20 The length of the third side must be greater than 8 inches and and 12 + 20 > x x > 8 32 > x, or x < 32 less than 32 inches. 5. 4 + 9 > 10 → 13 > 10 Yes 4 + 10 > 9 → 14 > 9 Yes 9 + 10 > 4 → 19 > 4 Yes yes; The sum of any two side lengths of a triangle is greater — . Using point A as the center, draw an 10. Draw a segment AB — arc with radius AB , then draw an arc with B as the center — and radius AB — to intersect the first arc on both sides of AB . Construct a segment from one arc intersection toward the other but stop at AB —. Label the arc-segment intersection as C — and the intersection with AB as G. △BGC is a right scalene triangle. C than the length of the third side. 6. no; The sum 8 + 9 = 17 is not greater than 18. A G B 7. no; The sum 5 + 7 = 12 is not greater than 12. 6.5 Exercises (pp. 344–346) Vocabulary and Core Concept Check 1. In an indirect proof, rather than proving a statement directly, you show that when the statement is false, it leads to a contradiction. 2. The longest side of a triangle is opposite the largest angle and the shortest side is opposite the smallest angle. Monitoring Progress and Modeling with Mathematics 3. Assume temporarily that WV = 7 inches. 4. Assume temporarily that xy is even. 5. Assume temporarily that ∠ B is a right angle. — 6. Assume temporarily that JM is not a median. 7. A and C; The angles of an equilateral triangle are always 60°. So, an equilateral triangle cannot be a right triangle. 8. B and C; If both ∠ X and ∠ Y have measures less than 30°, then their sum is less than 60°. Therefore, the sum of their measures cannot be 62°. 9. To construct a scalene triangle, draw a segment and label it —. Ensuring that AB — — — AC , BC , and AC are all different lengths, — draw an arc with center A and radius AB and an arc with — center C with radius CB . Where the two arcs intersect place point B. B A C — The largest angle is ∠ ABC and the opposite side, AC , is the longest side. The smallest angle is ∠ ACB and the opposite side, AB —, is the shortest side. The largest angle is ∠CGB because it is the right angle and — the opposite side, CB , is the longest side. The smallest angle — is ∠GCB and the opposite side, GB , is the shortest side. — — 11. The sides of △RST from smallest to largest are RT , TS , and — RS . So, by the Triangle Longer Side Theorem, the angles from smallest to largest are ∠ S, ∠ R, and ∠ T. — — 12. The sides of △JKL from smallest to largest are KL , JL , and — JK . So, by the Triangle Longer Side Theorem, the angles from smallest to largest are ∠ J, ∠ K, and ∠ L. 13. The angles of △ABC from smallest to largest are ∠ C, ∠ A, and ∠ B. So, by the Triangle Larger Angle Theorem, the — — — sides from shortest to longest are AB , BC , and AC . 14. The angles of △XYZ from smallest to largest are ∠ Z, ∠ X, and ∠ Y. So, by the Triangle Larger Angle Theorem, the — — — sides from shortest to longest are XY , ZY , and ZX . 15. m∠ M = 180° − (127° + 29°) = 24° The angles of △MNP from smallest to largest are ∠ M, ∠ P, and ∠ N. So, by the Triangle Larger Angle Theorem, the — — — sides from shortest to longest are PN , MN , and MP . 16. m∠ D = 180° − (90° + 33°) = 57° The angles of △DFG from smallest to largest are ∠ G, ∠ D, and ∠ F. So, by the Triangle Larger Angle Theorem, the — — — sides from shortest to longest are DF , GF , and GD . 17. x + 5 > 12 x>7 5 + 12 > x 17 > x or x < 17 The possible lengths of the third side are greater than 7 inches and less than 17 inches. 220 Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 28. Assume temporarily that the second group has 15 or more 18. x + 12 > 18 x>6 12 + 18 > x 30 > x or x < 30 The possible lengths of the third side are greater than 6 feet and less than 30 feet. students. Because the first group has 15 students, the total number of students in the class would be 30 students or more. Because the class has fewer than 30 students, the assumption must be false, and the second group must have fewer than 15 students. 29. C; m∠ U = 180° − (84° + 48°) = 180° − 132° = 48°, which indicates that △UTV is isosceles. By the Triangle Longer Side Theorem, UV > TV. 19. x + 24 > 40 x > 16 30. C and D; m∠ R = 180° − (65° + 56°) = 180° − 121° = 59°; 24 + 40 > x 64 > x or x < 64 The possible lengths of the third side are greater than 16 inches and less than 64 inches. 20. x + 25 > 25 By the Triangle Inequality Theorem, the order of the angles from smallest to largest is ∠ T, ∠ R, and ∠ S. The order of the sides from shortest to longest is RS < ST < RT ⇒ 8 < ST < RT. ST could possibly be 9 or 10, but not 7 or 8. 31. Given x>0 An odd number Prove An odd number is not divisible by 4. 25 + 25 > x 50 > x or x < 50 The possible lengths of the third side are greater than 0 meters and less than 50 meters. 21. 6 + 7 = 13 → 13 > 11 Yes 7 + 11 = 18 → 18 > 6 Yes 11 + 6 = 17 → 17 > 7 Yes Assume temporarily that an odd number is divisible by 4. Let the odd number be represented by 2y + 1 where y is a positive integer. Then there must be a positive integer x such that 4x = 2y + 1. However, when you divide each side of the equation by 4, you get x = —12 y + —14 , which is not an integer. So, the assumption must be false, and an odd number is not divisible by 4. 32. Given yes; The sum of any two side lengths of a triangle is greater than the length of the third side. 22. no; The sum 3 + 6 = 9 is not greater than 9. 23. no; The sum 28 + 17 = 45 is not greater than 46. 24. 35 + 120 = 155 → 155 > 125 Yes 120 + 125 = 255 → 255 > 35 Yes 125 + 35 = 160 → 160 > 120 Yes yes; The sum of any two side lengths of a triangle is greater than the length of the third side. 25. An angle that is not obtuse could be acute or right. Assume temporarily that ∠ A is not obtuse. — 26. Because 30° < 60° < 90° and 1 < √ 3 < 2, the longest side, which is 2 units long, should be across from the largest angle, which is the right angle. Prove △QRS, m∠ Q + m∠ R = 90° m∠ S = 90° Assume temporarily that in △QRS, m∠ Q + m∠ R = 90° and m∠ S ≠ 90°. By the Triangle Sum Theorem (Thm. 5.1), m∠ Q + m∠ R + m∠ S = 180°. Using the Substitution Property of Equality, 90° + m∠ S = 180°. So, m∠ S = 90° by the Subtraction Property of Equality, but this contradicts the given information. So, the assumption must be false, which proves that in △QRS, if m∠ Q + m∠ R = 90°, then m∠ S = 90°. 33. The right angle of a right triangle must always be the largest angle because the other two will have a sum of 90°. So, according to the Triangle Larger Angle Theorem (Thm. 6.10), because the right angle is larger than either of the other angles, the side opposite the right angle, which is the hypotenuse, will always have to be longer than either of the legs. 34. yes; If the sum of the lengths of the two shortest sides is greater than the length of the longest side, then the other two inequalities will also be true. 60° 1 2 —— 30° √3 27. Assume temporarily that your client committed the crime. Then your client had to be in Los Angeles, California, at the time of the crime. Security footage shows that your client was in New York at the time of the crime. Therefore, the assumption must be false, and the client must be innocent. Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 221 Chapter 6 35.a.The width of the river must be greater than 35 yards and — less than 50 yards. In △BCA, the width of the river, BA , — must be less than the length of CA , which is 50 yards, — because the measure of the angle opposite BA is less than — the measure of the angle opposite CA , which must be 50°. In △BDA, the width of the river, BA —, must be greater than the — length of DA , which is 35 yards, because the measure of — the angle opposite BA is greater than the measure of the — angle opposite DA , which must be 40°. b.You could measure from distances that are closer together. In order to do this, you would have to use angle measures that are closer to 45°. 36.a.By the side length requirements for a triangle, x < 489 + 565 = 1054 kilometers and x > 565 − 489 = 76 kilometers. b.Because ∠2 is the smallest angle, the distance between Granite Peak and Fort Peck Lake must be the shortest side of the triangle. So, the second inequality becomes x < 489 kilometers. 37. ∠WXY, ∠ Z, ∠ YXZ, ∠WYX and ∠ XYZ, ∠W; In △WXY, because WY < WX < YX, by the Triangle Longer Side Theorem (Thm. 6.9), m∠WXY < m∠WYX < m∠W. Similarly, in △XYZ, because XY < YZ < XZ, by the Triangle Longer Side Theorem (Thm. 6.9), m∠ Z < m∠ YXZ < m∠ XYZ. Because m∠WYX = m∠ XYZ and ∠W is the only angle greater than either of them, you know that ∠W is the largest angle. Because △WXY has the largest angle and one of the congruent angles, the remaining angle, ∠WXY, is the smallest. m∠ D + m∠ E + m∠ F = 180° 38. (x + 25)° + (2x − 4)° + 63° = 180° 3x = 96 x = 32 m∠D = 32 + 25 = 57° m∠E = 2 m∠F = 63° The order of the angles from least to greatest is 3x + 84 = 180 ⋅ 32 − 4 = 60° m∠ D < m∠ E < m∠ F. The order of the sides from least to greatest is EF < DF < DE. 39. By the Exterior Angle Theorem (Thm. 5.2), m∠1 = m∠ A + m∠B. Then by the Subtraction Property of Equality, m∠1 − m∠B = m∠ A. If you assume temporarily that m∠1 ≤ m∠B, then m∠ A ≤ 0. Because the measure of any angle in a triangle must be a positive number, the assumption must be false. So, m∠1 > m∠B. Similarly, by the Subtraction Property of Equality, m∠1 − m∠ A = m∠B. If you assume temporarily that m∠1 ≤ m∠ A, then m∠B ≤ 0. Because the measure of any angle in a triangle must be a positive number, the assumption must be false. So, m∠1 > m∠ A. 222 Geometry Worked-Out Solutions 40. JK + KL > JL x + 11 + 2x + 10 > 5x − 9 3x + 21 > 5x − 9 −2x > −30 x < 15 x + 11 + 5x − 9 > 2x + 10 4x > 8 x>2 2x + 10 + 5x − 9 > x + 11 7x + 1 > x + 11 6x > 10 The possible values for x are x > 2 and x < 15. 41. JK + JL > KL 6x + 2 > 2x + 10 KL + JL > JK x > — 10 = — 53 ≈ 1.667 6 UV + VT > TU 3x − 1 + 2x + 3 > 6x − 11 5x + 2 > 6x − 11 −x > −13 x < 13 6x − 11 + 2x + 3 > 3x − 1 TU + TV > UV 8x − 8 > 3x − 1 5x > 7 3x − 1 + 6x − 11 > 2x + 3 The possible values for x are x > 2 — and x < 13. 7 2 x > — 75 = 1 — 5 UV + TU > TV 9x − 12 > 2x + 3 7x > 15 1 x > — 15 = 2 — 7 7 1 42. The shortest route is along Washington Avenue. By the Triangle Inequality Theorem (Thm. 6.11), the length of Washington Avenue must be shorter than the sum of the lengths of Eighth Street and View Street, as well as the sum of the lengths of Hill Street and Seventh Street. Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 Prove m∠ BAC > m∠ C by the constructions in parts (a)–(c) because a triangle is formed only when the two sides formed by arc radii have — a sum that is greater than the length of side QR . B 1 A 2 3 D 46. Given △ABC C It is given that BC > AB and BD = BA. By the Base Angles Theorem (Thm. 5.6), m∠ 1 = m∠ 2. By the Angle Addition Postulate (Post. 1.4), m∠ BAC = m∠ 1 + m∠ 3. So, m∠ BAC > m∠ 1. Substituting m∠ 2 for m∠ 1 produces m∠ BAC > m∠ 2. By the Exterior Angle Theorem (Thm. 5.2), m∠ 2 = m∠ 3 + m∠ C. So, m∠ 2 > m∠ C. Finally, because m∠ BAC > m∠ 2 and m∠ 2 > m∠ C, you can conclude that m∠ BAC > m∠ C. 44. As an example, if the 24-centimeter string is divided into 10 centimeters, 10 centimeters, and 4 centimeters, the triangle is an acute isosceles triangle. 10 cm 10 cm 4 cm For a right scalene triangle, the sides could be 6 centimeters, 8 centimeters, and 10 centimeters. 10 cm 8 cm 6 cm 45.a. a b M Q R b. c d M R — yes; The arcs intersect in one point that lies on QR . They do not form a triangle because all three points are collinear. Prove AB + BC > AC, AC + BC > AB, and AB + AC > BC B 2 3 D 1 C A — Assume BC is longer than or the same length as each of — — the other sides, AB and AC . Then AB + BC > AC and AC + BC > AB. The proof for AB + AC > BC follows. STATEMENTS REASONS 1. △ABC 1. Given 2. Extend — AC to D so that — AB ≅ — AD . 2. Ruler Postulate (Post. 1.1) 3. AB = AD 3. Definition of segment congruence 4. AD + AC = DC 4. Segment Addition Postulate (Post. 1.2) 5. ∠ 1 ≅ ∠ 2 5. Base Angles Theorem (Thm. 5.6) 6. m∠ 1 = m∠ 2 6. Definition of angle congruence 7. m∠ DBC > m∠ 2 7. Protractor Postulate (Post. 1.3) 8. m∠ DBC > m∠ 1 8. Substitution Property 9. DC > BC 9. Triangle Larger Angle Theorem (Thm. 6.10) 10. AD + AC > BC 10. Substitution Property 11. AB + AC > BC 11. Substitution Property no; The arcs do not intersect. Q d. The Triangle Inequality Theorem (Thm. 6.11) is verified 43. Given BC > AB, BD = BA 47. The perimeter of △HGF must be greater than 4 and less than 24; Because of the Triangle Inequality Theorem (Thm. 6.11), FG must be greater than 2 and less than 8, GH must be greater than 1 and less than 7, and FH must be greater than 1 and less than 9. So, the perimeter must be greater than 2 + 1 + 1 = 4 and less than 8 + 7 + 9 = 24. c. f Q M R g yes; The arcs intersect in two points. Each point is a — vertex of a triangle with side QR . Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 223 Chapter 6 6.6 Monitoring Progress (pp. 349–350) Maintaining Mathematical Proficiency — — and BE is ∠ AEB. 48. The included angle between AE 1. PR = PSGiven — — PQ ≅ PQ Reflexive Property of Congruence (Thm. 2.1) — — 49. The included angle between AC and DC is ∠ ACD. — — 50. The included angle between AD and DC is ∠ ADC. m∠ QPR > m∠ QPSGiven — — 51. The included angle between CE and BE is ∠ BEC. 6.6 Explorations (p. 347) 2. 1.a.Check students’ work. b.Check students’ work. c.Check students’ work. d.AC ≅ DC , because all points on a circle are equidistant — — — — from the center; BC ≅ BC by the Reflexive Property of Congruence (Thm. 2.1). — — DB is 2.7 units, so AB > DB; m∠ ACB = 90° and e.As drawn, the length of AB is 3.6 units and the length of f. Sample answer: D AC BC 1. (4.75, 2.03) 2 3 2. (4.94, 2.5) 2 3. (5, 3) m∠ ACB m∠ BCD 3.61 2.68 90° 61.13° 3 3.61 3.16 90° 75.6° 2 3 3.61 3.61 90° 90° 4. (4.94, 3.5) 2 3 3.61 90° 104.45° 5. (3.85, 4.81) 2 3 3.61 4.89 90° 154.93° AB BD 4 g.If two sides of one triangle are congruent to two sides of another triangle, and the included angle of the first is larger than the included angle of the second, then the third side of the first is longer than the third side of the second. 2. If the included angle of one is larger than the included angle of the other, then the third side of the first is longer than the third side of the second. If the included angles are congruent, then you already know that the triangles are congruent by the SAS Congruence Theorem (Thm. 5.5). Therefore, the third sides are congruent because corresponding parts of congruent triangles are congruent. 3. Because the sides of the hinge do not change in length, the angle of the hinge can model the included angle and the distance between the opposite ends of the hinge can model the third side. When the hinge is open wider, the angle is larger and the ends of the hinge are farther apart. If the hinge is open less, the ends are closer together. 224 Hinge Theorem (Thm. 6.12) PR = PSGiven — — PQ ≅ PQ Reflexive Property of Congruence RQ < SQGiven m∠ RPQ < m∠ SPQConverse of the Hinge Theorem (Thm. 2.1) m∠ DCB = 61°, so m∠ ACB > m∠ DCB; yes; the results are as expected because the triangle with the longer third side has the larger angle opposite the third side. RQ > SQ — RQ is the longer segment. Geometry Worked-Out Solutions (Thm. 6.13) ∠ SPQ is the larger angle. 3. Assume temporarily that the third side of the first triangle with the larger included angle is not longer than the third side of the second triangle with the smaller included angle. This means the third side of the first triangle is equal to or shorter than the third side of the second triangle. 4. Group A: 135° Group B: 150° Group C: 180° − 40° = 140° Because 135° < 140° < 150°, Group C is closer to the camp than Group B, but not as close as Group A. 6.6 Exercises (pp. 351–352) Vocabulary and Core Concept Check 1. Theorem 6.12 refers to two angles with two pairs of sides that have the same measure, just like two hinges whose sides are the same length. Then the angle whose measure is greater is opposite a longer side, just like the ends of a hinge are farther apart when the hinge is open wider. — — — — 2. In △ABC and △DEF, AB ≅ DE , BC ≅ EF , and AC < DF. So, m∠ E > m∠ B by the Converse of the Hinge Theorem (Theorem 6.13). Monitoring Progress and Modeling with Mathematics 3. m∠ 1 > m∠ 2; By the Converse of the Hinge Theorem (Thm. 6.13), because ∠ 1 is the included angle in the triangle with the longer third side, its measure is greater than that of ∠ 2. 4. m∠ 1 < m∠ 2; By the Converse of the Hinge Theorem (Thm. 6.13), because ∠ 1 is the included angle in the triangle with the shorter third side, its measure is less than that of ∠ 2. 5. m∠ 1 = m∠ 2; The triangles are congruent by the SSS Congruence Theorem (Thm. 5.8). So, ∠ 1 ≅ ∠ 2 because corresponding parts of congruent triangles are congruent. Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 13. Your flight: 6. m∠ 1 > m∠ 2; By the Converse of the Hinge Theorem (Thm. 6.12), because ∠ 1 is the included angle in the triangle with the longer third side, its measure is greater than that of ∠ 2. — 7. AD > CD; By the Hinge Theorem (Thm. 6.12), because AD is the third side of the triangle with the larger included angle, —. it is longer than CD 50 miles 160° H 20° 100 miles Friend’s flight: — 8. MN < LK; By the Hinge Theorem (Thm. 6.12), because MN is the third side of the triangle with the smaller included —. angle, it is shorter than LK 50 miles 30° — 9. TR < UR; By the Hinge Theorem (Thm. 6.12), because TR is the third side of the triangle with the smaller included angle, —. it is shorter than UR — 150° 100 miles Because 160° >150°, the distance you flew is a greater distance than your friend flew by the Hinge Theorem (Thm. 6.12). 14. Your flight: 10. AC > DC; By the Hinge Theorem (Thm. 6.12), because AC is the third side of the triangle with the larger included angle, it —. is longer than DC 11. Given Prove — ≅ YZ —, m∠ WYZ > m∠ WYX XY W WZ > WX Z X STATEMENTS REASONS 1. 1. Given — XY ≅ — YZ 2. — WY ≅ — WY 210 miles Y 2. Reflexive Property of Congruence (Thm. 2.1) 110° 70° 80 miles 3. m∠ WYZ > m∠ WYX 3. Given 4. WZ > WX 4. Hinge Theorem (Thm. 6.12) 12. Given — ≅ DA —, DC < AB BC Prove m∠ BCA > m∠ DAC B A 210 miles D STATEMENTS BC ≅ — DA 1. — AC ≅ — AC 2. — Friend’s flight: C 50° REASONS 130° 1. Given 2. Reflexive Property of Congruence (Thm. 2.1) 3. DC < AB 3. Given 4. m∠ BCA > m∠ DAC 4. Converse of the Hinge Theorem (Thm. 6.13) 80 miles Because 130° > 110°, the distance your friend flew is a greater distance than the distance you flew by the Hinge Theorem (Thm. 6.12). 15. The measure of the included angle in △PSQ is greater than the measure of the included angle in △SQR; By the Hinge Theorem (Thm. 6.12), PQ > SR. 16. Given EF > ED and GD = GF, then m∠ EGF > m∠ DGE by the Converse of the Hinge Theorem (Thm. 6.13). Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 225 Chapter 6 17. The angle bisector of ∠ FEG will also pass through incenter H. 180° = 90°, Then, m∠ HEG + m∠ HFG + m∠ HGF = — 2 because they are each half of the measure of an angle of a triangle. By subtracting m∠ HEG from each side, you can conclude that m∠ HFG + m∠ HGF < 90°. Also, m∠ FHG + m∠ HFG + m∠ HGF = 180° by the Triangle Sum Theorem (Thm. 5.1). So, m∠ FHG > 90°, which means that m∠ FHG > m∠ HFG and m∠ FHG > m∠ HGF. So, FG > FH and FG > HG. — — — — — 18. Because NR is a median, PR ≅ QR . NR ≅ NR by the Reflexive Property of Congruence (Thm. 2.1). So, by the Converse of the Hinge Theorem (Thm. 6.13), m∠ NRQ > m∠ NRP. Because ∠ NRQ and ∠ NRP form a linear pair, they are supplementary. So, ∠ NRQ must be obtuse and ∠ NRP must be acute. 19. 180° − (27° + 102°) = 180° − 129° = 51° 110° > 51° 3x + 2 > x + 3 2x > 1 x > — 12 AD > BC 4x − 3 > 2x 2x > 3 x > — 32 — is the perpendicular bisector of AB . 21. Given CP Prove CA = CB C A P 24. The sum of the measures of the angles of a triangle in spherical geometry must be greater than 180°; The area of πr2 spherical △ABC = — (m∠ A + m∠B + m∠C − 180°), 180° where r is the radius of the sphere. Maintaining Mathematical Proficiency 25. x° + 115° + 27° = 180° x + 142 = 180 x = 38 27. 3x° = 180° 2x = 144 x = 72 28. x° = 44°+ 64° x = 60 x = 108 B 1. Let n be the stage, then the side length of the new triangles in each stage is 24 − n. So, the perimeter of each new triangle is ( 3 24 − n ). The number of new triangles is given by ( 3n − 1) . So, to find the perimeter of all the shaded triangles in each stage, start with the total from the previous stage and add ( 3 24 − n )( 3n − 1 ). The perimeter of the new triangles in stage 4 will be ( 3 24 − 4 )( 34 − 1 ) = 81. The total perimeter of the new triangles and old triangles is 81 + 114 = 195 units. ⋅ ⋅ or CA < CB. By the Reflexive Property of Congruence — — (Thm. 2.1), CP ≅ CP . Also, by the definition of — — perpendicular bisector, AP ≅ BP , ∠ APC is a right angle, and ∠ BPC is a right angle. So, by the Right Angles Congruence Theorem (Thm 2.3), ∠ APC ≅ ∠ BPC. If CA > CB, then by the Converse of the Hinge Theorem (Thm. 6.13), m∠ APC > m∠ BPC, which contradicts the fact that ∠ APC ≅ ∠ BPC. Also, if CA < CB, then by the Converse of the Hinge Theorem (Thm. 6.13), m∠ APC < m∠ BPC, which also contradicts the fact that ∠ APC ≅ ∠ BPC. So, the assumption must be false, and CA = CB. 22. By the Converse of the Hinge Theorem (Thm. 6.13), because 28 ft > 22 ft → AD > AB, then m∠ ACD > m∠ ACB. Geometry Worked-Out Solutions ⋅ 2. x + 5 > 12, x + 12 > 5, 5 + 12 > x; Because the length of the third side has to be a positive value, the inequality x + 12 > 5 will always be true. So, you do not have to consider this inequality in determining the possible values of x. Solve the other two inequalities to find that the length of the third side must be greater than 7 and less than 19. 3. If △ABC is an acute triangle, then m∠ BAC < m∠ BDC and the orthocenter D is inside the triangle. B Assume temporarily that CA ≠ CB. Then either CA > CB 226 26. 2x° + 36° = 180° 6.4–6.6 What Did You Learn? (p. 353) 20. By the Exterior Angle Theorem, m∠ ABD = m∠ BDC + m∠ C. So, m∠ ABD > m∠ BDC. 23. △ ABC is an obtuse triangle; If the altitudes intersect inside the triangle, then m∠ BAC will always be less than m∠ BDC — because they both intercept the same segment, CD . However, because m∠ BAC > m∠ BDC, ∠ A must be obtuse, and the altitudes must intersect outside of the triangle. D A C If △ABC is a right triangle, then m∠ BAC = m∠ BDC because the orthocenter D is on vertex A, where ∠ A is the right angle. B A D C Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 If △ABC is an obtuse triangle, then m∠ BAC > m∠ BDC and the orthocenter D is outside the triangle, where ∠ A is the obtuse angle. 5. Graph △LMN. 4 X(−2, 1) B y D(4, 3) 2 −2 A C −4 Chapter 6 Review (pp. 354–356) — ⊥ AC 1. DC = 20; Point B is equidistant from A and C, and BD . So, by the Converse of the Perpendicular Bisector Theorem (Thm. 6.2), DC = AD = 20. — — , and SP . So, 2. RS = 23; ∠ PQS ≅ ∠ RQS, SR ⊥ QR ⊥ QP by the Angle Bisector Theorem (Thm. 6.3), SR = SP. This means that 6x + 5 = 9x − 4, and the solution is x = 3. So, RS = 9(3) − 4 = 23. and FH 3. m∠ JFH = 47°; Point J is equidistant from FG . So, by the Converse of the Angle Bisector Theorem (Thm. 6.4), m∠ JFH = m∠ JFG = 47°. −6 −4 y U(0, −1) x −2 y=3 Z(6, −3) — midpoint of YZ = — , — The equation of the line perpendicular to YZ is x = 4. = — = −1 Slope of XY : m = — = — = — = — = 0 Slope of YZ : m = — ( 2 +2 6 −3 +2 (−3) ) 8 −6 = ( , ) = (4, −3) 2 2 — — — — — −4 −4 −3 − 1 2 − (−2) 2 + 2 4 — The slope of the line perpendicular to XY is m = 1. , midpoint of XY = — — −1 = 1 −1 = b ( −22+ 2 1 + 2(−3) ) 0 −2 = ( , ) = (0, −1) 2 2 — — — y = mx + b ⋅0 + b — The equation of the line perpendicular to XY through (0, −1) Intersection of x = 4 and y = x − 1: y=4−1=3 So, the coordinates of the circumcenter of △XYZ are (4, 3). is y = x − 1. 2 x = −3 Y(2, −3) x 4. Graph △TUV. 6 −3 − (−3) −3 + 3 0 6−2 4 4 — The slope of the line perpendicular to YZ is undefined. D 2 6. By the Incenter Theorem (Thm. 6.6), x = 5. −2 −4 T(−6, −5) −6 V(0, −5) ( 02 −1 +2 (−5) ) ( 20 −62 ) −6 + 0 −5 + (−5) — , midpoint of TV = ( ) = ( −62 , −102 ) 2 2 — = (0, −3) midpoint of UV = — , — = — , — The equation of the perpendicular bisector of UV through — — = (−3, −5) — — — its midpoint (0, −3) is y = −3, and the equation of the perpendicular bisector of TV — through its midpoint (−3, −5) is x = −3. The point of intersection of the two perpendicular bisectors is (−3, −3). So, the coordinates of the circumcenter of △TVU are (−3, −3). Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 227 Chapter 6 7. 8. y A(−10, 3) F(−6, 3) B(−4, 5) y 4 2 6 −2 6 8 x E(2, −2) G(5, −2) F(8, −2) 4 D(−4, 3) −4 2 E(−7, 2) −10 −8 −6 — −4 −2 C(4, −4) H(2, −5) −6 C(−4, 1) x −8 ( −4 +2 (−4) 5 +2 1 ) −8 6 = ( , ) = (−4, 3) 2 2 D(2, −8) — ( 2 +2 8 −2 +2 (−2) ) 10 −4 = ( , ) = (5, −2) 2 2 , — midpoint of BC = — midpoint of EF = — , — = 0. The slope of AD is —— = — = — = — = 2. The slope of DG is — The equation of AD through (−4, 3) is y = 3. , — midpoint of AC = —— 3 3 5−2 — = 1. The slope of BE is — = — = — −4 − (−7) −4 + 7 3 y = mx + b 2=1 2 = −7 + b 9=b — — — 3−3 −4 − (−10) — — 0 6 ( −10 +2 (−4) 3 +2 1 ) −14 4 = ( , ) = (−7, 2) 2 2 — — ⋅ (−7) + b — — — — −2 = 2 −2 = 10 + b −12 = b — The equation of DG through (5, −2) is y = 2x − 12. midpoint of ED = — , — = — = — = — . The slope of FH is — — ( 2 +2 2 −2 +2 (−8) ) 4 −10 = ( , = (2, −5) 2 2 ) — — — The centroid has the coordinates of the intersection of y = 3 −5 = — 2 + b y=x+9 3=x+9 −6 = x So, the coordinates of the centroid are (−6, 3). 228 Geometry Worked-Out Solutions −2 − (−5) 8−2 −2 + 5 6 3 6 1 2 y = mx + b 6 3 ⋅5 + b The equation of BE through (−7, 2) is y = x + 9. −2 + 8 3 y = mx + b and y = x + 9. −2 − (−8) 5−2 ⋅ 1 2 −5 = 1 + b −6 = b — The equation of FH through (2, −5) is y = — 2 x − 6. The centroid has the coordinates of the intersection of 2x − 12 = — x − 6 3x = 12 x=4 y = — 4 − 6 = 2 − 6 = −4 So, the coordinates of the centroid are (4, −4). 1 1 y = 2x − 12 and y = — 2 x − 6. 1 2 1 2x = — x + 6 2 1 2 2x = 2 — x + 2 6 2 4x = x + 12 ⋅ 1 2 ⋅ ⋅ ⋅ Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 9. 10. 2 5 y= x+4 y 6 y K(−8, 5) 4 G(1, 6) H(5, 6) L(−6, 3) y=x+5 2 J(3, 1) 4 x=3 6 x — 6−6 0 5−1 4 — The slope of the line perpendicular to GH through J(3, 1) is x = 3. = — = 0 slope of GH = — — 6−1 1−3 5 2 = − — slope of GJ = — The slope of the line perpendicular to GJ is — . y = mx + b 4=b — 2 5 ⋅ 2 6 = — 5 + b 5 6=2+b — The equation of the line perpendicular to GJ that passes through H(5, 6) is y = — x + 4. 2 5 2 x = −6 O(3, 5.2) M(0, 5) 2 The orthocenter is the intersection of x = 3 and y = — x + 4. 5 2 y = — 3 + 4 5 6 20 y = — + — 5 5 26 y = — 5 The orthocenter of △GHJ is inside the triangle with 26 coordinates 3, — . 5 ⋅ ( ) −4 x O(−6, −1) — 0 5−5 0 − (−8) 8 — The equation of the line perpendicular to KM through L(−3, 3) is x = −6. = 0 slope of KM = — = — — 5−3 −8 − (−6) 2 −8 + 6 2 −2 = −1 slope of KL = — = — = — The slope of the line perpendicular to KL is 1. y = mx + b — ⋅0 + b 5 =1 5=b y=x+5 — The equation of the line perpendicular to KL that passes The orthocenter is the intersection of x = −6 and y = x + 5. through M(0, 5) is y = x + 5. y=x+5 y = −6 + 5 y = −1 The orthocenter of △KLM is outside the triangle with coordinates (−6, −1). ( ) −12 12 = ( , ) = (−6, 6) 2 2 −6 + 0 8 + 4 −6 12 — , midpoint of AC = ( = , = (−3, 6) 2 2 ) ( 2 2) −6 + 0 4 + 4 −6 8 — , midpoint of BC = ( = , = (−3, 4) 2 2 ) ( 2 2) −6 + (−6) 8 + 4 — 11. midpoint of AB , — = — 2 2 The coordinates of the midsegments of △ABC are (−6, 6), — — — — — — — — — — (−3, 6), and (−3, 4). ( ) ( ) −3 + 1 1 + (−5) — , midpoint of DF = ( ) 2 2 −2 −4 = ( , ) = (−1, −2) 2 2 3 + 1 5 + (−5) — midpoint of EF = ( , ) = ( 42 , 02 ) = (2, 0) 2 2 −3 + 3 1 + 5 0 6 — , = — = (0, 3) 12. midpoint of DE = — , — — 2 2 2 2 The coordinates of the midsegments of △DEF are (0, 3), — — — — — — — — (−1, −2), and (2, 0). Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 229 Chapter 6 13. x + 4 > 8 4. WY = WZAngle Bisector Theorem (Thm. 6.3) x>4 6x + 2 = 9x − 13 −3x = −15 x=5 WY = 6x + 2 = 6 4+8>x The possible lengths for the third side of the triangle with 12 > x, or x < 12 sides 4 inches and 8 inches are 4 in. < x < 12 in. 14. x + 6 > 9 x>3 6+9>x The possible lengths for the third side of the triangle with 15 > x, or x < 15 sides 6 meters and 9 meters are 3 m < x < 15 m. 15. x + 11 > 18 x>7 11 + 18 > x 5. By the Incenter Theorem (Thm. 6.6), the incenter point is equidistant to each side of the triangle. Because WC = 20, BW = 20. 6. AB > CB; By the Hinge Theorem (Thm. 6.12) 7. m∠ 1 < m∠ 2; By the Converse of the Hinge Theorem (Thm. 6.13) 8. m∠ MNP < m∠ NPM; By the Triangle Larger Angle Theorem (Thm. 6.10) 9. The possible lengths for the third side of the triangle with y C(0, 6) 29 > x, or x < 29 y=2 x=2 4 circumcenter E(2, 2) D(1.3, 0.7) centroid sides 11 feet and 18 feet are 7 ft < x < 29 ft. 16. Given △XYZ, XY = 4, and XZ = 8 Prove YZ > 4 Assume temporarily that YZ > 4. Then it follows that either YZ < 4 or YZ = 4. If YZ < 4, then XY + YZ < XZ because 4 + YZ < 8 when YZ < 4. If YZ = 4, then XY + YZ = XZ because 4 + 4 = 8. Both conclusions contradict the Triangle Inequality Theorem (Thm. 6.11), which says that XY + YZ > XZ. So, the temporary assumption that YZ > 4 cannot be true. This proves that in △XYZ, if XY = 4 and XZ = 8, then YZ > 4. 17. Given that m∠ QRT > m∠SRT, by the Hinge Theorem (Thm. 6.12), QT > ST. 18. Given that QT > ST, by the Converse of the Hinge Theorem (Thm. 6.13), m∠ QRT > m∠SRT. Chapter 6 Test (p. 357) 1. By the Triangle Midsegment Theorem (Thm. 6.8), x = — 12 12 = 6. ⋅ 2. By the definition of midpoint, x = 9. RS = STPerpendicular Bisector Theorem (Thm. 6.1) 3x + 8 = 7x − 4 −4x = −12 x=3 ST = 7 3. ⋅ 3 − 4 = 21 − 4 = 17 −2 Geometry Worked-Out Solutions 4 A(0, −2) orthocenter 6 x B(4, −2) Circumcenter: midpoint of AB = — , — The line perpendicular to AB through (2, −2) is x = 2. midpoint of AC = — , — = 0, — = (0, 2) The line perpendicular to AC through (0, 2) is y = 2. The intersection of x = 2 and y = 2 is (2, 2). The coordinates of the circumcenter are (2, 2). Orthocenter: = — = undefined. The slope of AC is — ( 0 +2 4 −2 +2 (−2) ) 4 −4 = ( , ) = (2, −2) 2 2 — — — — — ( 02 6 + 2(−2) ) ( −42 ) — — 6 − (−2) 8 0−0 0 — The equation of the line perpendicular to AC through B(4, −2) is y = −2. −2 − (−2) 0 — = — = 0. The slope of AB is — 4−0 4 — The equation of the line perpendicular to AB through C(0, 6) The intersection of x = 0 and y = −2 is (0, −2). The coordinates of the orthocenter are (0, −2). Centroid: The slope of the line that contains B(4, −2) and the = — midpoint of AC , (0, 2), is — = −1. is x = 0. 230 ⋅ 5 + 2 = 32 — 4 2 − (−2) 0−4 −4 The equation of the line that contains B(4, −2) and the — midpoint of AC , (0, 2), is y = −x + 2. Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 6 The slope of the line that contains C(0, 6) and the midpoint = — = −4. of AB , (2, −2), is — — −2 − 6 −8 2−0 2 The equation of the line that contains C(0, 6) and the — midpoint of AB , (2, −2), is y = −4x + 6. The intersection of y = x + 2 and y = −4x + 6 is: −x + 2 = −4x + 6 3x = 4 4 x = — 3 4 6 2 4 = — y = − — + 2 = − — + — 3 3 3 3 ( 43 23 ) R Prove ∠R ≅ ∠P ≅ ∠Q Assume temporarily that △PQR is equilateral and P Q 11. By the Triangle Midsegment Theorem (Thm. 6.8), GH = — 12 FD. By the markings EG = GD. By the Segment Addition Postulate (Post. 1.2), EG + GD = ED. So, when you substitute EG for GD, you get EG + EG = ED, or 2(EG) = ED, which means that EG = — 12 ED. 1 So, the area of △GEH = — 2 bh 7 + 9 > x The possible lengths of Pine Avenue are 1 = — (EG)(GH) 2 1 1 = — — ED — 12 FD 2 2 1 = — (ED)(FD). 8 Note that the area of △DEF = — 12 bh = — 12 (ED)(FD). So, the area of △GEH = — 18 (ED)(FD) = — 14 — 12 (ED)(FD) ( )( ) [ ] = 12. 1.8 miles hiker 1 4 miles 140° 128° Visitor Center 4 miles 1.8 miles 52° Hiker one: 180° − 40° = 140° Hiker two: 180° − 52° = 128° Because 140° > 128°, the first hiker is farthest from the beach movie theater Chapter 6 Standards Assessment (pp. 358–359) 1. B; ( 1 +2 5 5 +2 2 ) ( 62 72 ) — = — = (3, 3.5) midpoint of MN = — , — , — The slope of the equation that contains the midpoint of MN — through L(3, 8): 8 − 3.5 4.5 = — = undefined slope = — 3−3 0 The equation of the line is x = 3. = — = (4, 5) midpoint of LN = — , — , — The slope of the equation that contains the midpoint of LN 14 A. — your house market — ( 3 +2 5 8 +2 2 ) ( 82 102 ) — through M(1, 5): hiker 2 40° 16 > x 15. To determine the placement of the market, construct the perpendicular bisectors of each side. Where they intersect is the location of the market. equiangular. Then it follows that m∠ P ≠ m∠ Q, m∠ Q ≠ m∠ R, or m∠ P ≠ m∠ R. By the contrapositive of the Base Angles Theorem (Thm. 5.6), if m∠ P ≠ m∠ Q, then PR ≠ QR, if m∠ Q ≠ m∠ R, then QP ≠ RP, and if m∠ P ≠ m∠ R, then PQ ≠ RQ. All three conclusions contradict the fact that △PQR is equilateral. So, the temporary conclusion must be false. This proves that if △PQR is equilateral, it must also be equiangular. 14. 9 miles; Because the path represents the shortest distance from the beach to Main Street, it must be perpendicular to Main Street, and you ended up at the midpoint between your house and the movie theater. So, the trail must be the perpendicular bisector of the portion of Main Street between your house and the movie theater. By the Perpendicular Bisector Theorem (Thm. 6.1), the beach must be the same distance from your house and the movie theater. So, Pine Avenue is the same length as the 9-mile portion of Hill Street between your house and the beach. x > 2 2 miles < x < 16 miles. . The coordinates of the centroid are — , — — ≅ QR — — PQ ≅ PR 10. Given 13. 7 + x > 9 5−5 4−1 0 3 = — = 0 slope = — The equation of the line is y = 5. The intersection of x = 3 and y = 5 is (3, 5). So, the coordinates of the centroid are (3, 5). visitor center, because the longer side is opposite the larger angle. Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 231 Chapter 6 2. G; By the Converse of the Perpendicular Bisector Theorem (Thm. 6.2), because point P is equidistant from points A and — B, it lies on the perpendicular bisector of AB . 3. 16; Because BD = DF and BE = EG, D is a midpoint of — and E is a midpoint of BG — — BF . So, by definition, DE is a midsegment of △BFG. So, by the Triangle Midsegment 1 Theorem (Thm. 6.8), DE = — FG. 2 1 DE = — 2 FG 4 = — 2 FG 2(4) = 2 — 2 FG 8 = FG Also, because BG = GC and BF = FA, G is a midpoint FG = — 2 AC 8 = — 2 AC 2(8) = 2 — 2 AC 16 = AC 8. C; In Step 1, the two points where arcs meet are each equidistant from A and B, so by the Converse of the Perpendicular Bisector Theorem (Thm. 6.2), the constructed — line is the perpendicular bisector of AB . Similarly, in Step 2 — the constructed line is the perpendicular bisector of BC . So, the point where the perpendicular bisectors intersect is the circumcenter, which is equidistant from each vertex. So, in Step 3, the circle is circumscribed about △ABC. 9. J; ∠ 4 forms a linear pair with ∠ 3, so ∠ 4 and ∠ 3 are supplementary. Also, because they are corresponding angles formed by parallel lines, ∠ 3 ≅ ∠ 7. So, ∠ 4 and ∠ 7 are supplementary but not necessarily congruent. So, statement J is not true. 1 ( ) 1 — — — of BC and F is a midpoint of BA . So, by definition, FG is a midsegment of △ABC. So, by the Triangle Midsegment 1 Theorem (Thm. 6.8), FG = — AC. 2 10. A; By the Converse of the Hinge Theorem (Thm. 6.13), m∠ JKM < m∠ LKM. Because m∠ LKM = 25°, the only possible measure for ∠ JKM is 20°. 1 1 ( ) 1 — So, the length of AC is 16 units. 4. A; Reflection in the x-axis (x, y) → (x, −y) : M(3, −1) → Z(3, 1). 5. G; By the Triangle Inequality Theorem (Thm. 6.11), you get x + 16 > 24, x + 24 > 16, and 16 + 24 > x. Disregard x + 24 > 16 because it has a negative solution, and all side lengths will have to be greater than a negative number. Solve x + 16 > 24 to obtain x > 8, and solve 16 + 24 > x to obtain 40 > x. So, the combined solution is 8 < x < 40. — — 6. A; Because YG is a perpendicular bisector of DF , you know that ∠ DEY and ∠ FEY are congruent right angles, and by the Perpendicular Bisector Theorem (Thm. 6.1), YD = YF. So, you can prove △DEY ≅ △FEY by the HL Congruence Theorem (Thm. 5.9). (Note: You can also use the Reflexive Property of Congruence (Thm. 2.1) and the SAS Congruence Theorem (Thm. 5.5).) 7. H; The transformation (x, y) → (3x, 3y) is a dilation because the image will be similar to the original but dilated by a factor of 3. (x, y) → (3x, y) is a horizontal stretch of the original. (x, y) → (−y, x) is a 90° rotation about the origin and represents a rigid motion, so the image is congruent to the original. (x, y) → (x + 3, y + 3) is a translation 3 units right and 3 units up and represents a rigid motion, so the image is congruent to the original. 232 Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved.