HPTER 5 Think & Discuss (p. 6). nswers may vary. Sample answer: Position may be the best position because he would have less space for the ball to pass him. He would also be more toward the middle of the area, so he wouldn t have as far to move to the left or right to intercept the ball.. The m x is approximately 38. n opponent could move closer to the goal to increase the shooting angle. Skill Review (p. 6). Sample answer:. Sample answer: Drawing onclusions. M is the perpendicular bisector of because M M with M between and and M on line D and M is perpendicular to since m 90.. Sample answers: G M F E D 3. Sample answer: 3. 4. 5. M 0, 4 0 0 3 4 0 m y y x x 6. The slope of the line perpendicular to is because. Lesson 5. Developing oncepts ctivity 5. (p. 63) Exploring the oncept. 3. Sample answer 3 4 9 6 5 5 M M 5. m M 90 0 4 0 P, 4 4, 4. Sample answer: M M 3 mm 6. Sample answer: 37 mm Point D Point E Point F Point G D 43 mm E 48 mm D 43 mm E 48 mm The distances from D to the endpoints and are equal, the distances from E to the endpoints and are equal, the distances from F to the endpoints and are equal, and the distances from G to the endpoints and are equal. 4. ny point on the perpendicular bisector has equal distances to the endpoints of the segments. 5. Guided Practice (p. 67). If D is on the perpendicular bisector of, then D is equidistant from and.. Point G must be on the angle bisector of HJK by Theorem 5.4. F 58 mm F 58 mm 3. D D 4. D and D are both right angles and are congruent. 5. because is on the perpendicular bisector of. 6. m LPM m NPM 7. The distance from M to PL is equal to the distance from M to PN. 5. Practice and pplications (pp. 68 7) 8. No; is not on the perpendicular bisector of because the measures of and are not equal. J G 67 mm G 67 mm H K G 80 Geometry
9. No; the diagram does not show that. 8. Statements Reasons 0. No; along with the information given, we would also need P P.. P. Given. No; since P is not equidistant from the sides of, P is not on the bisector of.. P P. Definition of congruent segments. No; since we do not know for sure that one of the distances given is a perpendicular distance. 3. P and P are right angles. 3. Definition of perpendicular lines 3. No; the diagram does not show that the segments with equal length are perpendicular segments. 4. P P 4. Reflexive Property for ongruence. 4. 5. 5. P P 5. SS ongruence Postulate 6. VT 8 7. SR 7 8. Point U must be on the perpendicular bisector SV. 9. NQ 0. Point M must be on the angle bisector JN... 3. D D 5 cm m XTV m TVX 90 (because triangle TVX is a right triangle) m VWU m UVW 90 because triangle UVW is a right triangle. because m VWU m VWX. 4. F 5. D 6. E 7. Given: P is on m. Prove: P. Statements Reasons. P is on line m.. Given. P P. y construction m TV 3090 m XTV 60 m VWX 5090 m VWX 40 3. P P 3. Reflexive Property of ongruence 4. P P 4. SSS ongruence Postulate 5. P P 5. orresponding parts of congruent triangles are congruent. 6. P 6. Theorem 3. D 3 cm 4 cm 4 cm 3 in. D is about.4 inches from each side of. D 6. 6. orresponding parts of congruent triangles are congruent. 7. 7. Definition of congruent segments 9. Statements Reasons. onstruct P at P. Perpendicular Postulate. P and P are. Definition of perpendicular right angles. lines 3. P and P are 3. Definition of right triangles right triangles. 4. 4. Definition of congruent 5. P P 5. Reflexive Property of ongruence 6. P P 6. HL ongruence Theorem 7. P P 7. orresponding parts of congruent triangles are congruent. 8. P is the perpendicular 8. Definition of perpendicular bisector of and is bisector on the perpendicular bisector of. 30. Statements Reasons. GJ is the perpendicular. Given bisector of HK.. GJ HK, HJ JK. Definition of perpencicular bisector of a segment 3. GH GK 3. Perpendicular isector Theorem 4. GH GK, HJ JK 4. Definition of congruent segments 5. GJ GJ, GM GM 5. Reflexive Property of ongruence 6. GJH GJK 6. SSS ongruence Theorem 7. GJH GJK 7. orresponding parts of congruent triangles are congruent. 8. GMH GMK 8. SS ongruence Theorem Theorem Geometry 8
3. The post is the perpendicular bisector between the ends of the wires. 3. Statements Reasons. D is in the interior of. Given.. D is equidistant from. Given and. 3. D D 3. Definition of equidistant 4. D, D 4. Definition of distance from a point to a line 5. D and D are 5. If two lines are, then they right angles form 4 right angles. 6. D D 6. Definition of right angles 7. D D 7. Reflexive Property of ongruence. 8. D D 8. HL ongruence Theorem 9. D D 9. orresponding parts of congruent triangles are congruent. 0. D bisects and 0. Definition of angle bisector point D is on the bisector of. 33. Line l is the perpendicular bisector of. 34. PG should be the angle bisector of P to give the goalie equal distances to travel on both sides. 35. The m P increases as the puck gets closer to the goal. This change makes it more difficult for the goalie because the goalie has a greater area to defend since the distances from goalie to the sides of P (the shooting angle) increase. 36. nswers may vary. Sample answer: D D D D 40 mm D D 48 mm This demonstrates the Perpendicular isector Theorem because D is on the perpendicular bisector of and D is equidistant from and. 37. a c. 38. 39. The perpendicular bisectors meet in one point. d. The fire station at should respond because it is closest to the house at x. WS YX because the product of their slopes is. WT YZ because the product of their slopes is. 3 3. WS 3 6 5 4 3 9 0 WT 5 6 4 3 9 0 40. YW bisects because the perpendicular distances XYZ from W to YS and YT are equal. We know these are perpendicular distances because in problem 38 it was shown that WS YX and WT YZ. 5. Mixed Review (p. 7) slope of WS m 5 4 3 6 slope of WT m 4 5 6 slope of YX m 8 6 3 4 3 3. X slope of YZ m 0 8 3 3 6 3 3 4. d cm r r 6 cm 4. r 3.4 6 37.68 cm 8 Geometry
43. r 0 5 5. Practice and pplications (pp. 75 78) 44. m 3.4 6 5. 6. 3.4 36 5 3.04 cm 5 45. 46. m 5 5 0 0 47. 9 4 5 48. 49. 50. 5. m m m x390 0 704x Lesson 5. 5 3 6 4 0 3 8 8 8 3 59 0 48 4x 6 48 8 8 0 3 0 Developing oncepts ctivity (p. 7). 4. Yes, all three bisectors intersect at the same point. onjecture: For any acute scalene triangle, the three perpendicular bisectors of the three sides will intersect at the same point. P P P 38 mm The three segments that are formed by connecting the vertices of the triangle to the point of intersection of the perpendicular bisectors of the sides have the same length. 5. Guided Practice (p. 75). If three or more points intersect at the same point, the lines are concurrent.. The incenter is made up of the words in and center. The incenter is the center for the circle that is in the triangle. The circumcenter is made up of the parts circum and center. ircum can be short for circumference which is the distance around the circle and can help us to remember it is the circle around the triangle. 3. G G 7 4. MK MJ 5 4 5 0 m 0 8 7 0 5. 8 7 8 7 6 40x 34 x 34 7. The perpendicular bisectors intersect outside the obtuse triangle. The perpendicular bisectors intersect at a point on the right triangle. 0. always. always. never 3. sometimes 4. DR SD 9 5. W W 0 6. J K KJ by the Pythagorean Theorem. 4 K 5 6 K 5 K 9 K 3 K K 3 7. Let the midpoint of PM be called point R. Then PR RM PM 48 4. PR QR PQ by the Pythagorean Theorem. 4 7 PQ 576 49 PQ QN PQ 5. 65 PQ 5 PQ The perpendicular bisectors intersect at a point inside an acute triangle. 8. and 9. Sample answer: The segments are congruent. This confirms Theorem 5.6. 8. The student s conclusion is false because D is not the point of intersection of the angle bisectors. D is the point of intersection of the perpendicular bisectors of the sides of the triangle. So D D D. 9. The student s conclusion is false because the angle bisectors of a triangle intersect in a point that is equidistant from the sides of the triangle, but MQ and MN are not necessarily distances to the sides; M is equidistant from JK, LK, and LJ. Geometry 83 D
0. To find the point that is equidistant from each location, draw the triangle, construct the perpendicular bisectors for each side, and the point of intersection is the point that is equidistant from each location.. Point H is the point of School intersection of the perpendicular bisectors. H So H is equidistant from each location. H would be the best location for the Office new home.. Statements Reasons., the bisectors of. Given,, and, DE, DF, DG. DE DG. D bisects, so D is equidistant from the sides of. 3. DE DF 3. D bisects, so D is equidistant from the sides of. 4. DF DG 4. Transitive property of equality 5. D is on the angle bisector 5. onverse of the ngle of. isector Theorem 6. The angle bisectors 6. Given and Steps, 3, 4 intersect and 5 at a point D that is equidistant from,, and. 3. She could construct the perpendicular bisectors to find the point that is equidistant from the vertices of the triangle. y doing so, she would see that the perpendicular bisectors do intersect at a point on the hypotenuse. Since the point on the hypotenuse would be the point of intersection of the perpendicular bisectors, then it would be equidistant from the vertices. 4. 5. y The radius is approximately (, 5) (4, ) (6, 3) ft. 6. ft 30 inches 30 in. 8 in. per year 3.75 years The mycelium is approximately 3.75 years old. x Factory 7. 8. m m m 80 E XT TW XW by the Pythagorean Theorem because T is the midpoint of XY. 9. The midpoint of is The midpoint of is The midpoint of is 00m m 80 XW WZ 3 XT 3 XT 44 69 0, 0 6 8, 6 0 0 8, 0 0 The slope of 6 0 0 6. The perpendicular bisector has slope because. m m 80 m m 80 m m 40 m D m D 40 m D m D m D 80 XT 5 XT 5 XY XT XY 5 XY 0 m D 4080, 6 30, 6 8, 0 m D 40 6, 3. 5, 3. 9, 0. y 3 6 y 3 The line y 5 is the equation of the perpendicular bisector of. The slope of 0 6 6 8 6. ONTINUED 84 Geometry
9. ONTINUED The perpendicular bisector of has slope because. y 3 5 y 3 5 The line y is the equation of the perpendicular bisector of. The slope of 0 0 so is 8 0 0 0, 8 horizontal. So the perpendicular bisector is the vertical line 9. 30. The lines y 5 and 9 intersect at the point 9, 3 because y 9 5 8 5 3. To show 9, 3 is also on y, plug the values in to see if it is true. y 3 9 3 3 Since 3 3is true, the point 9, 3 is on y. 3. Let P be the point 9, 3. P 9 0 3 0 9 3 8 9 90 9 0 3 0 P 9 3 6 3 9 9 8 90 9 0 3 0 P 9 8 3 0 9 3 8 9 90 9 0 3 0 Since P P P 3 0, P is equidistant from,, and. 5. Mixed Review (p. 78) 3. bh 33. 9 5 45.5 square units 34. j has slope because 3 3. The equation of j is y 3 3 3. 35. j has slope because. The equation of j is y 5. 36. j has slope because 3 3. y 8 3 y 8 3 3 The equation of j is y 3 5. 37. j has slope because 0 0 0. y 9 0 The equation of j is 38. There is enough information to prove DE by using the SS ongruent Postulate. 39. There is not enough information given to prove GJF GJH. One pair of congruent sides, one side congruent to itself, and one pair of congruent angles are given. ut the angles must be the included angles and they are not. 40. There is enough information given to prove PMN KML. One pair of congruent legs and one pair of congruent hypotenuses are given. The HL ongruence Theorem can be used to prove PMN KML. Lesson 5.3 3 y 4 3 y 4 3 3 y 3 3 3 y 6 7 y 6 7 y 5 3 y 3 5 y 9 0 y 9 0 5 y 0 56 5 5.3 Guided Practice (p. 8) bh y 0 56 5. 7 54 77 square units. The centroid of a triangle is the point where the three medians intersect.. The legs, KM and LM, of right KLM are also altitudes of KLM because KM is the perpendicular segment from K to LM and LM is the perpendicular segment from L to KM. Geometry 85
3. DG FG indicates that G is the midpoint of DF. Therefore, EG is a median of DEF. 4. EG DF indicates that EG is an altitude of DEF. 5. DEG FEG indicates the DEF is bisected. So EG is an angle bisector of DEF. 6. EG DF and DG FG indicates G is the midpoint of DF and EG is a perpendicular bisector of DEF, but it is also a median and an altitude. 7. DGE FGE indicates DEG FEG, DGE FGE, and DG FG. In this case EG is a perpendicular bisector, an angle bisector, a median, and an altitude of DEF. 5.3 Practice and pplications (pp. 8 84) 8. FH DH 9 9. EP 3 EH.. 3 8 EH EH HF EF Perimeter of DEF DE EF FD PH EH 4 3 PH EP 4 8 3. 4. Sample answer: 8 3 EH EH 9 EF 44 8 EF L 5 EF 5 EF is an acute triangle. D M DG 5 DH 7.5 5 9 5 5 8 48 units N 0. PH PE EH 5. Yes, they all met at the same point. It is labeled as point D. PH 8 PH 4 6. Sample answer: 7. 8. 9. 0. D 64 mm N 96 D 3 N 64 3 96 64 64 D 54 mm L 8 mm D 3 L 54 3 8 54 54 D 40 mm M 60 mm D 3 M 40 3 60 40 40 The distance from the centroid to a vertex is two thirds of the distance from that vertex to the midpoint of the opposite side. Q, PQ 5 5 0 6 6 PT 3 PQ PT 3 6 PT 4 0 6 0 36 36 The coordinates of T are 5, 6 4 5,. R 5 6, NT 5 6 NR 9 NT NR 3 6 9 3 3 3 0, 0 4, 4,. 8 M 5 3 4, 4, 8, 6 5, 0 86 Geometry
. 3. JP 5 7 6 0 4 4 6 0 4 5 5 JM 4 7 4 0 3 6 9 36 45 9 5 3 5 JP 3 JM 5 3 3 5 5 5 4. Sample answer: P P is the orthocenter of. 8. G and H are the same point. 9. When GH is measured, it is found that GH 0. Since GH 0, then G and H must be the same point; therefore the lines containing the three altitudes intersect at one point. 30. 3. 33. The measure of the angle between r and MP is approximately 0. 34. a. b. r S bh M P D D 9 R D D L D 5 44 D 5 D 8 0 9 90 square units c. This is in the drawing above. 5. Sample answer: E The orthocenter of EFG is point G. d. bh 90 90 E 5 E e. bh bh b h G F 90 7.5 E 6. Sample answer: K P is the orthocenter of KLM. E The length of the altitude is equal to twice the area divided by the base. L M P 7. E D G H F Geometry 87
35. Statements Reasons. is isosceles. Given D is a median to base.. D is the midpoint of.. Definition of median? 3. D D 3. Definition of midpoint? 4. 4. Definition of isosceles triangle 5. D D 5. Reflexive Property of ongruence 6. D D 6. SSS ongruence Postulate 7. D D 7. orresponding parts of congruent triangles are congruent. 8. D and D are a 8. Definition of linear pair linear pair. 9. D 9. If two lines intersect to form a linear pair of congruent angles, then the lines are perpendicular. 0. D is also an altitude. 0. Definition of altitude. 36. No, medians to the legs of an isosceles triangle are not also altitudes. It was the fact that the legs are congruent that made it possible to prove Exercise 35. One leg and the other side would have to be congruent as well for this to be true. 37. Yes, the medians of an equilateral triangle are also altitudes because the proof for the isosceles triangle could be used for the equilateral triangle. Yes, the medians would be contained in the angle bisectors. y looking at the proof in Exercise 35, it can be seen that the median was also the angle bisector since the two triangles are congruent. Yes the medians would be contained in the perpendicular bisectors because it was shown in Exercise 35 that the median was perpendicular to the side at the midpoint. 38. The median of an equilateral triangle is also a perpendicular bisector of a side, an altitude, and an angle bisector. 5.3 Mixed Review (p. 84) 39. The parallel line would also have slope. y 7 y 7 y 8 The equation of the line through P that is parallel to y 3 is y 8. 40. The parallel line would also have slope. y 8 3 y 8 3 y 8 6 y 4 The equation of the line through P that is parallel to y 3 is y 4. 4. The parallel line would also have slope 3. y 9 3 4 y 9 3 y 3 The equation of the line through P that is parallel to y 3 5 is y 3. 4. The parallel line would also have slope. y 4 y y x The equation of the line through P that is parallel to y is y x. 43. E H because you need the angles which do not have DF or GJ as a side. 44. F J because you need the angles which have EF or HJ as a side. 45. Sample answer: y (0, 9) (3, 0) x h 3 0 9 0 3 9 69 8 50 5 0 5 0 Quiz (p. 85). 4 9 3 5. 3y y 4 4 3 6 y 4 6 y 88 Geometry
3. 6 8 VT Extension 36 64 VT 4. VT VS 0 because the perpendicular bisectors intersect at a point equidistant from the vertices of the triangle. 5. The balancing point would be at point G because that is the centroid of the triangle. 5.3 Math and History (p. 85). You need to go to the post office (P), then the market (M), then the library (L) or in reverse order.. The goalie s position on the angle bisector optimizes the chance of blocking a scoring shot because the distance the goalie would have to travel to either side of the angle would be the same. The goalie would not have farther than that distance to either side of himself to block. Technology ctivity 5.3 (p. 86) Investigate 00 VT 0 VT. m F 4 m F 4 So F is the angle bisector of.. F was the point of intersection of the angle bisectors of angles and by construction. Since m F m F, F is the angle bisector of. So the three angle bisectors are concurrent. 3. G 57 mm and G 57 mm. This makes G a median also. 4. Since F was the point of intersection of two medians by construction. Since G G, G is the midpoint of and G is the third median of the triangle. Since F is on G by construction, F is on all three medians and the medians are concurrent. 5. D 7 mm F 08 mm D 7 F 08 3 Yes, D 3 F D 6. No, the quotient does not change because it is the F ratios based on measurements using the centroid. For any triangle in which the angle bisector is contained in the same ine as the median, the line will also contain an altitude and perpendicular bisector of the triangle. In this drawing D is the angle bisector of and D is the median of the triangle. So D D. Since D is the angle bisector, DE DF or DE DF. D D. Then DE DF by the HL ongruence Theorem. Therefore, DE DF or m DE m DF because corresponding parts of congruent triangles are congruent. lso FD ED by the HL ongruence Theorem. So, DF DE or m DF m DE because corresponding parts of congruent triangles are congruent. Then m DE m DF m DE m DE. So m D m D which are linear pairs. Therefore, D D and D D. This would mean that D would be an altitude and a perpendicular bisector, also. Lesson 5.4 5.4 Guided Practice (p. 90). In, if M is the midpoint of, N is the midpoint of, and P is the midpoint of, then MN, NP, and PN are midsegments of triangle.. It is convenient to position one of the sides of the triangle along the x-axis because some of the coordinates of two of the points will be zero and the side will be horizontal and have no slope. 3. JH DF 4. GH DE 5. EF EH 6. GH DE 7. 9. 0.6. DF DG 8 6 GJ EF. 0.6 F E Perimeter GJ JH GH 0.6 8 30.6 D 8. 4 JH DF 6 8 Geometry 89
0... LM 3. MN 4. 5. LN 0 0 MN 7 4 6. N LM 7. 0, 0 8 0, 8 4. 6 6 4 5.4 0 yds 54 yd 3 7 7 6 3 7 7 3 LM 3 7 9. The following pairs of angles are congruent: LN,, and NM because they are corresponding angles, as are NL,, and LM, ML MLN and MN LNM because they are alternate interior angles. So, by the Transitive Property of ongruence, LN,, NM, and LNM are congruent, as are NL,, LM, and NLM. Then, L, LMN, and MN are all congruent by the Third ngles Theorem and the Transitive Property of ongruence. 0. Sample answer:, 8, 4 5 5 4 9 5.4 9 8 LM 3 7 4 4 8 3 8 7 4 7 3 F D E 8., 6, 6 8 9 MN 6 8 3 9 3 8 8 4 6 8 6 4 8 4 8 6 Use the construction of the perpendicular bisectors to find the midpoints D, E, and F. DE, EF, DF are the midsegments for.. Slope of 3. 4. D 0 5, E 5 0 6, F 0 0, 6 Slope of DF 4 0 5 5 Since the slopes of and DF are equal, DF. DF 5 5 4 0 0 5 6 DF.5 DF So.5 4 6.5 6.5 5 8 5 64 89 DF and DF. y D (a, b) (a, b) (0, 0) F (c, 0) (c, 0) F 0 c, 0 0 Slope of DF a c b 0 a c b a c b 0 a c b a c b a c b ONTINUED 6 0 5 4 5 89 4 E(a c, b) 5, 0 5, 4 0, 8 8 5 89 4 x 4 8 5 5 c, 0 5, 0 5, 5, 4 89 c, 0 89 90 Geometry
4. ONTINUED Since the slopes of DF and are equal, DF. 5. 6. Slope of EF Since the slopes of EF and are equal, EF. DF a c b 0 a c b EF a c c b 0 a b a c b 0 a c b 4 a c 4b 4 a c b 4 a c b a c b a 0 b 0 a b 4a 4b 4 a b 4 a b a b DF a c b DF a c b DF EF a b EF a b EF L(, 3) y M(5, 9) a c c b 0 N(4, 4) Slope of LN 4 3 4 3 Draw a line through M with slope 3. Slope of MN 9 4 5 4 5 5 Draw a line through L with slope 5. x a b a 0 b 0 a b a b 7. 8. 9. Slope of LM 9 3 5 6 4 3 3 Draw a line through N with slope. The lines intersect at 0,, 8, 0, and, 8. y N(5, 4) Slope of LN 4 5 7 Draw a line through M with slope 3. Slope of MN 6 4 4 9 5 Draw a line through L with slope. Slope of LM 6 9 7 5 5 Draw a line through N with slope. The lines intersect at 3,,, 3, and 7, 9. GF D 8 D 6 D G G G G because G G. G 5 0 P D D 4 6 0 40 units SU PR PR 4 PR QR QU UR QR QU QU because QU UR. QR QU QR 9 QR 8 M(9, 6) L(7, ) x 3 3 P PR QR PQ 4 8 0 6 units Geometry 9
30. F Perimeter of is 4 times the perimeter of GHI. GH FD GH GH 4 HI DE HI HI 4 GI FE GI GI 4 I D H G E Perimeter of GHI GH HI GI 4 4 4 4 Perimeter of GHI 4 of perimeter of 4 times perimeter of GHI Perimeter of 3. The perimeter of the shaded triangle in Stage is because each side of the shaded triangle is of a side in the original triangle. The total perimeter of the shaded triangle in Stage is 4 4 4 4. The total perimeter of the shaded triangles in Stage 3 is 4 4 4 8 8 8 8 8 8 8 8 8 3 8. 3. The bottoms of the legs will be 60 inches apart. Since the cross bar attaches at the midpoints of the legs, the cross bar is a midsegment of the triangle formed by the two legs and the ground. Since the measure of the midsegments is half of the measure of third side, the measure of the cross bar, 30 inches, is half of the measure between the bottoms of the legs, 60 inches. 33. DE is a medsegment of, so D is the midpoint of and D D. y the Midsegment Theorem, DE and DE. ut F is the midpoint of, so F. Then by transitive property of equality and the definition of congruent segments, DE F. orresponding angles DE and are congruent, so DE DF by the SS ongruence Postulate. 34. In Exercise 33, it was shown that D D and DE F. So all that is left to show is that E DF. This can be done in the same manner that it was shown that DE F. y using the fact that DE is a midsegment, DF y using the fact that E is a midpoint of,. we can get E. Therefore, E DF or E DF and the triangles would be congruent by SSS ongruence Postulate. 35. Since PQ is closer to RS than MN, it must be longer than MN. Since MN is the midsegment, MN or MN RS 4 feet. So PQ cannot be 0 or feet long since it must be longer than MN which is feet long. PQ could be 4 feet long but not 4 feet long since PQ cannot equal RS. So, MN < PQ < RS, or < PQ < 4. 36. a. b. c. d. e. y (, 4) D m 4 3 (3, 7) E(3, ) (, ) y 5 4 y 5 8 y 3 slope of EF m 3 5 3 4 3 slope of FD m 3 5 4 4 The equation of the line containing has slope 3 and passes through D, 4. y 4 3 The equation of the line is y 4 3. The equation of the line containing has slope and passes through E 3,. y 3 The equation of the line is y 3. y 4 3 y 4 3 6 y 3 y 3 y 3 F (4, 5) (5, 3) x y ONTINUED 9 Geometry
f. To find, find the point of intersection of and 38. y 4 n m is the function 4 that gives the length of the 0 6 midsection at Stage n. From one stage to the next, the. has the equation y. has the equation y 3. y substitution, 3 3 5 Midsegment length 8 4 0 3 4 5 6 Stage n length is multiplied by. 37. So,,. To find, find the point of intersection of and. has the equation y. has the equation y 3. y substitution, 3 5 5 5 5 So, 5, 3. To find, find the point of intersection of and. has the equation y 3. has the equation y 3. y substitution, 3 3 3 5 5 5 3 y 3 y 3 3 y 9 7 So, 3, 7. 5 5 5 y 3 y 5 3 y 0 3 y 3 y 3 y 3 y 3 Stage n 0 3 4 5 Midsegment length 4 6 3.5 0.75 5.4 Mixed Review (p. 93) 39. 40. 4. 4. 43. 44. 45. 46. 3 4 ddition property of equality 3 3 46 3 33 Subtraction property of equality Division property of equality 8 7 8 8 ddition property of equality 6 8 Subtraction property of equality 3 Division property of equality 5 9 4 5 9 6 Subtraction property of equality 4 6 Subtraction property of equality 4 Division property of equality 4 4 4 7 Division property of equality 4 8 ddition property of equality Division property of equality 9 3 0 7 3 0 3 Division property of equality 3 7 Subtraction property of equality 7 3 Division property of equality 3 3 0 Subtraction property of equality 0 Division property of equality Subtraction property of equality 3 5 4 3 0 4 Distributive property 5 0 4 Simplify 5 6 Subtraction property of equality 6 5 Division property of equality Geometry 93
47. 3x80 34 80 46 3 48. 0 7 3880 7 6 80 7 9 7 49. 4x6 7 7 4 7 54 3 54 8 50. D D and D D because D, D, and D are angle bisectors and angle bisectors divide an angle into two congruent angles. 5. Point D is the incenter of because it is the intersection of the angle bisectors for. 5. DE DG DF because D is the point of intersection of the angle bisectors of and D is equidistant from the sides of the triangle. 53. DE E D DE 8 0 DE 64 00 DE 36 DE 6 ut DE DF because D is equidistant from the sides of. So, DF DE 6. Lesson 5.5 Technology ctivity 5.5 (p. 94) Investigate. The longest side is opposite the largest angle.. The shortest side is opposite the smallest angle. 3. The answers are the same. The longest side is opposite the largest angle. The shortest side is opposite the smallest angle. 4. The longest side will always be opposite the largest angle. The shortest side will always be opposite the smallest angle. The side with the middle length will be opposite the angle with the middle measure. Extension measure of smallest angle measure of largest angle The statement is false because the above example is a counterexample. 5.5 Guided Practice (p. 98) 4 0.5 8 length of shortest side 63 mm 0.656 length of longest side 96 mm. The inch side is opposite the smallest angle of 8, the 7 8 inches side is opposite the middle angle of 6, and the 8 inches side is opposite the largest angle of 90.. No, it is not possible to draw a triangle with side lengths of 5 inches, inches, and 8 inches because the sum of the lengths of any two sides must be greater than the length of the third side. ut 5 is not greater than 8. 3. m D m E m F 80 3m E 0380 m E 3580 m E 45 The smallest angle is D and the largest angle is F. 4. The shortest side is EF and the longest side is DE. 5. The distances between Guiuan and Masbate have to be between 65 99 66 miles and 65 99 64 miles (not inclusive). 5.5 Practice and pplications (pp. 98 30) 6. m m m 80 m 4780 m 380 m 67 is the shortest side because it is opposite the smallest angle. is the longest side because it is opposite the largest angle. 7. m R m S m T 80 m R 506580 m R 580 m R 65 RT is the shortest side. RS and ST are the longest sides RS ST. 8. m J m H 90 m J 3590 m J 55 JK is the shortest side. HJ is the longest side. 94 Geometry
9. is the smallest angle. is the largest angle. 5. 0. R is the smallest angle. Q is the largest angle.. H is the smallest angle. F is the largest angle.. 4.,, and 5. 6. DF, DE, and EF HJ, JG, and HG 7. L, K, and M 8. N, Q, and P 9. T, S, and R 0. 3. nswers may vary; sample answers are given. 0... 3. The following combinations of lengths will not produce triangles: 4 inches, 4 inches, and 0 inches; 3 inches, 5 inches, and 0 inches; and, inches, 7 inches, and 9 inches. 4. yzx 9030m F 80 m G m J m H 80 m HGJ 03580 5 in. 6 in. m D m E m F 80 0m F 80 8 in. 6 in. 7 in. 7 in. 4 in. 7 in. m G 5580 5 in. m F 60 6 in. 6 in. 8 in. 5 in. > 3 > 3 5 > 3 x > 3 7 x > 7 x < 7 m G 5 in. 8 in. 3. x > y x > z 8.5 in. 5.5 in. 4 in. > 4 > 3 6 > 3 x > 3 7 x > 7 x < 7 6. It is shorter to cut across the empty lot because the sum of the lengths of the two sidewalks is greater than the length of the diagonal across the lot. If the corner of Pleasant Street and Pine Street were labeled point, the corner of Pine Street and Union Street were labeled point, and the corner of Union Street and Oak Hill venue were labeled point, could be formed. We know >, or walking around the sidewalks is longer than walking through the lot. 7. The sides and angles could not be positioned as they are labeled; for example, the longest side is not opposite the largest angle. 8. No, a kitchen triangle cannot have side lengths of 9 feet, 3 feet, and 5 feet because 3 feet 5 feet 8 feet is not larger than 9 feet. 9. The boom is raised when the boom lines are shortened. 30. must be less than 00 50 50 feet. 3. Yes, when the boom is lowered and length of the boom lines,, is greater than 00 feet, then will be larger than. 3. The third inequality would be 4 > 0 and this is not helpful because x has to be positive and 4 is always greater than 0. 33. MJ JN, so MJN is a right triangle. The largest angle in a right triangle is the right angle, so m MJN > m MNJ, so MN > MJ. (If one angle of a triangle is larger than another, then the side opposite the larger angle is longer than the side opposite the smaller angle. Geometry 95
34. Statements Reasons.. Given. Extend to D such. Ruler Postulate that D. 3. D D 3. Segment ddition Postulate 4. 4. ase ngles Theorem 5. m D > m 5. Protractor Postulate 6. m D > m 6. Substitution property of equality 7. D < 7. If the 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. 8. D > 8. Substitution property of equality 9. m PD < m PD 9. Substitution property of equality 35. x > y since the side opposite the angle of x is longer than the side opposite the angle of y n 3 > n. 36. z is the measure of the exterior angle and xyz. 37. D 38. Statements Reasons. P plane M. Given. Let D be a point on. plane contains at least plane M distinct from. three noncollinear points 3. D P 3. Given 4. PD is a right angle. 4. If two lines are perpendicular, then they intersect to form four right angles. 5. PD is a right triangle. 5. Definition of right triangle. 6. PD is an acute angle 6. 7. m PD < 90 7. Definition of an acute angle 8. m PD 90 8. Definition of a right angle 9. m PD < m < PD 9. Substitution of equality 0. PD > P 0. 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. 5.5 Mixed Review (p. 30) 39. 4. nswers may vary. Sample answers are given. 39. The proof for Example on page 30 is a two-column proof. 40. The proof for Example on page 9 is a paragraph proof. 4. The proof for Example 3 on page 58 is a flow proof. 4. 5 and are corresponding angles. So are 5 and 9. 43. and 9 are vertical angles. 44. 6 and 3 are alternate interior angles. So are 6 and. 45. 7 and are alternate exterior angles. So are 7 and 0. 46. slope of LM m 3 4 slope of MN m slope of LN m 3 3 5 5 The line containing has slope 4 and passes through,. y 4 y 4 y 4 8 The line containing has slope and passes through 5 M, 3. y 4 7 y 3 5 y 3 5 4 5 y 9 5 5 The line containing has slope and passes through, N 3,. y 3 y 3 y 5 ONTINUED 3 3 4 4 96 Geometry
46. ONTINUED To find, find the point of intersection of and. y 4 7 y 9 5 5 4 7 9 5 5 4 54 5 5 8 54 5 5 3 y 4 7 y 4 3 7 y 7 y 5 The point has coordinates 3, 5. To find, find the point of intersection of and. y 4 7 y 5 4 7 5 4 9 9 9 y 4 7 y 4 7 y 4 7 y 3 The coordinates of are, 3. To find, find the point of intersection of and. 47. 5 y 9 5 5 y 5 9 5 5 5 5 4 38 5 4 63 9 63 7 y 5 y 7 5 y 7 5 y The coordinates of are 7,. slope of LM m 5 3 slope of MN m 0 6 4 slope of LN m 3 The line containing has slope and passes through L 3, 5. y 5 3 y 5 3 y 5 3 y 3 5 The line containing has slope and passes through 3 M,. y 5 3 5 0 3 6 3 5 3 3 y 5 3 y 5 0 3 3 y 5 6 3 3 ONTINUED Geometry 97
47. ONTINUED The line containing has slope 3 and passes through N 6, 0. y 0 3 6 y 3 6 y 3 8 To find, find the intersection of and. y 3 y 3 8 3 3 8 49 3 7 49 7 y 3 8 y 3 7 8 y 8 y 3 The coordinates of are 7, 3. To find, find the point of intersection of and. y 3 y 5 6 3 3 3 5 6 3 3 3 39 0 3 3 0 7 7 7 y 3 y y y 4 y 7 3 3 The coordinates of are, 7. 48. To find, find the intersection of and. The coordinates of are 5, 3. Slope of MN m 4 5 4 9 8 Slope of LN m 3 6 3 8 The line containing has slope 4 and passes through L 3, 6. y 6 4 3 y 6 4 The line containing has slope and passes through M 9, 5. y 5 9 y 5 9 The line containing has slope and passes through 6 N 8,. y 4 6 y 4 y 8 6 y 6 8 6 y 6 7 3 ONTINUED y 3 8 y 5 6 3 3 3 8 5 6 3 3 3 5 70 3 3 4 70 3 3 5 y 3 8 y 3 5 8 y 5 8 y 3 Slope of LM m 6 5 3 9 6 5 5 6 98 Geometry
48. ONTINUED To find, find the point of intersection of and 49.. y 4 6 y 6 7 3 4 6 6 7 3 4 5 6 3 5 5 6 3 y 4 6 y 4 6 y 8 6 y The coordinates of are,. To find, find the point of intersection of and. y 4 6 y 4 4 6 4 4 0 5 0 4 y 4 6 y 4 4 6 y 6 6 y 0 The coordinates of are 4, 0. To find, find the point of intersection of and. slope of LM m slope of MN m slope of LN m 3 LN is vertical. 4 undefined 0 The line containing has slope and passes through 3 L 3,. y 3 3 y 3 y 3 x The line containing is a vertical line passing through M 0, 4. 0 The line containing has slope and passes through 3 N 3, 6. y 6 3 3 y 6 3 y 3 8 To find, find the point of intersection of and. y 3 x y 3 8 4 3 0 4 6 0 3 6 3 3 3 3 3 y 6 7 3 y 4 6 7 4 3 35 6 3 5 35 6 3 4 y 4 y 4 4 y 0 3 3 8 4 3 8 6 y 3 x y 3 6 y 4 The coordinates of are 6, 4. ONTINUED The coordinates of are 4, 0. Geometry 99
49. ONTINUED To find, find the point of intersection of 8. x > 9 because 70 > 60. and. 9. 3 > 3 because 5 > 45. y 3 x 0 y 3 x y 3 0 y 0 The coordinates of are 0, 0. To find, find the point of intersection of and. y 3 8 0 y 3 8 y 0 8 3 y 0 8 y 8 The coordinates of are 0, 8. Lesson 5.6 5.6 Guided Practice (p. 305). n indirect proof might also be called a proof by contradiction because in an indirect proof, you prove that a statement is true by first assuming that its opposite is true. If this assumption leads to a contradiction, then you have proved that the original statement is true.. To use an indirect proof to show that two lines m and n are parallel, you would first make the assumption that lines m and n are not parallel. 3. m > m > 4. KL < NQ 5. D < FE 6. In, if you wanted to prove that >, you would use the two cases < and in an indirect proof. 5.6 Practice and pplications (pp. 305 307) 7. RS < TU 8. m m 9. m > m 0. XY > ZY. m m. m < m 3. > 4. UT > SV 5. m > m 6. The correct answer is because D D, D D, and > so by the onverse of the Hinge Theorem m 4 < m 5. 7. The correct answer is because D, D D, and m 3 < m 5 so by the Hinge Theorem > D. 3x > x > x > 0. 4 5 < 65 because < 4. 4x < 70 x < 7.5. Given that RS ST in. and ST 5 in., assume that RS 7 in.. Given MNP with Q the midpoint of NP, assume MQ is not a median. 3. Given with m m 90, assume m 90. 4. ssume that there are two points, P and Q, where m and n intersect. Then there are two lines (m and n) through points P and Q. ut this contradicts Postulate 5, which states that there is exactly one line through any two points. D It is false that m and n can intersect in two points, so they must intersect in exactly one point. 5. ase l: ssume EF < DF. 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, so m D m E. ut this contradicts the given information that m D > m E. ase : y the onverse of the ase ngles Theorem, m E m D. ut this contradicts the given information that m D > m E. Since both cases produce a contradiction, the assumption that EF is not greater than DF must be incorrect and EF > DF. 6. ssume m n Then m and n intersect in a point and a triangle is formed. m m m 3 80 by the Triangle Sum Theorem. Then m m 80 m 3 by the subtraction property of equality. ut m m 80 because and are supplementary. So 8080m 3 by substitution property of equality. Then m 3 0 by simplifying both sides. ut this is not possible; angle measures in a triangle cannot be zero. So the assumption that m n must be false. Therefore, m n. 00 Geometry
7. ase : ssume that RS > RT. Then m T > m S by 3. the onverse of the Hinge Theorem. ut Statements Reasons RUS RUT by the S ongruence. DE, EF,. Given Postulate, so S LT or m S m T. This is m > m DEF a contradiction. So RS > RT.. Let P be a point in the. Protractor Postulate ase : ssume RS < RT. Then m T < m S by the interior of such onverse of the Hinge Theorem. ut that m P m EFD RUS RUT by the S ongruence Postulate, so S T or m S m T. This 3. Let P be the point such 3. Ruler Postulate is a contradiction. So RS < RT. that P FD. Therefore RS RT and RST is an isosceles triangle. 4. P EFD 4. Definition of congruent angles 8. The path are described by two triangles in which two sides of one triangle are congruent to two sides of another triangle, but the included angle in your friend s triangle segments 5. P FD 5. Definition of congruent is larger than the included angle of your triangle, so the 6. P DEF 6. SS ongruence Postulate side representing the distance from the airport is longer in 7. P DEF 7. orresponding parts of your friend s triangle. P ED, P DF ongruent Triangles are 9. The paths are described by the two triangles in which two congruent sides of one triangle are congruent to two sides of another triangle, but the included angle in your friend s triangle 8. m P m DEF 8. Definition of congruent segments is larger than the included angle in your triangle, so the side representing the distance to the airport is longer in 9. m P m P 9. ngle ddition Postulate your friend s triangle. m 30. a. s ED increases, m ED increases because it is the 0. Let H be the angle 0. Protractor Postulate angle opposite ED. s ED increases, m D decreases because m ED bisector of P such that H is on. increases and D and ED are supplementary. b. s ED increases, D decreases because as ED. PH H. Definition of angle bisector increases m ED increases and m ED decreases making D decrease.. P. Transitive Property of congruence c. The cleaning arm illustrates the Hinge Theorem because the lengths of E and D remain constant 3. H H 3. Reflexive Property of congruence while m ED and ED change. So there are two triangles, 4. H PH 4. SS ongruence Postulate both being ED where two sides of the first tri- 5. H PH 5. orresponding Parts of angle are congruent to two sides of another triangle ongruent triangles are and the included angle of one triangle is larger than ongruent. the included angle of the other triangle, so the side opposite the larger angle is longer than the side opposite the smaller angle. 6. H PH 6. Definition of congruent segments 7. H H 7. Segment ddition Postulate 8. PH H 8. Substitution of equality 9. PH H > P 9. Triangular Inequality 0. > P 0. Substitution property of equality. P DF. Definition of congruent segments. > DF. Substitution property of equality Geometry 0
5.6 Mixed Review (p. 308) 3. isosceles 33. equilateral, equiangular, and isosceles 34. scalene 35. isosceles 36. equiangular, equilateral, and isosceles 37. isosceles 38. 39. 4. 4. T RU is the median, altitude, angle bisector, and perpendicular bisector. 5.6 Quiz (p. 308). FG E. If FG 8, then E 6. 3. If the perimeter of DE 4, then the perimeter of GHF 4. m L m M m Q 80 5. 6. 3 9 3x m 9 m 3 9 m 5 m m m 80 R m 54580 75m M 7480 LQ, LM, MQ m M m P m Q 80 m M 495080 MQ, MP, PQ m M m N m P 80 m M 487580 m M 380 m M 57 MP, NP, MN 3 3x m 9680 U m 84 S 3 x m M 4980 m M 3 m M 9980 m M 8 40. m 3 m 3 3 m 45 7. DE is longer than because two sides of are congruent to two sides of DEF and the included angle DFE is larger than included angle so DE >. 8. The nd Group is farther from the camp because the groups paths form two triangles with pairs of congruent sides and the included angle for the nd group is larger than the included angle for the st group. Review (pp. 30 3). If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.. If UR UT, then U must be on the perpendicular bisector SQ of RT. 3. If Q is equidistant from SR and ST, then Q is on the angle bisector of RST. 4. Let x abe the midpoint of ST. Then XT or XT ST 3 or XT 6. 5. KX XT KT 6 KT 44 56 KT 400 KT 0 KT ut KR KT 0 because K is equidistant from R, S, and T. W Y WY W 8 0 W 64 00 W 36 W 6 Since W is equidistant from the sides of XYZ, W W 6. 6. The special segments are angle bisectors and the point of concurrency is the incenter 7. The special segments are perpendicular bisectors and the point of concurrency is the circumcenter. 8. The special segments are medians and the point of concurrency is the centroid. 9. The special segments are the altitudes and the point of concurrency is the orthocenter. 0 Geometry
0. midpoint of midpoint of midpoint of m 3 0 0 4 3 4 The equation is m 3 0 0 3 The equation is The centroid is the point of intersection of the two equations. y 3 4 3 y 3 x 3 3 3 4 x 3 9 4 x 4 3 x y 3 x y 3 4 3 y XY 4 0, 0 0 4, 0, 0 XZ 0 0, 6 0 YZ 4 0, 0 6 4, 6, 3 y 0 3 4 4 y 3 4 4 y 3 4 3 y 0 3 0 y 3 x 0, 6 0, 3 The coordinates of the centroid of XYZ are 4 3,.. The coordinates of the orthocenter of XYZ are 0, 0 since XYZ is a right triangle and the two legs are also altitudes of XYZ.. The slope of The slope of The slope of LN m 3 3 4 6 The equation of the line containing HJ has slope and passes through M 8, 3 is y 3 8 y 3 8 y. The equation of the line containing HK has slope passes through 4, 3 is y 3 4 y 3 4 y. The equation of the line containing JK has slope 0 and passes through N 6, is y 0 6 y 0 y. H is the point of intersection of HJ and HK. The coordinates of H are 6, 5 J is the point of intersection of HJ and JK. y y 0 x 0 x y y 6 y y 6 y 5 LM m 3 3 8 4 0 4 0 MN m 3 8 6 The coordinates of J are 0,. ONTINUED Geometry 03
. ONTINUED K is the point of intersection of HK and JK. y y x The coordinates of K are,. 3. Let L be the midpoint of HJ, M be the midpoint of JK, and N be the midpoint of HK. The slope of LM 0 the slope of HK, so LM HK. The slope of MN the slope of HJ, so MN HJ. The slope of LN the slope of JK, so LN JK. 4. G G 9 5. RU UQ 9 G G RU UQ RQ 9 9 9 9 RQ 8 8 RQ GF D ST RQ D ST 8 4 D ST 9 P D D TU PQ P 8 4 TU 4 P 64 TU P ST TU SU P 9 0 P 3 6. The angle measurements in order from least to greatest are,, and. The side measurements in order from least to greatest are,, and. 7. The angle measurements in order from least to greatest are D, E, and F. The side measurements in order from least to greatest are EF, DF, and DE. 8. m J m H m G 80 m J 507080 m J 080 m J 60 The angle measurements in order from least to greatest are H, J, and G. The side measurements in order from least to greatest are GJ, GH, and HJ. 9. m L m K 90 m L 5590 m L 35 The angle measurements in order from least to greatest are L, K, and M. The side measurements in order from least to greatest are KM, LM, and KL. 0. The length of the third side must be less than the sum of the lengths of the other two sides. So the length of the third side must be less than 300 feet 00 00. So the maximum length of fencing needed is 600 feet 00 00 300 of fencing.. <. m < m 3. TU VS 4. In a MPQ, if M Q, then MPQ is not isosceles. 5. ssume has two right angles at and. Then m m 80 and, since m > 0, m m m > 80. This contradicts the Triangle Sum Theorem. Then the assumption that there is such a must be incorrect and no triangle has two right angles. hapter 5 Test (p. 33). If P is the circumcenter of RST, then PR, PS, and PT are always equal.. If D bisects, then D and D are sometimes congruent. 3. The incenter of a triangle never lies outside the triangle. 4. The length of a median of a triangle is sometimes equal to the length of a midsegment. 5. If M is the altitude to side of, then M is sometimes shorter than. 6. a. b. c. d. H 3 G H 3 H HG H 3 H 8 H 3 H 6 3 3 H 6 3 H 6 HG G H HE 3 E 0 H HE 3 HE H HE 3 HE 0 HE 3 HE 0 3 3 HE 0 3 HE 5 F F 9.9 9.9 9.8 6 8 H 36 64 H 00 H 04 Geometry
7. Point H is the centroid of the triangle. 8. G is a(n) median, perpendicular bisector, altitude, and angle bisector of. 9. EF and EF by the Midsegment Theorem. 0. m > m because the side opposite is longer than the side opposite.. To locate the pool so that its center is equidistant from the sidewalks, find the incenter of the triangle by constructing angle bisectors of each angle of the triangle and locating the point of intersection of the bisectors. This point will be equidistant from each sidewalk.. The converse of the Hinge Theorem guarantees that the angles between the legs get larger as the legs are spread apart. 3. The maximum distance between the end of two legs is 0 feet because the length of the third side of the triangle must be less than the sum of the lengths of the other two sides. 4. If m O > m O, then is longer than because two sides of one triangle are congruent to two sides in another triangle and the measure of the included angle of one triangle is larger than the measure of the included angle of the other triangle. 5. Statements Reasons.. Given.. Definition of congruent segments 3. is an isosceles 3. Definition of isosceles triangle triangle 4. 4. ase ngles Theorem 5. m m E 5. ngle ddition Postulate m E O 6. m m E m E 6. Substitution property of equality 7. m < m E 7. part is smaller than the whole. 8. E < E 8. Hinge Theorem 6. ssume m D. Then D because if two angles of a triangle are congruent, then the sides opposite them are congruent. So D by the definition of congruent segments. ut this contradicts the given statement that D. Therefore, the assumption must be false. So m D m. hapter 5 Standarized Test (pp. 34 35). 4 9 3. D 3. 9 4. The midpoint of FG is The centroid is two thirds of the distance from each vertex to the midpoint of the opposite side. ut MH. So H 3 MH The coordinates of are 7, 8 or 7, 3. E 5. KP 6, so PL 6 MK, so NP So P NP PL NL 6. 4 3 30 0 3 30 9 y 4 5y 6 9 y 5y y y 4 H 3 MH H 3 H 8 NP PL NL 6 NL 44 56 NL P 6 0 P 48 400 NL 0 NL M, 4, 7,. 7. The length of the side has to be between 8 6 or inches and 8 6 44 inches. ut it cannot be inches or 44 inches. 8. 9. xy90 because the sum of the measures of the acute angles of a right triangle is 90. Geometry 05
0. x > y because the side opposite G is longer than the side opposite H.. If y, then 45. ut x > y so x > 45.. The location of the point of intersection of the perpendicular bisectors is on GH because GHJ is a right triangle. 3. Let M be the midpoint of. M 0, 0 6 Let N be the midpoint of. N 8, 6 0 Let P be the midpoint of. P 0 8, 0 0 The slope of The equation of N is The slope of P m 6 0 9 6 3 The equation of P is y 0 9 The slope of The equation of M is y 4 9 The centroid is the point of intersection of N, P, and M. y 5 x y 8 8 5 9 5 8 0 y 5 x, 6 30, 6 8, 0 N m 3 0 5 0 y 0 0 5 y 5 x y 8 M m3 3 0 6 8 6, 3 5, 3 9, 0 3 5 5 3 y 0 8 4 4 4. The slope of m 0 0 8 0 0 8 0 The slope of the line perpendicular is undefined. So the line perpendicular to that passes through is the line. The slope of The slope of the line perpendicular to has slope because. The equation of the line perpendicular to and passing through 8, 0 is y 0 8 y 36. The slope of 6 m3 6 0 8 6. The slope of the line perpendicular to has slope because. The equation of the line parallel to and passing through 0, 0 is y 0 0 y x. The orthocenter is the point of intersection of M, N, and P. y 36 y 36 y 4 36 y The coordinates of the orthocenter are,. 5. a. The coordinates of the centroid are 0,. The coordinates of the orthocenter are,. Find the equation of the line passing through the centroid 0, and the orthocenter,, then show that the circumcenter 9, 3 is also on the line. slope m y 5 0 y 5 50 m 6 0 0 0 0 5 6 y 5 48 is the equation of the line Substitute the coordinates of the circumcenter into the equation of the line passing through the centroid and the orthocenter. ONTINUED y 0 5 y The coordinates of the centroid are 0,. 06 Geometry
5. ONTINUED 3 5 9 48 3 45 48 3 3 Since it is equal, the circumference is on the same line as the centroid and the orthocenter. Therefore, they are all collinear. b. The distance from the circumference to the centroid D is D. D 0 9 3 5 5 6 The distance from the circumcenter to the orthocenter P is P. P 9 3 3 5 9 5 34 9 6 3 6 Present Your Results Projects may vary. Extension The conjecture does not work for all four sided shapes. The following is an example for which it was not true. Investigation D 3 P 6 3 3 6 6 6 So the distance from the circumcenter to the centroid is one third the distance from the circumcenter to the orthocenter.. The lines are medians because they are the lines that contain the line segments whose endpoints are a vertex of the triangle and the midpoint of the opposite side.. The balancing point of the triangle is the centroid because it is the point of intersection of the medians. 3. nswers will vary. 4. onjecture: The balancing point of squares, rectangles, parallelograms, and rhombuses is the point of intersection of its diagonals. 5. nswer will vary. Sample answer: tested the conjecture by making more example shapes of each kind. The results were the same each time. The balance points were the points of intersection of the diagonals. Geometry 07