hapter, 7. Given: n Prove: () > () > () > One side, sa, is longer than or at least as long as each of the other sides. Then () and () are true. Proof of (): Statements Reasons. n. Given. Etend to so that >.. Ruler Postulate.. Segment ddition Postulate. >. ase ngles Theorem. m > m. Protractor Postulate 6. m > m 6. Substitution Propert of Equalit 7. > 7. 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. 8. > 8. Substitution Propert of Equalit 9. > 9. Substitution Propert of Equalit 8. a. Given: is a median of n. Prove: < ( ) Statements Reasons. is a median of. Given n... Segment ddition Postulate. > >. >. >. Triangle Inequalit Theorem. If a > b and c > d, then a c > b d. 6. > 6. Simplif.. Substitution Propert b. Given: X, Y, and Z are medians of n. Prove: X Y Z > ( ) Y M X Z Statements. X, Y, and Z are medians of n.. M X, M Z, M Y. M M >, M M >, M M >. X Y >, X Z >, Y Z >. (X Y Z) > 6. X Y Z > ( ) 7. X Y Z > ( ) ( ) 8. X Y Z > ( ) Florida Spiral Review 9.. Given Reasons. oncurrenc of Medians of a Triangle Theorem. Triangle Inequalit Theorem. Substitution. If a > b and c > d, then a c > b d. 6. Multiplication Propert of Inequalit 7. Rewrite right side as a sum. 8. If a > b and c is positive, then a > b c. Lesson.6.6 Guided Practice (pp. 0 ). the Hinge Theorem, RQ is longer than SQ.. the onverse of the Hinge Theorem, SPQ is larger than RPQ. opright Holt Mcougal 7. ( ) > 7. ivision Propert of Inequalit 0
hapter,. 7. m m Group. mi 0 You are given that two sides of one triangle are congruent to two sides of another triangle. The third side of each triangle is shared. Therefore, the two triangles are congruent b SSS. So, m m. 0 N Group mi mi. mi. mi mi 0 Group 8. m > m You are given that one side of one triangle is congruent to one side of another triangle. The side that is shared b each triangle is congruent to itself b the Refleive Propert. ecause > 9, b the onverse of the Hinge Theorem, m > m. 9. ; nklm and nkjm have two pairs of congruent The included angle of the triangle formed for Group is 808 08 08. the Hinge Theorem, Group is the farthest from camp, followed b Group, and finall Group which is the closest to camp.. The temporar assumption ou could make is 9. If ou substitute 9 and into, ou get 9 Þ which is a contradiction. Therefore, Þ 9.. The third side of the first is less than or equal to the third side of the second; ase : Third side of the first equals the third side of the second. ase : Third side of the first is less than the third side of the second. sides. Then because MJ 6 < 8 LM, ou know that m JKM < 8 b the onverse of the Hinge Theorem. 0. You are given that G > GF and ou know that EG > EG b the Refleive Propert. ecause EF > E, b the onverse of the Hinge Theorem, m FGE > m GE.. Suppose that the product is even.. Suppose that is a right angle.. Your stud partner has overlooked the possibilit that could be a right angle or a straight angle.. To use the Hinge Theorem the angle must be the included angle between the two pairs of congruent sides..6 Eercises (pp. ). The Hinge Theorem cannot be used to conclude that Skill Practice XW < XY because the two figures are not triangles.. n indirect proof is also called proof b contradiction because ou begin b assuming temporaril that the statement ou want to prove is false. You reason logicall until ou reach a contradiction.. The name Hinge Theorem is used because two sides and the included angle of a triangle can be thought of as a door with an included angle located at its hinge. s the door opens wider, the included angle gets larger making the third side of the triangle longer. 6. ( )8 < 668 < 6 < 0 7. 808 (08 78) 8 8 < 08, so >. >. > You are given that > and ou know that > b the Refleive Propert. ecause 68 > 8, m > m. So, b the Hinge Theorem, >. > 8. > >. MN < LK opright Holt Mcougal > You are given that JL > PM and JK > PN. ecause 8 < 8, m NPM < m JKL. So, b the Hinge Theorem, MN < LK.. TR < UR You are given that QR > SR and TQ > US. ecause 8 < 8, m TQR < m USR. So, b the Hinge Theorem, TR < UR. know that NR > NR b the Refleive Propert. You are given that NQ > NP. Therefore, b the onverse of the Hinge Theorem, m NRQ > m NRP. ecause NRQ and NRP are a linear pair, NRQ is obtuse and NRP must be acute. 6. m < m Two side of one triangle are congruent to two sides of the other triangle. ecause <, b the onverse of the Hinge Theorem, m < m. 9. ecause NR is a median of nnpq, QR > PR. You N Q R P LHGEFLSOL_c0.indd //09 :8: PM
hapter, 0.ecause m EFG m EGF m E 808, m EFG m EGF < 808. So, m HFG m HGF < 908. Then b the Triangle Sum Theorem, m FHG must be greater than 908. So FHG is larger than HFG or HGF which means that FG must be longer than FH or HG b Theorem.. F E H G. must be outside n. If is on n, then n must be a right n and and are the same verte because the altitudes from and meet at. This assumption implies m m. This contradicts the given statement that m > m. If is inside n, then m > m b Theorem., the Eterior ngle Theorem. Therefore, must be outside of n. is the orthocenter of the triangle, so n is an obtuse triangle. If m > 908 or m > 908, then m m b comparing n X and n Y, or n X and n Y. This contradicts the given statement that m > m. So, n is an obtuse triangle and m > 908. X Y Problem Solving. 0.8 mi First hiker 0 mi N Visitor center mi X Y 8 Second hiker.8 mi ecause 08 > 88, the first hiker is further from the visitor center b the Hinge Theorem.. Given: npqr is equilateral. Prove: npqr is equiangular. E. Temporaril assume that npqr is not equiangular.. That means that for some pair of vertices, sa P and Q, m P > m Q.. Then, b Theorem., ou can conclude that QR > PR.. ut this contradicts the given statement that npqr is equilateral.. The contradiction shows that the temporar assumption that npqr is not equiangular is false. This proves that npqr is equiangular.. Given: m > m E Prove: EF > F F E Proof: ssume temporaril that EF ò F. Then. it follows that either EF < F or EF F. ase : If EF < F, then m < m E b Theorem.0. This contradicts the given statement that m > m E. ase : If EF F, then EF > F. So, > E b the ase ngles Theorem. Therefore, m m E. This contradicts the given statement that m > m E. oth cases lead to contradictions, so the temporar assumption that EF ò F cannot be true. This proves that EF > F.. a. s KL increases, m LNK increases and m KNM decreases. b. ecause m KNM decreases as KL increases, KM decreases. c. Sample answer: In nknl, LN and KN have fied lengths. If m KNL is increased, a new nknl is formed with a longer side LK. This illustrates the Hinge Theorem. 6. Given: line k; point not on k; point on k such that k Prove: is the shortest segment from to k. Proof: ssume temporaril that is not the shortest segment from to k. This implies that there is a point on k such that is the shortest segment from to k. So, > and b Theorem.0, m > m. ecause k, m 908. So, m > 908. ut that means that m m > 808, which contradicts the Triangle Sum Theorem which guarantees that m m m 808. So, the temporar assumption that there is a point on k such that is the shortest segment from to k cannot be true. This proves that is the shortest segment from to k. opright Holt Mcougal
hapter, 7. Given: is divisible b. Statements Prove: is not an odd number. Reasons. PH H > P. Triangle Inequalit Theorem. > F. Substitution Proof: ecause is divisible b, n for some whole number n. So, multipling both sides b gives n. n n, can be divided b, which ecause implies is an even number. definition, an odd number cannot be divided b. Therefore, proving the contrapositive in this wa involves showing that the negation of the original conclusion leads to the negation of the original condition. This is similar to the indirect proof in Eample because there ou temporaril assume the negation of the original conclusion and then work backwards until ou reach a contradiction which happens to be the negative of the original condition. 8. Given: > E, > EF, m > m EF Florida Spiral Review 9. Each week manda rides her bike twice the distance the previous week. So, p 8, meaning manda will ride her bike 8 miles in week. 0. ; 0 0 0 P () () Prove: > F 6 0 0() H 0 F E The perimeter of the kite is 0 units. P Read To Go On? Quiz for..6 (p. ). No, because 6 must be greater than, so Statements. > E, > EF, m > m EF. onstruct P in the interior of so that P > FE and P > E. opright Holt Mcougal. np > n EF the Triangle Inequalit Theorem cannot be satisfied.. Given. onstruction >. 808 (798 88) 8. SS. H > PH. efinition of angle bisector 8 P R The sides from shortest to longest: QR, PQ, PR. > b the Hinge Theorem.. m > m b the onverse of the Hinge Theorem. 6. H > H 6. Refleive Propert 7. nh > nph 8. H > PH, P > F 7. SS 9. H PH, P F > The length of the third side must be greater than ards and less than ards. 79. onstruction 7 > Q. raw the angle bisector ###$ H of P through point H on.. > 7 Reasons s 8. orr. parts of > n are >. 9. efinition of congruent segments 0. H H 0. Segment ddition Postulate. PH H. Substitution Problem Solving onnections (p. 6). a. No, ou cannot conclude that ou and our friend live the same distance from the school if the path bisects the angle formed b Oak and Maple Streets because SS is not sufficient to prove triangles congruent. b. Yes, ou can conclude that ou and our friend live the same distance from the school if the path is perpendicular to irch Street b the Perpendicular isector Theorem. c. triangle is isosceles if its altitude and median are equal in length. ecause the path entrance is halfwa between houses it is a median. ecause it is perpendicular to irch Street, the path is an altitude. LHGEFLSOL_c0.indd //09 :8: PM
hapter,. ecause E, is a median. ecause 6 (6 8), G. Therefore, G is the centroid of the triangle b the oncurrenc of Medians of a Triangle Theorem. Then E is also a median and F FE. So, F 8.. No, the altitude labeled.7 centimeters can t be larger than the side labeled. centimeters because the altitude is a leg of a right triangle in which the side labeled. centimeters is the hpotenuse.. The angle between Gainesvill and atona each is 808 608 08 and the angle between Jacksonville and atona each is 808 08 08. ecause 08 < 08, b the Hinge Theorem Gainesville is closer to atona each.. a. l 6 > 6 > l l > 8 0 > l The length of the third side of the pen must be greater than 8 feet and less than 0 feet. b. 6 feet, feet, feet c. 6 ft 6 ft ft 6 ft 6 ft ft ft ft ft Onl the triangular pen that has side lengths of 6 feet, feet, and feet can accomodate a run of at least feet. The longest side in each of the other two sizes of pens is feet long, and there could not be a run longer than feet in either of those pens. 6. Sample answer: the Triangle Inequalit Theorem, E < 0. 0.6.0. You can use the Pthagorean Theorem to find the length of the hpotenuse of a right triangle with leg lengths of 0. mm and 0.6 mm. 0. 0.6 0.69 ø 0.7 mm. In this triangle, m F > 908, so b the Hinge Theorem, E > 0.7 mm. possible side length for E is 0.8 mm. hapter Review (pp. 9 6). perpendicular bisector is a segment, ra, line, or plane that is perpendicular to a segment at its midpoint.. To draw a circle that is circumscribed about a triangle, first draw at least two perpendicular bisectors of the triangle. The point of concurrenc is equidistant from the triangle s vertices, so it is the center of the circle that passes through all three vertices. raw the circle that passes through the vertices. The center of the circle is called the circumcenter. The radius of the circle is the distance from the circumcenter to a verte.. ; The incenter is the point of concurrenc of the angle bisectors of a triangle.. ; The centroid is the point of concurrenc of the medians of a triangle.. ; The orthocenter is the point of concurrenc of the altitudes of a triangle. 6. EF (7) 6 7. F () 90 E (90) 8. P(a, b) S(a, b) O(0, 0) T(a, 0) Q(a, 0) Slope of ST 0 b a a b a Slope of PO 0 b 0 a b a ecause the slopes of ST and PO are the same, ST i PO. 9. because @##$ bisects. b the Perpendicular isector Theorem. 0. 0 7 7. 6 6(). the oncurrenc of ngle isectors of a Triangle, R T. So.. c a b E 0 6 E 00 E E the oncurrenc of ngle isectors of a Triangle, G E. So,. opright Holt Mcougal
hapter,. 69> > > S R 0. 6 > 9 The length of the third side must be greater than meters and less than meters. M(, 0). > 0 T >8. Sides from smallest to largest b Theorem.: The distance from verte R(, 0) to midpoint M(, 0) is () 6 units. So, the centroid is (6) units to the right of R on RM. The coordinates of the centroid are (, 0), or (0, 0). S > The length of the third side must be greater than 8 feet and less than feet. midpoint M of ST : M(, 0). 0 > RQ, PR, PQ ngles from smallest to largest: P, Q, R. Sides from smallest to largest: LM, MN, LN ngles from smallest to largest b Theorem.0: N, L, M. 808 (908 78) 808 78 8 Sides from smallest to largest b Theorem.:,, T M(, ) R ngles from smallest to largest:,,. Two pairs of sides are congruent and >. So, b 6 midpoint M of RT : M, M(, ) the onverse of the Hinge Theorem, m > m. 6. ecause LK > NM, KLN > MNL, and LN > LN, nkln > nmnl b SS. Therefore, LM KN. The distance from verte S(, 6) to midpoint M(, ) 7. Given: Intersecting lines m and n is 6 units. So, the centroid is () units down from S on SM. Prove: The intersection of lines m and n is eactl one point. The coordinates of the centroid are (, 6 ), or (, ).. 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. 6. XQ XN. ut this contradicts Postulate, which states that through an two points there is eactl one line. QN XN. It is false that m and n can intersect in two points, so the must intersect in eactl one point. So XQ QN () 6 7. XM XY (7). hapter Test (p. 6) 8. Sample answer:. 0. F (6) opright Holt Mcougal. ecause E is the midpoint of and F is the midpoint of, the segment EF is a midsegment of n.. the Perpendicular isector Theorem, SV UV. 8 6 9. > 8 8> > > The length of the third side must be greater than inches and less than inches.. the ngle isector Theorem, PQ RQ. 6 9 9 LHGEFLSOL_c0.indd //09 :8:9 PM
hapter, 6. the onverse of the ngle isector Theorem, m HGJ m KGJ. 8. the concurrenc of Perpendicular isectors of a Triangle, the market is located at the intersection of the three perpendicular bisectors of the triangle formed b our house, the movie theater, and the beach. ( )8 ( )8 7 our house 7. Yes. T is equidistant from endpoints S and U of SU, so b the onverse of the Perpendicular isector Theorem, point T is on the perpendicular bisector of SU. c a b 8. 7 mi 9 mi movie theater beach market 6 76 9 7 the oncurrenc of ngle isectors of a Triangle Theorem, 7. 9. PS LS (6) ollege Entrance Eam Practice (p. 6).. ; ecause F is the midpoint fo, then E F F cm. So, the perimeter of n is PL PS LS P (6) () () PL 6 8 8 PL 8 0. TP TJ. JS JR. E. E 0 TJ RS JS JR 0. ; m R m S m T 808 m R 78 68 808 0 TJ m R 88 808 TP PJ TJ m R 8 0 PJ 0 PJ 0. No. The sum of the measures of an two sides of a triangle must be greater than the third side and 9 ò.. 8 6 ngles from smallest to largest b Theorem.0:,,. ecause the are the included angles between two pairs. LJK and JKM are the included angles between pairs of congruent sides. So, if MJ < LK, then LJK is larger than JKM b the onverse of the Hinge Theorem. 6. temporar assumption is RS 7. 7. 7 > 9 79> > 6 > The distance from the beach to the movie theater must be less than miles and greater than 6 miles. 6 opright Holt Mcougal of congruent sides, if m JKM > m LJK, then MJ is longer than LK b the Hinge Theorem. LHGEFLSOL_c0.indd 6 //09 :8: PM
hapter, hapter lgebra Review (p. 6) wins. a. losses Mastering the Standards (pp. 68 69). P(a, a) The team s win loss ratio is :. losses b. 6 total games The team s loss to total game ratio is :. length of mural. a. 6 height of mural The mural s length to height ratio is :. length of scale drawing. b. 6 length of mural The ratio is : 6. 8 8 0 females. 8 8 males The ratio of females to males in the choir is :. 0 7. 0.6 7 7 The increase is 60%. 0 6. 0. 0. 0.09 The decrease is 9%. 7 80 08 6. 0. 80 80 The decrease is %. 8 6 7. 0. 6 6 The increase is.%. 6 60 8.. The increase is 0%. 0 000 990 9. 0.00 000 000 The decrease is 0.%. 0. The new length is 00% % 0% of the original length. The new length is.0 7 ft 78 ft. Q(a, 0) O(0, 0) OP Ï (a 0) (a 0) Ïa a Ïa aï PQ Ï (a a) (0 a) Ï a a Ïa aï OQ Ï(a 0) (0 0) Ï a a nopq is an isosceles triangle. a0 Slope of OP a0 0a a Slope of PQ a a a Yes, nopq is a right triangle because the product of the slopes of OP and PQ is, so OP PQ.. the Midsegment Theorem, each side of the traffic triangle is times the parallel side of the middle traingle. Therefore, the perimeter of the traffic triangle is times the perimeter of the middle triangle. The traffic triangle has a perimeter of (0 feet), or 0 feet.. the Midsegment Theorem, the midsegment parallel to the floor would be (6 inches), or 8 inches. ecause the support is positioned five inches above where the midsegment would be, the length of the support is less than 8 inches.. It is the area of the original triangle. 8 0 6 6. The new time is 00% 6% 8% of the original time. The new time is 0.8 hours 7.8 hours. opright Holt Mcougal. The new amount is 00% 8% % of the original amount. The new amount is 0. $6,00 $7.. The new number of people is 00% 7.% 07.% of the original number. The new number of people is.07 80 people 86 people. Test Tackler (p. 67). Partial credit; The initial set-up of the problem is correct but no diagram is provided and the conclusion reached is incorrect.. Full credit; The eplanation is complete and valid, a correct diagram is provided, and the answer is correct. For eample, consider a triangle with sides of length 0,, and 6. You can see that the triangle formed b the midsegments is one of four congruent triangles in the interior of the original triangle.. The length of the base of an isosceles triangle with an 8 inch perimeter and a verte angle larger than 608 must be greater than 6 inches and less than 9 inches. n isosceles triangle with a verte angle of 608 would be equiangular and equilateral with sides of lengths 8 6 inches. Then if the verte angle is greater than 608, the base must be greater than 6 inches b Theorem.. lso, if the length of the base is, then the 8 length of each leg is.. No credit; The reasoning and the answer are incorrect. LHGEFLSOL_c0.indd 7 7 //09 :8: PM
hapter, the Triangle Inequalit Theorem,. M(, ) 8 8 > 8 > P(, ) 8 > 9> So, the base is shorter than 9 inches. The Hinge Theorem guarantees that the base must be longer than either side. 0 () 6 Midpoint of : M, M(, ) 6. The correct location for the booth is spot because the perpendicular bisectors of the triangle intersect at. So, b the oncurrenc of Perpendicular isectors of a Triangle Theorem, spot is equidistant from the three buildings. The median from verte must be a vertical segment because it passes through M(, ) and the centroid P(, ). the oncurrenc of Medians of a Triangle 7. n WP and nwp have two pairs of congruent sides and the included angles, WP and WP. Theorem, P M. This leads to P MP. m WP 808 68 88 ecause MP units, P () 6 units. So, is 6 units below P(, ) and its coordinates are (, 6) (, ). So, the -coordinate of the verte is. m WP 808 8 68 ecause m WP < m WP, P < P b the Hinge Theorem. So, the contestants should use point. 8. Sample answer: O c a b NG (6, ) (0, ) 69 NG NG 6 7 (8, ) 8 0 M(9, ) The perpendicular bisector of M is the set of points equidistant from and M. Points on one side of the perpendicular bisector are closer to than to M, and the points on the other side are closer to M than to. 9. Let M be the midpoint of RT. The triangles in the diagram are congruent b SS. ST TM + MS. TM +. 8.9 TM + 6.. TM. ø TM Perimeter of nrst ø.... ø.6 NG NE NG the oncurrenc of ngle isectors of a Triangle Theorem, NE NG.. a. 0 ft 7 ft 00 ft To find the location of the basket, draw the three angle bisectors of the triangle. The point where the angle bisectors intersect is equidistant from each line. b. The oncurrenc of ngle isectors of a Triangle Theorem verifies that the location is correct, because it states that the angle bisectors of a triangle intersect at a point that is equidistant from the sides of the triangle. 0. ; In an obtuse triangle, the circumcenter is outside the triangle.. a. (6, ) P (7, ) (, 0) (8, ). F; The two triangles have two pairs of congruent sides and included angles of 8 ñ 68. ecause 8 < 68, the length of the side opposite the 8 angle must be shorter than the length 8 of the side opposite the 68 angle b the Hinge Theorem. 8 8 6 midpoint of :, (7, ) opright Holt Mcougal. () slope of : 68 LHGEFLSOL_c0.indd 8 //09 :8:7 PM
hapter, b. Slope of perpendicular bisector of p slope c. of Slope of perpendicular bisector of Z (, ) Use the slope to find the intercept. m b J Slope intercept form (7) b Substitute coordinates of. 7 b Multipl. 7 b Subtract from each side. Use the slope and intercept to write an equation. K (, ) P 8 midpoint of JK:, (, ) The distance from verte P(, ) to (, ) is () 6 units. So, the centroid is (6) units up from P on P. The coordinates of the centroid Z are (, ), or Z(, ). c. Sample answer:, Mean of coordinates: When, () 0. 8 () Mean of coordinates: (, 0) Ï( 8) (0 ) Ï0 Yes, the relationship is true for njkp. The coordinates of the centroid are equal to the means of the coordinates and coordinates of the vertices. Ï( 6) (0 ) Ï0 This illustrates the Perpendicular isector Theorem: Point on the perpendicular bisector of is equidistant from the endpoints and. 6. a. K M(, ) J (, ) L 0 () midpoint of JL:, (, ) The distance from verte K(, 8) to (, ) is opright Holt Mcougal 8 () 9 units. So, the centroid is (9) 6 units down from K on K. The coordinates of the centroid M are (, 8 6), or (, ). 0 b. Mean of coordinates:, 8 () Mean of coordinates: The mean of the coordinates of the three vertices is the coordinate of the centroid and the mean of the coordinates of the three vertices is the coordinate of the centroid. LHGEFLSOL_c0.indd 9 9 //09 :8: PM