x = x = XY

Transcription

1 hapter hapter Maintaining Mathematical Proficienc (p. ) m. 8 d. 00 in.. and can be an real number, ; = ; no; bsolute value is never negative.. Vocabular and ore oncept heck (p. 8). ollinear points lie on the same line. oplanar points lie on the same plane.. Monitoring Progress and Modeling with Mathematics (pp. 8 0). Sample answer:,,, E 5. plane S 7. QW, line g 9. R, Q, S; Sample answer: T.. 5. E and E, E and E 7. Sample answer: P 9. Sample answer:. Sample answer:. Sample answer: N M ra ras 55. a. K, N b. Sample answer: plane JKL, plane JQN c. J, K, L, M, N, P, Q 57. sometimes; The point ma be on the line. 59. sometimes; The planes ma not intersect.. sometimes; The points ma be collinear.. sometimes; Lines in parallel planes do not intersect, and ma not be parallel.. Maintaining Mathematical Proficienc (p. 0) = 5 7. =. Vocabular and ore oncept heck (p. ). XY represents the segment XY, while XY represents the distance between points X and Y (the length of XY ).. Monitoring Progress and Modeling with Mathematics (pp. 8) , or about.. 97, or about and are not opposite ras because,, and are not collinear; and are opposite ras because,, and are collinear, and is between and. 7. J 9. Sample answer:. Sample answer:. E 5. point 7. segment 9. P, Q, R, S. K, L, M, N. L, M, Q, R 5. es; Use the point not on the line and two points on the line to draw the plane. 7. Three legs of the chair will meet on the floor to define a plane, but the point at the bottom of the fourth leg ma not be in the same plane. When the chair tips so that this leg is on the floor, the plane defined b this leg and the two legs closest to it now lies in the plane of the floor; no; Three points define a plane, so the legs of the three-legged chair will alwas meet in the flat plane of the floor. 9. ; The first two lines intersect at one point. The third line could intersect each of the first two lines. The fourth line can be drawn to intersect each of the first lines. Then the total is + + =. not congruent. The difference should have been taken; =.5 =.5. a. 88 mi b. about 50 mi/h 5. a. about 0. m; about 9. m b. about 8.9 m 7. a. + = ; = 5; RS = 0; ST = ; RT = b. 7 = 0; = ; RS = 0; ST = 0; RT = 0 c. + = + 0; = 7; RS = ; ST = ; RT = 7 d. + 0 = 8 ; = ; RS = 5; ST = 9; RT = 9. a. ft b. about 0. min c. few etra steps might be needed if other people are in the hall mi; If the round-trip distance is 7 miles, then the one-wa distance is.5 miles..5 7 = 9.5. =, =, = 9, =, =, = ; ; Two of the segments are units long. The other four are longer than that.

2 . Maintaining Mathematical Proficienc (p. 8) = 5. =. Vocabular and ore oncept heck (p. ). It bisects the segment.. Monitoring Progress and Modeling with Mathematics (pp. ). line k; 5. M; 7. M; 0 9. MN ;. M. E M (5, ) 5. (, 9 ) 7. (, ) 9. (8, 9). for a 5: ratio, should have used a = and b = 5; ( ) + 5(0) = QR = 7 MR = 8 5. House Librar School 5.7 km.85 km ( a + b ), c. location for lunch; The total distance traveled if ou return home is M + M + +. The total distance traveled if ou go to location for lunch is ecause < M, the second option involves less traveling.. cm. Maintaining Mathematical Proficienc (p. ) 5. ft, 0 ft 7. d, 0 d z F. a. L W b. about 0.7 mi c. about 7. mi. a. and b. (0, ), (, ), (, ) c. about 5.7 units, 0 square units 5. a. units, square units b. es; The sides are all the same length because each one is the hpotenuse of a right triangle with legs that are each units long. ecause the slopes of the lines of each side are either or, the are perpendicular. c. about. units, 8 square units; It is half of the area of the larger square. 7. =. Maintaining Mathematical Proficienc (p. ) 9. =. =. =.5 Vocabular and ore oncept heck (p. ). congruent.5 Monitoring Progress and Modeling with Mathematics (pp. ).,, 5., K, JKL (or LKJ) 7. HMK, KMN, HMN 9. 0 ; acute. 85 ; acute. The outer scale was used, but the inner scale should have been used because O passes through 0 on the inner scale; E,,,, , , 0 9., Vocabular and ore oncept heck (p. ). s. Monitoring Progress and Modeling with Mathematics (pp. ). quadrilateral; concave 5. pentagon; conve 7. units 9. about. units. about.9 units. 7.5 square units 5. 9 square units 7. about 9. units 9. about.7 units. square units. square units 5. The length should be 5 units; P = + w = (5) + () = ; The perimeter is units a. square units; square units; It is quadrupled. b. es; If ou double the side length and square it, then the new area will be = times as big.., 5., 7.,, , 5, 0. Subtract m from m to find m , a. + X b. + + = 9, 7, 9. a. acute b. acute c. acute d. right 5. a. Sample answer: (, ) b. Sample answer: (0, ) c. Sample answer: (, ) d. Sample answer: (, 0)

3 5. acute, right, or obtuse; The sum of the angles could be less than 90 (eample: = 50 ), equal to 90 (eample: = 90 ), or greater than 90 (eample: = 00 ). 55. Sample answer: You draw a segment, ra, or line in the interior of an angle so that the two angles created are congruent to each other; ngle bisectors and segment bisectors can be segments, ras, or lines, but onl a segment bisector can be a point. The two angles/segments created are congruent to each other, and their measures are each half the measure of the original angle/segment. 57. acute; It is likel that the angle with the horizontal is ver small because levels are tpicall used when something appears to be horizontal but still needs to be checked..5 Maintaining Mathematical Proficienc (p. ) 59. =. = 7. = 5. = 0. Vocabular and ore oncept heck (p. 5). djacent angles share a common ra, and are net to each other. Vertical angles form two pairs of opposite ras, and are across from each other.. Monitoring Progress and Maintaining Mathematical Proficienc (pp. 5 5). LJM, MJN 5. EGF, NJP m QRT = 7, m TRS =. m UVW =, m XYZ = and 5 7. es; The sides form two pairs of opposite ras. 9. 0, 0. 9, 8. The do not share a common ra, so the are not adjacent; and are adjacent ( + ) = 90; and 5. + ( 5 ) = 80; 7 and 7. alwas; linear pair forms a straight angle, which is sometimes; This is possible if the lines are perpendicular.. alwas; = 90. The measure of an obtuse angle is greater than 90. So, ou cannot add it to the measure of another angle and get a. 50, 0, 0 b. ; ecause all angles have supplements, the first paper can be an angle. Then there is a in chance of drawing its supplement. 7. es; ecause m KJL + = 90 and m MJN + = 90, it must be that m KJL + = m MJN +. Subtracting from each side of the equation results in the measures being equal. So, the angles are congruent. 9. a., (80 ), (80 ) b. The are alwas congruent; The are both supplementar to the same angle. So, their measures must be equal. 5. 7, 5 ; If two angles are complementar, then their sum is 90. If is one of the angles, then (90 ) is the complement. Write and solve the equation 90 = ( (90 )) + 7. The solution is = 5.. Maintaining Mathematical Proficienc (p. 5) 5. never; Integers are positive or negative whole numbers. Irrational numbers are decimals that never terminate and never repeat. 55. never; The whole numbers are positive or zero. 57. alwas; The set of integers includes all natural numbers and their opposites (and zero). 59. sometimes; Irrational numbers can be positive or negative. hapter Review (pp. 5 58). Sample answer: line h. Sample answer: XZ, YP. YX and YZ. P about about no 0. P = 7, M =. P = 9, M = 5. ( 0.5,.5). (, ). (, ) units, square units 7. about.9 units,.5 square units 8. 9, , hapter hapter Maintaining Mathematical Proficienc (p. ). a n = n ; a 50 = 97. a n = 7n ; a 50 = 80. a n = 0.n +.; a 50 =.. a n = n + ; a 50 = 7, or 8 5. a n = n + 0; a 50 = 70. a n = n + ; a 50 = 8 7. = 5 8. = + 9. = 0. = 7. = z. = z + +. no; The sequence does not have a common difference.. Vocabular and ore oncept heck (p. 7). a conditional statement and its contrapositive, as well as the converse and inverse of a conditional statement. Monitoring Progress and Modeling with Mathematics (pp. 7 7). If a polgon is a pentagon, then it has five sides. 5. If ou run, then ou are fast. 7. If =, then =. 9. If ou are in a band, then ou pla the drums.. If ou are registered, then ou are allowed to vote.

4 . The sk is not blue. 5. The ball is pink. 7. conditional: If two angles are supplementar, then the measures of the angles sum to 80 ; true converse: If the measures of two angles sum to 80, then the are supplementar; true inverse: If the two angles are not supplementar, then their measures do not sum to 80 ; true contrapositive: If the measures of two angles do not sum to 80, then the are not supplementar; true 9. conditional: If ou do our math homework, then ou will do well on the test; false converse: If ou do well on the test, then ou did our math homework; false inverse: If ou do not do our math homework, then ou will not do well on the test; false contrapositive: If ou do not do well on the test, then ou did not do our math homework; false. conditional: If it does not snow, then I will run outside; false converse: If I run outside, then it is not snowing; true inverse: If it snows, then I will not run outside; true contrapositive: If I do not run outside, then it is snowing; false. conditional: If 7 = 0, then = 9; true converse: If = 9, then 7 = 0; true inverse: If 7 0, then 9; true contrapositive: If 9, then 7 0; true 5. true; definition of right angle, the measure of the right angle shown is true; If angles form a linear pair, then the sum of the measures of their angles is point is the midpoint of a segment if and onl if it is the point that divides the segment into two congruent segments.. Two angles are adjacent angles if and onl if the share a common verte and side, but have no common interior points.. polgon has three sides if and onl if it is a triangle. 5. n angle is a right angle if and onl if it measures Taking four English courses is a requirement regardless of how man courses the student takes total, and the courses do not have to be taken simultaneousl; If students are in high school, then the will take four English courses before the graduate. 9. p q p p q. T T F T T F F T F T T T F F T F p q p q p q ( p q) T T F F T F T F F T T F F T T F F T F F T T T F. p q p q p T T F F T F F T F T T T F F T T 5. a. If a rock is igneous, then it is formed from the cooling of molten rock; If a rock is sedimentar, then it is formed from pieces of other rocks; If a rock is metamorphic, then it is formed b changing temperature, pressure, or chemistr. b. If a rock is formed from the cooling of molten rock, then it is igneous; true; ll rocks formed from cooling molten rock are called igneous. If a rock is formed from pieces of other rocks, then it is sedimentar; true; ll rocks formed from pieces of other rocks are called sedimentar. If a rock is formed b changing temperature, pressure, or chemistr, then it is metamorphic; true; ll rocks formed b changing temperature, pressure, or chemistr are called metamorphic. c. Sample answer: If a rock is not sedimentar, then it was not formed from pieces of other rocks; This is the inverse of one of the conditional statements in part (a). So, the converse of this statement will be the contrapositive of the conditional statement. ecause the contrapositive is equivalent to the conditional statement and the conditional statement was true, the contrapositive will also be true. 7. no; The contrapositive is equivalent to the original conditional statement. In order to write a conditional statement as a true biconditional statement, ou must know that the converse (or inverse) is true. 9. If ou tell the truth, then ou don t have to remember anthing. 5. If one is luck, then a solitar fantas can totall transform one million realities. 5. no; If 0 = +, then = is a false statement because = is also possible. The converse, however, of the original conditional statement is true. In order for a biconditional statement to be true, both the conditional statement and its converse must be true If toda is Februar 8, then tomorrow is March. 59. a. ats Lions If ou see a cat, then ou went to the zoo to see a lion; The original statement is true, because a lion is a tpe of cat, but the converse is false, because ou could see a cat without going to the zoo.

5 b. c. Helmet Sport If ou wear a helmet, then ou pla a sport; oth the original statement and the converse are false, because not all sports require helmets and sometimes helmets are worn for activities that are not considered a sport, such as construction work. Februar (8 or 9 das) Months with das Months with 0 das If this month is not Februar, then it has das; The original statement is true, because Februar never has das, but the converse is false, because a month that is not Februar could have 0 das.. Sample answer: If the are vegetarians, then the do not eat hamburgers.. Sample answer: slogan: This treadmill is a fat-burning machine! conditional statement: If ou use this treadmill, then ou will burn fat quickl.. Maintaining Mathematical Proficienc (p. 7) 5. add a square that connects the midpoints of the previousl added square; 7. add ; 5, 7 9.,,,...; 5,. Vocabular and ore oncept heck (p. 80). conjecture is an unproven statement that is based on observations. postulate is a rule that is accepted without proof.. Monitoring Progress and Modeling with Mathematics (pp. 80 8). The absolute value of each number in the list is greater than the absolute value of the previous number in the list, and the signs alternate from positive to negative;, 7 5. The list items are letters in backward alphabetical order; U, T 7. This is a sequence of regular polgons, each polgon having one more side than the previous polgon. 9. The product of an two even integers is an even integer. Sample answer: () = 8, () = 7, 8(0) = 80. The quotient of a number and its reciprocal is the square of that number. Sample answer: 9 9 = 9 9 = 9, = = ( ), 7 7 = 7 7 = ( 7 ). 5 = 5, 5 > 5 5. The could both be right angles. Then, neither are acute. 7. You passed the class. 9. not possible. not possible. If a figure is a rhombus, then the figure has two pairs of opposite sides that are parallel. 5. Law of Sllogism 7. Law of etachment 9. The sum of two odd integers is an even integer; Let m and n be integers. Then (m + ) and (n + ) are odd integers. (m + ) + (n + ) = m + n + = (m + n + ); (m + n + ) is divisible b and is therefore an even integer.. inductive reasoning; The conjecture is based on the assumption that a pattern, observed in specific cases, will continue.. deductive reasoning; Laws of nature and the Law of Sllogism were used to draw the conclusion. 5. The Law of etachment cannot be used because the hpothesis is not true; Sample answer: Using the Law of etachment, because a square is a rectangle, ou can conclude that a square has four sides. 7. Using inductive reasoning, ou can make a conjecture that male tigers weigh more than female tigers because this was true in all of the specific cases listed in the table. 9. n(n + ) = the sum of first n positive even integers. rgument ; This argument uses the Law of etachment to sa that when the hpothesis is met, the conclusion is true.. The value of is more than three times the value of ; = + ; Sample answer: If = 0, then = (0) + = ; If = 7, then = (7) + = a. true; ased on the Law of Sllogism, if ou went camping at Yellowstone, and Yellowstone is in Woming, then ou went camping in Woming. b. false; When ou go camping, ou go canoeing, but even though our friend alwas goes camping when ou do, he or she ma not choose to go canoeing with ou. c. true; It is known that if ou go on a hike, our friend goes with ou. It is also known that ou went on a hike. So, based on the Law of etachment, our friend went on a hike. d. false; It is known that ou and our friend went on a hike, but it is not known where. It is onl known that there is a -mile-long trail near where ou are camping.. Maintaining Mathematical Proficienc (p. 8) 7. Segment ddition Postulate (Post..) 9. Ruler Postulate (Post..). Vocabular and ore oncept heck (p. 87). three. Monitoring Progress and Modeling with Mathematics (pp ). Two Point Postulate (Post..) 5. Sample answer: Line q contains points J and K. 5

6 7. Sample answer: Through points K, H, and L, there is eactl one plane, which is plane M. 9.. Y P W m. es 5. no 7. es 9. es. In order to determine that M is the midpoint of or, the segments that would have to be marked as congruent are M and M or M and M, respectivel; ased on the diagram and markings, ou can assume and intersect at point M, such that M M and M M..,, F, H 5. Two Point Postulate (Post..) 7. a. If there are two points, then there eists eactl one line that passes through them. b. converse: If there eists eactl one line that passes through a given point or points, then there are two points; false; inverse: If there are not two points, then there is not eactl one line that passes through them; false; contrapositive: If there is not eactl one line that passes through a given point or points, then there are not two points; true 9. <. es; For eample, the ceiling and two walls of man rooms intersect in a point in the corner of the room.. Points E, F, and G must be collinear. The must be on the line that intersects plane P and plane Q; Points E, F, and G can be either collinear or noncollinear. P E F Q G. Maintaining Mathematical Proficienc (p. 88) 5. t = ; ddition Propert of Equalit 7. = ; Subtraction Propert of Equalit. Vocabular and ore oncept heck (p. 9). Refleive Propert of Equalit. Monitoring Progress and Modeling with Mathematics (p. 9 98). Subtraction Propert of Equalit; ddition Propert of Equalit; ivision Propert of Equalit 5. Equation Eplanation and Reason 5 0 = 0 Write the equation; Given 5 = 0 dd 0 to each side; ddition Propert of Equalit = ivide each side b 5; ivision Propert of Equalit X E F G V P Q 7. Equation Eplanation and Reason 8 = 0 Write the equation; Given 8 = 0 Subtract from each side; Subtraction Propert of Equalit = dd 8 to each side; ddition Propert of Equalit = ivide each side b ; ivision Propert of Equalit 9. Equation Eplanation and Reason 5( 0) = 0 Write the equation; Given 5 00 = 0 Multipl; istributive Propert 5 = 90 dd 00 to each side; ddition Propert of Equalit = ivide each side b 5; ivision Propert of Equalit. Equation Eplanation and Reason ( 5) = Write the equation; Given 0 = Multipl; istributive Propert = dd 0 to each side; ddition Propert of Equalit = ivide each side b ; ivision Propert of Equalit. Equation Eplanation and Reason (5 9) = ( + 7) Write the equation; Given 0 = Multipl on each side; istributive Propert = dd to each side; ddition Propert of Equalit = dd to each side; ddition Propert of Equalit = ivide each side b ; ivision Propert of Equalit 5. Equation Eplanation and Reason 5 + = 8 Write the equation; Given = Subtract 5 from each side; Subtraction Propert of Equalit 7. Equation Eplanation and Reason = Write the equation; Given = Subtract 0.5 from each side; Subtraction Propert of Equalit = ivide each side b ; ivision Propert of Equalit 9. Equation Eplanation and Reason = 0 + Write the equation; Given = 0 Subtract from each side; Subtraction Propert of Equalit = 0 + ivide each side b ; ivision Propert of Equalit. Equation Eplanation and Reason = πr Write the equation; Given π = r r = π ivide each side b π ; ivision Propert of Equalit Rewrite the equation; Smmetric Propert of Equalit

7 . Equation Eplanation and Reason S = 80(n ) Write the equation; Given S 80 = n ivide each side b 80; ivision Propert of Equalit S + = n dd to each side; ddition 80 Propert of Equalit S n = + Rewrite the equation; Smmetric 80 Propert of Equalit 5. Multiplication Propert of Equalit 7. Refleive Propert of Equalit 9. Refleive Propert of Equalit. Smmetric Propert of Equalit EF 7. XY GH 9. m = m. The Subtraction Propert of Equalit should be used to subtract from each side of the equation in order to get the second step. 7 = + Given = Subtraction Propert of Equalit = ivision Propert of Equalit. Equation Eplanation and Reason P = + w Write the equation; Given P w = Subtract w from each side; Subtraction Propert of Equalit P w = ivide each side b ; ivision Propert of Equalit = P w Rewrite the equation; Smmetric Propert of Equalit = m 5. Equation Eplanation and Reason m = m E Write the equation; Given m = m + m dd measures of adjacent angles; ngle ddition Postulate (Post..) m E = m + m dd measures of adjacent angles; ngle ddition Postulate (Post..) m = m + m Substitute m for m E; Substitution Propert of Equalit m + m = m + m Substitute m + m for m ; Substitution Propert of Equalit m = m Subtract m from each side; Subtraction Propert of Equalit 7. Transitive Propert of Equalit; ngle ddition Postulate (Post..); Transitive Propert of Equalit; m + m = m + m ; Subtraction Propert of Equalit 9. Equation Eplanation and Reason =, = = Marked in diagram; Given is equal to itself; Refleive Propert of Equalit + + = + + dd + to each side of = ; ddition Propert of Equalit + + = + + Substitute for and for ; Substitution Propert of Equalit 5. ZY = XW = 9 5.,, F 55. a. Equation Eplanation and Reason 9 5 b. c. = 5 ( F ) Write the equation; Given 9 5 = F Multipl each side b 9 5 ; 9 Multiplication Propert of Equalit + = F dd to each side; ddition Propert of Equalit egrees Fahrenheit ( F) F = 9 + Rewrite the equation; Smmetric 5 Propert of Equalit egrees elsius ( ) egrees Fahrenheit ( F) egrees elsius ( ) Yes, it is a linear function.. Maintaining Mathematical Proficienc (p. 98) 57. Segment ddition Postulate (Post..) 59. midpoint.5 Vocabular and ore oncept heck (p. 0). postulate is a rule that is accepted to be true without proof, but a theorem is a statement that can be proven..5 Monitoring Progress and Modeling with Mathematics (pp. 0 0). Given; ddition Propert of Equalit; PQ + QR = PR; Transitive Propert of Equalit 5. Transitive Propert of Segment ongruence (Thm..) 7. Smmetric Propert of ngle ongruence (Thm..) 9. Smmetric Propert of Segment ongruence (Thm..) 7

8 . STTEMENTS RESONS. segment eists with endpoints and.. equals the length of the segment with endpoints and.. Given. Ruler Postulate (Post..). =. Refleive Propert of Equalit.. STTEMENTS RESONS. GFH GHF. Given. efinition of congruent segments. m GFH = m GHF. efinition of congruent angles. EFG and GFH form a linear pair.. EFG and GFH are supplementar. 5. m EFG + m GFH = 80. m EFG + m GHF = EFG and GHF are supplementar.. Given (diagram). efinition of linear pair 5. efinition of supplementar angles. Substitution Propert of Equalit 7. efinition of supplementar angles 5. The Transitive Propert of Segment ongruence (Thm..) should have been used; ecause if MN LQ and LQ PN, then MN PN b the Transitive Propert of Segment ongruence (Thm..). 7. equiangular; ecause and, b the Transitive Propert of ngle ongruence (Thm..). ecause all three angles are congruent, the triangle is equiangular. (It is also equilateral and acute.) 9. The purpose of a proof is to ensure the truth of a statement with such certaint that the theorem or rule proved could be used as a justification in proving another statement or theorem. ecause inductive reasoning relies on observations about patterns in specific cases, the pattern ma not continue or ma change. So, the ideas cannot be used to prove ideas for the general case.. a. It is a right angle. b. STTEMENTS RESONS. m + m + m + m = 80. ngle ddition Postulate (Post..). (m + m ) = 80. istributive Propert. STTEMENTS RESONS. QR PQ, RS PQ,. Given QR = + 5, RS = 0. QR = PQ, RS = PQ. efinition of congruent segments. QR = RS. Transitive Propert of Equalit. + 5 = 0. Substitution Propert of Equalit = 0 5. ddition Propert of Equalit. 5 = 5. Subtraction Propert of Equalit 7. = 7. ivision Propert of Equalit.5 Maintaining Mathematical Proficienc (p. 0) 5.. Vocabular and ore oncept heck (p. ). ll right angles have the same measure, 90, and angles with the same measure are congruent.. Monitoring Progress and Modeling with Mathematics (pp. ). MSN PSQ b definition because the have the same measure; MSP PSR b the Right ngles ongruence Theorem (Thm..). The form a linear pair, which means the are supplementar b the Linear Pair Postulate (Post..8), and because one is a right angle, so is the other b the Subtraction Propert of Equalit; NSP QSR b the ongruent omplements Theorem (Thm..5) because the are complementar to congruent angles. 5. GML HMJ and GMH LMJ b the Vertical ngles ongruence Theorem (Thm..); GMK JMK b the Right ngles ongruence Theorem (Thm..). The form a linear pair, which means the are supplementar b the Linear Pair Postulate (Post..8), and because one is a right angle, so is the other b the Subtraction Propert of Equalit. 7. m = 7, m =, m = 7 9. m =, m =, m =. =, = 7. =, = 9 5. The epressions should have been set equal to each other because the represent vertical angles; ( + 5) = (9 + ) + 5 = = = 7. m + m = 90. ivision Propert of Equalit 8

9 7. Transitive Propert of ngle ongruence (Thm..); Transitive Propert of ngle ongruence (Thm..) STTEMENTS RESONS.. Given.,. Vertical ngles ongruence Theorem (Thm..).. Transitive Propert of ngle ongruence (Thm..).. Transitive Propert of ngle ongruence (Thm..) 9. complementar; m + m ; Transitive Propert of Equalit; m = m ; congruent angles STTEMENTS. and are complementar. and are complementar.. m + m = 90, m + m = 90. m + m = m + m RESONS. Given. efinition of complementar angles. Transitive Propert of Equalit. m = m. Subtraction Propert of Equalit efinition of congruent angles. ecause QRS and PSR are supplementar, m QRS + m PSR = 80 b the definition of supplementar angles. QRL and QRS form a linear pair and b definition are supplementar, which means that m QRL + m QRS = 80. So, b the Transitive Propert of Equalit, m QRS + m PSR = m QRL + m QRS, and b the Subtraction Propert of Equalit, m PSR = m QRL. So, b definition of congruent angles, PSR QRL, and b the Smmetric Propert of ngle ongruence (Thm..), QRL PSR.. STTEMENTS RESONS. E E. Given. m E = m E. efinition of congruent angles. m E = m E + m E. m E = m E + m E 5. m E = m E + m E. ngle ddition Postulate (Post..). Substitution Propert of Equalit 5. ngle ddition Postulate (Post..) 5. our friend; and are not vertical angles because the do not form two pairs of opposite ras. So, the Vertical ngles ongruence Theorem (Thm..) does not appl. 7. no; The converse would be: If two angles are supplementar, then the are a linear pair. This is false because angles can be supplementar without being adjacent ; 0 ; 50 ; 0. Maintaining Mathematical Proficienc (p. ). Sample answer:, I, and. Sample answer: plane and plane G 5. Sample answer:,, and hapter Review (pp. 8). conditional: If two lines intersect, then their intersection is a point. converse: If two lines intersect in a point, then the are intersecting lines. inverse: If two lines do not intersect, then the do not intersect in a point. contrapositive: If two lines do not intersect in a point, then the are not intersecting lines. biconditional: Two lines intersect if and onl if their intersection is a point.. conditional: If + 9 =, then =. converse: If =, then + 9 =. inverse: If + 9, then. contrapositive: If, then + 9. biconditional: + 9 = if and onl if =.. conditional: If angles are supplementar, then the sum to 80. converse: If angles sum to 80, then the are supplementar. inverse: If angles are not supplementar, then the do not sum to 80. contrapositive: If angles do not sum to 80, then the are not supplementar. biconditional: ngles are supplementar if and onl if the sum to 80.. conditional: If an angle is a right angle, then it measures 90. converse: If an angle measures 90, then it is a right angle. inverse: If an angle is not a right angle, then it does not measure 90. contrapositive: If an angle does not measure 90, then it is not a right angle. biconditional: n angle is a right angle if and onl if it measures The difference of an two odd integers is an even integer.. The product of an even and an odd integer is an even integer. 7. m = If =, then =. 9. es 0. es. no. no. Sample answer:. m E = m E. Transitive Propert of Equalit E 7. E E 7. efinition of congruent angles 9

10 . Sample answer: 5. Sample answer: X. Equation Eplanation and Reason 9 = 0 87 Write the equation; Given = 87 dd 0 to each side; ddition Propert of Equalit = dd to each side; ddition Propert of Equalit = ivide each side b ; ivision Propert of Equalit 7. Equation Eplanation and Reason 5 + = 7 + Write the equation; Given 8 + = Subtract 7 from each side; Subtraction Propert of Equalit 8 = 0 Subtract from each side; Subtraction Propert of Equalit = 5 ivide each side b 8; ivision Propert of Equalit 8. Equation Eplanation and Reason ( + 9) = 0 Write the equation; Given + 7 = 0 Multipl; istributive Propert = Subtract 7 from each side; Subtraction Propert of Equalit = Z P W K Y R ivide each side b ; ivision Propert of Equalit 9. Equation Eplanation and Reason 5 + ( ) = 5 Write the equation; Given 5 + = 5 Multipl; istributive Propert 9 = 5 ombine like terms; Simplif. 9 = 08 dd to each side; ddition Propert of Equalit = ivide each side b 9; ivision Propert of Equalit 0. Transitive Propert of Equalit. Refleive Propert of Equalit. Smmetric Propert of ngle ongruence (Thm..). Refleive Propert of ngle ongruence (Thm..). Transitive Propert of Equalit E 5. STTEMENTS RESONS. n angle with verte eists.. m equals the measure of the angle with verte.. Given. Protractor Postulate (Post..). m = m. Refleive Propert of Equalit.. efinition of congruent angles. Sample answer: m + m = 90 Given and are complementar. efinition of complementar angles and are complementar. Given hapter hapter Maintaining Mathematical Proficienc (p. ) ongruent omplements Theroem (Thm..5). = ( ). = 5 ( 5). = 7 ( ). = ( + 9) = 8( 7). + = ( + ) 7. = = + 9. = = 5. = 7. = + 9. when the point given is the -intercept; The slope and -intercept can be substituted for m and b respectivel without performing an calculations.. Vocabular and ore oncept heck (p. 9). skew. Monitoring Progress and Modeling with Mathematics (pp. 9 0). 5. F 7. MK and LS 9. no; The are intersecting lines.. and 5; and ; and 7; and 8. and 8; and 7 5. corresponding 7. consecutive interior 9. Lines that do not intersect could also be skew; If two coplanar lines do not intersect, then the are parallel.. a. true; The floor is level with the horizontal just like the ground. b. false; The lines intersect the plane of the ground, so the intersect certain lines of that plane. c. true; The balusters appear to be vertical, and the floor of the tree house is horizontal. So, the are perpendicular.. es; If the original two lines are parallel, and the transversal is perpendicular to both lines, then all eight angles are right angles. 5. HJG, FJ 7. F, HJ 9. no; The can both be in a plane that is slanted with respect to the horizontal.. Maintaining Mathematical Proficienc (p. 0). m =, m =, m = 59 0

11 . Vocabular and ore oncept heck (p. 5). oth theorems refer to two pairs of congruent angles that are formed when two parallel lines are cut b a transversal, and the angles that are congruent are on opposite sides of the transversal. However with the lternate Interior ngles Theorem (Thm..), the congruent angles lie between the parallel lines, and with the lternate Eterior ngles Theorem (Thm..), the congruent angles lie outside the parallel lines.. Monitoring Progress and Modeling with Mathematics (pp. 5 ). m = 7 b Vertical ngles ongruence Theorem (Thm..); m = 7 b lternate Eterior ngles Theorem (Thm..) 5. m = b lternate Interior ngles Theorem (Thm..); m = 58 b onsecutive Interior ngles Theorem (Thm..) 7. ; = 8 = 9. ; m 5 = ( 7) = = 80 = =. m = 00, m = 80, m = 00 ; ecause the 80 angle is a consecutive interior angle with both and, the are supplementar b the onsecutive Interior ngles Theorem (Thm..). ecause and are consecutive interior angles, the are supplementar b the onsecutive Interior ngles Theorem (Thm..).. In order to use the orresponding ngles Theorem (Thm..), the angles need to be formed b two parallel lines cut b a transversal, but none of the lines in this diagram appear to be parallel; 9 and 0 are corresponding angles. 5. t p. 9 0 = = 80; = 0, = 5. no; In order to make the shot, ou must hit the cue ball so that m = 5. The angle that is complementar to must have a measure of 5 because this angle is an alternate interior angle with the angle formed b the path of the cue ball and the vertical line drawn.. Maintaining Mathematical Proficienc (p. ) 5. If two angles are congruent, then the are vertical angles; false 7. If two angles are supplementar, then the form a linear pair; false. Vocabular and ore oncept heck (p. ). corresponding, alternate interior, alternate eterior. Monitoring Progress and Modeling with Mathematics (pp. ). = 0; Lines m and n are parallel when the marked corresponding angles are congruent. = 0 = 0 5. = 5; Lines m and n are parallel when the marked consecutive interior angles are supplementar. ( 5) + 50 = = 80 = 5 = 5 7. = 0; Lines m and n are parallel when the marked consecutive interior angles are supplementar. + = 80 = 80 = 0 9. n m p STTEMENTS q RESONS. p q. Given.. orresponding ngles Theorem (Thm..).. Vertical ngles ongruence Theorem (Thm..).. Transitive Propert of ongruence (Thm..) 7. m = 0 ; ecause the trees form parallel lines, and the rope is a transversal, the 7 angle and are consecutive interior angles. So, the are supplementar b the onsecutive Interior ngles Theorem (Thm..). 9. es; If two parallel lines are cut b a perpendicular transversal, then the consecutive interior angles will both be right angles.. q p It is given that. the Vertical ngles ongruence Theorem (Thm..),. Then b the Transitive Propert of ongruence (Thm..),. So, b the orresponding ngles Theorem (Thm..), p q.. es; lternate Interior ngles onverse (Thm..) 5. no 7. no 9. This diagram shows that vertical angles are alwas congruent. Lines a and b are not parallel unless =, and ou cannot assume that the are equal.. es; m E = 80 = 57 b the Linear Pair Postulate (Post..8). So, b definition, a pair of corresponding angles are congruent, which means that F b the orresponding ngles onverse (Thm..5).

12 . no; The marked angles are vertical angles. You do not know anthing about the angles formed b the intersection of F and E. 5. es; E. 0th ve. is parallel to E. 9th ve. b the orresponding ngles onverse (Thm..5). E. 9th ve. is parallel to E. 8th ve. b the lternate Eterior ngles onverse (Thm..7). E. 8th ve. is parallel to E. 7th ve. b the lternate Interior ngles onverse (Thm..). So, the are all parallel to each other b the Transitive Propert of Parallel Lines (Thm..9). 7. The two angles marked as 08 are corresponding angles. ecause the have the same measure, the are congruent to each other. So, m n b the orresponding ngles onverse (Thm..5). 9.,,, ; The orresponding ngles onverse (Thm..5) can be used because the angle marked at the intersection of line m and the transversal is a vertical angle with, and therefore congruent to, an angle that is corresponding with the other marked angle. The lternate Interior ngles onverse (Thm..) can be used because the angles that are marked as congruent are alternate interior angles. The lternate Eterior ngles onverse (Thm..7) can be used because the angles that are vertical with, and therefore congruent to, the marked angles are alternate eterior angles. The onsecutive Interior ngles onverse (Thm..8) can be used because each of the marked angles forms a linear pair with, and is therefore supplementar to, an angle that is a consecutive interior angle with the other marked angle.. two; Sample answer: 5, 7,, and 7 are supplementar. STTEMENTS 5. STTEMENTS RESONS.,. Given.. Vertical ngles ongruence Theorem (Thm..).. Transitive Propert of ongruence (Thm..).. Transitive Propert of ongruence (Thm..) 5. RESONS. m = 5, m = 5. Given. m + m = m + m. Refleive Propert of Equalit. m + m = Substitution Propert of Equalit. m + m = 80. Simplif. 5. and are supplementar. 5. efinition of supplementar angles. m n. onsecutive Interior ngles onverse (Thm.8) 5. lternate Interior ngles onverse (Thm..) 7. no; ased on the diagram b the lternate Interior ngles onverse (Thm..), but ou cannot be sure that. 9. a. p q r b. Given: p q, q r Prove: p r c. p STTEMENTS q r RESONS. p q, q r. Given.,. orresponding ngles Theorem (Thm..).. Transitive Propert of ongruence (Thm..). p r. orresponding ngles onverse (Thm..5). Maintaining Mathematical Proficienc (p. ). about.7.. Vocabular and ore oncept heck (p. 5). midpoint, right. Monitoring Progress and Modeling with Mathematics (pp. 5 5). about. units P m. In order to claim parallel lines b the Lines Perpendicular to a Transversal Theorem (Thm..), both lines must be marked as perpendicular to the transversal; Lines and z are perpendicular. 7. P m

13 . 5. g h ecause b definition, m = m. lso, b the Linear Pair Postulate (Post..8), m + m = 80. Then, b the Substitution Propert of Equalit, m + m = 80, and (m )= 80 b the istributive Propert. So, b the ivision Propert of Equalit, m = 90. Finall, g h b the definition of perpendicular lines. STTEMENTS RESONS. a b. Given. is a right angle.. efinition of perpendicular lines.. Vertical ngles ongruence Theorem (Thm..). m = 90. efinition of right angle 5. m = Transitive Propert of Equalit. and form a linear pair. 7. and are supplementar.. efinition of linear pair 7. Linear Pair Postulate (Post..8) 8. m + m = efinition of supplementar angles m = Transitive Propert of Equalit 0. m = Subtraction Propert of Equalit.. Vertical ngles ongruence Theorem (Thm..). m = 90. Transitive Propert of Equalit.,,, and are right angles.. efinition of right angle 7. none; The onl thing that can be concluded in this diagram is that v. In order to sa that lines are parallel, ou need to know something about both of the intersections between the transversal and the two lines. 9. m n, ecause m q and n q, lines m and n are parallel b the Lines Perpendicular to a Transversal Theorem (Thm..). The other lines ma or ma not be parallel.. n p; ecause k n and k p, lines n and p are parallel b the Lines Perpendicular to a Transversal Theorem (Thm..).. m = 90, m = 0, m = 0, m = 0, m 5 = 90 ; m = 90, because it is marked as a right angle. m = 90 0 = 0, because it is complementar to the 0 angle. m = 0, because it is a vertical angle with, and therefore congruent to, the 0 angle. m = 90 (0 + 0 ) = 0, because it forms a right angle with and the 0 angle. m 5 = 90, because it is a vertical angle with, and therefore congruent to,. 5. = 8 7.,,, E 9.. The lines segments that are perpendicular to the crosswalk require less paint, because the represent the shortest distance from one side of the crosswalk to the other.. about.5 units. Maintaining Mathematical Proficienc (p. 5) =. = 7.5 Vocabular and ore oncept heck (p. 59). directed.5 Monitoring Progress and Modeling with Mathematics (pp. 59 0) (, ) (0, 5). (7, 0.) 5. (.5,.5) 7. a c, b d 9. perpendicular; ecause m m = ( )( ) =, lines and are perpendicular b the Slopes of Perpendicular Lines Theorem (Thm..).. perpendicular; ecause m m = ( ) =, lines and are perpendicular b the Slopes of Perpendicular Lines Theorem (Thm..).

14 . ecause the slopes are opposites but not reciprocals, their product does not equal. Lines and are neither parallel nor perpendicular. 5. ( 5, 5 ) 7. It will be the same point. 9. ompare the slopes of the lines. The line whose slope has the greater absolute value is steeper.. no; m LM = 5, m LN = 7, and m MN = 9. None of these can pair up to make a product of, so none of the segments are perpendicular.. If and z, then b the Slopes of Parallel Lines Theorem (Thm..), m = m and m = m z. Therefore, b the Transitive Propert of Equalit, m = m z. So, b the Slopes of Parallel Lines Theorem (Thm..), z. 5. If lines and are horizontal, then b definition m = 0 and m = 0. So, b the Transitive Propert of Equalit, m = m. Therefore, b the Slopes of Parallel Lines Theorem (Thm..),..5 Maintaining Mathematical Proficienc (p. 0) 7. m = ; b = 9 9. m = ; b = 8. Vocabular and ore oncept heck (p. 5). perpendicular. Monitoring Progress and Modeling with Mathematics (pp. 5 ). = = 5. = = 5 = 7. = 9 = = + = 9 9. = = + 5 =. about. units. about 5. units 5. Parallel lines have the same slope, not the same -intercept. = +, (, ) = () + b = b The line = is parallel to the line = = 9. (0, ); = +. (, 0); = 9. a. p = 0t b. p = 0t + c. parallel; oth lines have a slope of es; If two lines have the same -intercept, then the intersect at that point. ut parallel lines do not intersect. 7. k = 9. about. units; The two lines have the same slope and are therefore parallel. So, the distance from a point on one line to the other line will be the same no matter which point is chosen. The line = is perpendicular to both lines and intersects = at (, ) and = + 5 at the origin. So, the distance between the lines is the same as the distance between these two points of intersection. ( 0) + ( 0).. a. d = b. Sample answer: Use a graphing calculator to graph d and find the minimum value. c. This method uses a variable point (, ) and a variable distance d, whereas the method in Eample uses eact points and equations; Sample answer: the method in Eample because it is more direct. Maintaining Mathematical Proficienc (p. ). 5 (0, ) 5. (, ) = 7 5 hapter Review (pp. 8 70). NR, MR, LQ, PQ. LM, JK, NP. JM, KL, KP, JN. plane JKP 5. = 5, = 5. =, = 7. =, = 9

15 8. =, = 7 9. = =. =. =. ; ecause z and z, lines and are parallel b the Lines Perpendicular to a Transversal Theorem (Thm..).. none; The onl thing that can be concluded in this diagram is that z and w. In order to sa that lines are parallel, ou need to know something about both of the intersections between the two lines and a transversal. 5. m n, a b; ecause a n and b n, lines a and b are parallel b the Lines Perpendicular to a Transversal Theorem (Thm..). ecause m a and n a, lines m and n are parallel b the Lines Perpendicular to a Transversal Theorem (Thm..). ecause b and n b, lines and n are parallel b the Lines Perpendicular to a Transversal Theorem (Thm..). ecause n and m n, lines and m are parallel b the Transitive Propert of Parallel Lines (Thm..9).. a b; ecause a n and b n, lines a and b are parallel b the Lines Perpendicular to a Transversal Theorem (Thm..). 7. undefined =. =. =. = + 5. about. units. about.7 units hapter hapter Maintaining Mathematical Proficienc (p. 75). reflection. rotation. dilation. translation 5. no; = 7 5, The sides are not proportional. 7. es; The corresponding angles are congruent and the corresponding side lengths are proportional. 7. es; The corresponding angles are congruent and the corresponding side lengths are proportional. 8. no; Squares have four right angles, so the corresponding angles are alwas congruent. ecause all four sides are congruent, the corresponding sides will alwas be proportional.. Vocabular and ore oncept heck (p. 8). is the preimage, and is the image.. Monitoring Progress and Modeling with Mathematics (pp. 8 8)., 7, E 9., 5. (, ) ( 5, + ). (, 0) 5. (5, ) E F F F 8 E F E P Q. P 8 X 8 Q R Z X R Y 8 0 X Z Y Z Y P Q. translation: (, ) ( + 5, + ), translation: (, ) ( 5, 5) 5. The quadrilateral should have been translated left and down; E H E H F G G F 7. a. The amoeba moves right 5 and down. b. about.8 mm c. about 0.5 mm/sec 9. r = 00, s = 8, t = 5, w = 5. E (, ), F (, 5), G (0, ). (, ) ( m, n); You must go back the same number of units in the opposite direction. 5. If a rigid motion is used to transform figure to figure, then b definition of rigid motion, ever part of figure is congruent to its corresponding part of figure. If another rigid motion is used to transform figure to figure, then b definition of rigid motion, ever part of figure is congruent to its corresponding part of figure. So, b the Transitive Propert of ongruence, ever part of figure is congruent to its corresponding part of figure. So b definition of rigid motion, the composition of two (or more) rigid motions is a rigid motion. 7. raw a rectangle. Then draw a translation of the rectangle. Net, connect each verte of the preimage with the corresponding verte in the image. Finall, make the hidden lines dashed. 9. es; ccording to the definition of translation, the segments connecting corresponding vertices will be congruent and parallel. lso, because a translation is a rigid motion, GH G H. So, the resulting figure is a parallelogram.. no; ecause the value of changes, ou are not adding the same amount to each -value. P Q R R 8 5

16 . Maintaining Mathematical Proficienc (p. 8). es 5. no Vocabular and ore oncept heck (p. 90). translation and reflection. Monitoring Progress and Modeling with Mathematics (pp. 90 9). -ais 5. neither 7. K K L K J J J L K J L L =. = L K J = = J L K 7. Reflect H in line n to obtain H. Then draw JH. Label the intersection of JH and n as K. ecause JH is the shortest distance between J and H and HK = H K, park at point K. 9. (5, 0). (, 0). = 5. m 7. Q = M M 9. = +. Maintaining Mathematical Proficienc (p. 9) Vocabular and ore oncept heck (p. 98). 70. Monitoring Progress and Modeling with Mathematics (pp ) P 5 J M F Q K J L G N G P N J F 7. T T T S S R S R R 7 5 L M K J S T R = R R T S T a. none b. c. OX d. none MOM S. X X X 5 Y Y Y

17 Y L X 9 8 M 7. es; Rotations of 90 and 80 about the center map the figure onto itself. 9. es; Rotations of 5, 90, 5, and 80 about the center map the figure onto itself.. no, es; 90, 80. es, no; one line of smmetr 5. The rule for a 70 rotation, (, ) (, ), should have been used instead of the rule for a reflection in the -ais; (, ) (, ), (, ) (, ) 7. X X L 5 N M M 5 N Y Y L 9. a. 90 : = +, 80 : = +, 70 : =, 0 ; = ; The slope of the line rotated 90 is the opposite reciprocal of the slope of the preimage, and the -intercept is equal to the -intercept of the preimage. The slope of the line rotated 80 is equal to the slope of the preimage, and the -intercepts of the image and preimage are opposites. The slope of the line rotated 70 is the opposite reciprocal of the slope of the preimage, and the -intercept is the opposite of the -intercept of the preimage. The equation of the line rotated 0 is the same as the equation of the preimage. b. es; ecause the coordinates of ever point change in the same wa with each rotation, the relationships described will be true for an equation with an slope and -intercept.. twice N. es; Sample answer: rectangle (that is not a square) is one eample of a figure that has 80 rotational smmetr, but not 90 rotational smmetr. 5. a. 5, n = b. 0, n = 7. X Y Z P 5 9. (, 0 ); (, 0 ); (, 00 ); The radius remains the same. The angle increases in conjunction with the rotation.. Maintaining Mathematical Proficienc (p. 00). and J, and K, and L, and M; and JK, and KL, and LM, and MJ. Vocabular and ore oncept heck (p. 08). congruent. Monitoring Progress and Modeling with Mathematics (pp. 08 0) Z. HJK QRS, EFG LMNP; HJK is a 90 rotation of QRS. EFG is translation 7 units right and units down of LMNP. 5. Sample answer: 80 rotation about the origin followed b a translation 5 units left and unit down 7. es; TUV is a translation units right of QRS. So, TUV QRS. 9. no; M and N are translated units right of their corresponding vertices, L and K, but P is translated onl unit right of its corresponding verte, J. So, this is not a rigid motion in translation 5 units right and a reflection in the -ais should have been used; is mapped to b a translation 5 units right, followed b a reflection in the -ais Reflect the figure in two parallel lines instead of translating the figure; The third line of reflection is perpendicular to the parallel lines. 5. never; ongruence transformations are rigid motions. 7. sometimes; Reflecting in = then = is not a rotation. Reflecting in the -ais then -ais is a rotation of no; The image on the screen is larger. X Y 7

18 . 9. L STTEMENTS. reflection in line l maps JK to J K, a reflection in line m maps J K to J K, and l m.. If KK intersects line l at L and line m at M, then L is the perpendicular bisector of KK, and M is the perpendicular bisector of K K.. KK is perpendicular to l and m, and KL = LK and K M = MK.. If d is the distance between l and m, then d = LM. 5. LM = LK + K M and KK = KL + LK + K M + MK RESONS. Given. efinition of reflection. efinition of perpendicular bisector. Ruler Postulate (Post..) 5. Segment ddition Postulate (Post..).. M M Not drawn to scale. U R Not drawn to scale. R R S L S N U T R S T N. KK = LK + LK + K M + K M. Substitution Propert of Equalit U T S 7. KK = ( LK + K M ) 7. istributive Propert 8. KK = ( LM ) 8. Substitution Propert of Equalit 9. KK = d 9. Transitive Propert of Equalit. 80 rotation; reflections: P (, ) P (, ) P (, ) and Q (, ) Q (, ) Q (, ) translation: P (, ) (, 5 ) (, ) and Q (, ) (, 5 ) Q (, ) 80 rotation P (, ) (, ) and Q (, ) (, ) 5. l m 5. U Not drawn to scale. Y Y Z 8 Z 8 X X T 7. V V. Maintaining Mathematical Proficienc (p. 0) 7. = 9. b =. n = %.5 Vocabular and ore oncept heck (p. ). P (k, k).5 Monitoring Progress and Modeling with Mathematics (pp. 8). 7 ; reduction 5. 5 ; reduction 7. M M Not drawn to scale. L P N L N W W 7 5 U U T T

19 R 8 U T S T U 8 R c. X W X Y W Z Y Z. The scale factor should be calculated b finding P P, not P P ; k = = 5. k = 5 ; = 7. k = ; = 9. k =. 00 mm. 90 mm 5. grasshopper, hone bee, and monarch butterfl; The scale factor for these three is k = 5. The scale factor for the black beetle is k = no; The scale factor for the shorter sides is 8 =, but the scale factor for the longer sides is 0 = 5. The scale factor for both sides has to be the same or the picture will be distorted. 9. = 5, = 5. original. original a. O = (O) b. O coincides with O. 7. k = 9. a. P = units, = square units b. X Y X 8 O O Y S 0 P = units, = square units; The perimeter of the dilated rectangle is the perimeter of the original rectangle. The area of the dilated rectangle is the area of the original rectangle. d. The perimeter changes b a factor of k. The area changes b a factor of k..5 Maintaining Mathematical Proficienc (p. 8) 5. (, 5 ), ( 0, 0 ), (, ) 5. ( 5, ), (, ), ( 0, ) 55. (, ), (, ), (, ). Vocabular and ore oncept heck (p. ). ongruent figures have the same size and shape. Similar figures have the same shape, but not necessaril the same size.. Monitoring Progress and Modeling with Mathematics (pp. ). F 5. H F H G G H H F G F F H G H G F G 5 H F G 5 F F F G H G H 5 F 8 G H G H G F H W W Z 8 P = 7 units, = 88 square units; The perimeter of the dilated rectangle is three times the perimeter of the original rectangle. The area of the dilated rectangle is nine times the area of the original rectangle. Z. F F H H G H G 8 F G. F 8 H H 8 F F G G H G 8 5. Sample answer: translation unit down and unit right followed b a dilation with center at E(, ) and a scale factor of 9