Answer Key
4.5Ā DilationsĀ Ā ExerciseĀ 1 Ā šš ā² ( šššš , šššš ) Ā Ā ExerciseĀ 2 Ā 60%Ā because thisĀ scaleĀ factorĀ isĀ lessĀ thanĀ 1.Ā Ā ExerciseĀ 3 Ā BecauseĀ š¶š¶šš ā² š¶š¶šš = 6 14 , Ā the scaleĀ factorĀ isĀ šš = 37 .Ā So the dilation isĀ a reduction.Ā Ā ExerciseĀ 4Ā BecauseĀ š¶š¶šš ā² š¶š¶šš = 24 9 ,Ā theĀ scaleĀ factorĀ isĀ šš = 83 .Ā SoĀ the dilationĀ isĀ an enlargement.Ā Ā ExerciseĀ 5 Ā BecauseĀ š¶š¶šš ā² š¶š¶šš = 9 15 ,the scaleĀ factorĀ isĀ šš = 35 .Ā So the dilation isĀ a reduction.Ā Ā ExerciseĀ 6 Ā BecauseĀ š¶š¶šš ā² š¶š¶šš = 28 8 ,Ā theĀ scaleĀ factorĀ isĀ šš = 72 .Ā So the dilationĀ isĀ an enlargement.Ā Ā ExerciseĀ 7 Ā Ā Ā ExerciseĀ 8Ā
Ā Ā ExerciseĀ 9Ā Ā Ā ExerciseĀ 10Ā
Ā Ā ExerciseĀ 11Ā Ā Ā ExerciseĀ 12Ā Ā Ā ExerciseĀ 13Ā
Ā Ā ExerciseĀ 14Ā Ā Ā ExerciseĀ 15Ā ( šš , šš ) ā (3 šš ,Ā 3 šš )Ā šš (6, ā 1) āĀ šš ā² (18, ā 3)Ā šš ( ā 2, ā 4) āĀ šš ā² ( ā 6, ā 12)Ā šš (1,2) āĀ šš ā² (3,6) Ā
Ā Ā ExerciseĀ 16 Ā ( šš , šš ) āĀ ļæ½ 65 šš , 65 ššļæ½ Ā š“š“ (0,5) āĀ š“š“ ā² (0,6)Ā šµšµ ( ā 10, ā 5) āĀ šµšµ ā² ( ā 12, ā 6)Ā š¶š¶ (5, ā 5) āĀ š¶š¶ ā² (6, ā 6) Ā Ā Ā ExerciseĀ 17 Ā ( šš , šš ) āĀ ļæ½ 23 šš , 23 ššļæ½ Ā šš (9, ā 3) āĀ šš ā² (6, ā 2)Ā šš (6,0) āĀ šš ā² (4,0)Ā šš (3,9) āĀ šš ā² (2,6)Ā šš (0,0) āĀ šš ā² (0,0) Ā
Ā Ā ExerciseĀ 18 Ā ( šš , šš ) ā (0.25 šš ,Ā 0.25 šš )Ā š½š½ (4,0) āĀ š½š½ ā² (1,0)Ā š¾š¾ ( ā 8,4) āĀ š¾š¾ ā² ( ā 2,1)Ā šæšæ (0, ā 4) āĀ šæšæ ā² (0, ā 1)Ā šš (12, ā 8) āĀ šš ā² (3, ā 2) Ā Ā Ā Ā ExerciseĀ 19 Ā ( šš , šš ) āĀ ļæ½ā 15 šš , ā 15 ššļæ½ Ā šµšµ ( ā 5, ā 10) āĀ šµšµ ā² (1,2)Ā š¶š¶ ( ā 10,15) āĀ š¶š¶ ā² (2, ā 3)Ā š·š· (0,5) āĀ š·š· ā² (0, ā 1) Ā
Ā Ā ExerciseĀ 20 Ā ( šš , šš ) ā ( ā 3 šš , ā 3 šš )Ā šæšæ (0,0) āĀ šæšæ ā² (0,0)Ā šš ( ā 4,1) āĀ šš ā² (12, ā 3)Ā šš ( ā 3, ā 6) āĀ šš ā² (9,18) Ā Ā Ā Ā
ExerciseĀ 21Ā ( šš , šš ) ā ( ā 4 šš , ā 4 šš )Ā š š ( ā 7, ā 1) āĀ š š ā² (28,4)Ā šš (2,5) āĀ šš ā² ( ā 8, ā 20)Ā šš ( ā 2, ā 3) āĀ šš ā² (8,12)Ā šš ( ā 3, ā 3) āĀ šš ā² (12,12) Ā Ā Ā Ā ExerciseĀ 22 Ā ( šš , šš ) ā ( ā 0.5 šš , ā 0.5 šš )Ā šš (8, ā 2) āĀ šš ā² ( ā 4,1)Ā šš (6,0) āĀ šš ā² ( ā 3,0)Ā šš ( ā 6,4) āĀ šš ā² (3, ā 2)Ā šš ( ā 2,2) āĀ šš ā² (1, ā 1) Ā Ā
Ā ExerciseĀ 23 Ā The scaleĀ factorĀ isĀ theĀ ratio ofĀ theĀ lengthĀ ofĀ the correspondingĀ sidesĀ ofĀ the image andĀ preimage.Ā šš = š¶š¶šš ā² š¶š¶šš = 3 12Ā = 14 Ā Ā ExerciseĀ 24Ā IfĀ P(x,y)Ā isĀ the preimage ofĀ a point,Ā then itsĀ image afterĀ dilation atĀ theĀ origin (0,0)Ā withĀ scaleĀ factorĀ kĀ isĀ theĀ pointĀ šš ā² ( šššš , šššš ). Ā Ā Therefore k=2.Ā Ā ExerciseĀ 25Ā šš = 15 9Ā = 53Ā šš = 35 šš Ā 53Ā = 35 šš Ā 5 šš =Ā 105Ā šš =Ā 21 Ā Ā ExerciseĀ 26Ā šš = 2814Ā =Ā 2Ā 2Ā = 12 šš Ā 2 šš =Ā 12Ā šš =Ā 6 Ā Ā ExerciseĀ 27Ā šš = 32Ā šš = šš 2Ā 32Ā = šš 2Ā 3Ā = šš Ā Ā ExerciseĀ 28Ā
šš = 7 28Ā = 14Ā šš = 4 šš Ā 14Ā = 4 šš Ā šš =Ā 16 Ā Ā ExerciseĀ 29Ā šš = 5 2.5Ā =Ā 2Ā šššš Ā šš = 7 3.5Ā =Ā 2 Ā Ā ExerciseĀ 30Ā šš = 10 8.5Ā = 100 85Ā = 2013 Ā Ā ExerciseĀ 31 Ā image length actualĀ length = šš Ā image length 60 =Ā 5Ā š¼š¼ššš¼š¼š¼š¼š¼š¼ Ā ššš¼š¼ššš¼š¼ššā Ā Ā =Ā Ā 300Ā šššš Ā Ā ExerciseĀ 32Ā image length actualĀ length = šš Ā image length 4.5 =Ā 10Ā š¼š¼ššš¼š¼š¼š¼š¼š¼ Ā ššš¼š¼ššš¼š¼ššā Ā Ā =Ā Ā 45Ā šššš Ā Ā ExerciseĀ 33Ā image length actualĀ length = šš Ā image length 47 =Ā 20Ā š¼š¼ššš¼š¼š¼š¼š¼š¼ Ā ššš¼š¼ššš¼š¼ššā Ā Ā =Ā Ā 980Ā šššš Ā Ā ExerciseĀ 34Ā image length actualĀ length = šš Ā image length 12 =Ā 15Ā š¼š¼ššš¼š¼š¼š¼š¼š¼ Ā ššš¼š¼ššš¼š¼ššā Ā Ā =Ā Ā 180Ā šššš Ā Ā ExerciseĀ 35Ā Grasshopper:Ā image length actualĀ length = 15 2Ā =Ā 7.5Ā BlackĀ beetle:Ā image length actualĀ length = 4.20.6Ā =Ā 7Ā
Honeybee:Ā image length actualĀ length = 75 16 5 8 =Ā 7.5Ā Monarch butterfly:Ā image length actualĀ length = 29.25 3.9Ā =Ā 7.5Ā Ā ThereforeĀ grasshoppers,Ā honeybeesĀ andĀ monarchĀ butterļæ½lies Ā wereĀ looked Ā atĀ using Ā theĀ sameĀ magnifyi Ā Ā Ā ExerciseĀ 36Ā BuildingĀ ā³ ABCĀ andĀ ā³ ABCĀ dilatationĀ ā³ A ā² B ā² C ā² .Ā WeĀ connectĀ each pointĀ toĀ itsĀ correspondingĀ image in orderĀ to constructĀ the dilation'sĀ center.Ā TheĀ centerĀ ofĀ dilationĀ isĀ where the three linesĀ converge.Ā Ā Ā ExerciseĀ 37Ā Small sides: 84 = šš 1 Ā Large sides:Ā 10 6 = šš 2 Ā Ā Because the scaleĀ factorĀ isĀ notĀ equal,Ā theĀ photo can'tĀ be dilatedĀ toĀ fitĀ theĀ frame.Ā Ā ExerciseĀ 38Ā image length actualĀ length = šš = 13Ā Hence,Ā theĀ largeĀ figure isĀ theĀ actualĀ figure and the otherĀ one isĀ theĀ dilatedĀ figure.Ā Ā ExerciseĀ 39Ā Since the largerĀ triangle isĀ the dilation ofĀ the smallerĀ triangle,Ā 62Ā = 2 šš +Ā 8 šš +Ā 1Ā Ā 3Ā = 2 šš +Ā 8 šš +Ā 1Ā Ā 3 šš +Ā 3Ā =Ā 2 šš +Ā 8Ā šš =Ā 6Ā Ā
Since Ā angle Ā measuresĀ areĀ preserved,Ā 3y ā 34Ā =Ā yĀ +Ā 16Ā 3y ā yĀ =Ā 16Ā +Ā 34Ā 2yĀ =Ā 50Ā yĀ =Ā 25 Ā Ā ExerciseĀ 40Ā ToĀ understand whyĀ a scaleĀ factorĀ ofĀ 2 isĀ equivalentĀ toĀ 200%,Ā we need toĀ firstĀ understandĀ whatĀ scale factorĀ meansĀ in mathematics.Ā AĀ scaleĀ factorĀ isĀ aĀ numberĀ thatĀ scales,Ā orĀ multiplies,Ā a quantityĀ orĀ measurementĀ byĀ a certain amount.Ā ForĀ example,Ā ifĀ we have aĀ line segmentĀ thatĀ isĀ 5Ā unitsĀ longĀ and we wantĀ to scale itĀ byĀ aĀ factorĀ ofĀ 2,Ā we wouldĀ multiplyĀ 5 byĀ 2 to getĀ aĀ newĀ length ofĀ 10Ā units.Ā Now,Ā let'sĀ considerĀ the percentage scale.Ā PercentagesĀ areĀ a wayĀ ofĀ expressingĀ aĀ numberĀ asĀ aĀ fraction ofĀ 100.Ā ForĀ example,Ā 50%Ā isĀ the same asĀ 50/100Ā orĀ 0.5.Ā To convertĀ a scaleĀ factorĀ toĀ aĀ percentage,Ā we can simplyĀ multiplyĀ itĀ byĀ 100.Ā ForĀ example,Ā ifĀ ourĀ scaleĀ factorĀ isĀ 2,Ā we canĀ multiplyĀ itĀ byĀ 100 to getĀ 200%,Ā which isĀ the sameĀ asĀ sayingĀ thatĀ we are scalingĀ ourĀ quantityĀ byĀ a factorĀ ofĀ 2.Ā InĀ mathematicalĀ notation,Ā we can expressĀ thisĀ as:Ā scaleĀ factorĀ ofĀ 2Ā =Ā 2Ā 200%=2Ć100Ā So,Ā aĀ scaleĀ factorĀ ofĀ 2 isĀ theĀ sameĀ asĀ 200%Ā because theyĀ both representĀ multiplyingĀ aĀ quantityĀ orĀ measurementĀ byĀ aĀ factorĀ ofĀ 2.Ā Ā ExerciseĀ 41Ā The originalĀ figure isĀ closerĀ toĀ theĀ centre ofĀ dilation.Ā Ā ExerciseĀ 42Ā The dilatedĀ figure isĀ closerĀ toĀ theĀ centre ofĀ dilation.Ā Ā ExerciseĀ 43Ā TheĀ originalĀ figure isĀ closerĀ toĀ theĀ centre ofĀ dilation.Ā Ā ExerciseĀ 44Ā The dilatedĀ figure isĀ closerĀ toĀ theĀ centre ofĀ dilation.Ā Ā ExerciseĀ 45 Ā a.Ā LengthĀ šš ā² š“š“ ā² Ā isĀ 2Ā timesĀ longerĀ thanĀ OA.Ā Ā b.Ā OAĀ andĀ šš ā² š“š“ ā² Ā are parallel.Ā Ā ExerciseĀ 46 Ā a.Ā LengthĀ š“š“ ā² šµšµ ā² Ā isĀ equalĀ toĀ halfĀ ofĀ theĀ lengthĀ ofĀ AB.Ā Ā b.Ā ABĀ andĀ š“š“ ā² šµšµ ā² Ā are parallel.Ā Ā ExerciseĀ 47 Ā image length actualĀ length = šš Ā 9 12 ā 12Ā = 9 144Ā = 1 16Ā HenceĀ theĀ scale factorĀ isĀ 1 16 . Ā Ā ExerciseĀ 48Ā
Yes!Ā WhenĀ theĀ scale factorĀ k=-1,Ā theĀ figure rotatesĀ 180 0 Ā butĀ the size andĀ shape don'tĀ change.Ā Ā ExerciseĀ 49 Ā a.Ā LengthĀ ofĀ the rectangleĀ =Ā 5-(-3)=8Ā and width ofĀ theĀ rectangle =3-(-1)=4Ā PerimeterĀ =Ā 2*8+2*4=24Ā unitsĀ Area =Ā 8*4=32Ā squareĀ unitsĀ Ā b.Ā LengthĀ ofĀ theĀ dilated rectangle =Ā 3*8=24Ā andĀ width ofĀ the dilated rectangle =3*4=12Ā PerimeterĀ =Ā 2*24+2*12=72Ā unitsĀ Area =Ā 24*12Ā =288Ā square unitsĀ Ā The perimeterĀ ofĀ theĀ dilated rectangle =3Ā (perimeterĀ ofĀ the initialĀ rectangle )Ā Area ofĀ the dilated rectangle =Ā 9Ā (areaĀ ofĀ theĀ initialĀ rectangle)Ā Ā c.Ā Length ofĀ the dilatedĀ rectangleĀ =Ā 84 =Ā 2 Ā and width ofĀ theĀ dilated rectangleĀ = 44 =Ā 1 Ā PerimeterĀ =Ā 2*2+2*1=6Ā unitsĀ Area =Ā 2*1Ā =2Ā square unitsĀ Ā The perimeterĀ ofĀ theĀ dilated rectangle =Ā (perimeterĀ ofĀ theĀ initialĀ rectangle )/4Ā Area ofĀ the dilated rectangle =Ā (area ofĀ the initialĀ rectangle)/16Ā Ā d.Ā The perimeterĀ ofĀ theĀ dilated rectangle =kĀ (perimeterĀ ofĀ theĀ initialĀ rectangle )Ā Ā Area ofĀ the dilated rectangle =Ā šš 2 Ā (areaĀ ofĀ the initialĀ rectangle)Ā where kĀ isĀ the scaleĀ factor.Ā Ā ExerciseĀ 50 Ā WhenĀ you reduceĀ a page andĀ place itĀ onĀ theĀ originalĀ page,Ā there isĀ alwaysĀ a pointĀ thatĀ isĀ in theĀ sameĀ placeĀ onĀ both pages.Ā ThisĀ isĀ because when you reduce the page,Ā allĀ ofĀ theĀ pointsĀ onĀ theĀ originalĀ page are scaled down byĀ the sameĀ factor.Ā Therefore,Ā theĀ relative distancesĀ betweenĀ theĀ pointsĀ on the originalĀ page are preserved in theĀ reducedĀ page.Ā ToĀ understand thisĀ conceptĀ mathematically,Ā let'sĀ considerĀ a pointĀ PĀ on theĀ originalĀ page with coordinatesĀ (x,y),Ā and let'sĀ assumeĀ thatĀ the page isĀ reduced byĀ aĀ factorĀ ofĀ r.Ā ThisĀ meansĀ thatĀ the coordinatesĀ ofĀ PĀ on the reduced page willĀ beĀ (rx,ry).Ā We canĀ seeĀ thatĀ theĀ relative distancesĀ between the pointsĀ on theĀ originalĀ page are preserved inĀ theĀ reduced page byĀ comparingĀ the distance between two pointsĀ on theĀ originalĀ page withĀ theĀ distance between the correspondingĀ pointsĀ on the reducedĀ page.Ā Let'sĀ considerĀ twoĀ pointsĀ PĀ and QĀ onĀ theĀ originalĀ page with coordinatesĀ ( šš 1 , šš 1 ) Ā andĀ ( šš 2 , šš 2 ) Ā respectively.Ā The distanceĀ betweenĀ theseĀ two pointsĀ canĀ be calculatedĀ usingĀ theĀ distanceĀ formula:Ā ļæ½ ( šš 2 āĀ šš 1 ) 2 +Ā ( šš 2 āĀ šš 1 ) 2 Ā Ā Now,Ā let's considerĀ theĀ correspondingĀ pointsĀ on the reduced page withĀ coordinatesĀ ļæ½šš š„š„ 1 , šš š¦š¦ 1 ļæ½ Ā andĀ ļæ½šš š„š„ 2 , šš š¦š¦ 2 ļæ½ .Ā Ā The distanceĀ betweenĀ theseĀ two pointsĀ canĀ also beĀ calculatedĀ usingĀ theĀ distanceĀ formula:Ā ļæ½ļæ½šš š„š„ 2 āĀ šš š„š„ 1 ļæ½ 2 + ļæ½šš š¦š¦ 2 āĀ šš š¦š¦ 1 ļæ½ 2 Ā Ā
WeĀ can simplifyĀ thisĀ expressionĀ usingĀ theĀ factĀ thatĀ šš š„š„ 1 = šš š„š„ šš Ā andĀ šš š¦š¦ 1 = šš š¦š¦ šš ,Ā and similarlyĀ forĀ the otherĀ point: �� š„š„ 2 āš„š„ 1 šš ļæ½ 2 + ļæ½ š¦š¦ 2 āš¦š¦ 1 šš ļæ½ 2 Ā WeĀ can seeĀ thatĀ theĀ expression on the right-handĀ side isĀ simplyĀ the originalĀ distanceĀ formulaĀ divided byĀ r.Ā ThisĀ meansĀ thatĀ the distanceĀ between theĀ two pointsĀ on the reducedĀ page isĀ scaledĀ down byĀ the sameĀ factorĀ asĀ theĀ page itself,Ā so the relative distancesĀ betweenĀ pointsĀ are preserved.Ā Ā Therefore,Ā we can conclude thatĀ there isĀ alwaysĀ a pointĀ thatĀ isĀ in theĀ sameĀ place onĀ bothĀ the originalĀ and reduced pages,Ā asĀ longĀ asĀ the reductionĀ isĀ done byĀ the sameĀ factorĀ inĀ both dimensions.Ā Ā ExerciseĀ 51 Ā š“š“ ā² (4,4), šµšµ ā² (4,12)Ā š¼š¼šššš Ā š¶š¶ ā² (10,4) Ā Ā Ā ExerciseĀ 52Ā ( šš , šš ) ā ( šš , ššĀ ā 4)Ā š“š“ (2, ā 1) āĀ š“š“ ā² (2, ā 5)Ā šµšµ (0,4) āĀ šµšµ ā² (0,0)Ā š¶š¶ ( ā 3,5) āĀ š¶š¶ ā² ( ā 3,1) Ā Ā ExerciseĀ 53Ā ( šš , šš ) ā ( ššĀ ā 1, šš +Ā 3)Ā š“š“ (2, ā 1) āĀ š“š“ ā² (1,2)Ā šµšµ (0,4) āĀ šµšµ ā² ( ā 1,7)Ā š¶š¶ ( ā 3,5) āĀ š¶š¶ ā² ( ā 4,8) Ā Ā ExerciseĀ 54Ā ( šš , šš ) ā ( šš +Ā 3, ššĀ ā 1)Ā š“š“ (2, ā 1) āĀ š“š“ ā² (5, ā 2)Ā šµšµ (0,4) āĀ šµšµ ā² (3,3)Ā š¶š¶ ( ā 3,5) āĀ š¶š¶ ā² (0,4) Ā Ā ExerciseĀ 55Ā
( šš , šš ) ā ( ššĀ ā 2, šš )Ā š“š“ (2, ā 1) āĀ š“š“ ā² (0, ā 1)Ā šµšµ (0,4) āĀ šµšµ ā² ( ā 2,4)Ā š¶š¶ ( ā 3,5) āĀ š¶š¶ ā² ( ā 4,5) Ā Ā ExerciseĀ 56Ā ( šš , šš ) ā ( šš +Ā 1, ššĀ ā 2)Ā š“š“ (2, ā 1) āĀ š“š“ ā² (3, ā 3)Ā šµšµ (0,4) āĀ šµšµ ā² (1,2)Ā š¶š¶ ( ā 3,5) āĀ š¶š¶ ā² ( ā 2,3) Ā Ā ExerciseĀ 57Ā ( šš , šš ) ā ( ššĀ ā 3, šš +Ā 1)Ā š“š“ (2, ā 1) āĀ š“š“ ā² ( ā 1,0)Ā šµšµ (0,4) āĀ šµšµ ā² ( ā 3,5)Ā š¶š¶ ( ā 3,5) āĀ š¶š¶ ā² ( ā 6,6) Ā
Chapter 4.5 Dilations
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