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NCERT Textual Exercises and Assignm
entsExErcisE 3.1
1. Givenherearesomefigures.
(i) (ii)
(iii) (iv)
(v) (vi)
(vii) (viii)
Classifyeachofthemonthebasisofthefollowing. (a) Simplecurve (b) Simpleclosedcurve (c) Polygon (d) Convexpolygon (e) Concavepolygon 2. Howmanydiagonalsdoeseachofthefollowinghave? (a) Aconvexquadrilateral (b) Aregularhexagon (c) Atriangle 3. Whatisthesumofthemeasuresoftheanglesofaconvexquadrilateral?Will
thispropertyholdifthequadrilateralisnotconvex?(Makeanon-convexquadrilateralandtry!)
4. Examinethetable.(Eachfigureisdividedintotrianglesandthesumoftheanglesdeducedfromthat.) Figure
Side 3 4 5 6Angle sum
180° 2 × 180° = (4 – 2) × 180°
3 × 180° = (5 – 2) × 180°
4 × 180° = (6 – 2) × 180°
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NCERT Textual Exercises and Assignm
ents Whatcanyousayabouttheanglesumofaconvexpolygonwithnumberof
sides? (a) 7 (b) 8 (c) 10 (d) n 5. Whatisaregularpolygon? Statethenameofaregularpolygonof (i) 3 sides (ii) 4 sides (iii) 6 sides 6. Findtheanglemeasurexinthefollowingfigures.
(a) (b)
(c) (d)
7.
(a) Findx + y + z (b) Findx + y + z + w
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NCERT Textual Exercises and Assignm
entsTest Yourself - UQ1
1. Findthevalueoftheunknownangleineachofthefollowingfigures:
(i) (ii)
(iii)
2. Whatisthesumofallexterioranglesofaconvexquadrilateral? 3. Whatissumofallinterioranglesofapentagon? 4. InaquadrilateralABCD,∠A = 150° and ∠B=∠C = ∠D,find∠B,∠C and
∠D. 5. Themeasuresofthreeanglesofaquadrilateralare39°,141°and13°.Find
thefourthangle. (a) 117° (b) 167° (c) 108° (d) 137°
ExErcisE 3.2 1. Findxinthefollowingfigures.
(a) (b)
2. Findthemeasureofeachexteriorangleofaregularpolygonof (i) 9 sides (ii) 15 sides
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NCERT Textual Exercises and Assignm
ents 3. Howmanysidesdoesaregularpolygonhaveifthemeasureofanexterior
angleis24°? 4. Howmanysidesdoesaregularpolygonhaveifeachofitsinteriorangles
is165°? 5. (a) Isitpossibletohavearegularpolygonwithmeasureofeachexterior
angleas22°? (b) Canitbeaninteriorangleofaregularpolygon?Why? 6. (a) Whatistheminimuminterioranglepossibleforaregularpolygon?
Why? (b) Whatisthemaximumexterioranglepossibleforaregularpolygon?
Test Yourself - UQ2 1. Aquadrilateralhasallfouranglesofthesamemeasure,whatisthemeasure
ofeachangle? 2. Twoanglesofaquadrilateralareofmeasure50°andtheothertwoangles
areequal.Whatisthemeasureofeachofthesetwoangles? 3. ABCDisaquadrilateral.AOandBOaretheanglebisectorsofangleAand
BwhichmeetatO.If∠C=70°,∠D=50°,find∠AOB. 4. Thevalueofyinthegivenfigureis:
(a) 89° (b) 91° (c) 101° (d) 79° 5. Findthemeasureof∠ADC.
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NCERT Textual Exercises and Assignm
ents (a) 37° (b) 47° (c) 133° (d) 43° 6. Whatisthevalueofa?
(a) 90° (b) 10° (c) 180° (d) 2° 7. Findthevaluesofx,y and zinthediagram.Givereasonswherevernecessary.
8. Inthegivendiagram,thevalueofx is
(a) 170° (b) 190° (c) 100° (d) 90° 9. Iftheanglesofaquadrilateralarex°,(2x+13)°,(3x+10)°,(x–6)°,findx.
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NCERT Textual Exercises and Assignm
entsExErcisE 3.3
1. Given a parallelogramABCD.Complete each statement alongwith thedefinitionorpropertyused.
(i) AD = ...... (ii) ∠DCB=...... (iii) OC=...... (iv) m∠DAB+m∠CDA = ...... 2. Considerthefollowingparallelograms.Findthevaluesoftheunknownsx,
y,z.
(i) (ii)
(iii) (iv)
(v)
3. CanaquadrilateralABCDbeaparallelogramif (i) ∠D + ∠B=180°? (ii) AB=DC=8cm,AD=4cmandBC=4.4cm? (iii) ∠A = 70° and ∠C=65°? 4. Drawaroughfigureofaquadrilateralthatisnotaparallelogrambuthas
exactlytwooppositeanglesofequalmeasure.
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NCERT Textual Exercises and Assignm
ents 5. Themeasuresof twoadjacent anglesof aparallelogramare in the ratio
3:2.Findthemeasureofeachoftheanglesoftheparallelogram. 6. Twoadjacentanglesofaparallelogramhaveequalmeasure.Findthemeasure
ofeachoftheanglesoftheparallelogram.
7. TheadjacentfigureHOPEisaparallelogram.Findtheanglemeasuresx,y and z.Statethepropertiesyouusetofindthem.
8. ThefollowingfiguresGUNSandRUNSareparallelograms.Findx and y. (Lengthsareincm)
(i) (ii)
9.
IntheabovefigurebothRISKandCLUEareparallelograms.Findthevalueof x.
10. Explainhowthisfigureisatrapezium.Whichofitstwosidesareparallel?(Fig3.32)
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NCERT Textual Exercises and Assignm
ents 11. Findm∠CinFig3.33ifAB DC|| .
12. Findthemeasureof∠P and ∠S if SP RQ|| inFig3.34.(Ifyoufindm∠R,istheremorethanonemethodtofindm∠P?)
Test Yourself - UQ3 1. Theanglesofaquadrilateralareintheratio2:3:5:8.Findthemeasure
ofeachofthefourangles. 2. ABCDisaparallelogram:If∠A=70°,calculate∠B,∠C and ∠D.
3. Theperimeterofaparallelogramis150cm.Oneofitssidesisgreaterthantheotherby25cm.Findthelengthsofallthesidesoftheparallelogram.
4. Theratiooftwosidesofaparallelogramis3:5,anditsperimeteris48cm.Findthesidesoftheparallelogram.
5. DiagonalsofaparallelogramABCDintersectat).XYcontainsO,andX,Yarepointsonoppositesidesoftheparallelogram.Givereasonsforeachofthefollowingstatements:
Maths VIII – Understanding Quadrilateral
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NCERT Textual Exercises and Assignm
ents (i) OB=OD (ii) ∠OBY=∠ODX (iii) ∠BOY=∠DOX (iv) DBOY≅ DDOX Now,stateifXYisbisectedatO. 6. ABCDisaparallelogram.CEbisects∠CandAFbisects∠A.Ineachofthe
following,ifthestatementistrue,giveareasonforthesame.
(i) ∠A = ∠C (ii) ∠FAB=1/2∠A (iii) ∠DCE=1/2∠C (iv) ∠FAB=∠DCE (v) ∠DCE=∠CEB (vi) ∠CEB=∠FAB (vii) CE||AF (viii) AE||FC 7. In a DABC,D,E,Farerespectively,themid-pointsofBC,CAandAB.If
thelengthsofsideAB,BCandCAare17cm,18cmand19cmrespectively,findtheperimeterofDDEF.
ExErcisE 3.4 1. StatewhetherTrueorFalse. (a) Allrectanglesaresquares (b) Allrhombusesareparallelograms (c) Allsquaresarerhombusesandalsorectangles (d) Allsquaresarenotparallelograms. (e) Allkitesarerhombuses. (f) Allrhombusesarekites.
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NCERT Textual Exercises and Assignm
ents (g) Allparallelogramsaretrapeziums. (h) Allsquaresaretrapeziums. 2. Identifyallthequadrilateralsthathave. (a) foursidesofequallength (b) fourrightangles 3. Explainhowasquareis. (i) aquadrilateral (ii) aparallelogram (iii) arhombus (iv) arectangle 4. Namethequadrilateralswhosediagonals. (i) bisecteachother (ii) areperpendicularbisectorsofeachother (iii) are equal 5. Explainwhyarectangleisaconvexquadrilateral. 6. ABCisaright-angledtriangleandOisthemidpointofthesideoppositeto
therightangle.ExplainwhyOisequidistantfromA,BandC.(Thedottedlinesaredrawnadditionallytohelpyou).
Test Yourself - UQ4 1. ABCDisarhombuswith∠ABC=126°.Determine∠ACD.
2. ABCDisasquare.Determine∠DCA.
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NCERT Textual Exercises and Assignm
ents 3. Thediagonalsofarhombusare6cmand8cm.Findthelengthofasideof
therhombus.
4. ABCDisarhombuswith∠ABC=56°.Determine∠CAD. 5. ABCDisatrapeziumandABEDisasquare.IfBE=EC,find:(a)∠BAE
(b) ∠ABC(c)WhatshapeisthefigureABCE? 6. ABCDisakiteand∠A = ∠C. If ∠CAD=70°,∠CBD=65°,find:(a)
∠BCD(b)∠ADC.
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NCERT Textual Exercises and Assignments
Exercise – 3.1 1. (a) Simplecurve
(1) (2) (5) (6) (7) (b) Simpleclosedcurve
(1) (2) (5) (6) (7) (c) Polygons
(1) (2) (4) (d) Convexpolygons
(1) (e) Concavepolygon
(1) (4) 2. (a) Aconvexquadrilateralhastwodiagonals. Here,ACandBDaretwodiagonals.
D C
A B (b) Aregularhexagonhas9diagonals. Here,diagonalsareAD,AE,BD,BE,FC,FB,AC,ECandFD.
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E D
C
BA
F
(c) Atrianglehasnodiagonal. 3. LetABCDisaconvexquadrilateral,thenwedrawadiagonalACwhichdividesthequadrilateral
intwotriangles. ∠A+B+∠C + ∠D = ∠1 + ∠6 + ∠5 + ∠4 + ∠3 + ∠2 = (∠1 + ∠2 + ∠3) + (∠4 + ∠5 + ∠6) =180°+180°[ByAnglesumpropertyoftriangle] Hence,thesumofmeasuresofthetrianglesofaconvexquadrilateralis360°
16
5
2 34
CD
B
A Yes,ifquadrilateralisnotconvexthen,thispropertywillalsobeapplied. LetABCDisanon-convexquadrilateralandjoinBD,whichalsodividesthequadrilateralin
twotriangles. Usinganglesumpropertyoftriangle, InDABD,
A1
D
C6
4
3B 2
5
∠1 + ∠2 + ∠3 = 180° ...(i) ∠4 + ∠5 + ∠6 = 180° ...(ii) Addingequation(i)and(ii), ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360° ⇒ ∠1 + ∠2 + (∠3 + ∠4) + ∠5 + ∠6 = 360° ⇒ ∠A + ∠B+∠C + ∠D = 360° Henceproved.4. (a) Whenn =7,then Angle sum of a polygon = (n – 2) × 180° = 5 × 180° = 900° (b) Whenn=8,then Angle sum of a polygon = (n – 2) × 180° = (8 – 2) × 180° = 6 × 180
=1080°
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(c) Whenn=10,then Angle sum of a polygon = (n – 2) × 180° = (10 – 2) ×180° = 8 × 180°= 1440° (d) Whenn = n, then Angle sum of a polygon = (n – 2) × 180 5. Aregularpolygon:Apolygonhavingallsidesofequallengthandtheinterioranglesofequal
sizeisknownasregularpolygon. (i) 3 sides Polygonhavingthreesidesiscalledatriangle. (ii) 4 sides Polygonhavingfoursidesiscalledaquadrilateral. (iii) 6 sides Polygonhavingsixsidesiscalledahexagon. 6. (a) Usinganglesumpropertyofaquadrilateral, 50° + 130° + 120° + x = 360° ⇒ 300° + x = 360° ⇒ x = 360° – 300 ⇒ x = 60° (b) Usinganglesumpropertyofaquadrilateral, 90° + 60° + 70° + x = 360° ⇒ 220° + x = 360° ⇒ x = 360° – 220 ⇒ x = 140° (c) Firstbaseinteriorangle=180°–70°=110° Secondbaseinteriorangle=180°–60°=120° \ Angle sum of a polygon = (n – 2) × 180° = (5 – 2) × 180° = 3 × 180° = 540° \ 30° + x + 110° + 120° + x = 540° ⇒ 260° + 2x = 540° ⇒ 2x = 280° ⇒ x = 140° (d) Angle sum of a polygon = (n – 2) × 180° = (5 – 2) × 180° = 3 × 180° = 540° \ x + x + x +x + x = 540° ⇒ 5x = 540° ⇒ x = 108° Henceeachinteriorangleis108°.
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7. (a) Sincesumoflinearpairangleis180° \ 90° + x = 180 ⇒ x = 180° – 90° = 90° and z + 30° = 180° ⇒ z = 180° – 30° = 150° Also y=90°+30°=120° [Exteriorangleproperty] \ x + y + x = 90° + 120° + 150° = 360° (b) Usinganglesumpropertyofaquadrilateral, 60° + 80° + 120° + n = 360° ⇒ 260° + n = 360° ⇒ n = 360° – 260° ⇒ n = 100° Sincesumoflinearpairanglesis180° \ w + 100 = 180° ....(i) x + 120° = 180° ....(ii) y + 80° = 180° ...(iii) z+60°=180° ...(vi) Addingeq.(i),(ii),(iii)and(iv) ⇒ x + y + z + w + 100° + 120° + 80° + 60° = 180° + 180° + 180° ⇒ x + y + z + w + 360° = 720° ⇒ x + y + z + w = 720° – 360° ⇒ x + y + z + w = 360°
Test Yourself - UQ1 1. (i) 80° (ii) 40° (iii) 233° 2. 360º 3. Sum=540º 4. 70°,70°,70° 5. 167°
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Exercise – 3.2ExErcisE 3.2 ExErcisE – 3.1 1. (a) Here,125°+m=180° [Linearpair] ⇒ m = 180° – 125° = 55°
and 125° + n=180° [Linearpair] ⇒ n = 180° – 125° = 55° Exterioranglex°=sumofoppositeinteriorangles \ x° = 55° + 55° = 110 (b) Sumofanglesofapentagon =(n – 2) × 180° = (5 × 2) × 180° = 3 × 180° = 540° Bylinearpairsofangles, ∠1 + 90° = 180° ...(i) ∠2 + 60° = 180° ...(ii) ∠3 + 90° = 180° ...(iii) ∠4+70°=180° ....(iv) ∠5 + x=180° ...(v) Addingequation(i),(ii),(iii),(iv)and(v) x + (∠1 + ∠2 + ∠3 + ∠4 + ∠5) + 310° = 900 ⇒ x + 540° + 310° = 900 ⇒ x + 850° = 900° ⇒ x = 900° – 850° = 50° 2. (i) Sum of angles of a regular polygon = (n – 2) × 180° = (9 – 2) × 180° = 7 × 180° = 1260°
Eachinteriorangle = = ° = °Sum of interior angles Number of sides
12609
140
(ii) Sumofexterioranglesofaregularpolygon=360°
Eachinteriorangle = = ° = °Sum of interior angles Number of sides
36015
24
3. Letnumberofsidesben. Sumofexterioranglesofaregularpolygon=360°
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Number of sides = = °°
=Sum of exterior angles Each interior angles
36024
15
Hence,theregularpolygonhas15sides. 4. Letnumberofsidesben. Exteriorangle=180°–165°=15°
Number of sides = = °°
=Sum of exterior angles Each interior angles
36015
24
Hence,theregularpolygonhas24sides. 5. (a) No.(since22isnotadivisorof360°) (b) No,(Becauseeachexteriorangleis180°–22°=158°,whichisnotadivisorof360°) 6. (a) Theequilateral trianglebeingaregularpolygonof3 sideshas the leastmeasureofan
interiorangleof60°. Sumofalltheanglesofatriangle=180° \ x + x + x = 180° ⇒ 3x= 180° ⇒ x = 60° (b) By(a),wecanobservethatthegreatestexteriorangleis180°–60°=120°.
Test Yourself - UQ2 1. 90° 2. 130° 3. 60° 4. 91° 5. 133° 6. 2° 7. 88°,68°,92° 8. 190° 9. x + 2x + 13 + 3x + 10 + x–1=360º 7x+17=360º 7x=360º–17 7x=343º
x = 343
7 x=49º
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Exercise – 3.3ExErcisE 3.2 ExErcisE – 3.1 1. (i) AD=BC[Sinceoppositesidesofaparallelogramareequal] (ii) ∠DCB=∠DAB[Sinceoppositeanglesofaparallelogramareequal] (iii) OC=OA[sincediagonalsofaparallelogrambisecteachother] (iv)m∠DAB+m∠CDA=180°[Adjacentanglesinaparallelogramaresupplementary] 2. (i) ∠B+∠C=180°[Adjacentanglesinaparallelogramaresupplementary]
⇒ 100° + x = 180° ⇒ x = 180° – 100° = 80° and z = x =80°[Sinceoppositeanglesofaparallelogramareequal]
Also y=100°[Sinceoppositeanglesofaparallelogramareequal] (ii) x+50°=180°[Adjacentanglesina||gmaresupplementary] ⇒ x = 180° – 50° = 130° ⇒ z = x =130°[Correspondingangles] (iii) x=90°[Verticallyoppositeangle] ⇒ y + x +30°=180°[Anglesumpropertyofatriangle] ⇒ y + 90° + 30° = 180° ⇒ y + 120 = 180° ⇒ y = 180° – 120° = 60° ⇒ z = y=60°[Alternateangles] (iv) z=80°[Correspondingangles]
⇒ x+80°=180°[Adjacentanglesina||gmaresupplementary] ⇒ x = 180° – 80° = 100° and y=80°[Oppositeanglesareequalina||gm]
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(v) y=112°[Oppositeanglesareequalina||gm]
⇒ y=112°[Oppositeanglesareequalina||gm] ⇒ 40° + y + x=180°[Anglesumpropertyofatriangle] ⇒ 40° + 112° + x = 180° ⇒ 152° + x = 180° ⇒ x = 180° – 152° = 28° and z = x =28°[alternateangles] 3. (i) ∠D + ∠B=180° Itcanbe,buthere,itneedsnottobe.
(ii) No,inthiscasebecauseonepairofoppositesidesareequalandanotherpairofoppositesides are unequal.
So,itisnotaparallellogram.
(iii) No. ∠A ≠ ∠C. Sinceoppositeanglesareequalinparallelogramandhereoppositeanglesarenotequalin
quadrilateralABCD. Thereforeitisnotaparallelogram.
4. ABCDisaquadrilateralinwhichangles∠A = ∠C = 110°. Therefore,itcouldbeakite.
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5. Lettwoadjacentanglesbe3x and 2x. Sincetheadjacentanglesinaparallelogramaresupplementary. \ 3x + 2x = 180° ⇒ 5x = 180°
⇒ x = ° = °1805
36
\ One angle = 3x = 3 × 36° = 108° And Anotherangle =2x = 2 × 36° = 72° 6. Leteachadjacentaanglebex. Sincetheadjacentanglesinaparallelogramaresupplementary. \ x + x = 180° ⇒ 2x = 180°
⇒ x = ° = °1802
90
Hence,eachadjacentangleis90°. 7. Here ∠HOP=180°–70°=110°[Angleoflinearpair] and ∠E=∠HOP[Oppositeanglesofa||gmareequal] ⇒ x = 110° ∠PHE=∠HPO[Alternateangles] \ y = 40°
Now ∠EHO=∠O=70°[Correspondingangles] ⇒ 40° + z = 70° ⇒ z = 70° – 40° = 30°
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8. (i) InparallelogramGUNS, GS=UN [Oppositesidesofparallelogramareequal]
⇒ 3x = 18 x = =183
6 cm
Also GU=SN [Oppositesidesofparallelogramareequal] ⇒ 3y – 1 = 26 ⇒ 3y = 26 + 1
⇒ 3y = 27 ⇒ y = =273
9 cm Hence,x=6cmandy=9cm. (ii) InparallelogramRUNS, y+7=20 [Diagonalsof||gmbisectseachother] ⇒ y =20–7=13cm And x + y = 16 ⇒ x + 13 = 16 ⇒ x = 16 – 3 ⇒ x=3cm Hence,x=3cmandy=13cm. 9. InparallelogramRISK, ∠RIS=∠K=120° [Oppositeanglesofa||gmareequal] ∠m+120°=180° [Linearpair] ⇒ ∠m = 180° – 120° = 60° And ∠ECl=∠L=70° [Correspondingangles]
⇒ m + n + ∠ECl=180° [Anglesumpropertyofatriangle] ⇒ 60° + n + 70° = 180° ⇒ 130° + n = 180° ⇒ n = 180° – 130° = 50° Also x = n =50° [Verticallyoppositeangles] 10. Here,∠M+∠L = 100° + 80° = 180°
[Sumofinterioroppositeangleis180°]
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\ NMandKLareparallel Hence,KLMNisatrapezium. 11. Here,∠B+∠C = 180°
AB||DC
\ 120 + m∠C = 180° ⇒ m∠C = 180° – 120° = 60° 12. Here,∠P + ∠Q=180° [Sumofco-interioranglesis180°] ⇒ ∠P + 130° = 180° ⇒ ∠P = 180° – 130° ⇒ ∠P = 50° \ ∠R=90° \ ∠S + 90° = 180° ⇒ ∠S= 180° – 90° ⇒ ∠S = 90° Yes,onemoremethodistheretofind∠P. ∠S + ∠R+∠Q + ∠P=360° [Anglesumpropertyofquadrilateral] ⇒ 90° + 90° + 130° + ∠P = 360° ⇒ 310° + ∠P = 360° ⇒ ∠P = 360° – 310° ⇒ ∠P = 50°.
Test Yourself - UQ3 1. 40°,60°,100°,160° 2. 110°,70°,110° 3. 25cm,50cm,25cm,50cm 4. 9cm,15cm,9cmand15cm 5.
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33
(i) Diagonalsofaparallelogrambisecteachother. (ii) Alternateinterioranglerareequal. (iii) Verticallyoppositeangles. (iv) A.A.S.congrencycriteria. 7. 27cm
Exercise – 3.4ExErcisE 3.2 ExErcisE – 3.1 1. (a) False.Since,allsidesofsquaresareequal. (b) True.Since,inrhombus,oppositeanglesareequalanddiagonalsintersectatmid-point. (c) True.Since,squareshavethesamepropertyofrhombusbutnotarectangle. (d) False.Since,allsquareshavethesamepropertyofparallelogram. (e) False.Since,allkitesdonothaveequalsides. (f) True.Since,allrhombuseshaveequalsidesanddiagonalsbisecteachother. (g) True.Since,trapeziumhasonlytwoparallelsides. (h) True.Since,allsquareshavealsotwoparalleltines. 2. (a) Rhombusandsquarehavesidesofequallength. (b) Squareandrectanglehavefourrightangles. 3. (i) AsquareIsaquadrilateral,ifithasfourunequallengthsofsides. (ii) Asquareisaparallelogram,sinceitcontainsbothpairsofoppositesidesequal. (iii)Asquareisalreadyarhombus.Since,ithasfourequalsidesanddiagonalsbisectat90to
eachother (iv)Asquareisaparallelogram,sincehavingeachadjacentanglearightangleandopposite
sides are equal. 4. (i) Ifdiagonalsofaquadrilateralbisecteachotherthenitisarhombus,parallelogram,rectangle
or square. (ii) Ifdiagonalsofaquadrilateralareperpendicularbisectorofeachother,thenitisarhombus
or square. (iii)Ifdiagonalsareequal,thenItisasquareorrectangle 5. ArectangleisaconvexquadrilateralsinceitsvertexareraisedandbothofItsdiagonalsliein
itsinterior. 6. Since,tworighttrianglesmakearectanglewhere0isequidistantpointfromA,B,CandD
because0 is themid-pointof the twodiagonalsofarectangle.SinceACandBDareequaldiagonalsandintersectatmid-pointSo,0istheequidistantfromA.B,CandD
Test Yourself - UQ4 1. 27° 2. 45°
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3. 5cm 4. 34 5. 45 6. 95°,40°
Maths VIII – Understanding Quadrilateral