E Zdd øãç ½ ø Ù ®Ý Ý Ä ÝÝ®¦ÄÃ ÄãÝ · (ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm? (iii) ∠A = 70° and ∠C = 65°? 4. Draw a rough figure of a quadrilateral

7

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°

Maths VIII – Understanding Quadrilateral

8


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


9


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


10


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.


11


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.


12


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.


13


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)


14


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:


15


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.


16


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.


17


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.


22

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.


23

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°


24

(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°.


25

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°


26

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°


27

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


28

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]


29

(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.


30

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°


31

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°]


32

\ 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.


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°


34

3. 5cm 4. 34 5. 45 6. 95°,40°


Documents

E Zdd øãç ½ ø Ù ®Ý Ý Ä ÝÝ®¦ÄÃ ÄãÝ · (ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm? (iii) ∠A = 70° and ∠C = 65°? 4. Draw a rough figure of a quadrilateral