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Q.1 In the adjoining figure AB\parallel CD Find the values of x \ and\  y . Give reasons.q1

\angle EGC=\angle DGH (Vertically opposite angles)

 \Rightarrow \angle DGH=2x

Hence y+2x=180   ------ (i)

\angle DGH=\angle GHA (Alternate interior angles)

\Rightarrow 2x+3x=180

\Rightarrow x=36^{\circ}

Substituting in (i) we get

y=180-72=108^{\circ}

Hence x=36^{\circ} \ and\ y=108^{\circ}

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Q.2 In each of the following figures, AB \parallel CO  Find. the value of x . Give reasons

i) Since AB\parallel CD q2

 \angle DGH+\angle BHG=180^{\circ} 

 4x-23+3x=180^{\circ} 

 7x=203

 x=29^{\circ} 

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ii)  Since AB\parallel CD , corresponding angles are equal thereforeq3

  2x+15=3x-20

  x=35^{\circ} 

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Q.3 In the adjoining figure,  AB \parallel CD are cut by a transversal, at  E \ and\  F respectively. If  \angle 2: \angle 1 = 5:4 , find the measure of each one of the marked angles.q4

Given  AB\parallel CD, \angle 2:\angle 1=5:4

Therefore  5x+4x=180

  x=20

  \therefore \angle 2=100^{\circ} \ \&\  \angle 1=80^{\circ}   

 Therefore

  \angle 3=\angle 1=80^{\circ}   (Vertically opposite angles)

  \angle 2=\angle 4=100^{\circ}   (Vertically opposite angles )

  \angle 4=\angle 6=100^{\circ}  (Alternate angles)

  \angle 3=\angle 5=80^{\circ}   (alternate angles)

 Similarly  \angle 5=\angle 7=80^{\circ}   \ and\  \angle 6=\angle 8=100^{\circ}

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Q.4 In the adjoining figure  AB \parallel  CD , Find the values of  x, \ y \ and\  z .q5

 \angle DPR=\angle PRQ (Alternate interior angles)

Therefore  80+3x+2x=180

  5x=100

  \Rightarrow x=20^{\circ}

Therefore  \angle QRP=y=60^{\circ}

  \angle QPR=2x=40^{\circ}

 Now   y+z=180 (angles on straight lines are complementary)

  z=180-60=120^{\circ}

  Hence \  x=20^{\circ}  ,\  y=60^{\circ}  , \ z=120^{\circ}

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Q.5 In each of the following figures,  AB \parallel CD  find, the value of  x  in each case

i) Given  AB\parallel CD q6

 Extend  AE and CD backwards

  \angle ABE+\angle EFC=180

  \angle EFC=76^{\circ}

  \angle FEC=180-x

  \angle ECF=64

 Therefore  76+180-x+64=180^{\circ}   \ or\  x+140^{\circ}

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ii)  AB\parallel CD q7

Draw a line parallel to  AB \ or\  CD passing through point E

 \angle CEF=\angle ECD (Alternate angles)

Therefore  x=35^{\circ}

Similarly  \angle BAE=\angle AEF (Alternate angles)

 Therefore  x=65^{\circ}

  x=x_1 +x_2=35+65=100^{\circ}

 Hence  x=100^{\circ}

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iii)  AB\parallel CD q8

Draw a line parallel to  AB \ or\  CD

Therefore

  x_1=35^{\circ} (Alternate angles)

  x_2=75^{\circ}  (Alternate angles)

 Hence  x=360-35-75

  x=250^{\circ}

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iv)  AB\parallel CD q9

Draw a line  XY \ parallel \ to \ AB \ and \ CD

 \angle ABE=\angle XMB (Corresponding angles)

  x_1=63^{\circ}

Similarly

  x_2=\angle MDC=180^{\circ}

Or  x_2=50^{\circ}

Hence  x=360-x_1-x_2

 =360-63-50

  =247^{\circ}

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v)  AB\parallel DC q10

  \angle AMP=\angle NMB

 Therefore

  \angle NMB=x+5+27=x+32

  \angle OND=\angle NMB (Corresponding angles)

  3(x-6) = x+32

  2x=32+18=50^{\circ}

 or  x=25^{\circ}

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vi)  AB\parallel CD q11

Sum of interior angles

  \angle BXY+\angle DYX=180^{\circ}

  2x-10+32+3x+38=180

  5x=180-60

  x=30^{\circ}

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Q.6 In the adjoining figure,  AB \parallel CD Find the values of  x,\  y,\  z q13

 AB \parallel CD

Therefore

 z+x=180^{\circ}  -----(i)

Also

  x+70+x=180

  2x=110

 or  x=55^{\circ}

Substituting in (i)

   z=180-55=125^{\circ}

Sum of angles of a triangle   = 180^{\circ}

Therefore  y + 90 +55 =180^{\circ}

  y=35^{\circ}

Q.7 In each of the following figures, AB \parallel CD . Find the values of  x,\  y,\  z .

i)  AB\parallel CD q14

 \angle ABO=\angle BCD (Alternate angles)

Therefore  z=80^{\circ}

Similarly  x=40^{\circ}

 \angle BAO=\angle ODC (Alternate angles)

Since

  x+80+y=180^{\circ}   (Angles of a triangle)

  y=180-80-40=60^{\circ}

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ii)  AB\parallel CD q15

  \angle CGH+\angle AHZ=180^{\circ}

  70+x=180^{\circ}

  x=110^{\circ}

  \angle HJZ=180-120=60^{\circ}

  \angle HJZ+\angle CGJ=180^{\circ}

  60+z+70=180^{\circ}

  z=180-130=50^{\circ}

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Q.8  In the given figure,  AB\parallel CD \ and\  EF\parallel GH  Find the values of  x, y, z, t and w .q16

 AB\parallel CD and EF\parallel GH

  x=60^{\circ}  (Vertically opposite angles )

  x=y (Alternate angles)

  y= 60^{\circ}

  \angle ZXY=180-110=70^{\circ}

 Therefore  \angle CYZ=180-70-60=50^{\circ}

 Therefore  z=180-50-60=70^{\circ}

  70+w=180^{\circ}

  w=110^{\circ}

  t+w=180^{\circ}   (angles on straight line)

  t=180-110=70^{\circ}

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Q.9 In the given figure,  AB\parallel CD, EF\parallel GH and JK\parallel LM  Find the values of   x, y, z, t and w .q17

 AB\parallel CD

  EF\parallel GH

  JK\parallel LM

  \angle TRS=\angle QSR (alternate angles)

 Therefore   w=70

  y+x=110 (sum of corresponding angles)

   x+t=110 (sum of angles of triangle)

   t=y (alternate angles)

   HG\parallel EF and CD is an intercept y=65^{\circ}

  Therefore  t=65^{\circ}  , x=45^{\circ}

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Q.10 In each of the following figures, find out for what value of  x will the lines I and m be parallel to each other?

i) For   l and m to be parallelq18

 3x-25=2x+5 (Corresponding angles)

  x=30^{\circ}

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ii) For  l and m to be parallelq19

  180-(2x-10)+180-(5x-20)=180^{\circ}

  180+10-2x-5x+20=0

  210=7x

  x=30^{\circ}

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iii) For   l and m to be parallelq20

  70+(3x+14)+(2x+6)=180 (corresponding angles)

  5x=90

  x=18^{\circ}

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iv) Given:  4y+3y+5y=180 (angles on a straight line )

  12y=180 q21

  y=15

 For  l and m to be parallel

  3y+5y=x (corresponding angles)

  8\times 15=x

 or  x=120^{\circ}

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Q.11 In the adjoining figure  AB\parallel CD and they cut the line.  PQ and QR at E, F and G, H respectively. Find the value of  x .

  AB\parallel CD q22

  \angle PEF=\angle PGH (corresponding angles)

 Therefore   \angle HGQ=180-85=95

  \angle QHG=180-125=55

 Hence  95+55+x=180

  x=30^{\circ}

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Q.12 In the adjoining figure:  AB\parallel BC\parallel CD .Find the value of  x .q23

 Given:  AB\parallel BC\parallel CD

  CE=180-135=45 (corresponding angles)

  \angle ABC=\angle BCD (alternate angles)

  70=x+45

  x=25^{\circ}

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Q.13 In the adjoining figure:  AB\parallel CD .Find the value of   x

 Given  AB\parallel CD q24

 Draw a line  l to AB and CD

  \angle BEX=\angle EXY (alternate angles)

  x1=20^{\circ}

 Similarly  \angle YXF=\angle XFD (alternate angles)

  x2=y

  \angle XEF=180-75-20=85^{\circ}   (Angles on a straight line)

  \angle DFE+\angle FEB=180^{\circ}

  25+y+105=180^{\circ}

  y=180-130=50^{\circ}

 Therefore  x2=y=50^{\circ}

 Hence x=x1+x2=70^{\circ} 

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Q.14 In the adjoining figure  AB\parallel CD and EF cuts them at G and H respectively.  GP and HQ are bisectors of  \angle AGH and \angle GHD respectively. Prove that $latex  GP \parallel HQ.q25

Given  AB\parallel CD

 GP is angle bisector of  \angle AGH

 HQ is angle bisector of  \angle GHD

 Because  AB\parallel CD

  2x=2y

  or   x=y

 Now EF intersect PG and HQ &s=1$

  \angle PGH=\angle QHG=x^{\circ}   (Alternate angles)

 Hence  PG\parallel HQ

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Q.15 In each of the following figures determine the values of:  x, y, z and t :

i) Using corresponding angle and alternate anglesq27

  t=67^{\circ}

  67+x=180

  x=113^{\circ}

  x+z=180^{\circ}

  z=180-113=67

  x+y=180^{\circ}

  y=180-113=67^{\circ}

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ii)  (3z-25)+(2y-10+=180 q28

  3z+2y=215  -----(i)

  2x+110=180

  x=35

  5t=2x= 70

  t=14

  3z-25+5\times 14=180

  3z=135

  Z=45^{\circ}

  3z+2y=215

  2y=215-135=80

  y=40

 Hence

  x=35^{\circ}  , y=40^{\circ}  , t=14^{\circ}  , z=45^{\circ}

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iii)  y+72=180  or  y=108^{\circ}   (corresponding angles)q29

  72+3z=180 or 3z=108 or z=36^{\circ}

  t+108=180 or t=72^{\circ}   (corresponding angle)

  132+x=180 or x=48^{\circ}  (corresponding angles)

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iv)   2y-11+5z=180 q30

  5z=125 or

  z=25

  4t+7=125

  t=118/4=27

  x+25=225 or x=90

  x+25+2y-11=180

  2y=180-90-14 or y=33

 Hence  x=90^{\circ}  , y=33^{\circ}  , z=25^{\circ}  and t=27^{\circ}

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Q.16 State, giving reasons, whether  AB\parallel CD or not. Given in (iii):  EF\parallel GH q27

i)  \angle EGD=\angle GHB=\angle AHF

  70=\angle GHB=70^{\circ}

 Hence

  AB\parallel CD

 Alternate angles are equal

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ii) Since  \angle AHG+\angle HGC \neq 180, AB is not parallel to CD q32

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iii) Given   EF\parallel GH q34

 Therefore  \angle AXF=\angle AYH=100^{\circ}

 But   \angle AYH+100 \neq 180^{\circ}

 Hence  AB is not parallel to CD

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iv)  \angle CXY=180-126=54^{\circ} q35

  \angle AYX=180-64=116^{\circ}

  \angle AYX+CXY=116+54=180^{\circ}

 Hence AB \parallel \  CD

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Q.17 In the given figure  AB\parallel CD and Prove that  AE\parallel \angle A= \angle C Prove that  AE\parallel CF q36

Given  AB\parallel CD,  \angle A=\angle C

 \angle A=\angle DXE=\angle C

Hence   \angle DXE=\angle XCF (Corresponding angles are equal)

 Hence  AE\parallel CF

 

ICSE Board: Suggested Books     ICSE Board:  Foundation Mathematics
Class 8: Reference Books               Class 8: NTSE Preparation
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