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

$\angle EGC=\angle DGH$(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 $AB\parallel CD$

$\angle DGH+\angle BHG=180^{\circ}$

$4x-23+3x=180^{\circ}$

$7x=203$

$x=29^{\circ}$

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

$2x+15=3x-20$

$x=35^{\circ}$

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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.

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 $x, \ y \ and\ z$.

$\angle DPR=\angle PRQ$ (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 $AB\parallel CD$

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) $AB\parallel CD$

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) $AB\parallel CD$

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) $AB\parallel CD$

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) $AB\parallel DC$

$\angle AMP=\angle NMB$

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) $AB\parallel CD$

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 $x,\ y,\ z$

$AB \parallel CD$

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) $AB\parallel CD$

$\angle ABO=\angle BCD$(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) $AB\parallel CD$

$\angle CGH+\angle AHZ=180^{\circ}$

$70+x=180^{\circ}$

$x=110^{\circ}$

$\angle HJZ=180-120=60^{\circ}$

$\angle HJZ+\angle CGJ=180^{\circ}$

$60+z+70=180^{\circ}$

$z=180-130=50^{\circ}$

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

$AB\parallel CD and EF\parallel GH$

$x=60^{\circ}$ (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 $x, y, z, t \ and \ w$.

$AB\parallel CD$

$EF\parallel GH$

$JK\parallel LM$

$\angle TRS=\angle QSR$ (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 $l \ and \ m$ be parallel to each other?

i) For $l \ and \ m$ to be parallel

$3x-25=2x+5$ (Corresponding angles)

$x=30^{\circ}$

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ii) For $l \ and \ m$ to be parallel

$180-(2x-10)+180-(5x-20)=180^{\circ}$

$180+10-2x-5x+20=0$

$210=7x$

$x=30^{\circ}$

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iii) For $l \ and \ m$ to be parallel

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

$5x=90$

$x=18^{\circ}$

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

$12y=180$

$y=15$

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 $x$.

$AB\parallel CD$

$\angle PEF=\angle PGH$ (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 $x$.

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 $AB\parallel CD$

Draw a line $l \ to \ AB \ and \ CD$

$\angle BEX=\angle EXY$(alternate angles)

$x_1=20^{\circ}$

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

$x_2=y$

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

$\angle DFE+\angle FEB=180^{\circ}$

$25+y+105=180^{\circ}$

$y=180-130=50^{\circ}$

Therefore $x_2=y=50^{\circ}$

Hence $x=x_1+x_2=70^{\circ}$

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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 $GP \parallel HQ$.

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 $EF \ intersect \ PG \ and \ HQ$

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

$t=67^{\circ}$

$67+x=180$

$x=113^{\circ}$

$x+z=180^{\circ}$

$z=180-113=67$

$x+y=180^{\circ}$

$y=180-113=67^{\circ}$

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ii) $(3z-25)+(2y-10+=180$

$3z+2y=215 -----(i)$

$2x+110=180$

$x=35$

$5t=2x= 70$

$t=14$

$3z-25+5\times 14=180$

$3z=135$

$Z=45^{\circ}$

$3z+2y=215$

$2y=215-135=80$

$y=40$

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)

$72+3z=180 \ or\ 3z=108\ or\ z=36^{\circ}$

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

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

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iv) $2y-11+5z=180$

$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): $EF\parallel GH$

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 $\angle AHG+\angle HGC \neq 180, \ AB \ is \ not \ parallel \ to \ CD$

iii) Given $EF\parallel GH$

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

$\angle AYX=180-64=116^{\circ}$

$\angle AYX+CXY=116+54=180^{\circ}$

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 $AE\parallel CF$

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

$\angle A=\angle DXE=\angle C$

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

Hence $AE\parallel CF$