Class 8: Lines and Angles – Exercise 28B – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

Answer:

$\angle EGC=\angle DGH$ (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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Question 2: In each of the following figures, $AB \parallel CO$ Find. the value of $x$ . Give reasons

Answer:

i) Since $AB\parallel CD$

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

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

$\Rightarrow 7x=203$

$\Rightarrow x=29^{\circ}$

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

$2x+15=3x-20$

$\Rightarrow x=35^{\circ}$

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Question 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.

Answer:

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

Therefore $5x+4x=180$

$\Rightarrow x=20$

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

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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Question 4: In the adjoining figure $AB \parallel CD$ , Find the values of $x, \ y \ and\ z$ .

Answer:

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

Therefore $80+3x+2x=180$

$\Rightarrow 5x=100$

$\Rightarrow x=20^{\circ}$

Therefore $\angle QRP=y=60^{\circ}$

$\Rightarrow \angle QPR=2x=40^{\circ}$

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

$\Rightarrow z=180-60=120^{\circ}$

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

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Question 5: In each of the following figures, $AB \parallel CD$ find, the value of $x$ in each case

Answer:

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

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 $x=100^{\circ}$

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 $x=360-35-75$

$x=250^{\circ}$

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 $x=360-x_1-x_2$

$=360-63-50$

$=247^{\circ}$

v) $AB\parallel DC$

$\angle AMP=\angle NMB$

Therefore

$\angle NMB=x+5+27=x+32$

$\angle OND=\angle NMB$ (Corresponding angles)

$3(x-6) = x+32$

$\Rightarrow 2x=32+18=50^{\circ}$

$\Rightarrow x=25^{\circ}$

vi) $AB\parallel CD$

Sum of interior angles

$\angle BXY+\angle DYX=180^{\circ}$

$2x-10+32+3x+38=180$

$\Rightarrow 5x=180-60$

$\Rightarrow x=30^{\circ}$

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Question 6: In the adjoining figure, $AB \parallel CD$ Find the values of $x,\ y,\ z$ q13

Answer:

$AB \parallel CD$

Therefore

$z+x=180^{\circ}$ … … … … … i)

Also

$x+70+x=180$

$2x=110$

$x=55^{\circ}$

Substituting in (i)

$z=180-55=125^{\circ}$

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

Therefore $y + 90 +55 =180^{\circ}$

$y=35^{\circ}$

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

Answer:

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)

$y=180-80-40=60^{\circ}$

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

Answer:

$AB\parallel CD and EF\parallel GH$

$x=60^{\circ}$ (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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Question 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

Answer:

$AB\parallel CD$

$EF\parallel GH$

$JK\parallel LM$

$\angle TRS=\angle QSR$ (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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Question 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? q18

Answer:

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

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

$x=30^{\circ}$

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

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

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

$210=7x$

$x=30^{\circ}$

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

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

$5x=90$

$x=18^{\circ}$

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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Question 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 $x$ .

Answer:

$AB\parallel CD$

$\angle PEF=\angle PGH$ (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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Question 12: In the adjoining figure: $AB\parallel BC\parallel CD$ .Find the value of $x$ . q23

Answer:

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

Answer:

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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Question 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 $GP \parallel HQ$ . q25