NCERT Solutions For Class 9 Maths Chapter 6

NCERT Solutions Class 9 Maths Lines and Angles

For any student it is important to be able to solve maths problems of every kind. By working on NCERT Solutions For Class 9 Maths Chapter 6 Lines and Angles, students will become familiar with concepts of lines and angles such as operations on angles when intersect, corresponding angles, alternate angles, interior angles when a transversal intersects two parallel lines, proving the sum of angles of a triangle is 180 and more. Thus download our NCERT Solutions For Class 9 Maths Chapter 6 pdf so that students can access the content from anywhere.

NCERT Solutions For Class 9 Maths Chapter 6 Exercises

Q1: Lines AB and CD intersect at 0. If AOC+BOE=70$\angle AOC +\angle BOE= 70^{\circ}$ and BOD=40$\angle BOD=40^{\circ}$, find  BOE$\angle BOE$ and reflex  COE$\angle COE$.

SOL:

Lines  ABandCDintersectatO$AB and CD intersect at O$.

AOC=BOD$∴ \angle AOC=\angle BOD$                           (Vertically Opposite Angles)

But BOD=40$\angle BOD=40^{\circ}$                              ………(i)

AOC=40$∴ \angle AOC=40^{\circ}$                    ……….(ii)

Now, AOC+BOE=7040+BOE=70$\angle AOC+\angle BOE=70^{\circ}\\ \Rightarrow 40^{\circ}+\angle BOE=70^{\circ}$

BOE=30$\Rightarrow \angle BOE=30^{\circ}$

Reflex COE=COD+BOD+BOE$\angle COE= \angle COD+\angle BOD+\angle BOE$

=COD40+30$\angle COD-40^{\circ}+30^{\circ}$                ………..(iii)     (using angle (i) and (ii)

COD=180$\angle COD=180^{\circ}$                                           (as it is a straight line)

Thus ,COE=180+40+30=270$\angle COE= 180^{\circ}+40^{\circ}+30^{\circ}=270^{\circ}$

Q2: In the figure, line XY$XY$ and MN$MN$ intersect at O$O$.

If POY=90$\angle POY=90^{\circ}$ and a:b=2:3$a:b=2:3$, find c.

Sol:

Ray OP$OP$ stands on line XY$XY$

POX+POY=180$∴ \angle POX+\angle POY=180^{\circ}$        (as is linear pair axiom)

Since POY=90$\angle POY=90^{\circ}$   (given)

POX=90$\Rightarrow \angle POX=90^{\circ}$ POM+XOM=90$\angle POM+\angle XOM=90^{\circ}$

a+b=90$\Rightarrow a+b=90^{\circ}$                 …………………….(i)

a:b=2:3ab=23ora2=b3=k$a:b=2:3\Rightarrow \frac{a}{b}=\frac{2}{3}\\ or \; \; \frac{a}{2}=\frac{b}{3}=k$ a=2k,b=3k$\Rightarrow a=2k, b=3k$

Substituting the values of a and b in (i), we get

2k+3k=90$2k+3k=90^{\circ}$ 5k=90k=18$\Rightarrow 5k=90^{\circ}\\ \Rightarrow k=18^{\circ}$

a=2k=2(18)=36b=3k=3(18)=54$∴ a=2k=2(18^{\circ})=36^{\circ}\\ b=3k=3(18^{\circ})=54^{\circ}$    …….(ii)

Ray OX$OX$ stands on line MN              (Linear pair axiom)

b+c=180$\Rightarrow b+c=180^{\circ}$ 54+c=180$\Rightarrow 54^{\circ}+c=180^{\circ}$

c=18054$\Rightarrow c=180^{\circ}- 54^{\circ}$        (using equation (ii)

c=126$\Rightarrow c=126^{\circ}$

Q3: Prove that PQS=PRT$\angle PQS=\angle PRT$ in the given isosceles triangle.

Sol:

Ray QP$QP$ stands on line ST$ST$.

PQS+PQR=180$∴ \angle PQS +\angle PQR=180^{\circ}$           ………….(i)

(as linear pair axiom ray QP$QP$ stands on line ST$ST$)

PQS+PQR=180$∴ \angle PQS+\angle PQR=180^{\circ}$           …………..(ii)

From (i) and (ii), we get

PQS+PQR=PRQ+PRT$\angle PQS+\angle PQR= \angle PRQ+\angle PRT$ PQS=PRT$\Rightarrow \angle PQS=\angle PRT$

(since PQR=PRQ$\angle PQR=\angle PRQ$)

Q4: In the given figure if x+y=w+z$x+y=w+z$ then prove that AOB$AOB$ is a line.

Sol:

x+y=w+z$x+y=w+z$                                  …………….(i)

We know, the sum of all the angles round a point is equal to 360$360^{\circ}$

So, x+y+w+z=360$x+y+w+z=360^{\circ}$

Putting the value of w+z in above equation we have,

x+y+x+y=360$x+y+x+y=360^{\circ}$ 2(x+y)=360$\Rightarrow 2(x+y)=360^{\circ}$ (x+y)=180$\Rightarrow (x+y)=180^{\circ}$

AOB$∴ AOB$ is a line.

Q5: In the figure POQ$POQ$  is a line. Ray QR$QR$ is a perpendicular to line PQ$PQ$. OS$OS$ is another ray lying between rays OPandOR$OP and OR$. Prove that ROS=12(QOSPOS)$\angle ROS=\boldsymbol{\frac{1}{2}}(\angle QOS-\angle POS)$.

Sol:

Given, ray QR$QR$ is perpendicular to line PQ$PQ$.

QOR=POR=90$∴ \angle QOR=\angle POR=90^{\circ}$                  ……………(i)

QOS=QOR+ROS$\angle QOS=\angle QOR+\angle ROS$                             …………….(ii)

QOS=PORROS$\angle QOS=\angle POR-\angle ROS$                               ……………(iii)

From equation (ii) and (iii), we have

QOSPOS=QOR+ROSPOR+ROS$\angle QOS-\angle POS=\angle QOR+\angle ROS-\angle POR+\angle ROS$ =(QORPOR)+2ROS$= (\angle QOR-\angle POR)+2\angle ROS$ =2ROS$=2\angle ROS$

(As we know QOR=POR$\angle QOR=\angle POR$ from equation (i) )

ROS=12(QOSPOS)$\Rightarrow \angle ROS=\frac{1}{2}(\angle QOS-\angle POS)$

Q6: It is given that XYZ=64$\angle XYZ=64^{\circ}$ and XY$XY$ is produced to point P$P$. Draw a figure from the given information.If ray YQ$YQ$ bisect ZYP$\angle ZYP$, find XYQ$\angle XYQ$ and reflex QYP$\angle QYP$.

Sol:

Given, ray YZ$YZ$ stands on line PX$PX$.

XYZ+ZYP=180$∴ \angle XYZ+\angle ZYP=180^{\circ}$

(As forms a linear pair axiom)

64+ZYP=180$\Rightarrow 64^{\circ}+\angle ZYP=180^{\circ}$

ZYP=116$\Rightarrow \angle ZYP=116^{\circ}$                                ……………(i)

Given ray YQ$YQ$ bisect ZYP$\angle ZYP$

PYQ=ZYQ=12ZYP=12(116)=58$∴\angle PYQ=\angle ZYQ=\frac{1}{2}\angle ZYP=\frac{1}{2}(116^{\circ})=58^{\circ}$                                  ……………..(ii)

reflexQYP=36058=302$∴ reflex \angle QYP=360^{\circ}-58^{\circ}=302^{\circ}$

Again ,XYQ=XYZ+ZYQ$\angle XYQ=\angle XYZ+\angle ZYQ$

=64+58$=64^{\circ}+58^{\circ}$                  (From eq (ii))

=122$=122^{\circ}$

Q7: In the given figure, find the values of xandy$x and y$ and then show that ABCD$AB\parallel CD$.

Sol:

Ray AE$AE$ stands on line GH$GH$

AEG+AEH=180$∴ \angle AEG+\angle AEH=180^{\circ}$

(as forms Linear pair axioms)

50+x=180$\Rightarrow 50^{\circ}+x=180^{\circ}$  (as AEG$\angle AEG$ is given)

x=18050=130$\Rightarrow x=180^{\circ}-50^{\circ}=130^{\circ}$     …….(i)

y=130$y=130^{\circ}$                           ……..(ii)    (vertically opposite angles)

From (i) and (ii) we have,

x=y$x=y$

But these angles are alternate interior angles and as they are equal.

So we can say that ABCD$AB\parallel CD$

Q8: In the figure, if ABCD$AB\parallel CD$, CDEF$CD\parallel EF$ and y:z=3:7$y:z=3:7$, find x$x$.

Sol:

Given ABCD$AB\parallel CD$, CDEF$CD\parallel EF$

ABEF$AB\parallel EF$

( as lines parallel to the same line are parallel to each other)

x=z$x=z$                    ………….(i)         (alternate interior angles)

x+y=180$x+y=180^{\circ}$ …………(ii)        (consecutive interior angles on the same side of the transversal GH$GH$ to parallel lines AB$AB$ and CD$CD$.

From (i) and (ii), we have

z+y=180$z+y=180^{\circ}$ y:z=3:7$y:z=3:7$

Now solving the ratios, we get

Sum of ratios =3+7=10

y=310×180=54$\Rightarrow y=\frac{3}{10}\times 180^{\circ}=54^{\circ}$ and

z=710×180=126$z=\frac{7}{10}\times 180^{\circ}=126^{\circ}$

As we know x=z$x=z$

x=z=126$∴ x=z=126^{\circ}$

Q9: In the figure, if ABCD$AB\parallel CD$, EFCD$EF\perp CD$ and GED=126$\angle GED=126^{\circ}$, find AGE,GEFandFGE.$\angle AGE, \angle GEF \; and \; \angle FGE.$

Sol:

(i) AGE=GED=126$\angle AGE=\angle GED=126^{\circ}$  (as forms alternate interior angles)

(ii) GED=GEF+FED=126$\angle GED=\angle GEF+\angle FED=126^{\circ}$

GEF+90=126$\Rightarrow \angle GEF+90^{\circ}=126^{\circ}$

(as given that EFCD$EF\perp CD$)

GEF=12690=36$\Rightarrow \angle GEF=126^{\circ}-90^{\circ}=36^{\circ}$

(iii) CEG+GED=180$\angle CEG+\angle GED=180^{\circ}$

And its given that GED=126$\angle GED=126^{\circ}$

CEG+126=180$\Rightarrow \angle CEG+126^{\circ}=180^{\circ}$ CEG=180126$\Rightarrow \angle CEG=180^{\circ}-126^{\circ}$ CEG=54$\Rightarrow \angle CEG=54^{\circ}$

FGE=CEG=54$\angle FGE=\angle CEG=54^{\circ}$        (alternate angles)

Q10: In the figure, if PQST$PQ\parallel ST$, PQR=110$\angle PQR=110^{\circ}$ and RST=130$\angle RST=130^{\circ}$, find QRS.$\angle QRS.$

[Hint:Draw a line parallel to ST$ST$ through point R$R$.

Sol:

From the hint we draw a line RU$RU$ parallel to ST$ST$ through point R$R$.

RST+SRU=180$\angle RST+\angle SRU=180^{\circ}$

(Sum of the consecutive interior angles on the same side of the transversal is 180$180^{\circ}$)

130+SRU=180$\Rightarrow 130^{\circ}+\angle SRU=180^{\circ}$

SRU=180130=50$\Rightarrow \angle SRU=180^{\circ}-130^{\circ}=50^{\circ}$       ………(i)

QRU=PQR=110$\angle QRU=\angle PQR=110^{\circ}$         (alternate interior angles)

QRS+SRU=110$\Rightarrow \angle QRS+\angle SRU=110^{\circ}$

QRS+50=110$\Rightarrow \angle QRS+50^{\circ}=110^{\circ}$     (using (i))

QRS=11050=60$\Rightarrow \angle QRS=110^{\circ}-50^{\circ}=60^{\circ}$

Q11: In the figure, if ABCD$AB\parallel CD$, APQ=50$\angle APQ=50^{\circ}$ and PRD=127$\angle PRD=127^{\circ}$, find xandy.$x and y.$

Sol:

From the figure we can see that

APQ=x=50$\angle APQ=x=50^{\circ}$     (as forms alternate interior angles)

PRD=x+y=127$\angle PRD=x+y=127^{\circ}$   (as we know that exterior angles of a triangle is equal to the sum of the two interior opposite  angles )

50+y=127$\Rightarrow 50^{\circ}+y=127^{\circ}$ y=12750=77$\Rightarrow y=127^{\circ}-50^{\circ}=77^{\circ}$

Q12 : In the Fig, PQ$PQ$ and RS$RS$ are two mirrors placed parallel to each other. An incident ray AB$AB$ strikes the mirror PQ$PQ$ at B$B$, the reflected ray moves along the path BC$BC$ and strikes the mirror RS$RS$ at C$C$ and again reflects back along CD$CD$. Prove that ABCD$AB\parallel CD$.

(Hint: Draw perpendiculars at A and B to the two plane mirrors. Recall that the angle of the incidence is equal to angle of reflection.)

Sol:

Construction: Draw ray BLPQ$BL\perp PQ$ and ray CMRS$CM\perp RS$.

BLPQ$BL\perp PQ$, CMRS$CM\perp RS$ and PQRS$PQ\parallel RS$.

BLCM$∴ BL\parallel CM$

LBC=MCB$\angle LBC=\angle MCB$     ……(i)     (Alternate interior angles)

ABL=LBC$\angle ABL=\angle LBC$ ………(ii)       (angle of incidence= angle of reflection)

MCB=MCD$\angle MCB=\angle MCD$  ……….(iii)       (angle of incidence = angle of reflection).

From (i), (ii) and (iii), we get ABL=MCD$\angle ABL=\angle MCD$….(iv)

Adding (i) and (iv), we get

LBC+ABL=MCB+MCD$\angle LBC+\angle ABL=\angle MCB+ \angle MCD$ ABC=BCD$\Rightarrow \angle ABC=\angle BCD$

But these are alternate interior angles and they are equal.

So, ABCD$AB\parallel CD$

Q13: In the figure, sides QPandRQ$QP\: \: and \: \: RQ$ are produced to point SandT$S\: \: and \: \: T$ respectively. If SRP=135andPQT=110$\angle SRP=135^{\circ}\; \; and \: \: \angle PQT=110^{\circ}$, find PRQ$\angle PRQ$.

Sol:

TR is a line So, PQT+PQR=180$\angle PQT+\angle PQR=180^{\circ}$

110+PQR=180$\Rightarrow 110^{\circ} +\angle PQR=180^{\circ}$

PQR=180110=70$\Rightarrow \angle PQR=180^{\circ}-110^{\circ} =70^{\circ}$   ……(i)

QS$QS$ is a line.

SPR+QRP=180$∴ \angle SPR+\angle QRP=180^{\circ}$ 135+QRP=180$\Rightarrow 135^{\circ}+\angle QRP=180^{\circ}$

QRP=180135=45$\Rightarrow \angle QRP=180^{\circ}-135^{\circ}=45^{\circ}$ …….(ii)

In PQR,PQR+QPR+PRQ=180$\bigtriangleup PQR, \angle PQR+\angle QPR+\angle PRQ=180^{\circ}$   (sum of all the interior angles of a traingle is 180$180^{\circ}$.

70+45+PRQ=180$\Rightarrow 70^{\circ}+45^{\circ}\angle+ PRQ=180^{\circ}$    (by equation (i) and (ii) )

115+PRQ=180$\Rightarrow 115^{\circ}+ PRQ=180^{\circ}$ PRQ=180115=65$\Rightarrow PRQ=180^{\circ}-115^{\circ}=65^{\circ}$

Q14 :  In the figure, X=62XYZ=54$\angle X=62^{\circ}\;\; \angle XYZ=54^{\circ}$. If YO$YO$ and ZO$ZO$ of  XYZandXZY$\angle XYZ\;\; and \;\; \angle XZY$ respectively of XYZ$\bigtriangleup XYZ$, find OZYandYOZ$\angle OZY \; \; and \angle YOZ$.

Sol:

In XYZ$\bigtriangleup XYZ$, XYZ+YZX+ZXY=180$\angle XYZ+ \angle YZX+ \angle ZXY=180^{\circ}$.       (Sum of interior angles of a triangle is 180$180^{\circ}$)

116+YZX=180$\Rightarrow 116^{\circ}+\angle YZX=180^{\circ}$

YZX=180116=64$\Rightarrow \angle YZX=180^{\circ}-116^{\circ}=64^{\circ}$  …..(i)

YO$YO$ is the bisector of XYZ$\angle XYZ$

XYO=OYZ=12XYZ=12(54)=27$\Rightarrow \angle XYO=\angle OYZ=\frac{1}{2}\angle XYZ=\frac{1}{2}(54^{\circ})=27^{\circ}$        …….(ii)

ZOisthebisectorofYZX$\Rightarrow ZO \; is \; the\; bisector\; of \;\angle YZX$

XZO=OZY=12YZX=12(64)=32$∴ \angle XZO =\angle OZY=\frac{1}{2}\angle YZX=\frac{1}{2}(64^{\circ})=32^{\circ}$     ……(iii)   (from equation (i))

In triangle OYZ$\bigtriangleup OYZ$, OYZ,OZYandYOZ=180$\angle OYZ, \; \;\angle OZY \; \; and \angle YOZ=180^{\circ}$     (sum of interior angle of a triangle is 180°)

27+32+YOZ=180$\Rightarrow 27^{\circ}+32^{\circ}+\angle YOZ=180^{\circ}$    ( using equation (i) and (ii))

59+YOZ=180$\Rightarrow 59^{\circ}+\angle YOZ=180^{\circ}$ YOZ=18059=121$\Rightarrow \angle YOZ=180^{\circ}-59^{\circ}=121^{\circ}$

Q15 : In the figure,AB=12DE$AB=\frac{1}{2}DE$,BAC=35andCDE=53$\angle BAC=35^{\circ} and \angle CDE=53^{\circ}$,find DCE$\angle DCE$.

Sol:

DEC=BAC=35$\angle DEC=\angle BAC=35^{\circ}$ ….(i)    (alternate interior angles)

CDE=53$\angle CDE=53^{\circ}$       ……..(ii)       (given)

In CDE,CDE+DEC+DCE=180°$\bigtriangleup CDE, \angle CDE+\angle DEC+\angle DCE=180°$    (as sum of interior angles of a triangle is 180$180^{\circ}$.

53+35+DCE=180$\Rightarrow 53^{\circ}+35^{\circ}+\angle DCE=180^{\circ}$ 88+DCE=180$\Rightarrow 88^{\circ}+\angle DCE=180^{\circ}$ DCE=18088=92$\Rightarrow \angle DCE=180^{\circ}-88^{\circ}=92^{\circ}$

Q16 : In the figure,if lines PQandRS$PQ\; \;and  RS$intersect at point T,$T,$such that PRT=40RPT=95,SQT=75,findSQT$\angle PRT=40^{\circ} \angle RPT=95^{\circ}, \angle SQT=75^{\circ} , \;\; find \angle SQT$.

Sol:

In In PRT,PTR+PRT+RPT=180°$\bigtriangleup PRT, \angle PTR+\angle PRT+\angle RPT=180°$    (as sum of interior angles of a triangle is 180°$180°$).

PTR+40+95=180$\Rightarrow \angle PTR+40^{\circ}+95^{\circ}=180^{\circ}$ PTR+135=180$\Rightarrow \angle PTR+135^{\circ}=180^{\circ}$ PTR=180135$\Rightarrow \angle PTR=180^{\circ}-135^{\circ}$ PTR=45$\Rightarrow \angle PTR=45^{\circ}$

QTR=PTR=45$\Rightarrow \angle QTR=\angle PTR=45^{\circ}$   (vertically opposite angles)

In TSQ,QTS+TSQ+SQT=180$\bigtriangleup TSQ, \angle QTS+\angle TSQ+\angle SQT=180^{\circ}$    (as sum of interior angles of a triangle is 180$180^{\circ}$.

45+75+SQT=180$\Rightarrow 45^{\circ}+75^{\circ}+\angle SQT=180^{\circ}$ 120+SQT=180$\Rightarrow 120^{\circ}+\angle SQT=180^{\circ}$ SQT=180120$\Rightarrow \angle SQT=180^{\circ}-120^{\circ}$ SQT=60$\Rightarrow \angle SQT=60^{\circ}$

Q17 :  In the figure, if PQPS,SQR=28andQRT=65$PQ\perp PS,\; \angle SQR=28^{\circ}\;\;and\; \; \angle QRT=65^{\circ}$,then find the value of xandy$x\;\; and\; \; y$.

Sol:

QRT=RQS+QSR$\angle QRT=\angle RQS+\angle QSR$ (exterior angle is equal to sum of the two opposite interior angles)

65=28+QSR$\Rightarrow 65^{\circ}=28^{\circ}+\angle QSR$ QSR=6528=37$\Rightarrow \angle QSR=65^{\circ}-28^{\circ}=37^{\circ}$

PQSP$PQ\perp SP$     (given)

QPS+PSR=180$\angle QPS+\angle PSR=180^{\circ}$     (the sum of consecutive interior angles on the same side of the transversal in 180$180^{\circ}$ )

90+PSR=180$\Rightarrow 90^{\circ}+\angle PSR=180^{\circ}$ PSR=18090=90$\Rightarrow \angle PSR=180^{\circ}-90^{\circ}=90^{\circ}$ PSQ+QSR=90$\Rightarrow \angle PSQ+\angle QSR=90^{\circ}$ y+37=90$\Rightarrow y+37^{\circ}=90^{\circ}$ Y=9037=53$\Rightarrow Y=90^{\circ}-37^{\circ}=53^{\circ}$

In PSQ,PSQ+QSP+QPS=180$\bigtriangleup PSQ, \angle PSQ+\angle QSP+\angle QPS=180^{\circ}$    (as sum of interior angles of a triangle is 180$180^{\circ}$.

x+y+90=180$\Rightarrow x+y+90{\circ}=180^{\circ}$ x+53+90=180$x+53^{\circ}+90^{\circ}=180^{\circ}$ x+143=180$x+143^{\circ}=180^{\circ}$ x=180143=37$x=180^{\circ}-143^{\circ}=37^{\circ}$

Q18 :  In the figure the side QRofPQR$QR \;\; of \bigtriangleup PQR$ is produced to a point S$S$. If the bisectors of PQRandPRS$\angle PQR\;\; and \angle PRS$ meet at point T$T$, then prove that QTR=12QPR$\angle QTR=\frac{1}{2}\angle QPR$.

Sol:

TRS$\angle TRS$ is an exterior angle of TQR$\bigtriangleup TQR$

TRS=TQR+QTR$\angle TRS=\angle TQR+\angle QTR$   …..(i)     (Since, the exterior angle is equal to the sum of the two interior opposite angles)

PRS$\angle PRS$ is an exterior angle of PQR$\bigtriangleup PQR$

PRS=PQR+QPR$\angle PRS=\angle PQR+\angle QPR$   ……(ii)     (since, the exterior angle is equal to the sum of the two interior opposite angles)

2TRS=2TQR+QPR$\Rightarrow 2\angle TRS=2\angle TQR+\angle QPR$   (QT$QT$ is the bisector of PQRandRT$\angle PQR \;and\; RT$ is the bisector of PRS$\angle PRS$

TRS$\Rightarrow \angle TRS$

2(TRSTQR)=QPR$\Rightarrow 2(\angle TRS-\angle TQR)=\angle QPR$   ……(iii)

From (i), TRSTQR=QTR$\angle TRS-\angle TQR=\angle QTR$     ……(iv)

From (iii) and (iv) , we obtain

2QTR=QPR$2\angle QTR=\angle QPR$ QTR=12QPR$\Rightarrow \angle QTR=\frac{1}{2}\angle QPR$