Ncert Solutions For Class 9 Maths Ex 6.1

Ncert Solutions For Class 9 Maths Chapter 6 Ex 6.1

Q1: Lines AB and CD intersect at 0. If AOC+BOE=70 and BOD=40, find  BOE and reflex  COE.

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SOL:

Lines  ABandCDintersectatO.

AOC=BOD                           (Vertically Opposite Angles)

But BOD=40                              ………(i)

AOC=40                    ……….(ii)

Now, AOC+BOE=7040+BOE=70

BOE=30

Reflex COE=COD+BOD+BOE

=COD40+30                ………..(iii)     (using angle (i) and (ii)

COD=180                                           (as it is a straight line)

Thus ,COE=180+40+30=270

 

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

If POY=90 and a:b=2:3, find c.

Sol:

Ray OP stands on line XY

POX+POY=180        (as is linear pair axiom)

Since POY=90   (given)

POX=90 POM+XOM=90

a+b=90                 …………………….(i)

a:b=2:3ab=23ora2=b3=k a=2k,b=3k

 

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

2k+3k=90 5k=90k=18

a=2k=2(18)=36b=3k=3(18)=54    …….(ii)

 

Ray OX stands on line MN              (Linear pair axiom)

b+c=180 54+c=180

c=18054        (using equation (ii)

c=126

 

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

Sol:

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Ray QP stands on line ST.

PQS+PQR=180           ………….(i)

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

PQS+PQR=180           …………..(ii)

From (i) and (ii), we get

PQS+PQR=PRQ+PRT PQS=PRT

(since PQR=PRQ)

 

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

Sol:

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

We know, the sum of all the angles round a point is equal to 360

So, x+y+w+z=360

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

x+y+x+y=360 2(x+y)=360 (x+y)=180

AOB is a line.

 

Q5: In the figure POQ  is a line. Ray QR is a perpendicular to line PQ. OS is another ray lying between rays OPandOR. Prove that ROS=12(QOSPOS).

Sol:

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Given, ray QR is perpendicular to line PQ.

QOR=POR=90                  ……………(i)

QOS=QOR+ROS                             …………….(ii)

QOS=PORROS                               ……………(iii)

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

QOSPOS=QOR+ROSPOR+ROS =(QORPOR)+2ROS =2ROS

(As we know QOR=POR from equation (i) )

ROS=12(QOSPOS)

 

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

Sol:

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Given, ray YZ stands on line PX.

XYZ+ZYP=180

(As forms a linear pair axiom)

64+ZYP=180

ZYP=116                                ……………(i)

Given ray YQ bisect ZYP

PYQ=ZYQ=12ZYP=12(116)=58                                  ……………..(ii)

reflexQYP=36058=302

Again ,XYQ=XYZ+ZYQ

=64+58                  (From eq (ii))

=122

 

Q7: In the given figure, find the values of xandy and then show that ABCD.

 

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Sol:

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Ray AE stands on line GH

AEG+AEH=180

(as forms Linear pair axioms)

50+x=180  (as AEG is given)

x=18050=130     …….(i)

y=130                           ……..(ii)    (vertically opposite angles)

From (i) and (ii) we have,

x=y

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

So we can say that ABCD

 

Q8: In the figure, if ABCD, CDEF and y:z=3:7, find x.

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Sol:

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Given ABCD, CDEF

ABEF

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

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

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

From (i) and (ii), we have

z+y=180 y:z=3:7

Now solving the ratios, we get

Sum of ratios =3+7=10

y=310×180=54 and

z=710×180=126

As we know x=z

x=z=126

 

Q9: In the figure, if ABCD, EFCD and GED=126, find AGE,GEFandFGE.

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Sol:

(i) AGE=GED=126  (as forms alternate interior angles)

(ii) GED=GEF+FED=126

GEF+90=126

(as given that EFCD)

GEF=12690=36

(iii) CEG+GED=180

And its given that GED=126

CEG+126=180 CEG=180126 CEG=54

FGE=CEG=54        (alternate angles)

 

Q10: In the figure, if PQST, PQR=110 and RST=130, find QRS.

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

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Sol:

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

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RST+SRU=180

(Sum of the consecutive interior angles on the same side of the transversal is 180)

130+SRU=180

SRU=180130=50       ………(i)

QRU=PQR=110         (alternate interior angles)

QRS+SRU=110

QRS+50=110     (using (i))

QRS=11050=60

 

Q11: In the figure, if ABCD, APQ=50 and PRD=127, find xandy.

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Sol:

From the figure we can see that

APQ=x=50     (as forms alternate interior angles)

PRD=x+y=127   (as we know that exterior angles of a triangle is equal to the sum of the two interior opposite  angles )

50+y=127 y=12750=77

 

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

(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.)

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 Sol:

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 Construction: Draw ray BLPQ and ray CMRS.

BLPQ, CMRS and PQRS.

BLCM

LBC=MCB     ……(i)     (Alternate interior angles)

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

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

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

Adding (i) and (iv), we get

LBC+ABL=MCB+MCD ABC=BCD

But these are alternate interior angles and they are equal.

So, ABCD

 

Q13: In the figure, sides QPandRQ are produced to point SandT respectively. If SRP=135andPQT=110, find PRQ.

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Sol:

TR is a line So, PQT+PQR=180

110+PQR=180

PQR=180110=70   ……(i)

QS is a line.

SPR+QRP=180 135+QRP=180

QRP=180135=45 …….(ii)

In PQR,PQR+QPR+PRQ=180   (sum of all the interior angles of a traingle is 180.

70+45+PRQ=180    (by equation (i) and (ii) )

115+PRQ=180 PRQ=180115=65

 

Q14 :  In the figure, X=62XYZ=54. If YO and ZO of  XYZandXZY respectively of XYZ, find OZYandYOZ.

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Sol:

In XYZ, XYZ+YZX+ZXY=180.       (Sum of interior angles of a triangle is 180)

116+YZX=180

YZX=180116=64  …..(i)

YO is the bisector of XYZ

XYO=OYZ=12XYZ=12(54)=27        …….(ii)

ZOisthebisectorofYZX

XZO=OZY=12YZX=12(64)=32     ……(iii)   (from equation (i))

 

In triangle OYZ, OYZ,OZYandYOZ=180     (sum of interior angle of a triangle is 180°)

27+32+YOZ=180    ( using equation (i) and (ii))

59+YOZ=180 YOZ=18059=121

 

Q15 : In the figure,AB=12DE,BAC=35andCDE=53,find DCE.

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Sol:

DEC=BAC=35 ….(i)    (alternate interior angles)

CDE=53       ……..(ii)       (given)

In CDE,CDE+DEC+DCE=180°    (as sum of interior angles of a triangle is 180.

53+35+DCE=180 88+DCE=180 DCE=18088=92

 

Q16 : In the figure,if lines PQandRSintersect at point T,such that PRT=40RPT=95,SQT=75,findSQT.

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Sol:

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

PTR+40+95=180 PTR+135=180 PTR=180135 PTR=45

QTR=PTR=45   (vertically opposite angles)

In TSQ,QTS+TSQ+SQT=180    (as sum of interior angles of a triangle is 180.

45+75+SQT=180 120+SQT=180 SQT=180120 SQT=60

 

Q17 :  In the figure, if PQPS,SQR=28andQRT=65,then find the value of xandy.

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Sol:

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

65=28+QSR QSR=6528=37

PQSP     (given)

QPS+PSR=180     (the sum of consecutive interior angles on the same side of the transversal in 180 )

90+PSR=180 PSR=18090=90 PSQ+QSR=90 y+37=90 Y=9037=53

 

In PSQ,PSQ+QSP+QPS=180    (as sum of interior angles of a triangle is 180.

x+y+90=180 x+53+90=180 x+143=180 x=180143=37

 

Q18 :  In the figure the side QRofPQR is produced to a point S. If the bisectors of PQRandPRS meet at point T, then prove that QTR=12QPR.

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Sol:

TRS is an exterior angle of TQR

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

PRS is an exterior angle of PQR

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

2TRS=2TQR+QPR   (QT is the bisector of PQRandRT is the bisector of PRS

TRS

2(TRSTQR)=QPR   ……(iii)

From (i), TRSTQR=QTR     ……(iv)

From (iii) and (iv) , we obtain

2QTR=QPR QTR=12QPR