NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles

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Written by Team Trustudies
Updated at 2021-02-12


NCERT solutions for class 9 Maths Chapter 6 Lines And Angles Exercise 6.1

Q1 ) In figure, lines AB and CD intersect at O. If ?AOC + ?BOE = 70? and ?BOD = 40?, find ?BOE and reflex ?COE.
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NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles


Answer :

Here, ?AOC and ?BOD are vertically opposite angles.

??AOC = ?BOD
? ?AOC = 40?
? ?BOD = 40?] ....(i)
It is given that,
?AOC+?BOE=70?

Hence, from Eq. (i),
?40?+?BOE=70?
??BOE=70??40?

? ?BOE = 30?

Now, by Linear pair axiom,
?AOC + ?COE + ?BOE = 180?
By substituting the values, we get
40? + ?COE + 30? = 180?
? ?COE = 180? - 40? - 30?
? ?COE = 110?

Now, so as to find the reflex angle,
?COE + reflex ?COE = 360?
? 110? + reflex ?COE = 360? .....(proved)
? reflex ?COE = 360? - 110?
? reflex ?COE = 250?

Q2 ) In figure, lines XY and MN intersect at O. If ?POY = 90? and a :b = 2:3. Find c.
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NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles


Answer :

It is given that, ?POY = 90?

Now, by Linear pair axiom,
?POY + ?POX = 180?
? ?POX = 180? - ?POY
? ?POX = 90?
So, we get, a + b = 90? ....(i)
It is also given that,
a :b = 2:3

Let a = 2k and b = 3k,
From eq. (i),
2k + 3k = 90?
? 5k = 90?
? k = 18?

So, a = 2 × 18? = 36?
and b = 3 × 18? = 54?
Again by Linear pair axiom,
?MOX + ?XON = 180?
? b + c = 180?
? c = 180? - 54?
? c = 126?

Q3 ) In figure,?PQR = ?PRQ, then prove that?PQS = ?PRT .
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NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles


Answer :

By Linear pair axiom,
?PQS + ?PQR = 180?...(i)

Similarly, again by Linear pair axiom,
?PRQ + ?PRT = 180? ...(ii)

Thus by Eq. (i) and (ii),we get,
?PQS + ?PQR = ?PRQ + ?PRT
As it is given, ?PQR = ?PRQ
? ?PQS + ?PRQ = ?PRQ + ?PRT
By cancelling equal terms,
we get, ?PQS = ?PRT
Hence, proved.

Q4 ) In figure, if x + y = w + z, then prove that AOB is a line.
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NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles


Answer :

Since, x,y,w and z are angles at a single point,
x + y + w + z = 360?

But it is given that, x + y = w + z

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

Hence, by converse of linear pair axiom, it is proved that AOB is a straight line.

Q5 ) 5.In figure below, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR.
Prove that ?ROS = 12 (?QOS - ?POS).
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NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles


Answer :

Since it is given that, OR is perpendicular to PQ, we have,?PQR = ?ROQ = 90?
Also, we can say that,
?POS + ?ROS = 90?
? ?ROS = 90? - ?POS ...(i)

Now, by adding, ?ROS on both the side, we get,

2?ROS = 90? - ?POS + ?ROS
? 2?ROS = (90? + ?ROS) - ?POS
? 2?ROS = ?QOS - ?POS
? ?ROS = 12 (?QOS - ?POS).
Hence, proved.

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



NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles


Answer :

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Given that, ray YQ bisects ?ZYP

Thus, ?ZYQ = ?QYP = 12 (?ZYP) ...(i)

By Linear pair axiom,
?XYZ + ?ZYQ + ?QYP = 180?

But, it is given that,

?XYZ = 64? ...(ii)

Therefore, from (i) and (ii), we get,

? 64? + ?ZYQ + ?ZYQ = 180?
? 2?ZYQ = 180? - 64?
? 2?ZYQ = 116?
? ?ZYQ = 58?

Now,
? ?XYQ = ?XYZ + ?ZYQ
? ?XYQ = 64? + 58? = 122?

Now,
?QYP + reflex?QYP = 360?
? 58? + reflex?QYP = 360?
? reflex?QYP = 302?

NCERT solutions for class 9 Maths Chapter 6 Lines And Angles Exercise 6.2

Q1 ) In figure, find the values of x and y and then show that AB ||CD.
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NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles


Answer :

? x + 50° = 180° .....(Linear pair)
? x = 130?
? y = 130? .....(vertically opposite pair)

Here, these are corresponding angles for lines AB and CD.

Hence, it is proved that AB || CD.

Q2 ) In figure, if AB || CD, CD || EF and y : z = 3:7 , find x.
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NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles


Answer :

It is given that, y : z = 3:7

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Let, y = 3k and z = 7k,

x = ?CHG (Corresponding angles) ....(i)
Z = ?CHG (Alternate angles) ....(ii)

From (i) and (ii), we get,
? x = z ....(iii)
Now, x + y = 180? ....(Internal angles on the same side of the transversal)
From eq.(iii),
z + y = 180?
? 3k + 7k = 180?
? 10k = 180?
? k = 18?

? y = 3 × 18? and z = 7 × 18?
? y = 54? and z = 126?
Hence, x = 126?

Q3 ) In figure, if AB || CD, EF is perpendicular to CD and ?GED = 126? , find ?AGE,?GEF and ?FGE
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NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles


Answer :

? ?AGE = ?GED ....(Alternate interior angles)

But it is given that,
? ?GED = 126?
? ?AGE = 126? ....(i)

Also, ?GEF + ?FED = 126?
? EF is perpendicular to CD,
? ?GEF + 90? = 126?
? ?GEF = 36?

Also, by linear pair axiom, we get,
?AGE + ?FGE = 180?
? 126? + ?FGE = 180? ....(from (i))
? ?FGE = 54?

Q4 ) In figure, if PQ || ST, ?PQR = 110? and ?RST = 130? , find ?QRS .
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NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles


Answer :

Construction : Draw a line parallel to ST through R.

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As it is given that,
PQ || ST,?PQR = 110? and ?RST = 130?
We can also say that, AB || PQ || ST

? ?PQR + ?QRA = 180?
....(interior angles on the same side of transversal)
? 110? + ?QRA = 180?
? ?QRA = 70?

? ?ARS = 130? ....(Alternate Interior angle)
As, ?RST = 130?
Now, so as to find ?QRS, We have,
?ARS = ?ARQ + ?QRS
? 130? = 70? + ?QRS
? ?QRS = 60?

Q5 ) In figure, if AB || CD, ?APQ = 50? and ?PRD = 127?, find x and y.
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NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles


Answer :

We have, AB || CD

?PQR = ?APQ ....(Alternate interior angles)

? x = 50? ....(i)(given, ?APQ = 50?))

Now, as we know that, Exterior angle is equal to sum of interior opposite angles of a triangle.
? (\angle{PQR}\) + ?QPR = 127?
from (i), we get,

? 50? + ?QPR = 127?
? y = 77?

Q6 ) In figure, 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 AB || CD.
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NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles


Answer :

Draw perpendiculars BE and CF on PQ and RS, respectively.
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Therefore, we can say that BE || CF
As angle of incidence = angle of reflection,
we get, ?a = ?b ....(i)
Similarly, we get, ?x = ?y ....(ii)

By Alternate angles theorem, ?b = ?y
Now, doubling the angles, we get,
2?b = 2?y
? ?b + ?b = ?y + ?y
from (i) and (ii),
?a + ?b = ?x + ?y

Hence, ?ABC = ?DCB

Thus by converse of alternate angles theorem,

we get, AB || CD
Hence, it is proved.

NCERT solutions for class 9 Maths Chapter 6 Lines And Angles Exercise 6.3

Q1 ) In figure, sides QP and RQ of ?PQR are produced to points S and T, respectively. If ?SPR = 135? and ?PQT = 110? , find ?PRQ.
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NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles


Answer :

By linear pair axiom,
? ?RPS + ?RPQ = 180?
? 135? + ?RPQ = 180?
? ?RPQ = 180? - 135?
?, ?RPQ = 45?

We also know that, Sum of interior opposite angles is equal to exterior angles.
? ?RPQ + ?PRQ = ?PQT
?45? + ?PRQ = 110?
??PRQ = 110? - 45?
? ?PRQ = 65?.

Q2 ) In figure, ?X = 60? , ?XYZ = 54? , if YO and ZO are the bisectors of ?XYZ and ?XZY respectively of ?XYZ , find ?OZY and ?YOZ.
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NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles


Answer :

In