# NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

## NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles - Free PDF

Class 9 Maths Chapter 6 focuses on lines and angles and the relationship between them. It introduces the basic types of angles formed when two lines intersect at various points. By solving the sums of Ch 6 Class 9 Maths students will get to learn the concepts of angle sum property of a triangle, pair of angles, vertically opposite angles, linear pair, etc. The NCERT Solutions for Class 9 Maths Ch 6 are available in PDF format on Vedantu. It will help students to learn the best problem-solving techniques for the sums of lines and angles. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 9 Science, Maths solutions, and solutions of other subjects that are available on Vedantu only.

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## Access NCERT Solutions for Class 9 Maths Chapter 6- Lines and Angles

### Question 1

In the given figure, lines AB and CD intersect at O. if $\angle \text{AOC+}\angle \text{BOE=7}{{\text{0}}^{\text{o}}}$ and $\angle \text{BOD=4}{{\text{0}}^{\text{o}}}$find $\angle \text{BOE}$ and reflex $\angle \text{COE}$

Ans:

AB is a straight line, OC and OE are rays from O.

We know that a straight line covers ${{180}^{\circ }}$

$\text{ }\Rightarrow \angle \text{AOC+}\angle \text{COE+}\angle \text{BOE=18}{{\text{0}}^{\circ }}$

By clubbing $\angle \text{AOC and }\angle \text{BOE}$ together we can rewrite the above equation as

$\Rightarrow \left( \angle \text{AOC+}\angle \text{BOE} \right)+\angle \text{COE=18}{{\text{0}}^{\circ }}$

Putting $\text{ }\angle \text{AOC+}\angle \text{BOE=7}{{\text{0}}^{\circ }}$

$\text{ }\Rightarrow \text{7}{{\text{0}}^{\circ }}+\angle \text{COE=18}{{\text{0}}^{\circ }}$

$\text{ }\Rightarrow \angle \text{COE=18}{{\text{0}}^{\circ }}\text{-}{{70}^{\circ }}$

$\text{ }\Rightarrow \angle \text{COE=11}{{\text{0}}^{\circ }}$

Hence  reflex $\text{ }\angle \text{COE=36}{{\text{0}}^{\circ }}\text{-11}{{\text{0}}^{\circ }}$

$\text{ reflex}\angle \text{COE=25}{{\text{0}}^{\circ }}\text{ }$

Similarly CD is a straight line, OB and OE are rays from O.

We know that a straight line covers ${{180}^{\circ }}$

$\text{ }\Rightarrow \angle B\text{OD+}\angle \text{COE+}\angle \text{BOE=18}{{\text{0}}^{\circ }}$

$\text{ }\Rightarrow \text{ }{{40}^{\circ }}\text{+}{{110}^{\circ }}\text{+}\angle \text{BOE=18}{{\text{0}}^{\circ }}$

$\text{ }\Rightarrow \text{ }\angle \text{BOE=18}{{\text{0}}^{\circ }}\text{-}\left( \text{4}{{\text{0}}^{\circ }}\text{+11}{{\text{0}}^{\circ }} \right)$

$\text{ }\Rightarrow \text{ }\angle \text{BOE=18}{{\text{0}}^{\circ }}\text{-}{{150}^{\circ }}$

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

Hence $\angle \text{BOE=}{{30}^{\circ }}$ and reflex $\angle \text{COE=25}{{\text{0}}^{\circ }}$

### Question 2

In the given figure, lines XY and MN intersect at O. If $\angle \text{POY=9}{{\text{0}}^{\text{o}}}$ and a: b = 2:3, find c.

Ans:

Let the common ratio between a and b be x.

$\therefore$$\text{a = 2x}$ , and $\text{b = 3x}$

XY is a straight line, OM and OP are rays from O.

We know that a straight line covers ${{180}^{\circ }}$

$\angle \text{XOM +}\angle \text{MOP +}\angle \text{POY = 18}{{\text{0}}^{\text{o}}}$

Putting values for $\angle \text{XOM=b and}\angle \text{MOP=a}$

$\Rightarrow \text{ b + a +}\angle \text{POY = 18}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ 3x + 2x +}\angle \text{POY = 18}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ 5x = 9}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ x = 1}{{\text{8}}^{\text{o}}}$

$\therefore \text{ a = 2x}$

$\Rightarrow \text{ a = 2}\times \text{1}{{\text{8}}^{\circ }}$

$\text{ = 3}{{\text{6}}^{\circ }}$

$\therefore \text{ b = 3x}$

$\Rightarrow \text{ b = 3}\times \text{1}{{\text{8}}^{\circ }}$

$\text{ = 5}{{\text{4}}^{\circ }}$

Similarly MN is a straight line, OX is a ray from O

$\text{ }\therefore \text{b+c=18}{{\text{0}}^{\circ }}$

$\text{5}{{\text{4}}^{\text{o}}}\text{ + c = 18}{{\text{0}}^{\text{o}}}$

$\text{ }c=\text{ }{{180}^{\circ }}\text{ }-\text{ }{{54}^{\circ }}$

$\text{ }c=\text{ 12}{{\text{6}}^{\circ }}$

### Question 3

In the given figure,$\angle \text{PQR=}\angle \text{PRQ}$ , then prove that $\angle \text{PQS=}\angle \text{PRT}$.

Ans:

ST is a straight line, QP is a line segment from Q in ST to any point P

by Linear Pair property

$\angle \text{PQS+}\angle \text{PQR=18}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ }\angle \text{PQR=18}{{\text{0}}^{\text{o}}}\text{-}\angle \text{PQS}$ ......($1$ )

Similarly

$\angle PRT+\angle PRQ={{180}^{\circ }}$

$\Rightarrow \text{ }\angle \text{PRQ=18}{{\text{0}}^{\text{o}}}\text{-}\angle \text{PRT}$ ......($2$ )

Now in the question it is given that $\angle \text{PQR=}\angle \text{PRQ}$

Therefore on equating equation ($1$ ) and ($2$ ) we get

$\text{18}{{\text{0}}^{\text{o}}}\text{-}\angle \text{PQS=18}{{\text{0}}^{\text{o}}}\text{-}\angle \text{PRT}$

$\Rightarrow \text{ }\angle \text{PQS=}\angle \text{PRT}$

Hence proved

### Question 4

In the given figure, if $\text{x+y=w+z}$ then prove that AOB is a line.

Ans:

It can be observed that,

Since there are ${{360}^{\circ }}$ around a point therefore we can write

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

It is given that,

$\text{x+y=w+z}$

Therefore writing$\text{x+y}$in place of $\text{w+z}$so that we can eliminate w and z, we get

$x+y+x+y={{360}^{\circ }}$

$2\left( x+y \right)={{360}^{\circ }}$

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

Since x and y form a linear pair, hence we can say that AOB is a line.

Hence proved

### Question 5

In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

$\angle \text{ROS=}\frac{\text{1}}{\text{2}}\left( \angle \text{QOS-}\angle \text{POS} \right)$

Ans:

Since $\text{OR}\bot \text{PQ}$  therefore

$\angle \text{POR=9}{{\text{0}}^{\text{o}}}$

$\angle POS+ROS={{90}^{\circ }}$

$\Rightarrow \text{ }\angle \text{ROS=9}{{\text{0}}^{\text{o}}}\text{-}\angle \text{POS}$ ......   ($1$ )

Similarly $\angle \text{QOR=9}{{\text{0}}^{\text{o}}}$ (Since $\text{OR }\!\!\hat{\ }\!\!\text{ PQ}$)

$\therefore \angle \text{QOS-}\angle \text{ROS=9}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ }\angle \text{ROS=}\angle \text{QOS-9}{{\text{0}}^{\text{o}}}$......($2$)

We can clearly see that on adding equation ($1$ ) and ($2$) ${{90}^{\circ }}$ get canceled out

$\Rightarrow \text{2}\angle \text{ROS=}\angle \text{QOS-}\angle \text{POS}$

Which can easily be written as

$\Rightarrow \angle \text{ROS=}\frac{\text{1}}{\text{2}}\left( \angle \text{ROS-}\angle \text{POS} \right)$

Hence proved

### Question 6

It is given that $\text{XYZ=6}{{\text{4}}^{\text{o}}}$ and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects $\text{ZYP}$, find $\text{XYQ}$ and reflex$\text{QYP}$.

Ans:

It is given that line YQ bisects $\angle \text{ZYP}$.

Hence, $\angle \text{QYP=}\angle \text{ZYQ}$

It can easily be understood that PX is a line, YQ and YZ being rays standing on it.

$\angle \text{XYZ+}\angle \text{ZYQ+}\angle \text{QYP=18}{{\text{0}}^{\text{o}}}$

From above relation $\angle \text{QYP=}\angle \text{ZYQ}$ we can write

$\text{6}{{\text{4}}^{\text{o}}}\text{+2}\angle \text{QYP=18}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ 2}\angle \text{QYP=18}{{\text{0}}^{\text{o}}}\text{-6}{{\text{4}}^{\text{o}}}$

$\Rightarrow \text{ }\angle \text{QYP=5}{{\text{8}}^{\circ }}$

Therefore $\angle \text{ZYQ=5}{{\text{8}}^{\circ }}$

Also Reflex $\angle \text{QYP=30}{{\text{2}}^{\circ }}$

Now we can write $\angle \text{XYQ}$ as below

$\angle \text{XYQ=}\angle \text{XYZ+}\angle \text{ZYQ}$

${{64}^{\circ }}+{{58}^{\circ }}={{122}^{\circ }}$

Therefore we found $\angle \text{XYQ=12}{{\text{2}}^{\circ }}$ and so the Reflex $\angle \text{QYP=30}{{\text{2}}^{\circ }}$

### Question 1

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

Ans:

From the figure it is clear that,

${{50}^{\circ }}+x={{180}^{\circ }}$ (Linear Pair)

$\Rightarrow \text{ x=13}{{\text{0}}^{\text{o}}}$......($1$ )

Also, from the figure it is clear that $y={{130}^{\circ }}$ (since vertically opposite angles are equal)

Similarly we can see that  x and y are alternate interior angles for lines AB and CD and so measures of these angles are equal to each other.

Therefore, line AB || CD.

Hence proved.

### Question 2

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

Ans:

It is given that AB$\parallel$ CD and CD $\parallel$ EF

$\therefore$ AB$\parallel$CD$\parallel$ EF (Lines parallel to other fixed line are parallel to each other)

It can easily be understood that

$\text{x=z}$  (since alternate interior angles are equal) ...... ($1$)

It is given that y: z = 3: 7

Without any loss of generality we can let $\text{y=3a}$and $\text{z=7a}$

Also, $\text{x+y=18}{{\text{0}}^{\text{o}}}$ (Co-interior angles together sum up to ${{180}^{\circ }}$)

From equation ($1$) we can write z in place of x as shown

$\text{z+y=18}{{\text{0}}^{\text{o}}}$

It can further written as shown

$7a+3a={{180}^{\circ }}$

$\Rightarrow \text{ 10a=18}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ a=1}{{\text{8}}^{\text{o}}}$

$\therefore \text{ x=7 }\!\!\times\!\!\text{ 1}{{\text{8}}^{\text{o}}}$

$\therefore \text{ x=12}{{\text{6}}^{\text{o}}}$

### Question 3

In the given figure, If AB$\parallel$CD, EF$\bot$ CD and$\angle \text{GED=12}{{\text{6}}^{\text{o}}}$ , find$\angle \text{AGE}$ , $\angle \text{GEF}$and $\angle \text{FGE}$.

Ans:

We are given that,

AB $\parallel$ CD and EF$\bot$ CD and

$\text{ }\angle \text{GED=12}{{\text{6}}^{\text{o}}}$

Which can be written as

$\Rightarrow \angle \text{GEF+}\angle \text{FED=12}{{\text{6}}^{\text{o}}}$

$\Rightarrow \text{ }\angle \text{GEF+9}{{\text{0}}^{\text{o}}}\text{=12}{{\text{6}}^{\text{o}}}$

Hence we can obtain $\text{GEF}$as shown below

$\Rightarrow \text{ }\angle \text{GEF=12}{{\text{6}}^{\text{o}}}\text{-9}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ }\angle \text{GEF=3}{{\text{6}}^{\circ }}$

$\angle \text{AGE=}\angle \text{GED=12}{{\text{6}}^{\text{o}}}$ (∠AGE and ∠GED are alternate interior angles)

But

$\angle \text{AGE+}\angle \text{FGE=18}{{\text{0}}^{\text{o}}}$ (because these form a linear pair)

$\Rightarrow \text{12}{{\text{6}}^{\text{o}}}\text{+}\angle \text{FGE=18}{{\text{0}}^{\text{o}}}$

$\Rightarrow \angle \text{FGE=18}{{\text{0}}^{\text{o}}}\text{-12}{{\text{6}}^{\text{o}}}$

$\Rightarrow \angle \text{FGE=5}{{\text{4}}^{\text{o}}}$

Hence, we found $\angle \text{AGE=12}{{\text{6}}^{\text{o}}}$, $\angle \text{GEF=3}{{\text{6}}^{\text{o}}}$ , $\angle \text{FGE=5}{{\text{4}}^{\text{o}}}$

### Question 4

In the given figure, if PQ $\parallel$ ST, $\angle \text{PQR=11}{{\text{0}}^{\text{o}}}$ and$\angle \text{RST=13}{{\text{0}}^{\text{o}}}$ , find $\angle \text{QRS}$.

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

Ans:

In this question we will have some construction of our own, we draw a line XY parallel to ST and so parallel to PQ passing through point R.

$\angle \text{PQR+}\angle \text{QRX=18}{{\text{0}}^{\text{o}}}$ (Co-interior angles on the same side of transversal QR together sum up to ${{180}^{\circ }}$)

$\therefore \text{11}{{\text{0}}^{\text{o}}}\text{+}\angle \text{QRX=18}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ }\angle \text{QRX=7}{{\text{0}}^{\text{o}}}$

Also,

$\angle \text{RST+}\angle \text{SRY=18}{{\text{0}}^{\circ }}$  (sum of Co-interior angles on the same side of transversal SR equals $\text{18}{{\text{0}}^{\circ }}$)

$\Rightarrow \text{ }\angle \text{SRY=18}{{\text{0}}^{\circ }}\text{-13}{{\text{0}}^{\circ }}$

$\Rightarrow \text{ }\angle \text{SRY=}{{50}^{\circ }}$

Now from the construction XY is a straight line. RQ and RS are rays from it.

$\angle \text{QRX+}\angle \text{QRS+}\angle \text{SRY=18}{{\text{0}}^{\circ }}$

$\Rightarrow \text{ 7}{{\text{0}}^{\circ }}+\angle QRS+{{50}^{\circ }}={{180}^{\circ }}$

Hence we found that

$\angle \text{QRS=6}{{\text{0}}^{\circ }}$

### Question 5

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

Ans:

$\angle APR=PRD$ (since alternate interior angles are equal)

$\therefore \text{5}{{\text{0}}^{\circ }}+y={{127}^{\circ }}$ (since $\angle APR=\angle \text{APQ+}\angle \text{PQR}$)

$\Rightarrow \text{y=7}{{\text{7}}^{\circ }}$

similarly,

$\angle \text{APQ=PQR}$ ∠APQ = ∠PQR (since alternate interior angles are equal)

$\therefore \text{x=5}{{\text{0}}^{\circ }}$

Therefore we found that $\text{x=5}{{\text{0}}^{\circ }}$ and $\text{y=7}{{\text{7}}^{\circ }}$

### Question 6

In the given 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 atC and again reflects back along CD. Prove that AB||CD.

Ans:

Let us construct BM$\bot$ PQ and CN$\bot$RS.

Since  PQ$\parallel$ RS, and so BM $\parallel$ CN

Therefore, CN and BM are two parallel lines and a transversal line BC cuts them at B and C respectively.

∠2 = ∠3......($1$) (since alternate interior angles are equal)

But, by laws of reflection in Physics

∠1 = ∠2 and ∠3 = ∠4

Now from equation ($1$)

∠1 = ∠2 = ∠3 = ∠4

Therefore

∠1 + ∠2 = ∠3 + ∠4

$\angle \text{ABC=}\sum \text{DCB}$

But, these are alternate interior angles.

$\therefore$ AB$\parallel$ CD

Hence  proved.

### Question 1

In the given figure, sides QP and RQ of $\Delta$PQR are produced to points S and T respectively. If ∠SPR = 135o and ∠PQT = 110o, find ∠PRQ.

Ans:

We are given that,

$\angle \text{SPR=13}{{\text{5}}^{\circ }}$ and$\angle \text{PQT=11}{{\text{0}}^{\circ }}$

Therefore we can write

$\text{ }\angle \text{SPR+}\angle \text{QPR=18}{{\text{0}}^{\circ }}$ (since Linear pair angles equals $\text{18}{{\text{0}}^{\circ }}$)

$\Rightarrow {{135}^{\circ }}+\angle \text{QPR=18}{{\text{0}}^{\circ }}$

$\therefore \text{QPR=4}{{\text{5}}^{\circ }}$

Similarly $\angle \text{PQT+PQR=18}{{\text{0}}^{\circ }}$ (Linear pair angles)

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

From here we get

$\angle \text{PQR=7}{{\text{0}}^{\circ }}$

Considering $\Delta$ PQR,

$\angle \text{QPR+}\angle \text{PQR+}\angle \text{PRQ}={{180}^{\circ }}$ (sum of all interior angles of a triangle is ${{180}^{\circ }}$)

$\Rightarrow {{45}^{\circ }}+{{70}^{\circ }}+\angle PRQ={{180}^{\circ }}$

Hence we found that

$\text{ }\angle \text{PRQ=6}{{\text{5}}^{\circ }}$

### Question 2

In the given figure,$\angle \text{X=6}{{\text{2}}^{\circ }}$, $\angle \text{XYZ=5}{{\text{4}}^{\circ }}$. If YO and ZO are the bisectors of $\angle \text{XYZ}$ and $\angle \text{XZY}$respectively of $\Delta$XYZ, find $\angle \text{OZY}$ and$\angle \text{YOZ}$.

Ans:

We know that  the sum of all interior angles of a triangle is${{180}^{\circ }}$, therefore, for $\Delta$ XYZ,

$\angle \text{X+}\angle \text{XYZ+XZY}$

$\Rightarrow {{62}^{\circ }}+{{54}^{\circ }}+\angle \text{XZY=18}{{\text{0}}^{\circ }}$

Hence we get

$\angle \text{XZY=6}{{\text{4}}^{\circ }}$

Similarly $\angle OZY={{27}^{\circ }}$ (since$\angle OZY=$$\frac{54}{2}$)

From $\Delta \text{OYZ}$ we obtain

$\angle OZY+\angle YOZ+\angle OZY=180$ ?(using angle sum property of a triangle)

$\Rightarrow {{27}^{\circ }}+\angle YOZ+{{32}^{\circ }}={{180}^{\circ }}$

From here we get

$\angle YOZ={{121}^{\circ }}$

Hence we found that, $\angle OZY={{27}^{\circ }}$and $\angle YOZ={{121}^{\circ }}$

### Question 3

In the given figure, if AB $\parallel$  DE, $\angle \text{BAC=3}{{\text{5}}^{\circ }}$ and$\angle \text{CDE=5}{{\text{3}}^{\circ }}$ , find $\angle DCE$.

Ans:

From the figure we can easily see AB $\parallel$  DE and AE is a transversal.

$\angle BAC=\angle CED$ (since alternate interior angles are equal)

$\angle \text{CED=3}{{\text{5}}^{\circ }}$

In $\Delta$CDE,

$\angle \text{CDE+}\angle \text{CED+}\angle \text{DCE=18}{{\text{0}}^{\circ }}$  (from the angle sum property of a triangle)

$\text{5}{{\text{3}}^{\circ }}+{{35}^{\circ }}+\angle DCE={{180}^{\circ }}$

From here we get

$\angle DCE={{92}^{\circ }}$

### Question 4

In the given figure, if lines PQ and RS intersect at point T, such that $\angle \text{PRT=4}{{\text{0}}^{\circ }}$,$\angle RPT={{95}^{\circ }}$and $\angle \text{TSQ=7}{{\text{5}}^{\circ }}$, find $\angle \text{STQ}$

Ans:

On applying angle sum property for $\Delta$PRT, we find

$\angle \text{PRT+}\angle \text{RPT+}\angle \text{PTR=18}{{\text{0}}^{\circ }}$

$\Rightarrow$ $\text{4}{{\text{0}}^{\circ }}+{{95}^{\circ }}+\angle PTR={{180}^{\circ }}$

From here we get

$\angle \text{PTR=4}{{\text{5}}^{\circ }}$

Now in $\Delta \text{STQ}$

$\angle STQ=\angle PTR$  (since they are vertically opposite angles)

$\therefore \angle STQ={{45}^{\circ }}$

Similarly on applying angle sum property for $\Delta$STQ, we find

$\angle STQ\text{+}\angle \text{SQT+}\angle \text{QST=18}{{\text{0}}^{\circ }}$

$\Rightarrow \text{ }{{45}^{\circ }}+\angle STQ+{{75}^{\circ }}={{180}^{\circ }}$

Hence we found that

$\angle \text{SQT=6}{{\text{0}}^{\circ }}$

### Question 5

In the given figure, if PQ $\bot$ PS, PQ $\parallel$ SR, $\angle \text{SQR=2}{{\text{8}}^{\circ }}$ and  $\angle \text{QRT=6}{{\text{5}}^{\circ }}$ then find the values of x and y.

Ans:

We are given that PQ$\parallel$ SR and QR is a transversal line.

$\therefore \angle \text{PQR=}\angle \text{QRT}$ (since alternate interior angles are equal)

$\Rightarrow x+{{28}^{\circ }}={{65}^{\circ }}$ (since $\angle \text{PQR=}\angle PQS+\angle SQR$)

From here we get

$\text{x=3}{{\text{7}}^{\circ }}$

By applying the angle sum property for $\Delta$SPQ, we find

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

${{90}^{\circ }}+{{37}^{\circ }}+y={{180}^{\circ }}$

From here we get

$\text{y=5}{{\text{3}}^{\circ }}$

Hence we found and $\text{x=3}{{\text{7}}^{\circ }}$and $\text{y=5}{{\text{3}}^{\circ }}$

### Question 6

In the given figure, the side QR of $\Delta$PQR is produced to a point S. If the bisectors of $\angle PQR$ and $\angle \text{PRS}$ meet at point T, then prove that$\angle \text{QTR=}\frac{1}{2}\angle QPR$.

Ans:

In $\Delta QTR$,we can easily observe ∠TRS is an exterior angle.

$\therefore \angle \text{QTR+}\angle \text{TQR=}\angle \text{TRS}$

$\Rightarrow \text{ }\angle \text{QTR=}\angle \text{TRS-}\angle \text{TQR}$...... ($1$)

In  $\Delta$PQR, we can easily observe $\angle \text{PRS}$ is an external angle.

$\therefore \angle QPR+\angle PQR=\angle PRS$

$\Rightarrow \angle QPR+2\angle TQR=2\angle TRS$ (Given QT and RT are angle bisectors) $\Rightarrow \text{ }\angle \text{QPR=2}\left( \angle TRS\text{-}\angle \text{TQR} \right)$

$\angle QPR=2\angle QTR$ (from equation $1$ )

Hence $\frac{1}{2}\angle QPR=\angle QTR$ proved.

## NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles - PDF Download

You can opt for Chapter 6 - Lines and Angles NCERT Solutions for Class 9 Maths PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

### NCERT Solutions For Class 9 Maths Chapter 6 Lines and Angles - Free PDF Download

Lines and Angles Class 9 Solutions are prepared by highly qualified and professional teachers at Vedantu. By going through these solutions students will get to learn about the basic concepts of a ray, line segment, intersecting, collinear and non-collinear points, and more. Apart from accurate solutions, students should go through all the formulas and the steps of solving the sums of this chapter. These NCERT Solutions help students enhance their mathematics problem-solving skills. These Maths Class 9 Chapter 6 Solutions provided are prepared as per the latest guidelines of CBSE so that students can secure good marks in the exam by following these solutions.

### NCERT Solutions Class 9 Maths Chapter 6 - Lines and Angles

Class 9th Maths Chapter 6 educates students about the difference between a line segment and a ray. A line with two endpoints refers to a line segment whereas a ray is a part of a line with one endpoint. Students will get to know that if three or more points lie on the same line, then they are referred to as collinear points. When two lines initiate from one point, then an angle is formed. Lines and angles have various real-life applications. From designing a building structure to finding the height of a tower or an aircraft’s location, angles are used everywhere.

Class 9 Maths NCERT Chapter 6 explains how ‘vertically opposite angles’ are formed and measure to be equal when two lines intersect each other. Students have to understand the proof of this theorem. A transversal intersects two or more parallel lines at diverse points. When a transversal intersects a pair of parallel lines, then different angles are formed, namely, alternate interior angles, exterior angles, corresponding angles, etc.

### Class 9 Maths Chapter 6 Lines and Angles - Weightage Marks

The total weightage of the Geometry unit in class 9 Maths is 22 marks, out of which, this chapter holds a weightage of 6 marks. The sections covered in class 9 Maths ch 6 are as follows.

Ex 6.1 Introduction.

Ex 6.2 Basic Terms.

Ex 6.3 Intersecting Lines and Non-Intersecting Lines.

Ex 6.4 Pairs of Angles.

Ex 6.5 Parallel Lines and a Transversal.

Ex 6.6 Lines Parallel to the Same Line are parallel and concepts.

Ex 6.7 Angle Sum Property in a Triangle.

NCERT Solutions Class 9 Maths Chapter 6 - Lines and Angles available on Vedantu will help students to understand these concepts easily.

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NCERT Solutions for Class 9 Maths Chapter 6, make a well-organized study material for students. It helps them have a clear idea of the lines and angles and related topics. The step-by-step solutions are prepared by professional experts having several years of experience. While studying and preparing for the final exams, these solutions play a vital role. Students can get the confidence to solve the exercise sums and get a grasp of the concepts. Some of the benefits of NCERT Class 9 Maths Chapter 6 Solutions are as follows.

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• The students can schedule their preparation from NCERT solutions class 9 Maths Chapter 6. With proper preparation, students can secure a good percentage.

1. How can NCERT Solutions for Class 9 Chapter 6 Maths Help in the Exam Preparation?

Answer: NCERT Solutions for Class 9 Chapter 6 make comprehensive study material and cover the concepts of the entire chapter to help students build a strong foundation. These solutions follow the latest CBSE curriculum and make students familiar with the types of questions being asked from the lines and angles chapter. Students can get in-depth knowledge about the topic, brief summary, solved questions, and more. Apart from reasoning and logical skills, these chapter solutions offer comprehensive learning to students. The difficult topics like types of angles, interior, and exterior angles are explained in a better way. Students can clear their tough concepts and can analyze their weak areas with the guidance of NCERT solutions class 6.

2. How are the Vertically Opposite Angles Equal when Two Lines Intersect each Other?

Answer: Let AB and CD be two lines intersecting at a single point, say, P. APC and BPD, and APD and BOC form two pairs of vertically opposite angles. Additionally, ray AP is perpendicular on line CD. It implies the sum of APC and APD is equal to 180o. Now, the sum of angles APD and BPD is equal to 180o. By using these two equations, it’s clear that the sum of angles APC and APD is equal to the sum of APD and BPD angles. It implies angle APC is equal to BPD. Similarly, it can be shown that angle APD is equal to angle BPC.