# NCERT Solutions for Class 9 Maths Chapter 6 – Lines And Angles Exercise 6.2

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## NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (Ex 6.2) Exercise 6.2

The Ex 6.2 Class 9 Maths NCERT Solutions provides students to enhance their analytical and Mathematical skills by practising the unanswered questions given in Exercise 6.2 Class 9 Solutions. Downloading and practising the sample questions from the NCERT Solutions, for example, NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 regulates aid to students as well as teachers.

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### Access NCERT Solutions for Class 9 Mathematics Chapter 6 – Lines and Angles

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### NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (Ex 6.2) Exercise 6.2

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In Exercise 6.2 of Class 9, Chapter Lines and Angles, students will get to know about the following:

1) Parallel Lines and a Transversal – Interior and Exterior Angles, Consecutive interior or Allied Angles or Co-Interior Angles, Axiom of Corresponding Angles, and usage of Theorems.

2) Lines Parallel to the Same Line.

The questions given in the Exercise 6.2 are in total six questions that questions students regarding:

• Finding the values of the mentioned integers
• Finding the angles
• Finding the angles with the help of drawing the parallel line and
• Proving whether the line is parallel to others or not.

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Q.1 In Fig. 6.28, find the values of x and y and then show that AB || CD.

Ans

$\begin{array}{l}\mathrm{x}+50\mathrm{°}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Linear}\text{pair of angles}\right]\\ ⇒\mathrm{x}=180\mathrm{°}-50\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=130\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}130\mathrm{°}=\mathrm{y}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Vertical}\text{opposite angles.}\right]\\ \mathrm{Since},\text{​ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}=\text{y}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}\parallel \text{CD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Alternate}\text{interior angles}\right]\end{array}$

Q.2 In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Ans

$\begin{array}{l}\mathrm{Since},\text{y:z}=\text{3:7}\\ \mathrm{Let}\text{y}=3\mathrm{a}\text{and z}=\text{7a}\\ \text{Since,}\mathrm{AB}\parallel \mathrm{CD}\text{and CD}\parallel \mathrm{EF}\\ \mathrm{so},\text{}\mathrm{AB}\parallel \mathrm{EF}\\ \mathrm{x}=\mathrm{z}\left[\mathrm{Alternate}\text{interior angles.}\right]\\ \because \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}\parallel \mathrm{CD}\\ \mathrm{x}+\mathrm{y}=180\mathrm{°}\left[\text{Cointerior angles}\right]\\ \mathrm{z}+\mathrm{y}=180\mathrm{°}\left[\because \mathrm{x}=\mathrm{z}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}7a}+3\mathrm{a}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}10\mathrm{a}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{180\mathrm{°}}{10}=18\mathrm{°}\\ \mathrm{So},\text{x}=\text{7a}\\ =\text{7}\left(18\mathrm{°}\right)\\ =126\mathrm{°}\end{array}$

Q.3 In Fig. 6.30, if AB||CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{AB}\parallel \mathrm{CD},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{EF}\perp \mathrm{CD}\text{and}\angle \text{GED=126°.}\\ \text{To find:}\angle \mathrm{A}\text{GE,}\angle \text{GEF and}\angle \mathrm{F}\text{GE}\\ \text{AB}\parallel \text{CD}\\ \angle \text{AGE}=\angle \text{GE}\mathrm{D}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=126\mathrm{°}\\ \angle \text{GED}=\angle \text{GEF}+\angle \text{FED}\\ \text{126°}=\angle \text{GEF}+90\mathrm{°}\\ \angle \text{GEF}=126\mathrm{°}-90\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=36\mathrm{°}\\ \angle \mathrm{A}\text{GE}+\angle \mathrm{F}\text{GE}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}126\mathrm{°}+\angle \mathrm{F}\text{GE}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{F}\text{GE}=180\mathrm{°}-126\mathrm{°}\\ =54\mathrm{°}\end{array}$

Q.4 In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{PQ}\parallel \mathrm{ST},\text{}\angle \text{PQR}=\text{11}0\mathrm{°},\text{\hspace{0.17em}\hspace{0.17em}}\angle \text{RST}=\text{13}0\mathrm{°}\\ \mathrm{To}\text{prove:}\angle \text{QRS}\\ \text{Construction:\hspace{0.17em}Draw line}\mathrm{l}\text{paralllel to ST through R.}\\ \text{Since, ST}\parallel \stackrel{↔}{\mathrm{l}}\text{, then}\\ \angle \text{\hspace{0.17em}}\mathrm{TSR}+\angle \text{\hspace{0.17em}}\mathrm{SRW}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Co}-\mathrm{interior}\text{​ angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}130\mathrm{°}+\angle \text{\hspace{0.17em}}\mathrm{SRW}=180\mathrm{°}\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{\hspace{0.17em}}\mathrm{SRW}=180\mathrm{°}-130\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=50\mathrm{°}\\ \mathrm{Since},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{PQ}\parallel \mathrm{ST}\text{and ST}\parallel \stackrel{↔}{\mathrm{l}}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{PQ}\parallel \stackrel{↔}{\mathrm{l}}\\ \angle \mathrm{PQR}=\angle \mathrm{SRW}\left[\mathrm{Alternate}\text{interior angles.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}110\mathrm{°}=\angle \mathrm{QRS}+\angle \mathrm{SRW}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\angle \mathrm{QRS}+50\mathrm{°}\\ \angle \mathrm{QRS}=110\mathrm{°}-50\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=60\mathrm{°}\end{array}$

Q.5 In Fig. 6.32, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{AB}\parallel \mathrm{CD},\text{\hspace{0.17em}\hspace{0.17em}}\angle \text{APQ}=\text{5}0\mathrm{°}\text{\hspace{0.17em}and}\mathrm{}\text{\hspace{0.17em}}\angle \text{PRD}=\text{127°}\\ \text{To find: x and y}\\ \text{Since, AB}\parallel \text{CD}\\ \text{So,}\angle \text{APR}=\angle \text{PRD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}50°}+\text{y}=127\mathrm{°}\\ ⇒\mathrm{y}=127\mathrm{°}-50\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=77\mathrm{°}\\ \text{And, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{APQ}=\angle \text{PQR}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}50\mathrm{°}=\mathrm{x}\\ \mathrm{Thus},\text{x}=\text{50° and y}=\text{77°.}\end{array}$

Q.6 In Fig. 6.33, 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.

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{PQ}\parallel \mathrm{RS},\text{\hspace{0.17em}AB is incident ray to PQ and BC is incident ray}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}to RS.}\\ \text{To Prove: AB}\parallel \text{CD}\\ \text{Construction: Draw MB}\perp \text{PQ and NC}\perp \text{RS.}\\ \text{Proof}:\text{}\angle \text{ABM}=\angle \mathrm{MBC}...\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{incident}\text{angle= reflected angle}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{BCN}=\angle \mathrm{NCD}...\left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{incident}\text{angle= reflected angle}\right]\end{array}$

$\begin{array}{l}\text{Since, perpendiculars to parallel lines are parallel.So,}\\ \text{\hspace{0.17em}\hspace{0.17em}MB}\parallel \text{NC}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{M}\text{BN}=\angle \mathrm{BCN}\\ \text{\hspace{0.17em}}\frac{1}{2}\angle \mathrm{A}\text{BC}=\frac{1}{2}\angle \mathrm{BCD}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{A}\text{BC}=\angle \mathrm{BCD}\\ \mathrm{Since},\text{alternate interior angles are equal, so}\\ \text{\hspace{0.17em}\hspace{0.17em}AB}\parallel \text{CD.}\end{array}$

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