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
The NCERT Class 9 Mathematics textbook is a comprehensive book that covers all the relevant topics and concepts in Mathematics. The NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2, is specially designed for students studying in Class 9 and has the syllabus prescribed by the NCERT. The NCERT textbook is heavily relied upon by students and teachers and is considered one of the most significant textbooks in the field of Mathematics. Class 9 Mathematics is important as it lays the foundation for more advanced Mathematics courses in high school and college. The textbook is easy to grasp and includes a wealth of practice problems to help students master the material. Therefore, if a student of Class 9 is searching for appropriate help for the intext Mathematics queries, they can explore the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 assisted by Extramarks. The NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2, are considered an important element in a student’s academic life and are also reviewed and encouraged by scholars and subjectmatter professionals.
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 CoInterior 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
$\begin{array}{l}\mathrm{x}+50\mathrm{\xb0}=180\mathrm{\xb0}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Linear}\text{pair of angles}\right]\\ \Rightarrow \mathrm{x}=180\mathrm{\xb0}50\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=130\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}130\mathrm{\xb0}=\mathrm{y}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Vertical}\text{opposite angles.}\right]\\ \mathrm{Since},\text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}=\text{y}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}\parallel \text{CD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Alternate}\text{interior angles}\right]\end{array}$
Q.2 In Fig. 6.29, if AB  CD, CD  EF and y : z = 3 : 7, find x.
Ans
$\begin{array}{l}\mathrm{Since},\text{y:z}=\text{3:7}\\ \mathrm{Let}\text{y}=3\mathrm{a}\text{and z}=\text{7a}\\ \text{Since,}\mathrm{AB}\parallel \mathrm{CD}\text{and CD}\parallel \mathrm{EF}\\ \mathrm{so},\text{}\mathrm{AB}\parallel \mathrm{EF}\\ \mathrm{x}=\mathrm{z}\left[\mathrm{Alternate}\text{interior angles.}\right]\\ \because \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}\parallel \mathrm{CD}\\ \mathrm{x}+\mathrm{y}=180\mathrm{\xb0}\left[\text{Cointerior angles}\right]\\ \mathrm{z}+\mathrm{y}=180\mathrm{\xb0}[\because \mathrm{x}=\mathrm{z}]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}7a}+3\mathrm{a}=180\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}10\mathrm{a}=180\mathrm{\xb0}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{180\mathrm{\xb0}}{10}=18\mathrm{\xb0}\\ \mathrm{So},\text{x}=\text{7a}\\ =\text{7}\left(18\mathrm{\xb0}\right)\\ =126\mathrm{\xb0}\end{array}$
Q.3 In Fig. 6.30, if ABCD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Ans
$\begin{array}{l}\mathrm{Given}:\mathrm{AB}\parallel \mathrm{CD},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{EF}\perp \mathrm{CD}\text{and}\angle \text{GED=126\xb0.}\\ \text{To find:}\angle \mathrm{A}\text{GE,}\angle \text{GEF and}\angle \mathrm{F}\text{GE}\\ \text{AB}\parallel \text{CD}\\ \angle \text{AGE}=\angle \text{GE}\mathrm{D}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=126\mathrm{\xb0}\\ \angle \text{GED}=\angle \text{GEF}+\angle \text{FED}\\ \text{126\xb0}=\angle \text{GEF}+90\mathrm{\xb0}\\ \angle \text{GEF}=126\mathrm{\xb0}90\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=36\mathrm{\xb0}\\ \angle \mathrm{A}\text{GE}+\angle \mathrm{F}\text{GE}=180\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}126\mathrm{\xb0}+\angle \mathrm{F}\text{GE}=180\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{F}\text{GE}=180\mathrm{\xb0}126\mathrm{\xb0}\\ =54\mathrm{\xb0}\end{array}$
Q.4 In Fig. 6.31, if PQ  ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
Ans
$\begin{array}{l}\mathrm{Given}:\mathrm{PQ}\parallel \mathrm{ST},\text{}\angle \text{PQR}=\text{11}0\mathrm{\xb0},\text{\hspace{0.17em}\hspace{0.17em}}\angle \text{RST}=\text{13}0\mathrm{\xb0}\\ \mathrm{To}\text{prove:}\angle \text{QRS}\\ \text{Construction:\hspace{0.17em}Draw line}\mathrm{l}\text{paralllel to ST through R.}\\ \text{Since, ST}\parallel \overleftrightarrow{\mathrm{l}}\text{, then}\\ \angle \text{\hspace{0.17em}}\mathrm{TSR}+\angle \text{\hspace{0.17em}}\mathrm{SRW}=180\mathrm{\xb0}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}[\mathrm{Co}\mathrm{interior}\text{ angles}]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}130\mathrm{\xb0}+\angle \text{\hspace{0.17em}}\mathrm{SRW}=180\mathrm{\xb0}\end{array}$
$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{\hspace{0.17em}}\mathrm{SRW}=180\mathrm{\xb0}130\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=50\mathrm{\xb0}\\ \mathrm{Since},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{PQ}\parallel \mathrm{ST}\text{and ST}\parallel \overleftrightarrow{\mathrm{l}}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{PQ}\parallel \overleftrightarrow{\mathrm{l}}\\ \angle \mathrm{PQR}=\angle \mathrm{SRW}\left[\mathrm{Alternate}\text{interior angles.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}110\mathrm{\xb0}=\angle \mathrm{QRS}+\angle \mathrm{SRW}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\angle \mathrm{QRS}+50\mathrm{\xb0}\\ \angle \mathrm{QRS}=110\mathrm{\xb0}50\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=60\mathrm{\xb0}\end{array}$
Q.5 In Fig. 6.32, if AB  CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
Ans
$\begin{array}{l}\mathrm{Given}:\mathrm{AB}\parallel \mathrm{CD},\text{\hspace{0.17em}\hspace{0.17em}}\angle \text{APQ}=\text{5}0\mathrm{\xb0}\text{\hspace{0.17em}and}\mathrm{}\text{\hspace{0.17em}}\angle \text{PRD}=\text{127\xb0}\\ \text{To find: x and y}\\ \text{Since, AB}\parallel \text{CD}\\ \text{So,}\angle \text{APR}=\angle \text{PRD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}50\xb0}+\text{y}=127\mathrm{\xb0}\\ \Rightarrow \mathrm{y}=127\mathrm{\xb0}50\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=77\mathrm{\xb0}\\ \text{And, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{APQ}=\angle \text{PQR}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}50\mathrm{\xb0}=\mathrm{x}\\ \mathrm{Thus},\text{x}=\text{50\xb0 and y}=\text{77\xb0.}\end{array}$
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
$\begin{array}{l}\mathrm{Given}:\mathrm{PQ}\parallel \mathrm{RS},\text{\hspace{0.17em}AB is incident ray to PQ and BC is incident ray}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}to RS.}\\ \text{To Prove: AB}\parallel \text{CD}\\ \text{Construction: Draw MB}\perp \text{PQ and NC}\perp \text{RS.}\\ \text{Proof}:\text{}\angle \text{ABM}=\angle \mathrm{MBC}...\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{incident}\text{angle= reflected angle}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{BCN}=\angle \mathrm{NCD}...\left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{incident}\text{angle= reflected angle}\right]\end{array}$
$\begin{array}{l}\text{Since, perpendiculars to parallel lines are parallel.So,}\\ \text{\hspace{0.17em}\hspace{0.17em}MB}\parallel \text{NC}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{M}\text{BN}=\angle \mathrm{BCN}\\ \text{\hspace{0.17em}}\frac{1}{2}\angle \mathrm{A}\text{BC}=\frac{1}{2}\angle \mathrm{BCD}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{A}\text{BC}=\angle \mathrm{BCD}\\ \mathrm{Since},\text{alternate interior angles are equal, so}\\ \text{\hspace{0.17em}\hspace{0.17em}AB}\parallel \text{CD.}\end{array}$
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