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

## NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (Ex 6.2) Exercise 6.2 Free PDF download of NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 (Ex 6.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 9 Maths Chapter 6 Lines and Angles Exercise 6.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise NCERT Book Solutions in your emails. Download NCERT maths class 9 at Vedantu.  Students can also avail of Class 9 Science from our website.

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Exercise: 6.2

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

Ans:

Proof:

It is observed that,

$x + {50^o} = {180^o}$ (Linear pair)

$x = {180^o} - {50^o}$

$x = {130^o}$   ---  (1)

Also, $y = {130^o}$ (vertically opposite angles)

As x and y are alternate interior angles for lines AB and CD and also measures of these angles are equal to each other, therefore, line AB $\parallel$CD.

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

Ans:

Given:

AB $\parallel$ CD, CD $\parallel$ EF and $y:z = 3:7$

Proof:

It is given that AB$\parallel$CD and CD$\parallel$EF.$$i.e, AB\parallel CD\parallel EF Lines parallel to the same line are parallel to each other. It is observed that x = z(Alternate interior angles) ---(1) It is given that y:z = 3:7 Let the common ratio between y and z. \therefore y = 3a and z = 7a Also, x + y = {180^o} Co-interior angles on the same side of the transversal. z + y = {180^o}  \Rightarrow 7a + 3a = {180^o}  \Rightarrow 10a = {180^o}  \Rightarrow a = {\dfrac{{180}}{{10}}^o}  \Rightarrow a = {18^o} To find x  \Rightarrow x = 7a  \Rightarrow x = 7 \times {18^o}  \Rightarrow x = {126^o} The value of x = {126^o}. 3. In the given figure, If AB \parallel CD, EF  \bot CD and \angle GED = {126^o}, find \angle AGE, \angle GEF and \angle FGE. (Image will be uploaded soon) Ans: Given: AB parallel to CD. EF perpendicular to CD. \angle GED = {126^o}. To find: \angle AGE, \angle GEF and \angle FGE. Proof: It is given that, AB \parallel CD, EF  \bot CD. \angle GED = {126^o} \angle GEF + \angle FED = {126^o} \angle GEF + {90^o} = {126^o} \angle GEF = {126^o} - {90^o} \angle GEF = {36^o}$$$As$\angle AGE$and$\angle GED$are alternate interior angles.$\angle AGE = \angle GED = {126^o}\angle AGE + \angle FGE = {180^o}{126^o} + \angle FGE = {180^o}\angle FGE = {180^o} - {126^o}\angle FGE = {54^o}$Hence$\angle GEF = {36^o}$,$\angle AGE = {126^o}$and$\angle FGE = {54^o}$. 4. In the given figure, if PQ$\parallel $ST,$\angle PQR = {110^o}$and$\angle RST = {130^o}$, find$\angle QRS$. [Hint: Draw a line parallel to ST through point R.] (Image will be uploaded soon) Ans: Given:$\angle PQR = {110^o}\angle RST = {130^o}$Proof: (Image will be uploaded soon) Let us draw a line XY parallel to ST and passing through point R.$\angle PQR + \angle QRX = {180^o}$. (Co-interior angles on the same side of transversal QR)${110^o} + \angle QRX = {180^o}\angle QRX = {180^o} - {110^o}\angle QRX = {70^o}$Also,$\angle RST + \angle SRY = {180^o}$(Co-interior angles on the same side of transversal SR)${130^o} + \angle SRY = {180^o}\angle SRY = {180^o} - {130^o}\angle SRY = {50^o}$XY is a straight line, RQ and RS stand on it.$\angle QRX + \angle QRS + \angle SRY = {180^o}$.${70^o} + \angle QRS + {50^o} = {180^o}\angle QRS = {180^o} - {70^o} - {50^o}\angle QRS = {180^o} - {120^o}\angle QRS = {60^o}$The value of$\angle QRS = {60^o}$. 5. In the given figure, if AB$\parallel $CD,$\angle APQ = {50^o}$and$\angle PRD = {127^o}$,find x and y. (Image will be uploaded soon) Ans: Given: AB parallel to CD$\angle APQ = {50^o}\angle PRD = {127^o}$To find: x and y Proof:$\angle APR = \angle PRD$(Alternate interior angles)${50^o} + y = {127^o}y = {127^o} - {50^o}y = {77^o}$Also,$\angle APQ = \angle PQR$(Alternate interior angles)$x = {50^o}$Hence$x = {50^o}$and$y = {77^o}$. 6. In the given figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD. (Image will be uploaded soon) Ans: (Image will be uploaded soon) Proof: Let us draw BM$ \bot $PQ and CN$ \bot $RS. PQ is parallel to RS$\therefore $BM parallel to CN BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.$\angle MBC = \angle BCN$(Alternate interior angles)$\angle 2 = \angle 3$However,$\angle 1 = \angle 2$and$\angle 3 = \angle 4$(By laws of reflection)$\angle 1 = \angle 2 = \angle 3 = \angle 4$Also,$\angle 1 + \angle 2 = \angle 3 + \angle 4\angle ABC = \angle DCB$However, these are alternate interior angles. Hence AB parallel to CD$\therefore AB\parallel CD\$.

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

Opting for the NCERT solutions for Ex 6.2 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 6.2 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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Q 1: Give an overview of the chapter Lines and angles.

Ans: Here, you will come across definitions and basic terms related to a line segment, collinear points, intersecting lines, ray, non collinear points and non intersecting lines. You shall also know about various pairs of angles such as linear pair of angles, the reflex angle, vertical opposite angle, complementary angle, adjacent angle and supplementary angles.

The chapter also gives you an idea of transversal and parallel lines with the help of the theorems to support over it. Here, the chapter also focuses upon of angle sum property of a triangle.

Q 2: What is the chapter mainly focused on?

Ans: The chapter mainly deals with:

• Points

• Line segments

• Ray

• Collinear points

• Non collinear points,

• Intersecting lines

• Concurrent lines

• Plane

• Right angle

• Acute angle

• Obtuse angle

• Straight angle

• Reflex angle

• Complete angle

• Equal angle

• Vertically opposite angle

• Bisector of an angle

• Parallel lines

• Linear pair of angles

• Transversal of the lines

• The angles that a transversal forms

• interior angles that stay on the same side of the transversal

• corresponding angles

• Consecutive interior angles

• Alternate interior angles

• Scalene triangle

• Isosceles triangle

• Equilateral triangle

• Acute triangle

• Right triangle

• Obtuse triangle

• Regular polygon

• Supplementary angles

• Complementary angles

• Parallel lines theorems

• Converse of theorems

• Triangle theorems

Q 3: What is a parallelogram triangle?

Ans: Inverse sides of a parallelogram are always parallel (by definition) and because of which they will never intersect. The area of a parallelogram is twice the area of a triangle designed by one of its diagonal sides. The area of a parallelogram is also considered to be equal to the magnitude of the vector across the product of two adjoining sides.

• Rhomboid – It is a quadrilateral whose opposite sides are adjacent and parallel sides are dissimilar, and whose angles are not always right angles.

• Rectangle – It is a parallelogram along with four angles of equal size (i.e., right angles).

• Rhombus – It is a parallelogram which consists of four sides with equal lengths.

• Square – It is a parallelogram with four sides of same and equal length and angles of identical and equal size (right angles).

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