Exercise 6.2 class 9

Table of Contents • • • • • • • • • • • • • NCERT Solutions for Class 9 Maths Exercise 6.2 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free. NCERT solutions for Class 9 Maths Lines and Angles NCERT Solutions for Class 9 Mathematics Lines and Angles 1. In the given figure, find the values of x and y and then show that AB || CD. Ans. We need to find the value of x and y in the figure given below and then prove that . From the figure, we can conclude that (Vertically opposite angles), and form a pair of linear pair. We know that the sum of linear pair of angles is . . From the figure, we can conclude that form a pair of alternate interior angles corresponding to the lines AB and CD. Therefore, we can conclude that . NCERT Solutions for Class 9 Maths Exercise 6.2 2. In the given figure, if AB || CD, CD || EF and y: z = 3: 7, find x. Ans. We are given that , and . We need to find the value of x in the figure given below. We know that lines parallel to the same line are also parallel to each other. We can conclude that . Let . We know that angles on same side of a transversal are supplementary. . (Alternate interior angles) Now There...

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. Exercise: 6.2 1. In the given figure, find the values of x and y and then show that AB $\parallel $CD. (Image will be uploaded soon) Ans: Proof: It is observed that, $x + $ . 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 alt...

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2 are part of • • • Board CBSE Textbook NCERT Class Class 9 Subject Maths Chapter Chapter 6 Chapter Name Lines and Angles Exercise Ex 6.2 Number of Questions Solved 6 Category NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2 Ex 6.2 Class 9 Maths Question 1. In the given figure, find the values of x and y and then show that AB||CD. Solution: Here, l is a straight line, So, x + 50° = 180° [by linear pair axiom] ⇒ x = 180° – 50° = 130° Now, line l and CD intersect each other at a point. So, y=130° [vertically opposite angles] Thus, x = y = 130° [alternate interior angles] Hence, AB || CD [by theorem 2] Hence proved. Ex 6.2 Class 9 Maths Question 2. In the given figure, if AB || CD, CD || EF and y :z = 3: 7, then find the value of x. Solution: Given, AB || CD and CD || EF ∴ AB ||EF [since, lines which are parallel to the same line, are parallel to each other] Then, ∠x= ∠z [alternate interior angles] … (i) Also, given that y : z = 3 :7 ⇒ y : x =3:7 [from eq. (i)] …(ii) Let ∠y = 3a and ∠x=7a Now, AB || CD ∠x + ∠y= 180° [since, interior angles are supplementary] ⇒ 7a +3a = 180° ⇒ 10a = 180° ⇒ a =\(\cfrac \) = 18° Hence, x = 7a = (7 x 18)° = 126° Ex 6.2 Class 9 Maths Question 3. In the given figure, if AB || CD, FE ⊥ CD, and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE. Solution: Given, AB || CD, EF ⊥ CD and GE is a transversal. .’. ∠AGE = ∠GED [alternate interior angles] ∠AGE = 126° [ ∠AGE =126° given] ...

NCERT Solutions Class 9 Maths Chapter 6 Exercise 6.2 Lines and Angles NCERT solutions for class 9 maths Chapter 6 exercise 6.2 Lines and Angles will introduce the students to the angles formed by a transversal intersecting a pair of parallel lines. These angles are named alternate interior angles, exterior angles Class 9 maths NCERT solutions Chapter 6 exercise 6.2 consists of a total of 6 questions focused on efficiently covering properties of these angles and their properties. With the practice of these NCERT solutions, students will be able to comprehend this knowledge easily. These solutions are available in a scrollable PDF format given below. ☛ Download NCERT Solutions Class 9 Maths Chapter 6 Exercise 6.2 Exercise 6.2 Class 9 Chapter 6 Download PDF More Exercises in Class 9 Maths Chapter 6 • • NCERT Solutions Class 9 Maths Chapter 6 Exercise 6.2 Tips By practicing the questions provided in The students should learn all the fundamental terms, theorems, definitions, and concepts to practice and solve the exercise questions. The Download Cuemath NCERT Solutions PDFs for free and start learning! Class 9 Maths NCERT Solutions Video Chapter 6 Exercise 6.2 NCERT Video Solutions for Class 9 Maths Chapter 6 Exercise 6.2

RD Sharma Solutions for Class 9 Mathematics Chapter 6 Exercise 6.2 Factorization of Polynomials are provided here. In this exercise, students will learn how to find the value of a polynomial. This study material helps students to master a concept by solving various questions listed in the exercise. To gain a better understanding of concepts, students can practise Previous Next Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 6 Factorization of Polynomials Exercise 6.2 Page Number 6.8 Exercise 6.2 Page No: 6.8 Question 1: If f(x) = 2x 3– 13x 2 + 17x + 12, find (i) f (2) (ii) f (-3) (iii) f(0) Solution: f(x) = 2x 3– 13x 2 + 17x + 12 (i) f(2) = 2(2) 3– 13(2) 2 + 17(2) + 12 = 2 x 8 – 13 x 4 + 17 x 2 + 12 = 16 – 52 + 34 + 12 = 62 – 52 = 10 (ii) f(-3) = 2(-3) 3– 13(-3) 2 + 17 x (-3) + 12 = 2 x (-27) – 13 x 9 + 17 x (-3) + 12 = -54 – 117 -51 + 12 = -222 + 12 = -210 (iii) f(0) = 2 x (0) 3– 13(0) 2 + 17 x 0 + 12 = 0-0 + 0+ 12 = 12 Question 2: Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases: (i) f(x) = 3x + 1, x = −1/3 (ii) f(x) = x 2– 1, x = 1,−1 (iii) g(x) = 3x 2– 2 , x = 2/√3 , −2/√3 (iv) p(x) = x 3– 6x 2 + 11x – 6 , x = 1, 2, 3 (v) f(x) = 5x –π, x = 4/5 (vi) f(x) = x 2 , x = 0 (vii) f(x) = lx + m, x = −m/l (viii) f(x) = 2x + 1, x = 1/2 Solution: (i) f(x) = 3x + 1, x = −1/3 f(x) = 3x + 1 Substitute x = −1/3 in f(x) f( −1/3) = 3(−1/3) + 1 = -1 + 1 = 0 Since, the result is 0, so x = −1/3 is the root of 3x...