Class 9 RD Sharma Solutions - Chapter 5 Factorisation of Algebraic Expressions- Exercise 5.4 Last Updated : 28 Mar, 2021 Summarize Comments Improve Suggest changes Share Like Article Like Report Factorize each of the following:Question 1. a3 + 8b3 + 64c3 – 24abc Solution: We know that a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) = a3 + 8b3 + 64c3 – 24abc = (a)3 + (2b)3 + (4c)3 – (3 × a × 2b × 4c) = (a + 2b + 4c) [(a)2 + (2b)2 + (4c)2 -(a × 2b)– (2b × 4c)– (4c × a)]2 = (a + 2b + 4c) (a2 + 4b2 + 16c2 – 2ab – 8bc – 4ca) Question 2. x3 – 8y3 + 27z3 + 18xyz Solution: We can simplify the given equation as : x3 – 8y3 + 27z3 + 18xyz = (x)3 + (-2y)3 + (3z)3 – ( 3 * x * (-2y) * (3 z)) = (x – y + 3z) (x2 + 4y2 + 9z2 + 2xy + 6yz – 3zx) Question 3. 27x3 – y3– z3 – 9xyz Solution: 27x3 - y3 - z3 - 9xyz = (3x)3 + (-y)3 + (-z)3 – (3 * 3x * (-y) (-z)) = (3x – y – z) [(3x)2 + (-y)2 + (-z)2 – (3x * (-y)) – ((-y) (-z))- (- z × 3x)] = (3x - y – z) (9x2 + y2 + z2 + 3xy – yz + 3zx) Question 4. (1/27)x3 - y3 + 125z3 + 5xyz Solution: =(1/3x)3 + -(y)3 +(5z)3 - 3*(1/3x) * (-y) * (5z) =((1/3x) - y +5z)[(1/3x)2 + (-y)2 + (5z)2 -((1/3x)*-y) - (-y * 5z) - (5z * (1/3x)) =((1/3x) - y +5z)[(1/9)x2 + y2 +25z2 + (1/3xy) + 5yz - (5/3xz)] Question 5. 8x3 + 27y3 – 216z3 + 108xyz Solution: 8x3 + 27y3 – 216z3 + 108xyz = (2x)3 + (3y)3 + (6z)3 – 3 × (2x) (3y) (-6z) = (2x + 3y – 6z) [(2x)2 + (3y)2 + (-6z)2 – (2x * 3y) – (3y * (-6z)) – ((-6z) * 2x)] = (2x + 3y – 6z) (4x2 + 9y2 + 36z2 – 6xy + 18yz + 12zx) Question 6. 125 + 8x3 – 27y3 + 90xy Solution: 125 + 8X3 – 27y3 + 90xy = (5)3 + (2x)3 + (-3y)3 – [3 * 5 * 2x * (-3y)] = (5 + 2x – 3y) [(5)2 + (2x)2 + (-3y)2 – (5 * 2x) – (2x * (-3y)) – ((-3y) * 5)] = (5 + 2x – 3y) (25 + 4x2 + 9y2– 10x + 6xy + 15y) Question 7. 8x3 – 125y3 + 180xy + 216 Solution: 8x3 – 125y3 + 180xy + 216 = (2x)3 + (-5y)3 + (6)3 – 3 * 2x *(-5y) * 6 = (2x – 5y + 6) [(2x)2 + (-5y)2 + (6)2 – 2x *(-5y) – (-5y) * 6 – 6 * 2x] = (2x - 5y + 6) (4x2 + 25y2 + 36 + 10xy + 30y – 12x) Question 8. Multiply: (i) x2 +y2 + z2 – xy + xz + yz by x + y – z (ii) x2 + 4y2 + z2 + 2xy + xz – 2yz by x- 2y-z (iii) x2 + 4y2 + 2xy – 3x + 6y + 9 by x – 2y + 3 (iv) 9x2 + 25y2 + 15xy + 12x – 20y + 16 by 3x – 5y + 4 Solution: (i) (x2 + y2 + z2 – xy + yz + zx) by (x + y – z) = x3 +y3 – z3 + 3xyz (ii) (x2 + 4y2 + z2 + 2xy + xz – 2yz) by (x – 2y – z) = (x -2y-z) [x2 + (-2y)2 + (-z)2 - (x * (- 2y)) – ((-2y)* (z)) – ((-z) (x))] = x3 + (-2y)3 + (-z)3 – 3x * (-2y) * (-z) = x3 – 8y3 – z3 – 6xyz (iii) x2 + 4y2 + 2xy – 3x + 6y + 9 by (x – 2y + 3) = (x – 2y + 3) (x2 + 4y2 + 9 + 2xy + 6y – 3x) = (x)3 + (-2y)3 + (3)3 – (3 * x * (-2y) x 3) = x3 – 8y3 + 27 + 18xy (iv) 9x2 + 25y2 + 15xy + 12x – 20y + 16 by (3x – 5y + 4) = (3x -5y + 4) [(3x)2 + (-5y)2 + (4)2 – 3x * (-5y) +(-5y x 4) + (4 × 3x)] = (3x)3 + (-5y)3 + (4)3 – 3 * 3x *(-5y) * 4 = 27x3 – 125y3 + 64 + 180xy Question 9. (3x – 2y)3 + (2y – 4z)3 + (4z – 3x)3 Solution: (3x – 2y)3 + (2y – 4z)3 + (4z – 3x)3 ∵ 3x – 2y + 2y – 4z + 4z – 3x = 0 ∴ (3x – 2y)3 + (2y – 4z)3 + (4z – 3x)3 = 3(3x – 2y) (2y – 4z) (4z – 3x) {∵ x3 + y3 + z3 = 3xyz if x + y + z = 0} Question 10. (2x – 3y)3 + (4z – 2x)3 + (3y – 4z)3 Solution: (2x – 3y)3 + (4z – 2x)3 + (3y – 4z)3 ∵ 2x – 3y + 4z – 2x + 3y – 4z = 0 ∴ (2x – 3y)3 + (4z – 2x)3 + (3y – 4z)3 = 3(2x – 3y) (4z – 2x) (3y – 4z) {∵ x3 + y3 + z3 = 3xyz if x + y + z = 0} Question 11. [(x/2)+y +(z/3)]3 + [(x/3) -(2y/3) +z]3 + [(-5x/6)-(y/3)-(4z/3)]3 Solution: [(x/2)+y +(z/3)]3 + [(x/3) -(2y/3) +z]3 + [(-5x/6)-(y/3)-(4z/3)]3 ∵ (x/2) + y +(z/3) +(x/3) -(2y/3) + z - (5x/6) -(y/3) - (4z/3) =0 ∴ [(x/2)+y +(z/3)]3 + [(x/3) -(2y/3) +z]3 + [(-5x/6)-(y/3)-(4z/3)]3 = 3[(x/2)+y +(z/3)] [(x/3) -(2y/3) +z] [(-5x/6)-(y/3)-(4z/3)] {∵ a3 + b3 + c3 = 3abc if a + b + c = 0} Question 12. (a – 3b)3 + (3b – c)3 + (c – a)3 Solution: (a- 3b)3 + (3b – c)3 + (c – a)3 ∵ a – 3b + 3b – c + c – a = 0 ∴ (a – 3b)3 + (3b – c)3 + (c – a)3 = 3(a – 3b) (3b – c) (c – a) {∵ a3 + b3 + c3 = 3abc if a + b + c = 0} Question 13. 2√2a3 + 3√3b3 + c3 - 3√6abc Solution: = (√2a)3 +(√3b)3 +c3 - 3 * √2a * √3b * c = (√2a + √3b +c)[(√2a)2 +(√3b)2 + c2 - (√2a * √3b) - (√3b * c) - (c * √2a) = (√2a + √3b +c)(2a2 + 3b2 + c2 - √6ab - √3bc - √2ca) Question 14. 3√3a3 - b3 - 5√5c3 - 3√15abc Solution: = (√3a)3 + (-b)3 +(-√5c)3 - 3*√3a* (-b) *(-√5c) = (√3a - b - √5c) [(√3a)2 + (-b)2 +(-√5c)2 - (√3a* -b) - (-b * (-√5c)) - (-√5c* √3a) = (√3a - b - √5c)(3a2 + b2 + 5c2 + √3ab - √5bc + √15ca) Question 15. 2√2 a3 + 16√2 b3 + c3 - 12abc Solution: =(√2a)3 + (2√2b)3 + c3 - (3 * √2a * 2√2b * c) =(√2a + 2√2b +c) [(√2a)2 + (2√2b)2 + c2 - (√2a* 2√2b) - (2√2b*c) - (c* √2a) =(√2a + 2√2b +c)[2a2 + 8b2 + c2 - 4ab - 2√2bc - √2ca] Question 16. Find the value of x3 + y3 – 12xy + 64, when x + y = - 4 Solution: x3 + y3 – 12xy + 64 x + y = -4 On Cubing both sides, x3 + y3 + 3 xy(x + y) = -64 Substitute the value of (x + y) ⇒ x3 + y3 + 3xy * (-4) = -64 ⇒ x3 + y3 – 12xy + 64 = 0 Comment More infoAdvertise with us Next Article Class 9 RD Sharma Solutions - Chapter 6 Factorisation of Polynomials- Exercise 6.1 D divyansheevarshney Follow Improve Article Tags : Mathematics School Learning Class 9 RD Sharma Solutions RD Sharma Class-9 Maths-Class-9 +2 More Similar Reads RD Sharma Class 9 Solutions RD Sharma Solutions for class 9 provides vast knowledge about the concepts through the chapter-wise solutions. These solutions help to solve problems of higher difficulty and to ensure students have a good practice of all types of questions that can be framed in the examination. Referring to the sol 10 min read Chapter 1: Number SystemsClass 9 RD Sharma Solutions - Chapter 1 Number System - Exercise 1.1Exercise 1.1 in Chapter 1 "Number System" typically deals with irrational numbers and their properties. This exercise often focuses on identifying irrational numbers, understanding their characteristics, and performing basic operations with them.RD Sharma's Class 9 Mathematics textbook is a comprehe 4 min read Class 9 RD Sharma Solutions - Chapter 1 Number System - Exercise 1.2Question 1. Express the following rational numbers as decimals: i) 42/100ii) 327/500 iii) 15/4 Solutions: (i) 42/100 Using long division method: So, 42/100 = 0.42 (ii) 327/500 Using long division method: So, 327/500 = 0.654 (iii) 15/4 Using long division method: So, 15/4 = 3.75 Question 2. Express t 2 min read Class 9 RD Sharma Solutions - Chapter 1 Number System - Exercise 1.3Question 1: Express each of the following decimals in the form p/q:(i) 0.39(ii) 0.750(iii) 2.15(iv) 7.010(v) 9.90(vi) 1.0001Solution:(i) Multiplying and Dividing the number by 1000.39 = 39/100(ii) Multiplying and Dividing the number by 10000.750 = 750/1000 = 3/4(iii) Multiplying and Dividing the num 2 min read Class 9 RD Sharma Solutions - Chapter 1 Number System - Exercise 1.4Chapter 1 of RD Sharma’s Class 9 Mathematics textbook covers the fundamentals of the Number System. Exercise 1.4 delves into specific problems and applications related to the various types of numbers including integers, rational numbers, and irrational numbers. The problems are designed to enhance s 12 min read Chapter 2: Exponents and Powers of Real NumbersClass 9 RD Sharma Solutions - Chapter 2 Exponents of Real Numbers- Exercise 2.1Chapter 2 of Class 9 RD Sharma focuses on the concept of exponents of real numbers. This chapter introduces students to the fundamental principles of the exponents and their application in simplifying mathematical expressions. Understanding exponents is crucial for grasping more advanced topics in a 9 min read Class 9 RD Sharma Solutions - Chapter 2 Exponents of Real Numbers- Exercise 2.2 | Set 1Question 1. Assuming that x, y, z are positive real numbers, simplify each of the following:(i) \left(\sqrt{x^{-3}}\right)^5 Solution: We have, = \left(\sqrt{x^{-3}}\right)^5 = \left(\sqrt{\frac{1}{x^{3}}}\right)^5 = \left(\frac{1}{x^{\frac{3}{2}}}\right)^5 = \frac{1}{x^{\frac{15}{2}}} (ii) \sqrt{x^ 7 min read Class 9 RD Sharma Solutions - Chapter 2 Exponents of Real Numbers- Exercise 2.2 | Set 2Set 2 of Exercise 2.2 in Chapter 2 typically delves deeper into the application of exponent laws. This set often introduces more complex problems that require students to combine multiple laws of exponents, work with negative and fractional exponents, and simplify more intricate expressions. The que 6 min read Chapter 3: RationalisationClass 9 RD Sharma Solutions - Chapter 3 Rationalisation- Exercise 3.1Question 1: Simplify each of the following: \\(i)\sqrt[3]{4} \times \sqrt[3]{16}\\ (ii)\frac{\sqrt[4]{1250}}{\sqrt[4]{2}} Solution: (i) Using the formula: \sqrt[n]{a}\times\sqrt[n]{b}= \sqrt[n]{a \times b}\\ Here, \\=\sqrt[3]{4\times 16}\\ =\sqrt[3]{64}\\ =\sqrt[3]{4^{3}}\\ =(4^{3})^{\frac{1}{3}}\\ 2 min read Class 9 RD Sharma Solutions - Chapter 3 Rationalisation- Exercise 3.2 | Set 1Question 1. Rationalise the denominator of each of the following (i-vii): (i)\frac{3}{\sqrt5} (ii)\frac{3}{2\sqrt5} (iii)\frac{1}{\sqrt{12}} (iv)\frac{\sqrt3}{\sqrt5} (v)\frac{\sqrt3+1}{\sqrt2} (vi)\frac{\sqrt2+\sqrt5}{3} (vii)\frac{3\sqrt{2}}{\sqrt5} Solution: (i) We know that rationalisation facto 7 min read Class 9 RD Sharma Solutions - Chapter 3 Rationalisation- Exercise 3.2 | Set 2Chapter 3 of RD Sharma's Class 9 mathematics textbook focuses on the concept of rationalization, a crucial algebraic technique used to simplify expressions containing surds or irrational numbers. Exercise 3.2 specifically deals with the rationalization of denominators, which is an essential skill fo 5 min read Chapter 4: Algebraic IdentitiesClass 9 RD Sharma Solutions - Chapter 4 Algebraic Identities- Exercise 4.1 | Set 1Chapter 4 of RD Sharma's Class 9 mathematics textbook delves into the crucial concept of Algebraic Identities, with Exercise 4.1 Set 1 specifically focusing on fundamental algebraic identities and their applications in various mathematical contexts. This section forms the cornerstone of algebraic ma 7 min read Class 9 RD Sharma Solutions - Chapter 4 Algebraic Identities- Exercise 4.1 | Set 2This Exercise question and its concept are repeatedly tested in JEE Mains with slightly difficult twists in the questions and also solid 3-mark questions in the Final Term Examination of Class 9. Check out the sample question paper CBSE and Previous JEE Mains paper for more such questions.Read More: 6 min read Class 9 RD Sharma Solutions - Chapter 4 Algebraic Identities- Exercise 4.2Question 1: Write the following in the expanded form: (i) (a + 2b + c)2 (ii) (2a − 3b − c)2 (iii) (−3x+y+z)2 (iv) (m+2n−5p)2 (v) (2+x−2y)2 (vi) (a2 +b2 +c2)2 (vii) (ab+bc+ca)2 (viii) (x/y+y/z+z/x)2 (ix) (a/bc + b/ac + c/ab)2 (x) (x+2y+4z)2 (xi) (2x−y+z)2 (xii) (−2x+3y+2z)2 Solution: By following the 5 min read Class 9 RD Sharma Solutions - Chapter 4 Algebraic Identities- Exercise 4.3 | Set 1Question 1. Find the cube of each of the following binomial expressions. (a) (1/x+y/3) (b) (3/x-2/x2) (c) (2x+3/x) (d) (4-1/3x) Solution: (a) (1/x+y/3) Given, (1/x+y/3)3 [(a+b)3=a3+b3+3ab(a+b)] we know that here, a = 1/x, b= y/3 By using formula: (1/x+y/3)3 = (1/x)3 + (y/3)3 + 3(1/x)(y/3)(1/x+y/3) = 5 min read Class 9 RD Sharma Solutions - Chapter 4 Algebraic Identities- Exercise 4.3 | Set 2Algebraic identities are equations that remain true for all values for the variables involved. It is employed to simplify expressions and tackle equations. In this Exercise going to use algebraic identities to solve quadratic equations. Solving Quadratic Equations using Algebraic Identities A quadra 5 min read Class 9 RD Sharma Solutions - Chapter 4 Algebraic Identities- Exercise 4.4IntroductionExercise 4.4 in Chapter 4 of RD Sharma's Class 9 mathematics textbook delves into the fascinating world of algebraic identities, a cornerstone of algebraic manipulation and problem-solving. This exercise serves as a crucial stepping stone in a student's mathematical journey, building upo 9 min read Chapter 5: Factorization of Algebraic ExpressionsClass 9 RD Sharma Solutions - Chapter 5 Factorisation of Algebraic Expressions- Exercise 5.1Chapter 5 of RD Sharma's Class 9 Mathematics textbook, specifically Exercise 5.1, delves into the critical topic of Factorisation of Algebraic Expressions. This exercise introduces students to fundamental factorisation techniques, including the common factor method, grouping method, factorisation of 8 min read Class 9 RD Sharma Solutions - Chapter 5 Factorisation of Algebraic Expressions- Exercise 5.2 | Set 1Factorize each of the following expressions:Question 1. p3+27 Solution: ⇒ p3+33 ⇒(p+3)(p2-3p-9) [ a3+b3=(a+b)(a2-ab+b2) ] Therefore, p3+27 = (p+3)(p2-3p-9) Question 2. y3+125 Solution: ⇒ y3+53 ⇒ (y+5)(y2-3y+25) [ a3+b3=(a+b)(a2-ab+b2) ] Therefore, y3+125 = (y+5)(y2-3y+25) Question 3. 1 - 27a3 Soluti 2 min read Class 9 RD Sharma Solutions - Chapter 5 Factorisation of Algebraic Expressions- Exercise 5.2 | Set 2Factorize each of the following expressions:Question 13. 8x2y3-x5 Solution: ⇒ x2((2y)3 - x3) ⇒ x2(2y-x)((2y)2+2xy+x2) [x3-y3=(x-y)(x2+xy+y2)] ⇒ x2(2y-x)(4y2+2xy+x2) Therefore, 8x3y3-x5 = x2(2y-x)(4y2+2xy+x2) Question 14. 1029 - 3x3 Solution: ⇒ 3 (343 - x3) ⇒ 3(73 - x3) [x3-y3=(x-y)(x2+xy+y2)] ⇒ 3 (7 2 min read Class 9 RD Sharma Solutions - Chapter 5 Factorisation of Algebraic Expressions - Exercise 5.3In this article, we will be going to solve the entire exercise 5.3 of RD Sharma's book. Factorization of algebraic expressions is the process of rewriting an expression as a product of its factors. This technique simplifies expressions, solves equations, and helps in various algebraic manipulations. 4 min read Class 9 RD Sharma Solutions - Chapter 5 Factorisation of Algebraic Expressions- Exercise 5.4Factorize each of the following:Question 1. a3 + 8b3 + 64c3 – 24abc Solution: We know that a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) = a3 + 8b3 + 64c3 – 24abc = (a)3 + (2b)3 + (4c)3 – (3 × a × 2b × 4c) = (a + 2b + 4c) [(a)2 + (2b)2 + (4c)2 -(a × 2b)– (2b × 4c)– (4c × a)]2 = (a 7 min read Chapter 6: Factorization of PolynomialsClass 9 RD Sharma Solutions - Chapter 6 Factorisation of Polynomials- Exercise 6.1Factorisation is a critical algebraic technique used to simplify polynomials and solve equations. In Class 9, RD Sharma's textbook introduces students to the fundamentals of factorization in Chapter 6. Exercise 6.1 focuses on applying these concepts to factorize polynomials effectively which is foun 4 min read Class 9 RD Sharma Solutions - Chapter 6 Factorisation of Polynomials- Exercise 6.2The Factorisation of polynomials is a fundamental algebraic skill that involves expressing a polynomial as a product of the simpler polynomials. In Class 9, students explore various techniques to factorize polynomials which helps in simplifying algebraic expressions and solving equations. Chapter 6 6 min read Class 9 RD Sharma Solutions - Chapter 6 Factorisation of Polynomials- Exercise 6.3The Factorisation of polynomials is a key algebraic skill taught in Class 9. This chapter focuses on breaking down polynomials into simpler components that when multiplied together give the original polynomial. Understanding factorization is crucial for solving algebraic equations and simplifying ex 5 min read Class 9 RD Sharma Solutions - Chapter 6 Factorisation of Polynomials- Exercise 6.4 | Set 1In Class 9 mathematics, the chapter on Factorisation of Polynomials is crucial for understanding algebraic expressions and their manipulation. Exercise 6.4 focuses on practical problems related to factorization enhancing students' skills in breaking down polynomials into simpler components. This exe 10 min read Class 9 RD Sharma Solutions - Chapter 6 Factorisation of Polynomials- Exercise 6.4 | Set 2In Class 9 mathematics, understanding the factorization of polynomials is crucial for solving quadratic equations and simplifying algebraic expressions. Factorisation involves expressing a polynomial as a product of its factors which simplifies the polynomial and aids in solving the various problems 15+ min read Class 9 RD Sharma Solutions - Chapter 6 Factorisation of Polynomials- Exercise 6.5 | Set 1In Class 9 mathematics, understanding the factorization of polynomials is crucial as it forms the foundation for the more advanced algebraic concepts. Chapter 6 of RD Sharma's textbook focuses on the various methods of factorizing polynomials helping the students simplify complex expressions and sol 12 min read Class 9 RD Sharma Solutions - Chapter 6 Factorisation of Polynomials- Exercise 6.5 | Set 2Factorisation is a key concept in algebra especially important for simplifying polynomial expressions. It involves breaking down a polynomial into a product of the simpler polynomials or factors. This process is crucial for solving polynomial equations and understanding polynomial functions in vario 8 min read Chapter 7: Introduction to Euclid’s GeometryClass 9 RD Sharma Solutions - Chapter 7 Introduction to Euclid’s Geometry- Exercise 7.1Question 1: Define the following terms.(i) Line segment(ii) Collinear points(iii) Parallel lines(iv) Intersecting lines(v) Concurrent lines(vi) Ray(vii) Half-line Solution: (i) A line segment is a one-dimensional line connecting two points. It is the shortest distance that is incident on both the po 3 min read Chapter 8: Lines and AnglesClass 9 RD Sharma Solutions - Chapter 8 Introduction to Lines and Angles- Exercise 8.1Question 1: Write the complement of each of the following angles: (i) 20° (ii) 35° (iii) 90° (iv) 77° (v) 30° Solution: (i) Given an angle 20° As we Studied in this Chapter, the sum of angle and its complement is 90 Therefore, its complement will be (90° - 20° = 70°) (ii) Given an angle 35° As we St 5 min read Class 9 RD Sharma Solutions - Chapter 8 Introduction to Lines and Angles- Exercise 8.2 | Set 1Chapter 8 of RD Sharma's Class 9 Mathematics textbook focuses on the "Introduction to Lines and Angles". This chapter introduces fundamental concepts of the geometry including the lines, angles and their properties. Exercise 8.2 | Set 1 presents a series of problems designed to deepen the understand 5 min read Class 9 RD Sharma Solutions - Chapter 8 Introduction to Lines and Angles- Exercise 8.2 | Set 2Question 11. In Figure, ACB is a line such that ∠DCA = 5x and ∠DCB = 4x. Find the value of x. Solution: It is given that ACB is a line in the figure given below. Thus, ∠ACD and ∠BCD form a linear pair. Therefore, their sum must be equal to 180°. Or we can say that ∠ACD + ∠BCD = 180° Also, ∠ACD = 4x 7 min read Class 9 RD Sharma Solutions - Chapter 8 Introduction to Lines and Angles- Exercise 8.3Question 1. In the below fig, lines l1, and l2 intersect at O, forming angles as shown in the figure. If x = 45°. Find the values of x, y, z, and u. Solution: Given that X = 45° Find: the value of Y, Z, and u z = x = 45° [Vertically opposite angles are equal] z + u = 180° [z and u are angles that ar 7 min read Class 9 RD Sharma Solutions - Chapter 8 Introduction to Lines and Angles- Exercise 8.4 | Set 1Question 1: In figure, AB, CD and ∠1 and ∠2 are in the ratio 3 : 2. Determine all angles from 1 to 8. Solution: Assume, ∠1 = 3x and ∠2 = 2x From the following figure: ∠1 and ∠2 are a linear pair of angles Thus, ∠1 + ∠2 = 180° 3x + 2x = 180° 5x = 180° x =\frac{180 }{ 5} x = 36° Hence, ∠1 = 3x = 108° 5 min read Class 9 RD Sharma Solutions - Chapter 8 Introduction to Lines and Angles- Exercise 8.4 | Set 2Question 11. In the given figure, lines AB and CD are parallel and P is any point as shown in the figure. Show that ∠ABP+ ∠CDP= ∠DPB. Solution: Here in the given figure: Given: AB || CD Now draw a line XY passing through point P and parallel to AB and CD. Here,XY || CD, thus, ∠CDP and ∠1are alternat 7 min read Class 9 RD Sharma Solutions - Chapter 8 Introduction to Lines and Angles- Exercise 8.4 | Set 3Question 21. In the given figure, transversal l intersects two linesmandn, ∠4 = 110° and ∠7 = 65°. Ism||n? Solution: The figure is given as follows: It is given that l is a transversal to lines m and n. Also, ∠4 = 110° and ∠7 = 65° We need to check whether m || n or not. We have ∠7 = 65°. Also, ∠7 a 7 min read Chapter 9: Triangle and Its AnglesClass 9 RD Sharma Solutions - Chapter 9 Triangles and its Angles- Exercise 9.1Question 1: In a triangle ΔABC, if ∠A = 55° ∠B = 40°, find ∠C. Solution: Given: A = 55° and B = 40° Theorem Used: Sum of all angles of a triangle is 180°. From the theorem we can write that: ∠A + ∠B + ∠C = 180° 55° + 40° + ∠C = 180° //Putting the values of A and B. 95° + ∠C = 180° ∠C = 180° - 95° ∠C 9 min read Class 9 RD Sharma Solutions - Chapter 9 Triangles and its Angles- Exercise 9.2Triangles are fundamental geometric shapes in Euclidean geometry, consisting of three straight sides and three angles. The sum of the interior angles in any triangle is always 180°. This is a fundamental property used in many geometric proofs and problems.In this article, we will be going to solve t 8 min read Like