Class 10 RD Sharma Solutions - Chapter 11 Constructions - Exercise 11.2 | Set 1 Last Updated : 10 Feb, 2023 Comments Improve Suggest changes Like Article Like Report Question 1. Construct a triangle of sides 4 cm, 5 cm, and 6 cm and then a triangle similar to it whose sides are (2/3) of the corresponding sides of it. Solution: Follow these steps for the construction: Step 1: Construct a line segment BC of 5 cm. Step 2: From the centre B take the radius of 4 cm and from the centre C take the radius 6 cm, construct arcs bisecting each other at the point A. Step 3: Connect the lines AB and AC. After that we will have ABC as the triangle. Step 4: Construct a ray BX creating an acute angle with the line BC and break off 3 equal parts creating BB1 = B1B2 = B2B3. Step 5: Further connect B3C. Step 6: Construct B’C’ parallel to B3C and C’A’ parallel to CA Therefore, We have the required triangle ΔA’BC’. Question 2. Construct a triangle similar to a given ΔABC such that each of its sides is (5/7)th of the corresponding sides of ΔABC. It is given that AB = 5 cm, BC = 7 cm and ∠ABC = 50°. Solution: Follow these steps for the construction: Step 1: Construct a line segment BC of 7 cm. Step 2: Construct a ray BX an angle of 50° and break off BA = 5 cm. Step 3: Connect AC. After that we have ABC is the triangle. Step 4: Construct a ray BY creating an acute angle with BC and break off 7 equal parts creating BB = B1B2 = B2B3 = B3B4 = B4Bs = B5B6 = B6B7 Step 5: Now connect B7 and C Step 6: Construct B5C’ parallel to B7C and C’A’ parallel to CA. Therefore, We have the required triangle ΔA’BC’. Question 3. Construct a triangle similar to a given ΔABC such that each of its sides is (2/3)rd of the corresponding sides of ΔABC. It is given that BC = 6 cm, ∠B = 50° and ∠C = 60°. Solution: Follow these steps for the construction: Step 1: Construct a line segment BC of 6 cm. Step 2: Construct a ray BX creating an angle of 50° and CY creating 60° Along BC which bisect each other at A. After that, ABC is the triangle. Step 3: From B, Construct one more ray BZ creating an acute angle below BC and intersect 3 equal parts, creating BB1 = B1B2 = B2B2. Step 4: Connect B3C. Step 5: From B2, Construct B2C’ parallel to B3C and C’A’ parallel to CA. Therefore, We have the required triangle ΔA’BC’. Question 4. Construct a ΔABC in which BC = 6 cm, AB = 4 cm and AC = 5 cm. Draw a triangle similar to ΔABC with its sides equal to 3/4th of the corresponding sides of ΔABC. Solution: Follow these steps for the construction: Step 1: Construct a line segment BC of 6 cm. Step 2: Along centre B and radius 4 cm and Along centre C and radius 5 cm, Construct arcs’ bisecting each other at A. Step 3: Connect AB and AC. After that ABC is the triangle, Step 4: Construct a ray BX creating an acute angle along BC and break off 4 equal parts creating BB1= B1B2 = B2B3 = B3B4. Step 5: Connect B4 and C. Step 6: From B3C Construct C3C’ parallel to B4C and from C’, Step 7: Construct C’A’ parallel to CA. Therefore, We have the required triangle ΔA’BC’. Question 5. Construct a triangle Along sides 5 cm, 6 cm, and 7 cm and then another triangle whose sides are (7/5)th of the corresponding sides of the first triangle. Give the justification of the construction. Solution: Follow these steps for the construction: Step 1: Construct a line segment BC of 5 cm. Step 2: Along centre B and radius 6 cm and Along centre C and radius 7 cm, Construct arcs bisecting each other at A. Step 3: Connect AB and AC. After that ABC is the triangle. Step 4: Construct a ray BX creating an acute angle along BC and break off 7 equal parts creating BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7. Step 5: Connect B5 and C. Step 6: From B7, Construct B7C’ parallel to B5C and C’A’ parallel CA. Therefore, We have the required triangle ΔA’BC’. Question 6. Construct a right triangle ABC in which AC = AB = 4.5 cm and ∠A = 90°. Draw a triangle similar to ΔABC with its sides equal to (5/4)th of the corresponding sides of ΔABC. Solution: Follow these steps for the construction: Step 1: Construct a line segment AB of 4.5 cm. Step 2: At A, Construct a ray AX perpendicular to AB and break off AC = AB = 4.5 cm. Step 3: Connect BC. After that ABC is the triangle. Step 4: Construct a ray AY creating an acute angle along AB and break off 5 equal parts creating AA1 = A1A2 = A2A3 = A3A4 = A4A5 Step 5: Connect A4 and B. Step 6: From 45, Construct 45B’ parallel to A4B and B’C’ parallel to BC. Therefore, We have the required triangle ΔAB’C’. Question 7. Construct a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle. Solution: Follow these steps for the construction: Step 1: Construct a line segment BC of 5 cm. Step 2: At B, Construct perpendicular BX and break off BA = 4 cm. Step 3: Connect AC , After that ABC is the triangle Step 4: Construct a ray BY creating an acute angle along BC, and break off 5 equal parts creating BB1 = B1B2 = B2B3 = B3B4 = B4B5. Step 5: Connect B3 and C. Step 6: From B5, Construct B5C’ parallel to B3C and C’A’ parallel to CA. Therefore, We have the required triangle ΔA’BC’. Question 8. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 3/2 times the corresponding sides of the isosceles triangle. Solution: Follow these steps for the construction: Step 1: Construct a line segment BC of 8 cm. Step 2: Construct its perpendicular bisector DX and break off DA = 4 cm. Step 3: Connect AB and AC. After that ABC is the triangle. Step 4: Construct a ray DY creating an acute angle along OA and break off 3 equal parts creating DD1 = D1D2 = D2D3 = D3D4. Step 4: Connect D2. Step 5: Construct D3A’ parallel to D2A and A’B’ parallel to AB meeting BC at C’ and B’ respectively. Therefore, We have the required triangle ΔB’A’C’. Question 9. Draw a ΔABC with BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are (3/4)th of the corresponding sides of the ΔABC. Solution: Follow these steps for the construction: Step 1: Construct a line segment BC of 6 cm. Step 2: At B, Construct a ray BX creating an angle of 60° Along BC and break off BA of 5 cm. Step 3: Connect AC. After that ABC is the triangle. Step 4: Construct a ray BY creating an acute angle along BC and break off 4 equal parts creating BB1= B1B2 = B2B3=B3B4. Step 5: Connect B4 and C. Step 6: From B3, Construct B3C’ parallel to B4C and C’A’ parallel to CA. Therefore, We have the required triangle ΔA’BC’. Question 10. Construct a triangle similar to ΔABC in which AB = 4.6 cm, BC = 5.1 cm, ∠A = 60° Along scale factor 4 : 5. Solution: Follow these steps for the construction: Step 1: Construct a line segment AB of 4.6 cm. Step 2: At A, Construct a ray AX creating an angle of 60°. Step 3: Along centre B and radius 5.1 cm. Construct an arc bisecting AX at C. Step 4: Connect BC. After that ABC is the triangle. Step 5: From A, Construct a ray AX creating an acute angle along AB and break off 5 equal parts creating AA1 = A1A2 = A2A3 = A3A4=A4A5. Step 6: Connect A4 and B. Step 7: From A5, ConstructA5B’ parallel to A4B and B’C’ parallel to BC. Therefore, We have the required triangle ΔC’AB’. Comment More infoAdvertise with us Next Article Class 10 RD Sharma Solutions - Chapter 11 Constructions - Exercise 11.2 | Set 1 Y yashchuahan Follow Improve Article Tags : Mathematics School Learning Class 10 RD Sharma Solutions RD Sharma Class-10 Maths-Class-10 +2 More Similar Reads RD Sharma Class 10 Solutions RD Sharma Class 10 Solutions offer excellent reference material for students, enabling them to develop a firm understanding of the concepts covered. in each chapter of the textbook. As Class 10 mathematics is categorized into various crucial topics such as Algebra, Geometry, and Trigonometry, which 9 min read Chapter 1: Real NumbersClass 10 RD Sharma Solutions - Chapter 1 Real Numbers - Exercise 1.1 | Set 1In Class 10 Mathematics, Chapter 1 of RD Sharma's textbook deals with Real Numbers. This chapter lays the foundation for understanding the different types of numbers and their properties. 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This chapter is fundamental as it introduces the concept of the real numbers which form the basis for understanding more complex mathematical id 15+ min read Class 10 RD Sharma Solutions- Chapter 1 Real Numbers - Exercise 1.2 | Set 2In this article, we will explore the solutions to Exercise 1.2 from Chapter 1: Real Numbers of RD Sharma's Class 10 Mathematics textbook. The exercise focuses on the fundamental concepts related to real numbers including Euclid’s Division Lemma Fundamental Theorem of Arithmetic and their application 10 min read Class 10 RD Sharma Solutions- Chapter 1 Real Numbers - Exercise 1.3In Chapter 1 of Class 10 Mathematics, RD Sharma explores the topic of "Real Numbers". This chapter delves into various properties of the real numbers including their classification and how they can be expressed as rational and irrational numbers. 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This chapter is foundational laying the groundwork for understanding different types of numbers and their properties. It covers the concept of the real numbers, their classification, and key operations performed with the 6 min read Chapter 2: PolynomialsClass 10 RD Sharma Solutions- Chapter 2 Polynomials - Exercise 2.1 | Set 1The Polynomials are fundamental algebraic expressions consisting of variables raised to the whole-number powers and combined using addition, subtraction, and multiplication. They form the basis of many concepts in algebra including solving equations, graphing functions, and mathematical modeling. In 15+ min read Class 10 RD Sharma Solutions- Chapter 2 Polynomials - Exercise 2.1 | Set 2In this section, we delve into Chapter 2 of the Class 10 RD Sharma textbook, focusing on Polynomials. Exercise 2.1 is designed to help students understand the fundamental concepts of polynomials, including their types, properties, and various operations performed on them.Class 10 RD Sharma Solutions 12 min read Class 10 RD Sharma Solutions- Chapter 2 Polynomials - Exercise 2.2In this article, we will explore the solutions to Exercise 2.2 from Chapter 2 of RD Sharma's Class 10 Mathematics textbook which focuses on the "Polynomials". This chapter is fundamental for understanding the basics of polynomials their properties and how to manipulate them. By solving Exercise 2.2 7 min read Class 10 RD Sharma Solutions - Chapter 2 Polynomials - Exercise 2.3Chapter 2 of RD Sharma's Class 10 Mathematics book delves into the polynomials a fundamental algebraic concept. This chapter introduces polynomials, their types and their properties laying the groundwork for the solving polynomial equations. Exercise 2.3 focuses on the solving polynomial equations u 11 min read Chapter 3: Pair of Linear Equations in Two VariablesClass 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.1Question 1. Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a rig on the items in the stall, and if the ring covers any object completely you get it). The number of times she played Hoopla is half the number of rides she h 7 min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.2 | Set 1Question 1. Solve the following of equation graphicallyx + y = 32x + 5y = 12 Solution: Given that, 2x + 5y = 12 and x + y = 3 We have, x + y = 3, When y = 0, we get x = 3 When x = 0, we get y = 3 So, the following table giving points on the line x + y = 3: x03y30 Now, 2 + 5y = 12 y = (12 - 2x)/5 Whe 11 min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.2 | Set 2In Chapter 3 of RD Sharma's Class 10 Mathematics, students explore the concept of solving pairs of linear equations in two variables. Exercise 3.2 | Set 2 presents various problems designed to reinforce these concepts through practical application. This article comprehensively addresses each problem 14 min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.2 | Set 3Question 23. Solve the following system of linear equations graphically and shade the region between the two lines and the x-axis.(i) 2x + 3y = 12 and x - y = 1 Solution: Given that, 2x + 3y = 12 and x - y = 1 Now, 2x + 3y = 12 x = (12-3y)/2 When y = 2, we get x = 3 When y = 4, we get x = 0 So, the 15+ min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.3 | Set 1Solve the following system of equations:Question 1. 11x + 15y + 23 = 0 and 7x – 2y – 20 = 0 Solution: 11x +15y + 23 = 0 …………………………. (i) 7x – 2y – 20 = 0 …………………………….. (ii) From (ii) 2y = 7x – 20 ⇒ y = (7x −20)/2 ……………………………… (iii) Putting y in (i) we get, ⇒ 11x + 15((7x−20) / 2) + 23 = 0 ⇒ 11x + (10 9 min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.3 | Set 2Solve the following system of equations:Question 14. 0.5x + 0.7y = 0.74 and 0.3x + 0.5y = 0.5 Solution: 0.5x + 0.7y = 0.74……………………… (i) 0.3x – 0.5y = 0.5 ………………………….. (ii) Multiply LHS and RHS by 100 in (i) and (ii) 50x +70y = 74 ……………………….. (iii) 30x + 50y = 50 …………………………… (iv) From (iii) 50x = 74 9 min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.4 | Set 1Solve each of the following systems of equations by the method of cross multiplication.Question 1. x + 2y + 1 = 0 and 2x – 3y – 12 = 0 Solution: Given that, x + 2y + 1 = 0 2x - 3y - 12 = 0 On comparing both the equation with the general form we get a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = −3, c2 = -12 N 11 min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.4 | Set 2Chapter 3 of RD Sharma’s Class 10 Mathematics book delves into the topic of Pair of Linear Equations in Two Variables. This chapter is fundamental in understanding how to solve systems of linear equations and their applications. Exercise 3.4 | Set 2 provides the practice problems to strengthen stude 14 min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.5 | Set 1Question 1. In each of the following systems of equations determine whether the system has a unique solution, no solution, or infinite solutions. In case there is a unique solution, find it from 1 to 4:x − 3y − 3 = 0, 3x − 9y − 2 = 0.Solution: Given that,x − 3y − 3 = 0 ...(1) 3x − 9y − 2 = 0 ...(2) 9 min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.5 | Set 2Chapter 3 of RD Sharma's Class 10 Mathematics textbook deals with the "Pair of Linear Equations in the Two Variables." This chapter explores methods for solving systems of linear equations involving the two variables. Understanding these methods is crucial for solving the problems in algebra and geo 11 min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.5 | Set 3Question 27. For what value of a, the following system of the equation has no solution:ax + 3y = a − 312x + ay = a Solution: Given that, ax + 3y = a − 3 ...(1) 12x + ay = a ...(2) So, the given equations are in the form of: a1x + b1y − c1 = 0 ...(3) a2x + b2y − c2 = 0 ...(4) On comparing eq (1) with 15+ min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.6 | Set 1Question 1. 5 pens and 6 pencils together cost ₹ 9 and 3 pens and 2 pencils cost ₹ 5. Find the cost of 1 pen and 1 pencil. Solution: Let the cost of 1 pen be Rs. x and the cost of 1 pencil be Rs. y Now, 5 pens and 6 pencils cost Rs. 9 => 5x + 6y = 9 (1) And, 3 pens and 2 pencils cost Rs. 5 => 9 min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.6 | Set 2Linear equations in two variables form a critical part of algebraic problems in mathematics. These equations which represent lines on a graph provide the foundation for solving various real-world problems involving relationships between the two variables. Chapter 3 of RD Sharma's Class 10 mathematic 7 min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.7In Class 10 mathematics, solving linear equations is a fundamental concept that underpins many advanced topics. Chapter 3 of RD Sharma's textbook titled "Pair of Linear Equations in Two Variables" explores methods to solve systems of equations involving the two variables. Understanding these concept 14 min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.8In this section, we explore Chapter 3 of the Class 10 RD Sharma textbook, which focuses on Pairs of Linear Equations in Two Variables. Exercise 3.8 is designed to help students understand the methods for solving pairs of linear equations, enhancing their problem-solving skills in algebra.Class 10 RD 9 min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.9In Class 10 Mathematics, Chapter 3 focuses on Pair of Linear Equations in Two Variables. This chapter deals with the solving systems of the equations where each equation is linear and has two variables. The importance of this topic lies in its applications in various fields such as economics, engine 11 min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.10Question 1. Points A and B are 70km. apart on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7hrs, but if they travel towards each other, they meet in one hour. Find the speed of two cars. Solution: Let’s assume that th 15+ min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.11 | Set 1Question 1. If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however the length is reduced by 1 unit and the breadth increased by 2 units, the area increases by 33 square units. Find the area of the rectangle. Solution: Let’s 15 min read Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.11 | Set 2In this article, we will discuss the solutions to Exercise 3.11 Set 2 from Chapter 3 of RD Sharma's Class 10 Mathematics textbook which focuses on the "Pair of Linear Equations in the Two Variables". This chapter is vital as it introduces students to methods for solving systems of linear equations a 11 min read Chapter 4: TrianglesClass 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.1Question 1. (Fill in the blanks using the correct word given in brackets: (i) All circles are _____________ (congruent, similar).(ii) All squares are _____________ (similar, congruent).(iii) All _____________ triangles are similar (isosceles, equilaterals) :(iv) Two triangles are similar, if their c 2 min read Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.2Question 1: In a ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC.(i) If AD = 6 cm, DB = 9 cm and AE = 8 cm, Find AC. Solution: Given: Δ ABC where Length of side AD = 6 cm, DB = 9 cm, AE = 8 cm Also, DE ∥ BC, To find : Length of side AC By using Thales Theorem we get, 15 min read Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.3Problem 1: In a ∆ABC, AD is the bisector of ∠A, meeting side BC at D.(i) If BD = 2.5 cm, AB = 5 cm and AC = 4.2 cm, find DC Solution: Given: Length of side BD = 2.5 cm, AB = 5 cm, and AC = 4.2 cm. To find: Length of side DC In Δ ABC, AD is the bisector of ∠A, meeting side BC at D. Since, AD is ∠A bi 11 min read Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.4Question 1. (i) In fig., if AB || CD, find the value of x. Solution: Given, AB∥ CD. To find the value of x. Now, AO/ CO = BO/ DO [Diagonals of a parallelogram bisect each other] ⇒ 4/ (4x – 2) = (x +1)/ (2x + 4) 4(2x + 4) = (4x – 2)(x +1) 8x + 16 = x(4x – 2) + 1(4x – 2) 8x + 16 = 4x2 – 2x + 4x – 2 -4 3 min read Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.5 | Set 1The Triangles are fundamental geometric shapes characterized by three sides, three angles and three vertices. They are categorized based on their side lengths and angles leading to the various types such as the equilateral, isosceles, and scalene triangles. The study of the triangles includes unders 6 min read Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.5 | Set 2Chapter 4 of the Class 10 RD Sharma Mathematics textbook, "Triangles," covers the properties and theorems related to triangles, including similarity, congruence, and the Pythagorean theorem. Exercise 4.5 focuses on applying these properties to solve problems related to triangles.RD Sharma Solutions 10 min read Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.6 | Set 1This exercise contains questions regarding the similarity of triangles. Questions in this exercise are based on the concepts of properties of two similar triangles. The RD Sharma Solutions Class 10 provides all solutions required for a quick preparation for exams. The following questions and their s 7 min read Class 10 RD Sharma Mathematics Solutions - Chapter 4 Triangles - Exercise 4.6 | Set 2Chapter 4 of RD Sharma's Class 10 mathematics book focuses on "Triangles" which is a fundamental concept in the geometry. This chapter deals with the various properties and theorems related to triangles such as the congruence similarity and Pythagorean theorem. Understanding these concepts is essent 9 min read Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.7 | Set 1Question 1. If the sides of a triangle are 3 cm, 4 cm, and 6 cm long, determine whether the triangle is a right-angled triangle.Solution: According to the question The sides of triangle are: AB = 3 cm BC = 4 cm AC = 6 cm According to Pythagoras Theorem:  AB2 = 32 = 9 BC2 = 42 = 16 AC2 = 62 = 36 Sinc 11 min read Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.7 | Set 2Question 15. Each side of a rhombus is 10 cm. If one of its diagonals is 16 cm, find the length of the other diagonal. Solution: Given, In rhombus ABCD, diagonals AC and BD bisect each other at O at right angles Each side = 10 cm and one diagonal AC = 16 cm ∴ AO = OC = 16/2 = 8 cm Now in ∆AOB, By us 11 min read Chapter 5: Trigonometric RatiosClass 10 RD Sharma Solutions - Chapter 5 Trigonometric Ratios - Exercise 5.1 | Set 1Chapter 5 of the Class 10 RD Sharma Mathematics textbook, "Trigonometric Ratios," introduces students to the basic trigonometric functions and their applications. Exercise 5.1 focuses on solving problems involving the fundamental trigonometric ratios of angles in a right-angled triangle.RD Sharma So 9 min read Class 10 RD Sharma Solutions - Chapter 5 Trigonometric Ratios - Exercise 5.2 | Set 2Evaluate each of the following(14-19)Question 14. \frac{sin30°-sin90°+2cos0°}{tan30°tan60°} Solution: Given:\frac{sin30°-sin90°+2cos0°}{tan30°tan60°} -(1) Putting the values of sin 30° = 1/2, tan 30° = 1/√3, tan 60° = √3, sin 90° = cos 0° = 1 in eq(1) = \frac{(\frac{1}{2})-(1)+2(1)}{(\sqrt3)(\frac{1 4 min read Class 10 RD Sharma Solutions - Chapter 5 Trigonometric Ratios - Exercise 5.1 | Set 3Trigonometry is a fundamental branch of mathematics that deals with the relationships between the angles and sides of triangles. Chapter 5 of RD Sharma's Class 10 textbook focuses on the Trigonometric Ratios which are the foundation of trigonometry. In this chapter, students will learn to apply thes 5 min read Class 10 RD Sharma Solutions - Chapter 5 Trigonometric Ratios - Exercise 5.2 | Set 1Evaluate each of the following(1-13)Question 1. sin 45° sin 30° + cos 45° cos 30° Solution: Given: sin 45° sin 30° + cos 45° cos 30° -(1) Putting the values of sin 45° = cos 45°= 1/√2, sin 30° = 1/2, cos 30° = √3/2 in eq(1) = (1/√2)(1/2) + (1/√2})(√3/2) = (1/2√2) + (√3/2√2) = (1 + √3)/2√2 Question 2 4 min read Class 10 RD Sharma Solutions - Chapter 5 Trigonometric Ratios - Exercise 5.3 | Set 1Question 1. Evaluate the following:(i) sin 20°/cos 70° Solution: Given: sin 20°/cos 70° = sin(90° − 70°)/cos 70° = cos 70°/cos 70° -(∵ sin (90° - θ) = cos θ) = 1 Hence, sin 20°/cos 70° = 1 (ii) cos 19°/sin 71° Solution: Given: cos 19°/sin 71° = cos(90° − 71°)/sin 71° = sin 71°/sin 71° -(∵ cos (90° - 8 min read Class 10 RD Sharma Solutions - Chapter 5 Trigonometric Ratios - Exercise 5.3 | Set 2Question 8. Prove the following:(i) sin θ sin (90° - θ) - cosθ cos (90° - θ) = 0 Solution: We have to prove that sin θ sin (90° - θ) - cosθ cos (90° - θ) = 0 Taking LHS = sin θ sin (90° - θ) - cosθ cos (90° - θ) -(∵ sin (90° - θ) = cos θ) = sin θ cosθ - cosθ sinθ = 0 LHS = RHS Hence proved (ii) \fra 9 min read Chapter 6: Trigonometric IdentitiesClass 10 RD Sharma Solutions - Chapter 6 Trigonometric Identities - Exercise 6.1 | Set 1Prove the following trigonometric identities:Question 1. (1 – cos2 A) cosec2 A = 1 Solution: We have, L.H.S. = (1 – cos2 A) cosec2 A By using the identity, sin2 A + cos2 A = 1, we get, = (sin2 A) (cosec2 A) = sin2 A × (1/sin2 A) = 1 = R.H.S. Hence proved. Question 2. (1 + cot2 A) sin2 A = 1 Solution 7 min read Class 10 RD Sharma Solutions - Chapter 6 Trigonometric Identities - Exercise 6.1 | Set 2Prove the following trigonometric identities:Question 29. \frac{1+secθ}{secθ}=\frac{sin^2θ}{1-cosθ} Solution: We have, L.H.S. = \frac{1+secθ}{secθ} = \frac{1+\frac{1}{cosθ}}{\frac{1}{cosθ}} = \frac{cosθ+1}{cosθ}×{\frac{cosθ}{1}} = 1 + cos θ = \frac{(1+cosθ)(1-cosθ)}{1-cosθ} = \frac{1-cos^2θ}{1-cosθ} 7 min read Class 10 RD Sharma Solutions - Chapter 6 Trigonometric Identities - Exercise 6.1 | Set 3The Trigonometric identities are fundamental formulas in trigonometry that relate the angles and sides of the triangles. These identities simplify trigonometric expressions and solve the equations involving trigonometric functions. In Class 10 RD Sharma's Chapter 6 on the Trigonometric Identities Ex 13 min read Class 10 RD Sharma Solutions - Chapter 6 Trigonometric Identities - Exercise 6.2In Chapter 6 of RD Sharma's Class 10 Mathematics textbook, we explore Trigonometric Identities a fundamental topic in trigonometry. Exercise 6.2 focuses on applying various trigonometric identities to solve problems and simplify expressions. Understanding these identities is crucial for solving comp 8 min read Chapter 7: StatisticsClass 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.1 | Set 1Problem 1: Calculate the mean for the following distribution:x:56789f:4 814113 Solution: x ffx542068487 1498811889327 N = 40 ∑ fx = 281 We know that, Mean = ∑fx/ N = 281/40 = 7.025 Problem 2: Find the mean of the following data :x:19212325272931f:13151618161513 Solution: x ffx19132472115315231636825 4 min read Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.1 | Set 2Problem 11: Five coins were simultaneously tossed 1000 times and at each toss, the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4, and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.No. of heads per tossNo. of tosses 03811442 3 min read Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.2Question1: The number of telephone calls received at an exchange per interval for 250 successive one-minute intervals are given in the following frequency table:Number of calls (x) 0 1 2 3 4 5 6 Number of intervals (f) 15 24 29 46 54 43 39 Compute the mean number of calls per interval. Solution: Let 6 min read Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.3 | Set 1Chapter 7 of RD Sharma’s Class 10 textbook focuses on Statistics a fundamental area in mathematics that deals with data collection, analysis, interpretation, and presentation. Exercise 7.3 | Set 1 of this chapter covers the various problems designed to enhance students' understanding and application 8 min read Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.3 | Set 2Question 14. Find the mean of the following frequency distribution:Class interval:25 – 2930 – 3435 – 3940 – 4445 – 4950 – 5455 – 59Frequency:1422166534 Solution: Let’s consider the assumed mean (A) = 42 Class intervalMid – value xidi = xi – 42ui = (xi – 42)/5fifiui25 – 2927-15-314-4230 – 3432-10-222 13 min read Class 10 RD Sharma Solution - Chapter 7 Statistics - Exercise 7.4 | Set 1In this article, we will explore the solutions to Exercise 7.4 from Chapter 7 of RD Sharma's Class 10 Mathematics textbook which focuses on Statistics. This chapter is crucial as it lays the foundation for understanding how data is collected, organized, analyzed, and interpreted in the various field 8 min read Class 10 RD Sharma Solution - Chapter 7 Statistics - Exercise 7.4 | Set 2Question 11. An incomplete distribution is given below :Variable10-2020-3030-4040-5050-6060-7070-80Frequency1230-65-2518You are given that the median value is 46 and the total number of items is 230. (i) Using the median formula fill up missing frequencies. (ii) Calculate the AM of the completed dis 9 min read Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.5 | Set 1Statistics is a branch of mathematics dealing with data collection, analysis, interpretation, and presentation. It provides methods for summarizing and drawing conclusions from the data which is essential in various fields such as economics, sociology, and science. Chapter 7 of RD Sharma's Class 10 9 min read Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.5 | Set 2Question 11. Find the mean, median, and mode of the following data:Classes0-5050-100100-150150-200200-250250-300300-350Frequency2356531 Solution: Let mean (A) = 175 Classes Class Marks (x) Frequency (f) c.f. di = x - A A = 175 fi * di0-502522-150-30050-1007535-100-300100-150125510-50-250150-200175-A 8 min read Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.6Chapter 7 of RD Sharma's Class 10 textbook focuses on Statistics a branch of mathematics that deals with the collection, analysis, interpretation, and presentation of data. This chapter is crucial for understanding how to summarize and make sense of numerical data in various contexts. The aim is to 8 min read Chapter 8: Quadratic EquationsClass 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.1Question 1. Which of the following are quadratic equations?(i) x2 + 6x - 4 = 0 Solution: As we know that, quadratic equation is in form of: ax2 + bx + c = 0 Here, given equation is x2 + 6x - 4 =0 in quadratic equation. So, It is a quadratic equations. (ii) √3x2 − 2x + ½= 0 Solution: As we know that, 12 min read Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.2Quadratic equations are a fundamental concept in algebra that involves a polynomial of degree two. They play a crucial role in various fields of mathematics, physics, engineering, and many other disciplines. Understanding quadratic equations is essential for solving problems related to motion, area, 5 min read Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.3 | Set 1Solve the following quadratic equations by factorizationQuestion 1. (x - 4) (x + 2) = 0 Solution: We have equation, (x - 4) (x + 2) = 0 Implies that either x - 4 = 0 or x + 2 = 0 Therefore, roots of the equation are 4 or -2. Question 2. (2x + 3) (3x - 7) = 0 Solution: We have equation, (2x + 3) (3x 8 min read Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.3 | Set 2The Quadratic equations are fundamental to algebra and appear frequently in various mathematical contexts. They play a crucial role in understanding polynomial functions and solving real-world problems. In this article, we will delve into Exercise 8.3 | Set 2 from RD Sharma’s Class 10 Mathematics bo 10 min read Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.4Question 1: Find the roots of the following quadratic (if they exist) by the method of completing the square: x^2-4\sqrt{2x}+6=0 . Solution: Given: x^2-4\sqrt{2x}+6=0 We have to make the equation a perfect square. => x^2-4\sqrt{2x} = -6 => x^2 - 2\times2\sqrt{2x} + (2\sqrt{2})^2 = -6 + (2\sqrt 5 min read Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.5Question 1. Find the discriminant of the following quadratic equations:(i) 2x2 - 5x + 3 = 0 Solution: Given quadratic equation: 2x2 - 5x + 3 = 0 ....(1) As we know that the general form of quadratic equation is ax2 + bx + c = 0 ....(2) On comparing eq(1) and (2), we get Here, a = 2, b = -5 and c = 3 15+ min read Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.6 | Set 1Question 1. Determine the nature of the roots of following quadratic equations : (i) 2x² – 3x + 5 = 0 (ii) 2x² – 6x + 3 = 0 (iii) 3/5 x² – 2/3 x + 1 = 0 (iv) 3x² – 4√3 x + 4 = 0 (v) 3x² – 2√6 x + 2 = 0 Solution: (i) 2x² – 3x + 5 = 0 Here a=2, b=-3, c=5 D=b2-4ac=(-3)2-4*2*5 =-9-40=-31 D<0 Roots ar 9 min read Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.6 | Set 2Question 11. If – 5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal-roots, find the value of k. Solution: 2x² + px – 15 = 0 -5 is its one root It will satisfy it 2(-5)2+p(-5)-15=0 2*25-5p-15=0 50-15-5p=0 -5p+35=0 -5p=-35 p=-35/-5=7 Now in 6 min read Class 10 RD Sharma Solutions- Chapter 8 Quadratic Equations - Exercise 8.7 | Set 1Question 1: Find two consecutive numbers whose squares have the sum 85. Solution: Let first number is = x ⇒ Second number = (x+1) Now according to given condition— ⇒ Sum of squares of the numbers = 85 ⇒ x2 + (x+1)2 = 85 ⇒ x2 + x2 + 2x + 1 = 85 [ because (a+b)2 = a2 + 2ab + b2] ⇒ 2x2 + 2x + 1 - 85 = 8 min read Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.7 | Set 2Question 11: The sum of a number and its square is 63/4 , find the numbers. Solution: Let the number is = x so it's square is = x2 Now according to condition- ⇒(number)+(number)2=63/4 ⇒ x+x2 = 63/4 ⇒ x2+x-63/4=0 Multiplying by 4- ⇒ 4x2+4x-63=0 ⇒ 4x2 +(18-14)x - 63 =0 [because 63*4=252 so 18*14=252 5 min read Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.8Chapter 8 of RD Sharma's Class 10 textbook focuses on the Quadratic Equations an essential topic in mathematics that lays the foundation for understanding more advanced algebraic concepts. In this chapter, students will learn how to solve quadratic equations using various methods such as factorizati 15+ min read Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.9The Quadratic equations are a fundamental topic in algebra often introduced at the Class 10 level. They are equations of the form ax2 +bx+c=0 where a, b, and c are constants and x is the variable. Understanding quadratic equations is crucial as they form the basis for many mathematical concepts and 9 min read Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.10Question 1. The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides. Solution: Let the length of other two sides of the triangle be x and y. Therefore, according to the question, x - y = 5 or, x = 5 min read Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.11Question 1: The perimeter of the rectangular field is 82 m and its area is 400 m2. Find the breadth of the rectangle? Solution: Given: Perimeter = 82 m and its area = 400 m2 Let the breadth of the rectangle be 'b' m. As we know, Perimeter of a rectangle = 2×(length + breadth) 82 = 2×(length + b) 41 6 min read Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.12Question 1: A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work. Solution: Let us assume that B takes 'x' days to complete a piece of work. So, B’s 1 day work = 1/x Now, A takes 3 min read Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.13Question 1. A piece of cloth costs ₹ 35. If the piece were 4 m longer and each meter costs ₹ 1 less, the cost would remain unchanged. How long is the piece? Solution: Let us considered the length of piece of cloth = x m Given: The total cost = ₹ 35 So, the cost of 1 m cloth = ₹ 35/x According to the 11 min read Chapter 9: Arithmetic ProgressionsClass 10 RD Sharma Solutions - Chapter 9 Arithmetic Progression Exercise 9.1Problem 1: Write the first terms of each of the following sequences whose nth term are:(i) an = 3n + 2 Solution: Given: an = 3n + 2 By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence a1 = (3 × 1) + 2 = 3 + 2 = 5 a2 = (3 × 2) + 2 = 6 + 2 = 8 a3 = (3 × 3) + 2 = 9 + 2 = 11 a4 = (3 × 4) 7 min read Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.2Problem 1: Show that the sequence defined by an = 5n – 7 is an A.P., find its common difference. Solution: Given: an = 5n – 7 Now putting n = 1, 2, 3, 4,5 we get, a1 = 5.1 – 7 = 5 – 7 = -2 a2 = 5.2 – 7 = 10 – 7 = 3 a3 = 5.3 – 7 = 15 – 7 = 8 a4 = 5.4 – 7 = 20 – 7 = 13 We can see that, a2 – a1 = 3 – ( 7 min read Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.3Problem 1: For the following arithmetic progressions write the first term a and the common difference d:(i) -5, -1, 3, 7, ………… Solution: Given sequence is -5, -1, 3, 7, ………… ∴ First term is -5 And, common difference = a2 - a1 = -1 - (-5) = 4 ∴ Common difference is 4 (ii) 1/5, 3/5, 5/5, 7/5, …… Solut 10 min read Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.4 | Set 1In arithmetic progressions (AP), the terms follow a consistent pattern where each term is derived from the previous one by adding a fixed value, known as the common difference. To analyze various aspects of APs, such as finding specific terms or determining the number of terms, certain formulas and 15+ min read Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.4 | Set 2Question 14. The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference. Solution: Let’s assume that the first term and the common difference of the A.P to be a and d respectively. Given that, 4th term of an A.P. 10 min read Class 10 RD Sharma Solutions- Chapter 9 Arithmetic Progressions - Exercise 9.5Question 1: Find the value of x for which (8x + 4), (6x – 2) and (2x + 7) are in A.P. Solution: Since (8x + 4), (6x – 2) and (2x + 7) are in A.P. Now we know that condition for three number being in A.P.- ⇒2*(Middle Term)=(First Term)+(Last Term) ⇒2(6x-2)=(8x+4)+(2x+7) ⇒12x-4=10x+11 ⇒(12x-10x)=11+4 5 min read Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.6 | Set 1Class 10 RD Sharma Solutions are comprehensive guides to the exercises and problems presented in the RD Sharma Mathematics textbook, which is a popular reference for students studying mathematics at the 10th class level. These solutions provide step-by-step explanations and answers to the questions 15+ min read Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.6 | Set 2Question 25. In an A.P. the first term is 22, nth term is –11 and the sum of first n term is 66. Find n and the d, the common difference. Solution: Given A.P. has first term(a) = 22, nth term(an) = –11 and sum(Sn) = 123. Now by using the formula of sum of n terms of an A.P. Sn = n[a + an] / 2 => 15+ min read Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.6 | Set 3Question 49. Find the sum of first n odd natural numbers. Solution: First odd natural numbers are 1, 3, 5, 7, . . .2n – 1. First term(a) = 1, common difference(d) = 3 – 1 = 2 and nth term(an) = 2n – 1. Now by using the formula of sum of n terms of an A.P. Sn = n[a + an] / 2 So, = n[1 + 2n – 1] / 2 = 15+ min read Like