# RD Sharma Class 9 Solution

## RD Sharma Class 9 Solution Mathematics is the basic building block of all other subjects or fields. Without maths nothing is possible and hence to build your basics concepts, more examples with solutions of class 9 RD Sharma solutions are the best solutions ever. In this RD Sharma solutions, most difficult numerical are solved with easy methods with detailed explanation. We know that, “practice makes man perfect” and hence in this solutions most of practice examples are solved based on the different topics given in syllabus. It is very helpful to revise the syllabus for students before exam as it involves step by step solutions with valid reasons also. If you are referring this RD Sharma solutions then definitely you can score best marks in Mathematics in grade 10 also. Because of all above reasons it is better to refer RD Sharma solutions for practicing and knowing different methods how to solve the numerical through different ways.

### Chapter 1: Number System

In this chapter of RD Sharma solutions, all about rational numbers and irrational numbers is included. Mainly it includes how to find the rational numbers between any two numbers, how to express rational numbers in decimal form. Also it includes the methods how to identify the given numbers are rational or not. Mainly, the important topic with example explained here is how to represent the rational numbers on number line accurately. And at last one method is given how to visualise any numbers on the number line using successive magnification method. Many examples are solved here to understand in detail.

### Chapter 2: Exponents of Real Numbers

In this chapter of RD Sharma solutions exponential, their properties with examples are explained basically. Some important formulae involved here are given below:

• (am)n = am*n
• (a*b)n= a3*bn
• a(m+n)=am*an
• 1/am = a-m
• (a/b)n= an/bn
• a0= 1

Most of the complicated examples based on this properties are solved step by step in this RD Sharma solutions. How to simplify any bigger or complicated exponents easily using exponential properties is explained in detail here. This practice examples given in RD Sharma solutions are more sufficient to understand all the properties of exponents.

### Chapter 3: Rationalisation

In this chapter of RD Sharma solutions, exponential properties based examples are given and which are solved mainly by the method of Rationalisation. Many different types of examples are given to understand the properties in detail. Rationalisation method is used only when there is rational number is present in the denominator. Rationalisation method is the easiest method to simplify given number. This RD Sharma solutions involves examples from basic to some applied part also. So over all this practice questions may completely clear all the doubts of students about exponential.

### Chapter 4: Algebraic Identities

• In this chapter of RD Sharma solutions, algebraic identities, their properties and how to solve given questions using these identities is explained in detail. Some of the important identities given here are as below:
• (a – b)2 = a2 – 2ab + b2
• (a + b)2 = a2 + 2ab + b2
• (a + b)*(a – b) = a2 – b2
• (a +b)3 = a3 + b3 + 3ab*(a + b)
• (a – b)3 = a3 – b3 – 3ab*(a – b)
• And also, how to simplify given complicated algebraic expressions using identities is given here with solutions of examples. Some examples are like how to expand the given algebraic expression, how to find the product of given algebraic expression is also explained here in detail.

Chapter 4 Solution

### Chapter 5: Factorisation of Algebraic Expressions

In this chapter of RD Sharma solutions, the properties of Algebraic Identities explained in previous chapter are taken to factorise the given algebraic expression. Many examples are solved here step by step to understand more easily. Here, also to factorise the given algebraic expression some algebraic identities are used effectively. So because of extra examples solved in this RD Sharma solutions students may clarify their doubts easily through practice only.

### Chapter 6: Factorisation of polynomials

In this chapter of RD Sharma solutions, polynomials, definition, types of polynomials, properties of polynomials and examples based on it is explained basically. How to identify the coefficient, degree and type of given polynomial is explained through different examples. Polynomials and their classification, also examples based on actual division of polynomials are solved here. Many examples are based on how to identify the factors of given polynomial.

Thus, in this RD Sharma solutions all the basic concepts which are needed in grade 10 about polynomials are cleared by giving solutions of different examples.

### Chapter 7: Introduction to Euclid’s Geometry

In this chapter of RD Sharma solutions, all about basic concepts of geometry like line, line segment, collinear points, parallel lines, intersecting lines, concurrent lines, ray are explained with diagram. Also, through various conditions these geometrical concepts are explained here by giving meaningful diagrams which build basic concepts of geometry.

### Chapter 8: Lines and Angles

In this chapter of RD Sharma solutions, examples based on the concepts like angles, types of angles are given with the help of diagram. The concept included hare are like acute angle, obtuse angle, right angle, supplementary angles, complementary angles, adjacent angles and their properties also. In this RD Sharma solutions with the help of diagram mostly the solutions are given which helps to students to understand more easily. Various concepts based examples with solutions are given here so that students build their knowledge about angels which is necessary in higher classes also.

### Chapter 9: Triangles and it’s Angles

In this RD Sharma solutions, the examples based on types of triangles, properties of triangles are solved by easy and time saving methods. If some information about triangles like exterior or interior angles is given then how to find other properties is explained through various examples. This chapter solutions are explained and clarified with the help of well labelled diagram also which helps to understand more and easily.

### Chapter 10: Congruent Triangles

RD Sharma solutions are the best solution for geometry chapters also. Because in this solutions with diagram all the examples are solved. Which triangles are said to be congruent is given with help of solutions of many questions. The view of students seeing towards such geometrical problem is enhanced here through more practice questions.In this RD Sharma solutions the examples are solved which includes the basics from previous chapter also. Let’s practice the question from RD Sharma to build your confidence in geometry.

### Chapter 11: Co-ordinate Geometry

In this chapter of RD Sharma solutions, problems are solved which are based on plotting the coordinates in the Cartesian plane and also how to find the coordinates of given points which are already plotted on Cartesian plane are explained. Easy and simple methods are given here which are easily understandable.

### Chapter 12: Heron’s Formula

In this solutions, the numerical are solved of finding area of a triangle if there sides of a triangle are given by Heron’s formula. The Heron’s formula is given by

A(∆) = √[s(s-a)*(s-b)*(s-c)]

Where, a, b and c are the lengths of three sides of a triangle and s is the semi perimeter. While solving the examples most of the properties of geometrical figures like rhombus, trapezium, quadrilaterals, triangles are explained. Also, how to use Pythagoras theorem is also given through many practice questions.

### Chapter 13: Linear equations in Two variables

We already know that this chapter is the basic of algebra, and hence in this RD Sharma solutions all examples from basic to complicated are solved by giving valid reasons. If two equations are given how to find their solutions, how to find nature of the lines and many thing are covered through this solutions. Again some equations are given from that how to plot graph of such lines js explained with colourful graphs here which may clear all doubts of students about graphical representation. Also, how to locate given equations on number line and in Cartesian plane is also explained Graphically which builds your confidence about plotting any equation in Cartesian plane. Many practice questions are solved here also. Go through this solutions and be perfect in plotting equations on graphs.

In this chapter of RD Sharma solutions, problems based on different types quadrilaterals, their properties are solved to improve concepts of different quadrilaterals. Inside quadrilaterals, some triangle are so there so how to solve that problem with given data is given in more precise manner. Most of the examples in which mid point theorem is used to explain better. Also congruence properties of triangles are used to solve given n examples.

### Chapter 15: Area of Parallelogram and Triangles

In this chapter of RD Sharma solutions, problems based on area of Parallelogram and Triangles are solved more efficiently. Mostly, here properties of triangles and quadrilaterals are used to find the areas directly from the given conditions. This are the best solution for geometry ever. Let’s solve examples and enhance your knowledge of geometry.

### Chapter 16: Circles

In this chapter of RD Sharma solutions, problems related to circles, conceptual examples are solved with colourful diagrams. Also, here diagrammatic explanation is given clearly and many practice examples are solved with valid reasons. If you are going through this RD Sharma solutions it will confirm that all concepts will be cleared only through this solutions.

### Chapter 17: Construction

In this exercise solutions, examples are solved like bisecting an angle, how to draw angle of given measure accurately, how to draw a perpendicular bisector, how to form a triangle of given measurement is explained step by step only in this RD Sharma solutions. While drawing such diagrams you need such stepwise guidance with basics is only given here in this RD Sharma solutions.

### Chapter 18: Surface Areas and Volume of Cuboid and Cone

In this chapter of RD Sharma solutions, examples based on finding the area and volume of Cuboid and Cone are given. Some formulae which involved here are as given below:

• Lateral surface area of cube = 4*(side)2
• Total surface area of a cube = 6*(side)2
• Volume of a cube= (side)3
• Lateral surface area of Cuboid = 2*(length + breadth)*height
• Total surface area of a cuboid = 2*(lb + bh + lh)
• Volume of a cuboid= l*b*h

Many exercise are given in which different types of examples are solved to build your basic about cube, cuboid and their properties. Practice these questions so that idea about cube and cuboid get cleared totally.

### Chapter 19: Surface Area and volume of right circular cylinder

In this chapter of RD Sharma solutions, problems based on how to find the area and volume of right circular cylinder are given. The three dimensional geometrical shapes that is right circular cylinder and it’s properties are explained by solving many questions here. Some basic formulae are given here which are as given below:

• Curved surface area of a cylinder = 2*πrh
• Total surface area of a cylinder= 2πr*(h + r)
• Volume of a cylinder = π*r2*h

Here examples of cylinder like cylindrical tank, cylindrical pipe, cylindrical pillar, are solved though different types which makes easy to understand. Many practice questions are given based on volume of cylinder also which are related to daily life examples and it’s applied part also.

### Chapter 20: Surface area and volume of right circular cone

In this RD Sharma solutions, the examples of finding area and volume of right circular cone are given. If we know the basic formulae then we can directly find the volume and area of right circular cone. Some formulae which are given here are:

• Curved surface area of right circular cone= π*rl
• Total surface area of a cone = πr*(l + r)
• Volume of a cone = 1/3*π*r2*h

So many practice questions are given here with valid solutions which may increase your confidence to solve daily life examples or you can directly apply this tricks of finding area and volume of right circular cone.

### Chapter 21 : Surface Area and Volume of a Sphere

In this RD Sharma solutions, problems based on surface area and volume of a Sphere are given which involves mostly daily life examples. These type of examples and their practice may increase your confidence to solve hot type questions also. Hemisphere, sphere and their properties are explained by giving more and more examples with valid solutions also. Here the direct formulae used are as given below:

Surface area of sphere = 4πr2

Volume of a Sphere= 4/3*πr3

Surface area of hemisphere= 2πr2

Total surface area of hemisphere= 3πr2

### Chapter 22: Tabular representation of statistical data

In this RD Sharma solutions, exercise are based on the basic concepts of statistics, and the characteristics of statistics also. It includes primary data, secondary data, group data, variate class interval, class size, class mark, frequency, class limits like concepts. Examples given are based on statistical data from which frequency table is prepared. So you get all basic clarity about statistics through this exercise.

### Chapter 23: Graphical representation of Statistical Data

In this type of exercise solutions, problems based on statistical data are given which are represented Graphically with the help of symbols or colourful pictures which makes more interesting and easy to understand also. Bar graphs with particular shades so that you easily distinguish it from other. Proper labelling and solutions of each example is explained with the help of graphs. Histogram examples are also solved here effectively.

### Chapter 24: Measure of Central Tendency

In this type of solutions of RD Sharma, if some data is given from that how to find mean directly with short tricks is also explained. In some examples distribution table is given already from that here mean is determined correctly. Also, if data and mean is given then how to find the missing value is described very well in this exercise. RD Sharma solutions, are the best solution for such type of problems in which median, mode, examples are also solved. More and more practice questions are also solved with valid solutions.

Mean= (sum of all quantities)/ total number of quantities

Median = [n/2th value + (n/2 + 1)th value]/ 2

Mode= 3(median) – 2(mode)

### Chapter 25: Probability

In this chapter exercise solutions, how to find probability if some outcomes are given is explained clearly. Most important examples with their solutions are given in this RD Sharma solutions. Basics about probability are cleared here through solutions.