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## RD Sharma Solutions for class 9

RD Sharma Class 9 Solutions PDF Text book Solutions is based on the latest syllabus prescribed as per the CCE guidelines by CBSE. RD Sharma Class 9 maths textbook is in accordance with the latest syllabus prescribed by CBSE.  Here all solutions to the questions in RD sharma Class 9 textbook is given in a detailed and step by step explation to help students to clear all their doubts.

## RD Sharma Solutions for CBSE Class 9 Maths Chapters

Chapter 1 – Number System

This chapter acquaints us with the number system including rational and irrational numbers, presenting numbers on the number line, rationalising and exponents. The methods and tricks explained through the solutions help students greatly.

Simply put, a number system is a way to represent numbers. We are used to using the base-10 number system, which is also called decimal. Other common number systems include base-16 (hexadecimal), base-8 (octal), and base-2 (binary).

Chapter 2 – Exponents of Real Numbers

This chapter explains topics like integral exponents of a real number, laws of integral exponents and rational exponents of a real number, all solved by our in-house academic experts.

Exponents are used to show repeated multiplication of a number by itself. For example, 7 × 7 × 7 can be represented as 73. Here, the exponent is '3' which stands for the number of times the number 7 is multiplied. 7 is the base here which is the actual number that is getting multiplied.

Chapter 3 – Rationalisation

This chapter provides solutions to identities and rationalisation of the denominator. The examples that are illustrated in the textbook solutions will immensely help kids.

Chapter 4 – Algebraic Identities

This chapter goes deep into the topic of identities, identity for the square of a trinomial, identity for the cube of a binomial, sum and difference of cubes and one more identity.

An algebraic identity is an equality that holds for any values of its variables. For example, the identity ( x + y ) 2 = x 2 + 2 x y + y 2 (x+y)^2 = x^2 + 2xy + y^2 (x+y)2=x2+2xy+y2 holds for all values of x and y.

Chapter 5 – Factorisation of Algebraic Expressions

This chapter explains the concepts of factorisation, factorisation as a sum or difference of two cubes, factorisation using the formulae for the cube of a binomial and factorisation of algebraic expressions.

A number or quantity that when multiplied with another number produces a given number or expression. For example, the factors of 12 are 1, 2, 3, 4, 6 and 12. Any number can be expressed in the form of its factors as shown

Factorise the expression 5(x - 2) - y(2 - x).

(x + 2)(y - 5)

(x - 2)(y - 5)

(x - 2)(5 + y)

(x + 2)(y + 5)

Chapter 6 – Factorisation of Polynomials

Learn topics like zeros of a polynomial, remainder theorem, factor theorem and factorisation of polynomials using the factor theorem.

Trinomials can be factored using a process called factoring by grouping. ... The sum of cubes and the difference of cubes can be factored using equations. Polynomials containing fractional and negative exponents can be factored by pulling out a GCF.

Sum of cubes: a3+b3=(a+b)(a2−ab+b2)a3+b3...

Difference of cubes: a3−b3=(a−b)(a2+ab+b2)a...

Difference of squares: a2−b2=(a+b)(a−b)a2−b...

Chapter 7 – Introduction to Euclid’s Geometry

In this chapter, students learn about axioms and theorems, incidence properties, parallel and intersecting lines, line segment, length axioms and plane. The solutions are explained in a manner that it becomes easy for students to understand.

Euclid's postulates were : Postulate 1 : A straight line may be drawn from any one point to any other point. Postulate 2 :A terminated line can be produced indefinitely. Postulate 3 : A circle can be drawn with any centre and any radius. Postulate 4 : All right angles are equal to one another.

Chapter 8 – Lines and Angles

Here, the discussion of different types of angles and their relations with parallel lines are discussed. The explanations are drafted by in-house experts to facilitate understanding.

Linear Pair of Angles: When the sum of two adjacent angles is 180°, then they are called a linear pair of angles. (i) If a ray stands on a line, then the sum of two adjacent angles so formed is 180°. (ii) If the sum of two adjacent angles is 180°, then a ray stands on a line (that is the non-common arms form a line).

Chapter 9 – Triangle and its Angles

Triangles and the concepts associated with it are discussed here. This chapter covers types of angles basis angles and sides along interior and exterior angles.

A triangle has three sides, three angles, and three vertices. The sum of all internal angles of a triangle is always equal to 180°. This is called the angle sum property of a triangle. ... Any exterior angle of the triangle is equal to the sum of its interior opposite angles.

Chapter 10 – Congruent Triangles

The primary objective of this chapter is to introduce important concepts like congruence of line segments, angles, triangles, congruence criteria and inequality relations.

Triangles that have exactly the same size and shape are called congruent triangles. The symbol for congruent is ≅. Two triangles are congruent when the three sides and the three angles of one triangle have the same measurements as three sides and three angles of another triangle.

Chapter 11 – Co-ordinate Geometry

Learn about the rectangular or Cartesian co-ordinates of a point and plotting of points. All the exercises mentioned in the textbook are illustrated properly by academic experts.

Coordinate Geometry (or the analytic geometry) describes the link between geometry and algebra through graphs involving curves and lines. It provides geometric aspects in Algebra and enables them to solve geometric problems.

Chapter 12 – Heron’s Formula

Heron’s formula and its applications are explained at length in this chapter. Our experienced experts explain problem-solving for this chapter using the best possible methods.

For a quadrilateral, when one of its diagonal value and the sides are given, the area can be calculated by splitting the given quadrilateral into two triangles and use the Heron's formula. Example :A park, in the shape of a quadrilateral ABCD, has ∠C=90?, AB = 9 cm, BC = 12 cm, CD = 5 cm and AD = 8 cm.

Chapter 13 – Linear Equations in Two Variables

Linear equations are discussed in this chapter. The topics of linear equations in two variables, solution of a linear equation, graph of a linear equation in two variables and equations of lines parallel to the X-axis and Y-axis have been explained.

Summary. An equation of the form ax + by + c = 0, where a, b and c are real numbers, such that a and b are not both zero, is called a linear equation in two variables. The equation is called linear because the equation is of the first degree. A solution is an ordered pair of real numbers which satisfies the equation.

This chapter explains important concepts like quadrilateral, types of quadrilaterals, angle sum property, properties of a parallelogram, rectangle, rhombus and square and useful facts about the triangle.

Square Rectangle
All sides are equal Yes No
Opposite sides are parallel Yes Yes
Opposite sides are equal Yes Yes
All the angles are of the same measure Yes Yes

Chapter 15 – Areas of Parallelograms and Triangles

This chapter explains figures on the same base and between the same parallels, polygonal regions and area axioms. Our academic experts have also used some simple formulas to help students remember the concept.

Parallelograms on the same base (or equal base) and between the same parallels are equal in area. Triangles on the same base (or equal bases) and between the same parallels are equal in area. Area of a parallelogram = base × height. A median of a triangle divides it into two triangles of equal area.

Chapter 16 – Circles

This chapter focuses on circles and all important concepts around circles. It deals with arcs of a circle, chord and segment of a circle, congruence of circles and arcs, some results on equal chords, arcs and angles subtended by them and cyclic quadrilaterals.

Circles are geometric figures whose points all lie the same distance from a given point, the circle's center. They are not polygons, because they are not made up of segments. Points that lie in the same line, like those in a segment, are never equidistant (an equal distance) from a single point.

Chapter 17 – Constructions

In this chapter, students discover basic constructions, construction of standard angles and construction of triangles. Everything is explained in a detailed manner to assist students in understanding the concepts better.

The most-used straightedge and compass constructions include:

• Constructing the perpendicular bisector from a segment.
• Finding the midpoint of a segment.
• Drawing a perpendicular line from a point to a line.
• Bisecting an angle.
• Mirroring a point in a line.
• Constructing a line through a point tangent to a circle.

Chapter 18 – Surface Area and Volume of a Cuboid and a Cube

In this chapter, students learn the concepts of the units of measurement of area and volume, cuboid and cube, surface area and lateral surface area of a cuboid and a cube and volume of a cuboid.

Cube and Cuboid Formulas

Cube Cuboid
Total Surface Area = 6(side)2 Total Surface area = 2 (Length x Breadth + breadth x height + Length x height)
Lateral Surface Area = 4 (Side)2 Lateral Surface area = 2 height(length + breadth)
Volume of cube = (Side)3 Volume of the cuboid = (length × breadth × height)

Chapter 19 – Surface Area and Volume of a Right Circular Cylinder

This chapter focuses on problems and formulas used in determining the surface area and volume of a right circular cylinder.

The radius and height of a right cylinder are given as 5 m and 6.5 m respectively. Find the volume and total surface area of the right cylinder. Hence, the volume of the given right cylinder is 510.25 cubic m. Hence, the total surface area of the given right cylinder is 361.1 m2.

Chapter 20 – Surface Area and Volume of a Right Circular Cone

The primary concepts of this chapter are right circular cone and its surface area and volume. Students can easily understand the solutions created by our academic experts and score better marks.

Right Circular Cone Formula

For a right circular cone of radius 'r', height 'h' and slant height 'l', we have; Curved surface area of right circular cone = π r l. Total surface area of a right circular cone = π(r + l) r. Volume of a right circular cone = 1/3π r2 h.

Chapter 21 – Surface Area and Volume of a Sphere

A sphere is a three-dimensional object which has volume and surface area. Sphere, section of a sphere by a plane, surface area and volume of a sphere are discussed here.

A sphere is a solid figure that exists in three-dimensional space. Spheres consist of all the points that are equidistant from a center point. Every point on the sphere is the distance of the radius from the center. ... When thinking about spheres, we will be working with surface area and volume.

Chapter 22 – Tabular Representation of Statistical Data

Students understand the steps that are needed to represent data in the tabular form. They will also learn more about statistics, statistical data, presentation of data, frequency distribution, construction of a frequency distribution table and cumulative frequency distribution.

What is Tabular Presentation of Data? It is a table that helps to represent even a large amount of data in an engaging, easy to read, and coordinated manner. The data is arranged in rows and columns. This is one of the most popularly used forms of presentation of data as data tables are simple to prepare and read.

Chapter 23 – Graphical Representation of Statistical Data

The graphical representation of data, the bar graph, the histogram and the method of constructing a frequency polygon have been explained in this chapter.

Graphic representation is another way of analysing numerical data. A graph is a sort of chart through which statistical data are represented in the form of lines or curves drawn across the coordinated points plotted on its surface. Graphs enable us in studying the cause and effect relationship between two variables.

Chapter 24 – Measures of Central Tendency

Through this chapter, students learn basic concepts like measures of central tendency, the arithmetic mean of grouped data and the median.

A measure of central tendency is a single value that attempts to describe a set of data by identifying the central position within that set of data. The mean (often called the average) is most likely the measure of central tendency that you are most familiar with, but there are others, such as the median and the mode.

Chapter 25 – Probability

This is an important chapter as it is required in higher classes also. This chapter discusses probability and operations at length.

The probability of an event is the ratio between the number of outcomes in the event set and the number of possible outcomes in the sample space. The probability of the union of two events is calculated using: P(A∪B)=P(A)+P(B)−P(A∩B). Mutually exclusive events are two events that cannot occur at the same time.

## Some questions are asked by class 9 students

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Aspirants are advised to check our website studentcare.com to get all chapters RD Sharma Class 9 Solutions for free.

4. Is solving RD Sharma Class 9 Solutions enough for board exam preparation?

Why should I study from RD Sharma solutions?

RD Sharma solutions provide answers to each and every textbook question, which is a good way to get acquainted with the chapter.

Will CBSE Class 9 Maths help me in other exams?

The topics and concepts that you will study now will come in handy in other competitive exams such as JEE, IAS and NDA.

Is it important to practise Maths?

Since Maths is an application-based subject, it cannot be memorised. Therefore, it is essential that you practise Maths every day, especially the difficult chapters.

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If you have any doubts regarding any topic, you can leave your doubt in the ‘Doubts and Solutions’ section. Your query will be answered within 24 hours by our experts.

Why are RD Sharma Solutions important?

RD Sharma is one of the greatest Mathematics professors largely followed by CBSE students as his numerical skills are fantastic. And if you are looking for the best solutions for the RD Sharma Mathematics textbook for CBSE Class 9, then our RD Sharma textbook solutions will be ideal for you. Student care provides RD Sharma solutions which will help you study all the topics in detail. In CBSE Class 9, there are chapters like Rationalisation, Polynomials, Lines & Angles and several others which are tricky and take a lot of time to understand.

Mathematics is a scoring subject, and effective preparation is a must. CBSE Class 9 is important for students, as they will start learning new concepts and applications in Maths. RD Sharma textbooks are beneficial, as they are created according to the latest norms of the CBSE syllabus. So, referring to RD Sharma textbook solutions can help CBSE Class 9 students in scoring more marks.

Question
A coin is tossed 1000 times with the following frequencies
Head : 455, Tail : 545.
Compute the probability for each event.
Solution:
Total number of events (m) 1000
(i) Possible events (m) 455
∴Probability P(A) = mn = 4551000
= 91200 = 0.455
(ii) Possible events (m) = 545
∴Probability P(A) = mn = 5451000 = 109200 = 0.545

Question
Two coins are tossed simultaneously 500 times with the following frequencies of different
outcomes:
One tail : 290 times
Find the probability of occurrence of each of these events.
Solution:
Two coins are tossed together simultaneously 500 times
∴ Total outcomes (n) 500
(i) 2 heads coming (m) = 95 times
∴Probability P(A) = mn
= No.ofpossibleeventsTotalnumberofevents
= 95500 = 19100 = 0.19
(ii) One tail (m) = 290 times
∴Probability P(A) = mn = 290500 = 5801000 = 58100 = 0.58
(iii) No head (m) = 115 times
∴Probability P(A) = mn = 115500 = 23100 = 0.23

Question
Three coins are tossed simultaneously 1oo times with the following frequencies of different outcomes:

If the three coins are simultaneously tossed again, compute the probability of:
(iii) at least one head coming up.
(iv) getting more heads than tails.
(v) getting more tails than heads.
Solution:
Three coins are tossed simultaneously 100 times
Total out comes (n) = 100

(i) Probability of 2 heads coming up (m) = 36
Probability P(A) = mn = 36100 = 0.36
(ii) Probability of 3 heads (m) = 12
ProbabilityP(A)= mn = 12100 = 0.12
(iii) Probability of at least one head coming up (m) = 38 + 36 + 12 = 86
Probability P(A) = mn = 86100 = 0.86
(iv) Probability of getting more heads than tails (m) = 36 + 12 = 48
Probability P(A) = mn = 48100 = 0.48
(v) Getting more tails than heads (m) = 14 + 38 = 52
Probability P(A) = mn = 52100 = 0.52

Question
Is zero a rational number? Can you write it P in the form pq , where p and q are integers and q ≠ 0? [NCERT]
Solution:
Yes, zero is a rational number e.g. Question
Find five rational numbers between 1 and 2. [NCERT]
Solution:
We know that one rational number between two numbers a and b = a+b2
Therefore one rational number between 1 and 2   Question
Find six rational numbers between 3 and 4. [NCERT]
Solution:
One rational number between 3 and 4   Question
Find five rational numbers between 35 and 45
Solution: Question
Are the following statements true or false?
(i) Every whole number is a natural number. [NCERT]
(ii) Every integer is a rational number.
(iii) Every rational number is an integer.
(iv) Every natural number is a whole number,
(v) Every integer is a whole number.
(vi) Every rational number is a whole number.
Solution:
(i) False, as 0 is not a natural number.
(ii) True.
(iii) False, as 12, 13 etc. are not integers.
(iv) True.
(v) False, ? negative natural numbers are not whole numbers.
(vi) False, ? proper fraction are not whole numbers

Question
Simplify the following: Solution:  Question
If a = 3 and b =-2, find the values of:
(i) aa+ bb
(ii) ab + ba
(iii) (a+b)ab
Solution: Question
Prove that: Solution:  Question
Simplify each of the following: Solution:  Question
Simplify the following expressions:
(i)  (4 + 7–√) (3 + 2–√)
(ii) (3 + 3–√ )(5- 2–√)
(iii) (5–√ -2)(3–√ – 5–√)
Solution: Question
Evaluate each of the following using identities:
(i) (2x –1x)2
(ii)  (2x + y) (2x – y)
(iii) (a2b – b2a)2
(iv) (a – 0.1) (a + 0.1)
(v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2)
Solution:  Question
Evaluate each of the following using identities:
(i) (399)2
(ii) (0.98)2
(iii) 991 x 1009
(iv) 117 x 83
Solution: Question
Define a trial.
Solution:
When we perform an experiment, it is called a trial of the experiment.

Question
Define an elementary event.
Solution:
An outcome of a trial of an experiment is called an elementary event.

Question
Define an event.
Solution:
An event association to a random experiment is said to occur in a trial.

Question
Define probability of an event.
Solution:
In n trials of a random experiment if an event A happens m times, then probability of happening
of A is given by P(A) = mn

Question
A bag contains 4 white balls and some red balls. If the probability of drawing a white ball from the bag is 25, find the number of red balls in the bag
Solution:
No. of white balls = 4
Let number of red balls = x
Then total number of balls (n) = 4 white + x red = (4 + x) balls

Question
Which one of the following is a correct statement?
(a) Decimal expansion of a rational number is terminating
(b) Decimal expansion of a rational number is non-terminating
(c) Decimal expansion of an irrational number is terminating
(d) Decimal expansion of an irrational number is non-terminating and non-repeating
Solution:
Decimal expansion of an irrational number is non-terminating and non-repeating . (d)

Question
Which one of the following statements is true?
(a) The sum of two irrational numbers is always an irrational-number
(b) The sum of two irrational numbers is always a rational number
(c) The sum of two irrational numbers may be a rational number or an irrational number
(d) The sum of two irrational numbers is always an integer
Solution:
The sum of two irrational numbers may be a rational number or an irrational number (c)

Question
Write (625)–1/4 in decimal form.
Solution: Question
State the product law of exponents:
Solution:
xm x xn = xm +n

Question
Write the value of (2 + 3–√ ) (2 – 3–√).
Solution:
(2+ 3–√ )(2- 3–√ ) = (2)2-(3–√ )2
{? (a + b) (a – b) = a2 – b2}
= 4-3=1

Question
Write the reciprocal of 5 + 2–√.
Solution: Question
2a2 + 26–√ ab +3b2
Solution:
2a2 + 26–√  ab +3 b2
= (2–√ a)2+ 2–√ a x 3–√ b+ (3–√ b)2
= (2–√a + 3–√ b)2

Question
a2 + b2 + 2(ab + bc + ca)
Solution:
a2 + b2 + 2(ab + bc + ca)
= a2 + b2 + 2 ab + 2 bc + 2 ca
= (a + b)2 + 2c(b + a)
= (a + b)2 + 2c(a + b)
= (a + b) (a + b + 2c)

Question
4(x – y)2 – 12(x -y) (x + y) + 9(x + y)2
Solution:
4(x – y)2 – 12(x – y) (x + y) + 9(x + y)2
= [2(x – y)2 + 2 x 2(x – y) x 3(x + y) + [3 (x+y]2        {? a2 + b2 + 2 abc = (a + b)2}
= [2(x – y) + 3(x + y)]2
= (2x-2y + 3x + 3y)2
= (5x + y)2

Question
a2 – b2 + 2bc – c2
Solution:
a2 – b2 + 2bc – c2
= a2 – (b2 – 2bc + c2)                                           {? a2 + b2 – 2abc = (a – b)2}
= a2 – (b – c)2
= (a)2 – (b – c)2          {? a2 – b2 = (a + b) (a – b)}
= (a + b – c) (a – b + c)

Question
xy9 – yx9
Solution:
xy9 – yx9 = xy(y8 – x8)
= -xy(x8 – y8)
= -xy[(x4)2 – (y4)2]
= -xy (x4 + y4) (x4 – y4)                                         {? a2-b2 = (a + b) (a – b)}
= -xy (x4 + y4) {(x2)2 – (y2)2}
= -xy(x4 + y4) (x2 + y2) (x2 – y2)
= -xy (x4 +y4) (x2 + y2) (x + y) (x -y)
= -xy(x – y) (x + y) (x2 + y2) (x4 + y4)

Question
Write the degrees of each of the following polynomials:
(i) 7x3 + 4x2 – 3x + 12
(ii) 12 – x + 2x3
(iii) 5y – 2–√
(iv) 7
(v) 0
Solution:
(i) Degree of the polynomial 7x3 + 4x2 – 3x + 12 is 3
(ii) Degree of the polynomial 12 – x + 2x3 is 3
(iii) Degree of the polynomial 5y – 2–√is 1
(iv) Degree of the polynomial 7 is 0
(v) Degree of the polynomial 0 is 0 undefined.

Question
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer: Solution:
(i) 3x2 – 4x + 15,
(ii) y2 + 23–√ are polynomial is one variable. Others are not polynomial or polynomials in one variable.

Question
Write the coefficient of x2 in each of the following: Solution:
Coefficient of x2,
in (i) is 7
in (ii) is 0 as there is no term of x2 i.e. 0 x2 Question
Write the degrees of each of the following polynomials:
(i) 7x3 + 4x2 – 3x + 12
(ii) 12 – x + 2x3
(iii) 5y – 2–√
(iv) 7
(v) 0
Solution:
(i) Degree of the polynomial 7x3 + 4x2 – 3x + 12 is 3
(ii) Degree of the polynomial 12 – x + 2x3 is 3
(iii) Degree of the polynomial 5y – 2–√is 1
(iv) Degree of the polynomial 7 is 0
(v) Degree of the polynomial 0 is 0 undefined.

Question
(i) x + x2 + 4
(ii) 3x – 2
(iii) 2x + x2
(iv) 3y
(v) t2 + 1
(v) 7t4 + 4t3 + 3t – 2
Solution:
(i)  x + x2 + 4 It is a quadratic polynomial.
(ii) 3x – 2 : It is a linear polynomial.
(iii) 2x + x2: It is a quadratic polynomial.
(iv) 3y It is a linear polynomial.
(v) t2+ 1 It is a quadratic polynomial.
(vi) 7t4 + 4t3 + 3t – 2 It is a biquadratic polynomial.

Question
Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) -2x + 3y = 12
(ii) x-y2 -5 = 0
(iii) 2x + 3y = 9.35
(iv) 3x = -7y
(v) 2x + 3 = 0
(vi) y – 5 = 0
(vii) 4 = 3x
(viii) y = x2
Solution:
(i) -2x + 3y = 12
⇒  -2x + 3y – 12 = 0
Here a -2, b = 3, c = – 12
(ii) x – y2 -5 = 0
Here a = 1, b =12 ,c = -5
(iii) 2x + 3y = 9.35
⇒  2x + 3y – 9.35 = 0
Here a = 2, b = 3, c = – 9.35
(iv) 3x = -7y
⇒  3x + 7y + 0 = 0
Here a = 3, b = 7,c = 0
(v) 2x + 3 = 0
⇒ 2x + 0y + 3 = 0
Here a = 2, b = 0, c = 3
(vi) y-5 = 0 ⇒ ox+y-5 = 0
Here a = 0, b = 1, c = -5
(vii) 4 = 3x
⇒ 3x – 4 = 0
⇒ 3x + 0y – 4 = 0
Here a = 3, b = 0, c = -4
(Viii) y= x2
⇒  x2 – y+ 0 = 0
⇒  x-2y + 0 = 0
Here a = 1, y = -2, c = 0

Question
Plot the following points on the graph paper:
(i)  (2, 5)
(ii) (4, -3)
(iii) (-5, -7)
(iv) (7, -4)
(v) (-3, 2)
(vi) (7, 0)
(vii) (-4, 0)
(viii) (0, 7)
(ix) (0, -4)
(x) (0, 0)
Solution:
The given points have been plotted on the graph as given below:

Question
Write the coordinates of each of the following points marked in the graph paper.

Solution:
The co-ordinates of the points given in the graph are A (3, 1), B (6, 0), C (0, 6), D (-3, 0), E (-4, 3), F (-2, -4), G (0, -5), H (3, -6), P (7, -3).

Question 1.
In a ?ABC, if ∠A = 55°, ∠B = 40°, find ∠C.
Solution:
? Sum of three angles of a triangle is 180°
∴ In ?ABC, ∠A = 55°, ∠B = 40°
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle) ⇒ 55° + 40° + ∠C = 180°
⇒ 95° + ∠C = 180°
∴ ∠C= 180° -95° = 85°

Question
If the angles of a triangle are in the ratio 1:2:3, determine three angles.
Solution:
Ratio in three angles of a triangle =1:2:3
Let first angle = x
Then second angle = 2x
and third angle = 3x
∴ x + 2x + 3x = 180° (Sum of angles of a triangle)
⇒6x = 180°
⇒x = 180?6  = 30°
∴ First angle = x = 30°
Second angle = 2x = 2 x 30° = 60°
and third angle = 3x = 3 x 30° = 90°
∴ Angles are 30°, 60°, 90°

Question
Three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angle.
Solution:
Sum of four angles of a quadrilateral = 360°
Three angles are = 110°, 50° and 40°
∴ Fourth angle = 360° – (110° + 50° + 40°)
= 360° – 200° = 160°

Question
In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1 : 2 : 4 : 5. Find the measures of each angle of the quadrilateral.
Solution:
Sum of angles of a quadrilateral ABCD = 360°
Ratio in angles = 1 : 2 : 4 : 5
Let first angle = x
Second angle = 2x
Third angle = 4x
and fourth angle = 5x
∴ x + 2x + 4x + 5x = 360°
⇒ 12x = 360° ⇒ x = 360?12  = 30°
∴ First angle = 30°
Second angle = 30° x 2 = 60°
Third angle = 30° x 4 = 120°
Fourth angle = 30° x 5 = 150°

1. If a Cubiod has length l, breadth b, and height h, then
• Perimetof Cubiod = 4(l + b + h)
• Surface area of the cubiod = 2(lb+bh+lh)
• Lateral surface area of the cubiod = 2(l+b)xh
•
• Volume of the Cubiod = lbh
2. If the length of each edge of a cube is l, then
• Perimeter of the cube = 12l = 12 (Edge)
• Surface area of the Cube = 6l2 = 6 (Edge)2
• Lateal surface area of the cube = 4l2 =4(Edge)2

Volume of the cube =l3 = (Edge)3

Question
Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height. [NCERT]
Solution:
Curved surface area of a cylinder = 4.4 m2 Question
In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. [NCERT]
Solution:
Diameter of the pipe = 5 cm Question
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of 12.50 per m2. [NCERT]
Solution:
Diameter of cylindrical pillar = 50 cm Question
It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same? [NCERT]
Solution:
Height of cylinder (h) = 1 m = 100 cm
Diameter of box = 140 cm  Question
If the heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively. Find the mean height.
Solution:
Heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm

Question
Find the mean of 994, 996, 998, 1002 and 1000.
Solution:
Mean of 994, 996, 998, 1002 and 1000

Question
Find the mean of first five natural numbers.
Solution:
First five natural numbers are 1, 2, 3, 4, 5
∴ Mean = x¯¯¯=1+2+3+4+55=155 = 3

Question
Find the mean all factors of 10.
Solution:
Factors of 10 = 1, 2, 5, 10
∴ Mean = x¯¯¯=1+2+5+104=184 = 4.5

Question
A coin is tossed 1000 times with the following frequencies
Head : 455, Tail : 545.
Compute the probability for each event.
Solution:
Total number of events (m) 1000
(i) Possible events (m) 455
∴Probability P(A) = mn = 4551000
= 91200 = 0.455
(ii) Possible events (m) = 545
∴Probability P(A) = mn = 5451000 = 109200 = 0.545

Question
Two coins are tossed simultaneously 500 times with the following frequencies of different
outcomes:
One tail : 290 times
Find the probability of occurrence of each of these events.
Solution:
Two coins are tossed together simultaneously 500 times
∴ Total outcomes (n) 500
(i) 2 heads coming (m) = 95 times
∴Probability P(A) = mn
= No.ofpossibleeventsTotalnumberofevents
= 95500 = 19100 = 0.19
(ii) One tail (m) = 290 times
∴Probability P(A) = mn = 290500 = 5801000 = 58100 = 0.58
(iii) No head (m) = 115 times
∴Probability P(A) = mn = 115500 = 23100 = 0.23