Bhagya

JAC Jharkhand Board Class 12th History Solutions in Hindi & English Medium

JAC Board Class 12th History Solutions in Hindi Medium

• Chapter 1 ईंटें, मनके तथा अस्थियाँ : हड़प्पा सभ्यता
• Chapter 2 राजा, किसान और नगर : आरंभिक राज्य और अर्थव्यवस्थाएँ
• Chapter 3 बंधुत्व, जाति तथा वर्ग : आरंभिक समाज
• Chapter 4 विचारक, विश्वास और इमारतें : सांस्कृतिक विकास
• Chapter 5 यात्रियों के नज़रिए : समाज के बारे में उनकी समझ
• Chapter 6 भक्ति-सूफी परंपराएँ : धार्मिक विश्वासों में बदलाव और श्रद्धा ग्रंथ
• Chapter 7 एक साम्राज्य की राजधानी : विजयनगर
• Chapter 8 किसान, ज़मींदार और राज्य : कृषि समाज और मुगल साम्राज्य
• Chapter 9 शासक और इतिवृत्त : मुगल दरबार
• Chapter 10 उपनिवेशवाद और देहात : सरकारी अभिलेखों का अध्ययन
• Chapter 11 विद्रोही और राज : 1857 का आंदोलन और उसके व्याख्यान
• Chapter 12 औपनिवेशिक शहर : नगर-योजना, स्थापत्य
• Chapter 13 महात्मा गांधी और राष्ट्रीय आंदोलन : सविनय अवज्ञा और उससे आगे
• Chapter 14 विभाजन को समझना : राजनीति, स्मृति, अनुभव
• Chapter 15 संविधान का निर्माण : एक नए युग की शुरुआत

JAC Board Class 12th History Solutions in English Medium

• Chapter 1 Bricks, Beads and Bones: The Harappan Civilisation
• Chapter 2 Kings, Farmers and Towns: Early States and Economies
• Chapter 3 Kinship, Caste and Class: Early Societies
• Chapter 4 Thinkers, Beliefs and Buildings: Cultural Developments
• Chapter 5 Through the Eyes of Travellers: Perceptions of Society
• Chapter 6 Bhakti-Sufi Traditions: Changes in Religious Beliefs and Devotional Texts
• Chapter 7 An Imperial Capital: Vijayanagara
• Chapter 8 Peasants, Zamindars and the State: Agrarian Society and the Mughal Empire
• Chapter 9 Kings and Chronicles: The Mughal Courts
• Chapter 10 Colonialism and the Countryside: Exploring Official Archives
• Chapter 11 Rebels and the Raj: 1857 Revolt and its Representations
• Chapter 12 Colonial Cities: Urbanisation, Planning and Architecture
• Chapter 13 Mahatma Gandhi and The Nationalist Movement: Civil Disobedience and Beyond
• Chapter 14 Understanding Partition: Politics, Memories, Experiences
• Chapter 15 Framing the Constitution: The Beginning of a New Era

Students should go through these JAC Class 9 Maths Notes Chapter 2 Polynomials will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 2 Polynomials

Polynomials:
An algebraic expression f(x) of the form f(x) = a₀ + a₁x + a₂x² + …… + a_nxⁿ, where a₀, a₁, a₂ ……, a_n are real numbers and all the index of x’ are nonnegative integers is called a polynomial in x.
→ Degree of a Polynomial: Highest Index of x in algebraic expression is called the degree of the polynomial, here a₀, a₁x, a₂x² ….. a_nxⁿ, are called the terms of the polynomial and a₀, a₁, a₂, …… a_n are called various coefficients of the polynomial f(x).
Note: A polynomial in x is said to be in standard form when the terms are written either in increasing order or decreasing order of the indices of x in various terms.

→ Different Types of Polynomials: Generally, we divide the polynomials in the following categories.
→ Based on degrees:
There are four types of polynomials based on degrees. These are listed below:

• Linear Polynomials: A polynomial of degree one is called a linear polynomial. The general form of linear polynomial is ax + b, where a and b are any real constant and a ≠ 0.
• Quadratic Polynomials: A polynomial of degree two is called a quadratic polynomial. The general form of a quadratic polynomial is ax² + bx + c, where a ≠ 0, a, b, c ∈ R.
• Cubic Polynomials: A polynomial of degree three is called a cubic polynomial. The general form of a cubic polynomial is ax³ + bx² + cx + d, where a ≠ 0 and a, b, c, d ∈ R.
• Biquadratic (or quadric) Polynomials: A polynomial of degree four is called a biquadratic (quadric) polynomial. The general form of a biquadratic polynomial is ax⁴ + bx³ + cx² + dx + e, where a ≠ 0 and a, b, c, d, e are real numbers.

Note: A polynomial of degree five or more than five does not have any particular name. Such a polynomial usually called a polynomial of degree five or six or ….etc.

→ Based on number of terms:
There are three types of polynomials based on number of terms. These are as follow:

• Monomial: A polynomial is said to be monomial if it has only one term. e.g. x, 9x², 5x³ all are monomials.
• Binomial: A polynomial is said to be binomial if it contains only two terms e.g. 2x² + 3x, \(\sqrt{3}\)x + 5x³, -8x³ + 3, all are binomials.
• Trinomial: A polynomial is said to be a trinomial if it contains only three terms.e.g. 3x³ – 8x + \(\frac{1}{2}\), \(\sqrt{7}\) x¹⁰ + 8x⁴ – 3x², 5 – 7x + 8x⁹, are all trinomials.

Note: A polynomial having four or more than four terms does not have particular name. These are simply called polynomials.

→ Zero degree polynomial: Any non-zero number (constant) is regarded as polynomial of degree zero or zero degree polynomial. i.e. f(x) = a. where a ≠ 0 is a zero degree polynomial, since we can write f(x) = a, as f(x) = ax⁰.

→ Zero polynomial: A polynomial whose all coefficients are zero is called as zero polynomial i.e. f(x) = 0, we cannot determine the degree of zero polynomial.

Algebraic Identities:
An identity is an equality which is true for all values of the variables.
Some important identities are:
(i) (a + b)² = a² + 2ab + b²
(ii) (a – b)² = a² – 2ab + b²
(iii) a² – b² = (a + b)(a – b)
(iv) a³ + b³ = (a + b)(a² – ab + b²)
(v) a³ – b³ = (a – b)(a² + ab + b²)
(vi) (a + b)³ = a³ + b³ + 3ab (a + b)
(vii) (a – b)³ = a³ – b³ – 3ab (a – b)
(viii) a⁴ + a²b² + b⁴ = (a² + ab + b²)(a² – ab + b²)
(ix) a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ac)

Special case: if a + b + c = 0 then a³ + b³ + c³ = 3abc.
Other Important Identities
(i) a² + b² = (a + b)² – 2ab,
if a + b and ab are given
(ii) a² + b² = (a – b)² + 2ab
if a – b and ab are given
(iii) a + b = \(\sqrt{(a-b)^2+4 a b}\)
if a – b and ab are given
(iv) a – b = \(\sqrt{(a+b)^2-4 a b}\)
if a + b and ab are given

Factors Of A Polynomial:
→ If a polynomial f(x) can be written as a product of two or more other polynomials f₁(x), f₂(x), f₃(x)…. then each of the polynomials f₁(x), f₂(x), f₃(x)….. is called a factor of polynomial f(x). The method of finding the factors of a polynomial is called factorisation.

Zeroes Of A Polynomial:
→ A real number α is a zero of polynomial f(x) = a_nxⁿ + a_n-1x^n-1 + a_n-2x^n-2 + ….. +a₁x + a₀, if f(α) = 0. i.e. a_nαⁿ + a_n-1α^n-1 + a_n-2α^n-2+ ….. + a₁α + a₀ = 0.
For example x = 3 is a zero of the polynomial f(x) = x³ – 6x² + 11x – 6, because f(3) = (3)³ – 6(3)² + 11(3) – 6 = 27 – 54 + 33 – 6 = 0.
but x = -2 is not a zero of the above mentioned polynomial,
∵ f(-2) = (-2)³ – 6(-2)² + 11(-2) – 6
f(-2) = -8 – 24 – 22 – 6
f(-2) = -60 ≠ 0.

→ Value of a Polynomial: The value of a polynomial f(x) at x = a is obtained by substituting x a in the given polynomial and is denoted by f(a). Eg if f(x) = 2x³ – 13x² + 17x + 12 then its value at x = 1 is
f(1) = 2(1)³ – 13(1)² + 17(1) + 12
= 2 – 13 + 17 + 12 = 18.

Remainder Theorem:
Let ‘p(x)’ be any polynomial of degree greater than or equal to one and ‘a’ be any real number and if p(x) is divided by (x – a). then the remainder is equal to p(a). Let q(x) be the quotient and r(x) be the remainder when p(x) is divided by (x – a), then
Dividend = Divisor × Quotient + Remainder
∴ p(x) = (x – a) × q(x) + [r(x) or r], where r(x) = 0 or degree of r(x) < degree of (x – a). But (x – 2) is a polynomial of degree 1 and a polynomial of degree less than 1 is a constant. Therefore, either r(x) = 0 or r(x) = Constant. Let r(x) = r, then p(x) = (x – a)q(x) + r.
Putting x = a in above equation, p(a)
p(a) = (a – a)q(a) + r = 0 × q(a) + r
p(a) = 0 + r
⇒ p(a) = r
This shows that the remainder is p(a) when p(x) is divided by (x – a).
Remark: If a polynomial p(x) is divided by (x + a),(ax – b), (ax + b), (b – ax) then the remainder is the value of p(x) at x.
= \(-a, \frac{b}{a},-\frac{b}{a}, \frac{b}{a} \text { i.e. } p(-a)\)
\(p\left(\frac{b}{a}\right), p\left(-\frac{b}{a}\right), p\left(\frac{b}{a}\right)\) respectively.

Factor Theorem:
Let ‘p(x)’ be a polynomial of degree greater than or equal to 1 and ‘a’ be a real number such that p(a) = 0, then (x – a) is a factor of p(x). Conversely, if(x – a) is a factor of p(x). then p(a) = 0.

Factorisation Of A Quadratic Polynomial:
→ For factorisation of a quadratic expression ax² + bx + c where a ≠ 0, there are two methods.
→ By Method of Completion of Square:
In the form ax² + bx + c where a ≠ 0, firstly we take ‘a’ common in the whole expression then factorise by converting the expression \(a\left\{x^2+\frac{b}{a} x+\frac{c}{a}\right\}\) as the difference of two squares, which is

→ By Splitting the Middle Term:
→ x² + lx + m = x² + (a + b)x + ab
Where l = a + b and m = ab, such that a and b are real numbers
= x² + ax + bx + ab
= x (x + a) + b (x + a)
= (x + a) (x + b)
Method: We express l as the sum of two such numbers whose product is m.

→ ax² + bx + c = prx² + (ps + qr)x + qs
where b = ps + qr, a = pr, c = qs
so that (ps) (gr) (pr) (qs) = ac
∴ prx² + (ps + qr)x + qs
= prx² + psx + qrx + qs
= px (rx + s) + q(rx + s)
= (px + q) (rx + x)
Method: We express b as the sum of two such numbers whose product is ac.

→ Integral Root Theorem:
If f(x) is a polynomial with integral coefficient and the leading coefficient is 1, then any integral root of f(x) is a factor of the constant term. Thus if f(x) = x³ – 6x² + 11x – 6 has an Integral root, then it is one of the factors of 6 which are ±1, ±2, ±3, ±6.
Now in fact,
f(1) = (1)³ – 6(1)² + 11(1) – 6 = 1 – 6 + 11 – 6 = 0
f(2) = (2)³ – 6(2)² + 11(2) – 6
= 8 – 24 + 22 – 6 = 0
f(3) = (3)³ – 6(3)² + 11(3) – 6
27 – 54 + 33 – 6 = 0
Therefore Integral roots of f(x) are 1, 2, 3.

→ Rational Root Theorem:
Let \(\frac{b}{c}\) be a rational fraction in lowest terms. If \(\frac{b}{c}\) is a rational root of the polynomial f(x) = a_nxⁿ + a_n-1x^n-1 +…+ a₁x + a₀, an ≠ 0 with integral coefficients, then b is a factor of constant term a₀, and C is a factor of the leading coefficient a_n.
For example: If \(\frac{b}{c}\) is a rational root of the polynomial f(x) = 6x³ + 5x² – 3x – 2, then the values of b are limited to the factors of -2, which are ±1, ±2 and the values of care limited to the factors of 6, which are ±1, ±2, ±3, ±6. Hence, the possible rational roots of f(x) are ±1, ±2, \(\pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm \frac{2}{3}\). In fact -1 is an integral root and \(\frac{2}{3}\), –\(\frac{1}{2}\) are the rational roots of f(x) = 6x³ + 5x² – 3x – 2.
Note: (i) n^th degree polynomial can have at most n real roots.
→ Finding a zero of polynomial f(x) means solving the polynomial equation f(x) = 0. It follows from the above discussion that if f(x) = ax + b, a ≠ 0 is a linear polynomial, then it has only one zero given by
f(x) = 0 i.e. f(x) = ax + b = 0
⇒ ax = -b
⇒ x = –\(\frac{b}{a}\)
Thus, x = –\(\frac{b}{a}\) is the only zero of f(x) = ax + b.
→ If a polynomial of degree n has more than n zeros then all the coefficients of powers of x including constant term of polynomial are zero.

Students should go through these JAC Class 9 Maths Notes Chapter 8 Quadrilaterals will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 8 Quadrilaterals

Quadrilateral
A quadrilateral is a closed figure obtained by joining four points (with no three points collinear) in an order.
→ Since, ‘quad’ means ‘four’ and ‘lateral’ is for ‘sides therefore quadrilateral means a figure bounded by four sides’
→ Every quadrilateral has:
(A) Four vertices
(B) Four sides
(C) Four angles and
(D) Two diagonals.
→ A diagonal is a line segment obtained on joining the opposite vertices.

Sum of the Angles of a Quadrilateral:

Consider a quadrilateral ABCD as shown in figure. Join A and C to get the diagonal AC which divides the quadrilateral ABCD into two triangles ABC and ADC.
We know the sum of the angles of each triangle is 180°
∴ In ΔABC; ∠CAB + ∠B + ∠BCA = 180°
and In ΔADC; ∠DAC + ∠D + ∠DCA = 180°
On adding, we get:
(∠CAB + ∠DAC) + ∠B + ∠D + (∠BCA + ∠DCA) = 180° + 180°
⇒ ∠A + ∠B + ∠D + ∠C = 360°
Thus, the sum of the angles of a quadrilateral is 360°.

Types of Quadrilaterals:
→ Trapezium: It is a quadrilateral in which one pair of opposite sides are parallel and one pair is unparallel. In the quadrilateral ABCD, drawn alongside, sides AB and DC are parallel and AD and BC are unparallel therefore it is a trapezium

→ Parallelogram: It is a quadrilateral in which both the pairs of opposite sides are equal and parallel. The figure shows a quadrilateral ABCD in which AB is parallel and equal to DC and AD is parallel and equal to BC, therefore ABCD is a parallelogram.

Here, (A) ∠A = ∠C and ∠B = ∠D
(B) AB = CD and AD = BC
(C) AB || CD and AD || BC
→ Rectangle: It is a parallelogram whose each angle is 90°.
(a) ∠A + ∠B = 90° + 90° = 180°
⇒ AD || BC, also AD = BC.
(b) ∠B + ∠C = 90° + 90° = 180°
⇒ AB || DC, also AB = DC.
(c) Diagonals AC and BD are equal.

Rectangle ABCD is also a parallelogram.
→ Rhombus: It is a also parallelogram whose all the sides are equal and diagonals are perpendicular to each other. The figure shows a parallelogram ABCD in which AB = BC = CD = DA; AC ⊥ BD.; therefore it is a rhombus.

→ Square: It is a parallelogram whose all the sides are equal and each angle is 90°. Also, diagonals are equal and perpendicular to each other. The figure shows a parallelogram ABCD in which AB = BC = CD = DA, ∠A = ∠B = ∠C = ∠D = 90°, AC ⊥ BD and AC = BD, therefore ABCD is a square.

→ Kite: It is not parallelogram in which two pairs of adjacent sides are equal The figure shows a quadrilateral ABCD in which adjacent sides AB and AD are equal i.e. AB = AD and also the other pair of adjacent sides are equal i.e., BC = CD; therefore it is a kite or kite shaped figure.

Remarks:

• Square, rectangle and rhombus are all parallelograms.
• Kite and trapezium are not parallelograms.
• A square is a rectangle.
• A square is a rhombus.
• A parallelogram is a trapezium.

Parallelogram Theorems
A parallelogram is a quadrilateral in which both the pairs of opposite sides are equal and parallel.

Theorem 1.
A diagonal of a parallelogram divides the parallelogram into two congruent triangles.
Proof:
Given: A parallelogram ABCD.
To Prove: A diagonal divides the parallelogram
into two congruent triangles i.e., if diagonal AC is drawn then ΔABC ≅ ΔCDA and if diagonal BD is drawn
then ΔABD ≅ ΔCDB.
Construction: Join A and C
Proof: Since, ABCD is a parallelogram
AB || DC and AD || BC
In ΔABC and ΔCDA
∠BAC = ∠DCA [Alternate angles]
∠BCA = ∠DAC [Alternate angles]
And, AC = AC [Common side]
∴ ΔABC ≅ ΔCDA [By ASA]
Similarly, we can prove that
ΔABD ≅ ΔCDB

Theorem 2.
In a parallelogram, opposite sides are equal.

Proof:
Given: A parallelogram ABCD in which
AB || DC and AD || BC.
To Prove: Opposite sides are equal i.e.. AB = DC and AD = BC
Construction: Join A and C
Proof: In ΔABC and ΔCDA
∠BAC = ∠DCA [Alternate angles]
∠BCA = ∠DAC [Alternate angles]
AC = AC [Common]
∴ ΔABC ≅ ΔCDA [By ASA]
⇒ AB = DC and AD = BC [By CPCT]
Hence, proved.

Theorem 3.
If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Proof:
Given: A quadrilateral ABCD in which AB = DC and AD = BC.
To Prove: ABCD is a parallelogram ie, AB || DC and AD || BC
Construction: Join A and C
Proof: In ΔABC and ΔCDA
AB = DC [Given]
AD = BC [Given]
And AC = AC [Common]
∴ ΔABC ≅ ΔCDA [By SSS]
⇒ ∠1 = ∠3 [By CPCT]
And ∠2 = ∠4 [By CPCT]
But these are alternate angles and whenever alternate angles are equal, the lines are parallel.
∴ AB || DC and AD || BC
⇒ ABCD is a parallelogram.
Hence, proved.

Theorem 4.
In a parallelogram, opposite angles are equal.

Proof:
Given: A parallelogram ABCD in which AB || DC and AD || BC.
To Prove: Opposite angles are equal i.e. ∠A = ∠C and ∠B = ∠D
Construction: Draw diagonal AC.
Proof: In ΔABC and ΔCDA
∠BAC = ∠DCA [Alternate angles]
∠BCA = ∠DAC [Alternate angles]
AC = AC [Common]
∴ ΔABC ≅ ΔCDA [By ASA]
⇒ ∠B = ∠D [By CPCT]
Similarly, we can prove that
∠A = ∠C Hence, proved.

Theorem 5.
If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
Proof:
Given: A quadrilateral ABCD in which opposite angles are equal. i.e., ∠A = ∠C and ∠B = ∠D
To prove: ABCD is a parallelogram i.e.,
AB || DC and AD || BC
Proof: Since the sum of the angles of quadrilateral is 360°
⇒ ∠A + ∠B + ∠C+ ∠D = 360°
⇒ ∠A + ∠D + ∠A + ∠D = 360°
[∵ ∠A = ∠C and ∠B = ∠D]
⇒ 2∠A + 2∠D = 360°
⇒ ∠A + ∠D = 180°
[∵ The sum of interior angles on the same side of transversal AB is 180°]
⇒ AB || DC
Similarly, ∠A + ∠B + ∠C + ∠D = 360°
⇒ ∠A + ∠B + ∠A + ∠B = 360°
[∵ ∠A = ∠C and ∠B = ∠D]
⇒ 2∠A + 2∠B = 360°
⇒ ∠A + ∠B = 180°
[∵ The sum of interior angles on the same side of transversal AB is 180°]
∴ AD || BC
So, AB || DC and AD || BC
⇒ ABCD is a parallelogram.
Hence, proved.

Theorem 6.
The diagonal of a parallelogram bisect each other.

Proof:
Given: A parallelogram ABCD. Its diagonals AC and BD intersect each other at point O..
To Prove: Diagonals AC and BD bisect each other i.e., OA = OC and OB = OD.
Proof: In ΔAOB and ΔCOD
∵ AB || DC and BD is a transversal.
∴ ∠ABO = ∠CDO [Alternate angles]
∵ AB || DC and AC is a transversal line.
∴ ∠BAO = ∠DCO [Alternate angles]
And, AB = DC
⇒ ΔAOB ≅ ΔCOD [By ASA]
⇒ OA = OC and OB = OD [By CPCT]
Hence, proved.

Theorem 7.
If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Proof:
Given: A quadrilateral ABCD whose diagonals AC and BD bisect each other at point O.
i.e., OA = OC and OB = OD
To prove: ABCD is a parallelogram i.e..
AB || DC and AD || BC.
Proof: In ΔAOB and ΔCOD
OA = OC [Given]
OB = OD [Given]
And, ∠AOB = ∠COD [Vertically opposite angles]
⇒ ΔAOB ≅ ΔCOD [By SAS]
⇒ ∠1 = ∠2 [By CPCT]
But these are alternate angles and whenever alternate angles are equal, the lines are parallel.
∴ AB is parallel to DC ie., AB || DC Similarly,
ΔAOD ≅ ΔCOB [By SAS]
⇒ ∠3 = ∠4
But these are also alternate angles
⇒ AD || BC
AB || DC and AD || BC
⇒ ABCD is parallelogram.
Hence, proved.

Theorem 8.
A quadrilateral is a parallelogram, if a pair of opposite sides is equal and parallel.

Proof:
Given: A quadrilateral ABCD in which AB = DC and AB || DC.
To Prove: ABCD is a parallelogram, i.e. AB || DC and AD || BC.
Construction: Join A and C.
Proof: Since AB is parallel to DC and AC is transversal
∠BAC = ∠DCA [Alternate angles]
AB = DC [Given]
And AC = AC [Common]
⇒ ΔBAC ≅ ΔDCA [By SAS]
⇒ ∠BCA = ∠DAC [By CPCT]
But these are alternate angles and whenever alternate angles are equal, the lines are parallel.
⇒ AD || BC
Now, AB || DC (given) and AD || BC
[Proved above]
⇒ ABCD is a parallelogram
Hence, proved.

Remarks:
In order to prove that given quadrilateral is parallelogram, we can prove any one of the following.

• Opposite angles of the quadrilateral are equal, or
• Diagonals of the quadrilateral bisect each other, or
• A pair of opposite sides is parallel and is of equal length, or
• Opposite sides are equal.
• Every diagonal divides the parallelogram into two congruent triangles.

Mid-Point Theorem
Statement: In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side and is half of it.

Given: A triangle ABC in which P is the mid-point of side AB and Q is the mid-point of side AC.
To Prove: PQ is parallel to BC and is half of it
i.e., PQ || BC and PQ = \(\frac{1}{2}\)BC
Construction: Produce PQ upto point such that PQ = QR. Join R and C.
Proof: In ΔAPQ and ΔCRQ
PQ = QR [By construction]
AQ = QC [Given]
And, ∠AQP = ∠CQR [Vertically opposite angles]
⇒ ΔAPQ ≅ ΔCRQ [By SAS]
⇒ AP = CR [By CPCT]
And, ∠APQ = ∠CRQ [By CPCT]
But, ∠APQ and ∠CRQ are alternate angles and we know, whenever the alternate angles are equal, the lines are parallel.
⇒ AP || CR
⇒ AB || CR
⇒ BP || CR
Given, P is mid-point of AB
⇒ AP = BP
⇒ CR = BP [As, AP = CR]
Now, BP = CR and BP || CR
⇒ BCRP is a parallelogram.
[When any pair of opposite sides are equal and parallel, the quadrilateral is a parallelogram]
BCRP is a parallelogram and opposite sides of a parallelogram are equal and parallel.
∴ PR = BC and PR || BC
Since, PQ = QR
⇒ PQ = \(\frac{1}{2}\)PR = \(\frac{1}{2}\)BC [AS, PR = BC]
Also, PQ || BC [As, PR || BC]
∴ PQ || BC and PQ = \(\frac{1}{2}\)BC
Hence, proved.

Converse of the Mid-Point Theorem
Statement: The line drawn through the midpoint of one side of a triangle parallel to the another side bisects the third side.

Given: A triangle ABC in which is the midpoint of side AB and PQ is parallel to BC.
To prove: PQ bisects the third side AC i.e., AQ = QC.
Construction: Through C, draw CR parallel to BA, which meets PQ produced at point R.
Proof: Since, PQ || BC i.e., PR || BC [Given]
CR || BA i.e., CR || BP [By construction]
∴ Opposite sides of quadrilateral PBCR are parallel.
⇒ PBCR is a parallelogram
⇒ BP = CR
Also, BP = AP [As Pis mid-point of AB]
∴ CR = AP (As CR = BP)
Now, AB || CR and AC is transversal, ∠PAQ = ∠ROQ [Alternate angles]
Also, AB || CR and PR is transversal, ∠APQ = ∠CRQ [Alternate angles]
In ΔAPQ and ΔCRQ
CR = AP, ∠PAQ = ∠RCQ and ∠APQ = ∠CRQ
⇒ ΔAPQ ≅ ΔCRO [By ASA]
⇒ AQ = QC Hence, proved.

Students should go through these JAC Class 9 Maths Notes Chapter 5 Introduction to Euclid’s Geometry will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 5 Introduction to Euclid’s Geometry

Introduction:
The credit for introducing geometrical concepts goes to the distinguished Greek mathematician ‘Euclid’ who is known as the “Father of Geometry” and the word geometry comes from the Geek words ‘geo’ which means “Earth’ and ‘Metreon’ which means ‘measure’.

Basic Concepts In Geometry:
A point, a ‘line’ and a plane are the basic concepts to be used in geometry.
→ Axioms:
The statement that is taken to be true without proof, to serve as a premise for further reasoning and arguments, are called axioms.

Euclid’s Definitions:

• A point is that which has no part.
• A line is breadthless length.
• The ends of a line segment are points.
• A straight line is that which has length only.
• A surface is that which has length and breadth only.
• The edges of surface are lines.
• A plane surface is a surface which lies evenly with the straight lines on itself.

Euclid’s Five Postulates:
→ A straight line may be drawn from any one point to any other point.
→ A terminated line or a line segment can be produced infinitely.

→ A circle can be drawn with any centre and of any radius.
→ All right angles are equal to one another.
→ If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced infinitely, meet on that side on which the sum of angles is less than two right angles.

Important Axioms:
→ A line is the collection of infinite number of points.
→ Through a given point, infinite lines can be drawn.

→ Given two distinct points, there is one and only one line that contains both the points.
→ If P is a point not lying on a line l, then one and only one line can be drawn through P which is parallel to l.

→ Two distinct lines cannot have more than one point in common.

→ Two lines which are both parallel to the same line, are parallel to each other. i.e. IF l || n, m || n ⇒ l || m.

Some Important Definitions:
→ Collinear points: Three or more points are said to be collinear if there is a line which contains all of them.

→ Concurrent lines: Three or more lines are said to be concurrent if there is a point which lies on all of them.

→ Intersecting lines: Two lines are intersecting if they have a common point. The common point is called the “point of intersection”.

→ Parallel lines: Two lines I and m in a plane are said to be parallel lines if they do not have a common point.

→ Line segment: Given two points A and B on a line l, the connected part (segment) of the line with end points at A and B is called the line segment AB.

→ Interior point of a line segment: A point R is called an interior point of a line segment PQ if R lies between Pand O but Ris neither P nor Q.

→ Congruence of line segment: Two line segments AB and CD are congruent if trace copy of one can be superposed on the other so as to cover it completely and exactly in this case we write AB ≅ CD. In other words we can say two lines are congruent if their lengths are same.
→ Distance between two points: The distance between two points P and Q is the length of the line segment PO.
→ Ray: Directed line segment is called a ray. If AB is a ray, then it is denoted by \(\overrightarrow{\mathrm{AB}}\). Point A is called initial point of ray.

→ Opposite rays: Two rays AB and AC are said to be opposite rays if they are collinear and point A is the only common point of the two rays and A lies in between B and C.

Theorem 1.
If l, m, n are lines in the same plane such that l intersects m and n || m, then l also intersects n.
Answer:
Given: Three lines l, m, n in the same plane such that intersects m and n || m.
To prove: Lines land n are intersecting lines.

Proof: Let l and n be non intersecting lines. Then. l || n. But, n || m [Given]
∴ l || n and n || m
⇒ l || m
⇒ l and m are non-intersecting lines.
This is a contradiction to the hypothesis that I and m are intersecting lines. So our supposition is wrong.
Hence, l intersects line n.

Theorem 2.
If lines AB, AC, AD and AE are parallel to a line l, then points A, B, C, D and E are collinear.
Answer:
Given: Lines AB, AC, AD and AE are parallel to a line l.
To prove: A, B, C, D, E are collinear.
Proof: Since AB, AC, AD and AE are all parallel to a line l. Therefore point A is outside l and lines AB, AC, AD, AE are drawn through A and each line is parallel to l.
But by parallel lines axiom, one and only one line can be drawn through the point A outside a line l and parallel to it.
This is possible only when A, B, C, D and E all lie on the same line. Hence, A, B, C, D and E are collinear.

JAC Jharkhand Board Class 9th Maths Solutions in Hindi & English Medium

JAC Board Class 9th Maths Solutions in English Medium

JAC Class 9 Maths Chapter 1 Number Systems

• Chapter 1 Number Systems Ex 1.1
• Chapter 1 Number Systems Ex 1.2
• Chapter 1 Number Systems Ex 1.3
• Chapter 1 Number Systems Ex 1.4
• Chapter 1 Number Systems Ex 1.5
• Chapter 1 Number Systems Ex 1.6

JAC Class 9 Maths Chapter 2 Polynomials

• Chapter 2 Polynomials Ex 2.1
• Chapter 2 Polynomials Ex 2.2
• Chapter 2 Polynomials Ex 2.3
• Chapter 2 Polynomials Ex 2.4
• Chapter 2 Polynomials Ex 2.5

JAC Class 9 Maths Chapter 3 Coordinate Geometry

• Chapter 3 Coordinate Geometry Ex 3.1
• Chapter 3 Coordinate Geometry Ex 3.2
• Chapter 3 Coordinate Geometry Ex 3.3

JAC Class 9 Maths Chapter 4 Linear Equations in Two Variables

• Chapter 4 Linear Equations in Two Variables Ex 4.1
• Chapter 4 Linear Equations in Two Variables Ex 4.2
• Chapter 4 Linear Equations in Two Variables Ex 4.3
• Chapter 4 Linear Equations in Two Variables Ex 4.4

JAC Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry

• Chapter 5 Introduction to Euclid’s Geometry Ex 5.1
• Chapter 5 Introduction to Euclid’s Geometry Ex 5.2

JAC Class 9 Maths Chapter 6 Lines and Angles

• Chapter 6 Lines and Angles Ex 6.1
• Chapter 6 Lines and Angles Ex 6.2
• Chapter 6 Lines and Angles Ex 6.3

JAC Class 9 Maths Chapter 7 Triangles

• Chapter 7 Triangles Ex 7.1
• Chapter 7 Triangles Ex 7.2
• Chapter 7 Triangles Ex 7.3
• Chapter 7 Triangles Ex 7.4
• Chapter 7 Triangles Ex 7.5

JAC Class 9 Maths Chapter 8 Quadrilaterals

• Chapter 8 Quadrilaterals Ex 8.1
• Chapter 8 Quadrilaterals Ex 8.2

JAC Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

• Chapter 9 Areas of Parallelograms and Triangles Ex 9.1
• Chapter 9 Areas of Parallelograms and Triangles Ex 9.2
• Chapter 9 Areas of Parallelograms and Triangles Ex 9.3
• Chapter 9 Areas of Parallelograms and Triangles Ex 9.4

JAC Class 9 Maths Chapter 10 Circles

• Chapter 10 Circles Ex 10.1
• Chapter 10 Circles Ex 10.2
• Chapter 10 Circles Ex 10.3
• Chapter 10 Circles Ex 10.4
• Chapter 10 Circles Ex 10.5
• Chapter 10 Circles Ex 10.6

JAC Class 9 Maths Chapter 11 Constructions

• Chapter 11 Constructions Ex 11.1
• Chapter 11 Constructions Ex 11.2

JAC Class 9 Maths Chapter 12 Heron’s Formula

• Chapter 12 Heron’s Formula Ex 12.1
• Chapter 12 Heron’s Formula Ex 12.2

JAC Class 9 Maths Chapter 13 Surface Areas and Volumes

• Chapter 13 Surface Areas and Volumes Ex 13.1
• Chapter 13 Surface Areas and Volumes Ex 13.2
• Chapter 13 Surface Areas and Volumes Ex 13.3
• Chapter 13 Surface Areas and Volumes Ex 13.4
• Chapter 13 Surface Areas and Volumes Ex 13.5
• Chapter 13 Surface Areas and Volumes Ex 13.6
• Chapter 13 Surface Areas and Volumes Ex 13.7
• Chapter 13 Surface Areas and Volumes Ex 13.8
• Chapter 13 Surface Areas and Volumes Ex 13.9

JAC Class 9 Maths Chapter 14 Statistics

• Chapter 14 Statistics Ex 14.1
• Chapter 14 Statistics Ex 14.2
• Chapter 14 Statistics Ex 14.3
• Chapter 14 Statistics Ex 14.4

JAC Class 9 Maths Chapter 15 Probability

• Chapter 15 Probability Ex 15.1

JAC Board Class 9th Maths Solutions in Hindi Medium

JAC Class 9 Maths Chapter 1 संख्या पद्धति

• Chapter 1 संख्या पद्धति Ex 1.1
• Chapter 1 संख्या पद्धति Ex 1.2
• Chapter 1 संख्या पद्धति Ex 1.3
• Chapter 1 संख्या पद्धति Ex 1.4
• Chapter 1 संख्या पद्धति Ex 1.5
• Chapter 1 संख्या पद्धति Ex 1.6

JAC Class 9 Maths Chapter 2 बहुपद

• Chapter 2 बहुपद Ex 2.1
• Chapter 2 बहुपद Ex 2.2
• Chapter 2 बहुपद Ex 2.3
• Chapter 2 बहुपद Ex 2.4
• Chapter 2 बहुपद Ex 2.5

JAC Class 9 Maths Chapter 3 निर्देशांक ज्यामिति

• Chapter 3 निर्देशांक ज्यामिति Ex 3.1
• Chapter 3 निर्देशांक ज्यामिति Ex 3.2
• Chapter 3 निर्देशांक ज्यामिति Ex 3.3

JAC Class 9 Maths Chapter 4 दो चरों वाले रैखिक समीकरण

• Chapter 4 दो चरों वाले रैखिक समीकरण Ex 4.1
• Chapter 4 दो चरों वाले रैखिक समीकरण Ex 4.2
• Chapter 4 दो चरों वाले रैखिक समीकरण Ex 4.3
• Chapter 4 दो चरों वाले रैखिक समीकरण Ex 4.4

JAC Class 9 Maths Chapter 5 युक्लिड के ज्यामिति का परिचय

• Chapter 5 युक्लिड के ज्यामिति का परिचय Ex 5.1
• Chapter 5 युक्लिड के ज्यामिति का परिचय Ex 5.2

JAC Class 9 Maths Chapter 6 रेखाएँ और कोण

• Chapter 6 रेखाएँ और कोण Ex 6.1
• Chapter 6 रेखाएँ और कोण Ex 6.2
• Chapter 6 रेखाएँ और कोण Ex 6.3

JAC Class 9 Maths Chapter 7 त्रिभुज

• Chapter 7 त्रिभुज Ex 7.1
• Chapter 7 त्रिभुज Ex 7.2
• Chapter 7 त्रिभुज Ex 7.3
• Chapter 7 त्रिभुज Ex 7.4
• Chapter 7 त्रिभुज Ex 7.5

JAC Class 9 Maths Chapter 8 चतुर्भुज

• Chapter 8 चतुर्भुज Ex 8.1
• Chapter 8 चतुर्भुज Ex 8.2

JAC Class 9 Maths Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल

• Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Ex 9.1
• Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Ex 9.2
• Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Ex 9.3
• Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Ex 9.4

JAC Class 9 Maths Chapter 10 वृत्त

• Chapter 10 वृत्त Ex 10.1
• Chapter 10 वृत्त Ex 10.2
• Chapter 10 वृत्त Ex 10.3
• Chapter 10 वृत्त Ex 10.4
• Chapter 10 वृत्त Ex 10.5
• Chapter 10 वृत्त Ex 10.6

JAC Class 9 Maths Chapter 11 रचनाएँ

• Chapter 11 रचनाएँ Ex 11.1
• Chapter 11 रचनाएँ Ex 11.2

JAC Class 9 Maths Chapter 12 हीरोन का सूत्र

• Chapter 12 हीरोन का सूत्र Ex 12.1
• Chapter 12 हीरोन का सूत्र Ex 12.2

JAC Class 9 Maths Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन

• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.1
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.2
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.3
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.4
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.5
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.6
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.7
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.8
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.9

JAC Class 9 Maths Chapter 14 सांख्यिकी

• Chapter 14 सांख्यिकी Ex 14.1
• Chapter 14 सांख्यिकी Ex 14.2
• Chapter 14 सांख्यिकी Ex 14.3
• Chapter 14 सांख्यिकी Ex 14.4

JAC Class 9 Maths Chapter 15 प्रायिकता

• Chapter 15 प्रायिकता Ex 15.1

Students should go through these JAC Class 9 Maths Notes Chapter 7 Triangles will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 7 Triangles

Triangle
A plane figure bounded by three lines in a plane is called a triangle. Every triangle has three sides and three angels. If ABC is any triangle then AB, BC and CA are three sides and ∠A, ∠B and ∠C are three angles.

Types of Triangles
→ On the basis of sides we have three types of triangles:

• Scalene triangle – A triangle whose no two sides are equal is called a scalene triangle.
• Isosceles triangle – A triangle having two sides equal is called an isosceles triangle.
• Equilateral triangle – A triangle in which all sides are equal is called an equilateral triangle.

→ On the basis of angles we have three types of triangles:

• Right triangle – A triangle in which any one angle is a right angle (= 90°) is called right triangle.
• Acute triangle – A triangle in which all angles are acute (0° >angle >90°) is called an acute triangle.
• Obtuse (90° < angle < 180° ) triangle – A triangle in which any one angle is obtuse is called an obtuse triangle.

Congruent Figures
The figures are called congruent if they have same shape and same size. In other words, two figures are called congruent if they are having equal length, width and height.

In the above figures {Fig. (i) and Fig. (ii)} both are equal in length, width and height, so these are congruent figures.

Congruent Triangles
Two triangles are congruent if and only if one of them can be made to superimposed on the other, so as to cover it exactly.

If two triangles ΔABC and ΔDEF are congruent then there exist a one to one correspondence between their vertices and sides. i.e. we get following six equalities.
∠A = ∠D, ∠B = ∠E, ∠C = ∠F and AB = DE, BC = EF, AC = DF.
If ΔABC and ΔDEF are congruent under one to one correspondence A ↔ D, B ↔ E, C ↔ F then we write ΔABC ≅ ΔDEF We cannot write it as ΔABC ≅ ΔDFE or ΔABC ≅ ΔEDF or in other forms because ΔABC ≅ ΔDFE have following one-one correspondence A ↔ D, B ↔ F, C ↔ E.
Hence, we can say that ‘two triangles are congruent if and only if there exists a oneone correspondence between their vertices such that the corresponding sides and the corresponding angles of the two triangles are equal.

Sufficient Conditions for Congruence of two Triangles
→ SAS Congruence Criterion:

Two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding sides and the included angle of the other triangle.

→ ASA Congruence Criterion:

Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.

→ AAS Congruence Criterion:
If any two angles and a non included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.

→ SSS Congruence Criterion:
Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle.

→ RHS Congruence Criterion:
Two right angled triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and one side of the other triangle.

→ Congruence Relation in the Set of all Triangles:
By the definition of congruence of two triangles, we have following results.

• Every triangle is congruent to itself i.e., ΔABC ≅ ΔABC
• If ΔABC ≅ ΔDEF then ΔDEF ≅ ΔABC
• If ΔABC ≅ ΔDEF and ΔDEF ≅ ΔΡQR then ΔΑΒC ≅ ΔΡQR

NOTE: If two triangles are congruent then their corresponding sides and angles are also congruent by CPCT (corresponding parts of congruent triangles are also congruent).

Theorem 1.
Angles opposite to equal sides of an isosceles triangle are equal.

Given:
ΔABC in which AB = AC
To Prove: ∠B = ∠C
Construction: We draw the bisector AD of ∠A which meets BC in D.
Proof: In ΔABD and ΔACD, we have
AB = AC [Given]
∠BAD = ∠CAD [∵ AD is bisector of ∠A]
And, AD = AD [Common side]
∴ By SAS criterion of congruence, we have
ΔΑΒD ≅ ΔΑCD
⇒ ∠B = ∠C [by CPCT]
Hence, proved.

Theorem 2.
If two angles of a triangle are equal, then sides opposite to them are also equal.
Given: ΔABC in which ∠B = ∠C
To Prove: AB = AC
Construction: We draw the bisector of ∠A

which meets BC in D.
Proof: In ΔABD and ΔACD, we have
∠B = ∠C [Given]
∠BAD = ∠CAD [∵ AD is bisector of ∠A]
AD = AD [Common side]
∴ By AAS criterion of congruence, we get
ΔΑΒD ≅ ΔΑCD
⇒ AB = AC [By CPCT] Hence, proved.

Theorem 3.
If the bisector of the vertical angle bisects the base of the triangle, then the triangle is isosceles.

Given: ΔABC in which AD is the bisector of ∠A meeting BC in D such that BD = CD
To Prove: ΔABC is an isosceles triangle.
Construction: We produce AD to E such that AD = DE and join EC
Proof: In ΔADB and ΔEDC, we have
AD = DE [By construction]
∠ADB = ∠CDE [Vertically opposite angles]
BD = DC [Given]
∴ By SAS criterion of congruence, we get
ΔADR ≅ ΔEDC ⇒ AB = EC ……(i)
And, ∠BAD = ∠CED [By CPCT]
But, ∠BAD = ∠CAD
∴ ∠CAD = ∠CED
⇒ AC = EC [Sides opposite to equal angles are equal]
⇒ AC = AB [By eq. (i)] Hence, proved

Some Inequality Relations In A Triangle
→ If two sides of a triangle are unequal, then the longer side has greater angle opposite to it, i.e., if in any ΔABC, AB > AC then ∠C > ∠B.
→ In a triangle the greater angle has the longer side opposite to it, ie, if in any ΔABC, ∠A > ∠B then BC > AC.
→ The sum of any two sides of a triangle is always greater than the third side, i.e., in any ΔABC, AB + BC > AC, BC + CA > AB and AC + AB > BC.
→ Of all the line segments that can be drawn to a given line, from a point, not lying on it, the perpendicular line segment is the shortest.

P is any point not lying on line l, PM ⊥ l then PM < PN.
→ The difference of any two sides of a triangle is less than the third side, i.e., in any ΔABC, AB – BC < AC, BC – CA < AB and AC – AB < BC.

Students should go through these JAC Class 9 Maths Notes Chapter 3 Coordinate Geometry will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 3 Coordinate Geometry

Co-Ordinate System:
In two dimensional coordinate geometry, we generally use two types of coordinate systems.

• Cartesian or Rectangular coordinate system.
• Polar coordinate system.

In cartesian coordinate system we represent any point by ordered pair (x, y) where x and y are called x and y coordinate of that point respectively.
In polar coordinate system we represent any point by ordered pair (r, θ) where is called radius vector and ‘θ’ is called vectorial angle of that point, which will be studied in higher classes.

Cartesian Coordinate System:
→ Rectangular Coordinate Axes:
Let XX’ and YY’ are two lines such that XX’ is horizontal and YY’ is vertical lines in the same plane and they intersect each other at O. This intersecting point is called origin Now choose a convenient unit of length and starting from origin as zero, mark off a number scale on the horizontal line XX’, positive to the right of origin O and negative to the left of origin O. Also mark off the same scale on the vertical line YY’, positive upwards and negative downwards of the origin. The line XX’ is called X-axis and the line YY’ is known as Y-axis and the two lines taken together are called the coordinate axes.

→ Quadrants:
The coordinates axes XX’ and YY’ divide the plane of graph paper into four parts XY, X’Y, X’Y’ and XY’. These four parts are called the quadrants. The parts XY, X’Y, X’Y’ and XY’ are known as the first second, third and fourth quadrants respectively.

→ Cartesian Coordinates of a Point:
Let -axis and y-axis be the coordinate axes and P be any point in the plane. To find the position of P with respect of x-axis and y-axis, we draw two perpendicular line segment from P on both coordinate axes.

Let PM and PN be the perpendiculars on x-axis and y-axis resepectively. The length of the line segment OM is called the x-coordinate or abscissa of point P. Similarly the length of line segment ON is called the y-coordinate or ordinate of point P.

Let OM = x and ON = y. The position of the point P in the plane with respect to the coordinate axes is represented by the ordered pair (x, y). The ordered pair (x, y) is called the coordinates of point P. “Thus, for a given point, the abscissa and ordinate are the distances of the given point from y-axis and x-axis respectively”.

The above system of coordinating of ordered pair (x, y) with every point in plane is called the Rectangular or Cartesian coordinate system.

Cartesian coordinate system

→ Convention of Signs:
As discussed earlier that regions XOY, Χ’ΟΥ, Χ’ΟΥ’ and ΧΟΥ’ are known as the first second, third and fourth quadrants respectively. The ray OX is taken as positive X-axis, OX’ as negative x-axis, OY as positive y-axis and OY as negative y-axis. Thus we have.
In first quadrant: x > 0, y > 0
In second quadrant: x < 0, y > 0
In third quadrant: x < 0, y < 0
In fourth quadrant: x > 0, y < 0

→ Points on Axis:
If point P lies on x-axis then clearly its distance from x-axis will be zero, therefore we can say that its ordinate will be zero. In general, if any point lies on x-axis then its y-coordinate will be zero. Similarly if any point Q lies on y-axis, then its distance from y-axis will be zero therefore we can say its x-coordinate will be zero. In general, if any point lies on y-axis then its x-coordinate will be zero.

→ Plotting of Points:
In order to plot the points in a plane, we may use the following algorithm.
Step I: Draw two mutually perpendicular lines on the graph paper, one horizontal and other vertical.
Step II: Mark their intersection point as O (origin).
Step III: Choose a suitable scale on X-axis and Y-axis and mark the points on both the axes.
Step IV: Obtain the coordinates of the point which is to be plotted. Let the point be P(a, b). To plot this point start from the origin and |a| units move along OX, OX’ according as ‘a’ is positive or negative respectively. Suppose we arrive at point M. From point M move vertically upward or downward |b| units according as ‘b’ is positive or negative respectively The point where we arrive finally is the required point P(a, b).

Distance Between Two Points:
→ If there are two points A (x₁, y₁) and B(x₂, y₂) on the XY plane, the distance between them is given by
AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

Students should go through these JAC Class 9 Maths Notes Chapter 14 Statistics will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 14 Statistics

Introduction:
The branch of science known as Statistics has been used in India from ancient times. Statistics deals with collection of numerical facts ie, data, their classification and tabulation and their interpretation. In statistics we shall try to study, in detail about collection, classification and tabulation of such data.
→ Importance of Data: Expressing facts with the helps of data is of great importance in our day-to-day life. For example, instead of saying that India has a large population it is more appropriate to say that the population of India, based on the census of 2001 is more than one billion.

→ Collection of Data: On the basis of methods of collection, data can be divided into two categories:
(i) Primary data: Data which are collected for the first time by the statistical investigator or with help of his workers is called primary data. For example if an investigator wants to study the condition of the workers working in a factory then for this he collects some data like their monthly income, expenditure, number of brothers, sisters, etc.

(ii) Secondary data: Data already collected by a person or a society and may be available in published or unpublished form is known as secondary data. Secondary should be carefully used. Such data is generally obtained from the following two sources.

• Published sources
• Unpublished sources

→ Classification of Data: When the data is compiled the same form and order in which it is collected, it is known as Raw Data. It is also called Crude Data. For example, the marks obtained by 20 students of class X in English out of 10 marks are as follow:
7. 4, 9, 5, 8, 9, 6, 7, 9, 2, 0 3, 7, 6, 2, 1, 9, 8, 3, 8
(i) Geographical basis: Here, the data is classified on the basis of place or region. For example, the production of food grains of different states is shown in the following table:

(ii) Chronological classification data’s classification is based on hour, day, week, month or year, then it is called chronological classification. For example, the population of India in different years is shown in following table:

(iii) Qualitative basis: When the data is classified into different groups on the basis of their descriptive qualities and properties such a classification is known as descriptive or qualitative classification. Since the attributes cannot be measured directly they are counted on the basis of presence or absence of qualities. For example intelligence, literacy, unemployment, honesty etc. The following table shows classification on the basis of sex and employment.

(iv) Quantitative basis: If facts are such that they can be measured physically e.g, marks obtained, height, weight, age, income, expenditure etc, such facts are known as variable values. If such facts are kept into classes then it is called classification according to quantitative or class intervals.

Definitions:
→ Variate: The numerical quantity whose value varies is called a variate, generally a variate is represented by x. There are two types of variate.
(i) Discrete variate: Its magnitude is fixed. For example, the number of teachers in different branches of a institute are 30, 35, 40 etc.
(ii) Continuous variate: Its magnitude is not fixed. It is expressed in groups like 10 – 20, 20 – 30, … etc.
→ Range: The difference between the maximum and the minimum values of the variable x is called range.
→ Class frequency: in each class, the number of times a data is repeated is known as its class frequency.
→ Class limits: The lowest and the highest value of the class are known as lower and upper limits respectively of that class.
→ Classmark: The average of the lower and the upper limits of a class is called the mid value or the class mark of that class. It is generally denoted by x.
If x is the mid value and his the class size, then the class limits are \(\left(x-\frac{h}{2}, x+\frac{h}{2}\right)\).

Example:
The mid values of a distribution are 54, 64, 74, 84 and 94. Find the class size and class limits.
Solution:
The class size is the difference of two consecutive class marks, therefore class size (h) = 64 – 54 = 10.
Here the mid values are given and the class size is 10. So, class limits are:

Therefore, class limits are 49 – 59, 59 – 69, 69 – 79, 79 – 89, and 89 – 99.

Frequency Distribution:
The marks scored by 30 students of IX class of a school in the first test of Mathematics out of 50 marks are as follows:

The number of times a mark is repeated is called its frequency. It is denoted by f.

Above type of frequency distribution is called ungrouped frequency distribution. Although this representation of data is shorter than representation of raw data, but from the angle of comparison and analysis it is quite bit. So to reduce the frequency distribution, it can be classified into groups in following ways and it is called grouped frequency distribution.

(a) Kinds of Frequency Distribution: Statistical methods like comparison, decision taken etc. depends on frequency distribution. Frequency distribution are of three types.
(i) Individual frequency distribution: Here each item or original price of unit is written separately. In this category, frequency of each variable is one.

Example:
Total marks obtained by 10 students in a class.

(ii) Discrete frequency distribution: When number of terms is large and variable are discrete, i.e., variate can accept some particular values only under finite limits and is repeated then it is called discrete frequency distribution. For example, the wages of employees and their numbers is shown in following table.

The above table shows ungrouped frequency distribution and same facts can be written in grouped frequency as follows:

Note: If variable is repeated in individual distribution then it can be converted into discrete frequency distribution.
(iii) Continuous frequency distribution:
When number of terms is large and variate is continuous, i.e. variate can accept all values under finite limits and they are repeated then it is called continuous frequency distribution. For example age of students in a school is shown in the following table:

Classes can be made mainly by two methods:
(i) Exclusive series: In this method upper limit of the previous class and lower limit of the next class is same. In this method the value of upper limit in a class is not considered in the same class, it is considered in the next class.

(ii) Inclusive series: In this method value of upper and lower limit are both contained in same class. In this method the upper limit of class and lower limit of next class are not same. Some time the value is not a whole number, it is a fraction or in decimals and lies in between the two intervals then in such situation the class interval can be constructed as follows:

Cumulative Frequency:
→ Discrete frequency distribution:
From the table of discrete frequency distribution, it can be identified that number of employees whose monthly income is 4000 or how many employees of monthly income 11000 are there. But if we want to know how many employees whose monthly income is upto 11000, then we should add 10, 8, 5 and 7 i.e., number of employees whose monthly income is upto 11000 is 10 + 8 + 5 + 7 = 30. Here we add all previous frequency and get cumulative frequency. It will be more clear from the following table:

→ Continuous frequency distribution: In (a) part, we obtained cumulative frequency for discrete series. Similarly, cumulative frequency table can be made from continuous frequency distribution also.
For example, for table:

Graphical Representation Of Data:
(i) Bar graphs
(ii) Histograms
(iii) Frequency polygons
(iv) Frequency curves
(v) Cumulative frequency curves or Ogives
(vi) Pie Diagrams

(i) Bar Graphs. A bar graph is a graph that present categorical data with rectangular bars with heights or lengths proportional to the values that they represent.

Example:
A family with monthly income of ₹ 20,000 had planned the following expenditure per month under various heads: Draw bar graph for the data giyen below:

Solution:

To draw a bar graph, class intervals are marked along x-axis on a suitable scale. Frequencies are marked along y-axis on a suitable scale, such that the areas of drawn rectangles are proportional to corresponding frequencies.

(ii) Histogram: Histogram is rectangular representation of grouped and continuous frequency distribution in which class intervals are taken as base and height of rectangles are proportional to corresponding frequencies.

Now we shall study construction of histo grams related with four different kinds of frequency distributions.

• When frequency distribution is grouped and continuous and class intervals are also equal.
• When frequency distribution is grouped and continuous but class interval are not equal.
• When frequency distribution is grouped but not continuous.
• When frequency distribution is ungrouped and middle points of the distribution are given.

Now we try to make the above facts clear with some examples.

Example:
Draw a histogram of the following frequency distribution.

Solution:
Here frequency distribution is grouped and continuous and class intervals are also equal. So mark the class intervals on the x-axis i.e., age in year (scale 1 cm = 5 year). Mark frequency i.e., number of students (scale 1 cm = 25 students) on the y-axis.

Example:
The weekly wages of workers of a factory are given in the following table. Draw histogram for it.

Solution:
Here frequency distribution is grouped and continuous but class intervals are not same. Under such circumstances the following method is used to find heights of rectangle so that heights are proportional to frequencies the least.
(i) Write the least class size (h), here h = 500.
(ii) Redefine the frequencies of classes by using following formula.
Redefined frequency of class = \(\frac{\mathrm{h}}{\text { class size }}\) × frequency of class interval.
So, here the redefined frequency table is obtained as follows:

Now mark class interval on x-axis (scale 1 cm = 500) and no of workers on y-axis (scale 1 cm = 5). On the basis of redefined frequency distribution, construct rectangles A, B, C, D and E.

This is the required histogram of the given frequency distribution.

(a) Difference between Bar Graph and Histogram

• In histogram there is no gap in between consecutive rectangles as in bar graph.
• The width of the bar is significant in histogram. In bar graph, width is not important at all.
• In histogram the areas of rectangles are proportional to the frequency, however if the class size of the classes are equal then heights of the rectangle are proportional to the frequencies.

(iii) Frequency polygon: A frequency polygon is also a form of a graphical representation of frequency distribution Frequency polygon can be constructed in two ways:

• With the help of histogram.
• Without the help of histogram.

Following procedure is useful to draw a frequency polygon with the help of histogram.

• Construct the histogram for the given frequency distribution.
• Find the middle point of each upper horizontal line of the rectangle.
• Join these middle points of the successive rectangles by straight lines.
• Join the middle point of the initial rectangle with the middle point of the previous expected class interval on the x-axis.
• Join the middle point of the last rectangle with the middle point of the next expected class interval on the x-axis.

Example:
For the following frequency distribution, draw a histogram and construct a frequency polygon with it.

Solution:
The given frequency distribution is grouped and continuous, so we construct a histogram by the method given earlier Join the middle points P, Q, R, S, T of upper horizontal line of each rectangles A, B, C, D, E by straight lines.

Example:
Draw a frequency polygon of the following frequency distribution table.

Solution:
Given frequency distribution is grouped and continuous. So we construct a histogram by using earlier method. Join the middle points P, Q, R, S, T, U, V, W, X, Y of upper horizontal lines of each rectangle A, B, C, D, E, F, G, H, I, J by straight line in successions.

Example:
Draw a frequency polygon of the following frequency distribution.

Solution:
Here frequency distribution is grouped and continuous so here we obtain following table on the basis of class.

Now taking suitable scale on graph mark the points (5, 15), (15, 12), (25, 10) (35, 4), (45, 11), (55, 14).

Measures Of Central Tendency:
The commonly used measure of central tendency are:
(a) Mean,
(b) Median,
(c) Mode

(a) Mean: The mean of a number of observations is the sum of the values of all the observations divided by the total number of observations. It is denoted by the symbol \(\overline{\mathrm{x}}\), read as x bar.
(i) Properties of mean:
→ If a constant real number ‘a’ is added to each of the observations, then new mean will be \(\overline{\mathrm{x}}\) + a.
→ If a constant real number ‘a’ is subtracted from each of the observations, then new mean will be \(\overline{\mathrm{x}}\) – a.
→ If a constant real number ‘a’ is multiplied with each of the observations, then new mean will be \(\overline{\mathrm{x}}\).
→ If each of the observation is divided by a constant no ‘a’ then new mean will be \(\frac{\overline{\mathrm{x}}}{\mathrm{a}}\).

(ii) Mean of ungrouped data: If x₁, x₂, x₃, ….., x_n are n values (or observations) then A.M. (Arithmetic mean) is

i.e. product of mean and no. of items gives sum of observations.

Example:
Find the mean of the factors of 10.
Solution:
Factors of 10 are 1, 2, 5 and 10.
\(\overline{\mathrm{x}}\) = \(\frac{1+2+5+10}{4}=\frac{18}{4}\) = 4.5

Example:
If the mean of 6, 4, 7, P and 10 is 8, find P.
Solution:
8 = \(\frac{6+4+7+P+10}{5}\)
⇒ P = 13

(iii) Method for Mean of ungrouped frequency distribution:

Then, mean \(\overline{\mathrm{x}}\) = \(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}\)

(iv) Method for Mean of grouped frequency distribution:
Example:
Direct Method: For finding mean

\(\overline{\mathrm{x}}\) = \(\frac{\sum f_i x_i}{\sum f_i}=\frac{1430}{40}\) = 35.75

(v) Combined Mean:
\(\overline{\mathrm{x}}\) = \(\frac{\mathrm{n}_1 \overline{\mathrm{x}}_1+\mathrm{n}_2 \overline{\mathrm{x}}_2+\ldots \ldots .}{\mathrm{n}_1+\mathrm{n}_2+\ldots \ldots}\)

(vi) Uses of Arithmetic Mean

• It is used for calculating average marks obtained by a student.
• It is extensively used in practical statistics.
• It is used to obtain estimates.
• It is used by businessmen to find out profit per unit article, output per machine, average monthly income and expenditure etc.

(b) Median: Median of a distribution is the value of the variable which divides the distribution into two equal parts.
(i) Median of ungrouped data:

• Arrange the data in ascending or descending order.
  Count the no. of observations (Let there be ‘n’ observations)
  If n is odd then median = value of \(\left(\frac{\mathrm{n}+1}{2}\right)^{\mathrm{th}}\) observation.
  If n is even then median = value of mean of \(\left(\frac{n}{2}\right)^{\text {th }}\) observation and \(\left(\frac{\mathrm{n}}{2}+1\right)^{\mathrm{th}}\) observation.

Example:
Find the median of the following values:
37, 31, 42, 43, 16, 25, 39, 45, 32
Solution:
Arranging the data in ascending order, we have
25, 31, 32, 37, 39, 42, 43, 45, 46
Here the number of observations, n
= 9 (odd)
∴ Median
= Value of \(\left(\frac{9+1}{2}\right)^{\text {th }}\) observation
= Value of 5th observation = 39.

Example:
The median of the observation 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 arranged in ascending order is 24. Find the value of x.
Solution:
Here, the number of observations, n = 10.
Since n is even, therefore
Median

(ii) Uses of Median:
(A) Median is the only average to be used while dealing with qualitative data which cannot be measured quantitatively but can be arranged in ascending or descending order of magnitude.
(B) It is used for determining the typical value in problems concerning wages, distribution of wealth etc.

(c) Mode:
(i) Mode of ungrouped data (By inspection only): Arrange the data in an array and then count the frequencies of each variate. The variate having maximum frequency is the mode.

Example: Find the mode of the following array of an individual series of scores 7, 10, 12, 12, 12, 11, 13, 13, 17.

∴ Mode is 12
(ii) Uses of Mode: Mode is the average to be used to find the ideal size, e.g., in business forecasting, in manufacture of ready-made garments, shoes etc.

Empirical Relation Between Mode, Median And Mean:
Mode = 3 Median – 2 Mean
Range:
The range is the difference between the highest and lowest scores of a distribution. It is the simplest measure of dispersion. It gives a rough idea of dispersion. This measure is useful for ungrouped data.
(a) Coefficient of the Range:
If R and h are the lowest and highest scores in a distribution then the coefficient of the Range = \(\frac{\mathrm{h}-\mathrm{R}}{\mathrm{h}+\mathrm{R}}\)

Example: Find the range of the following distribution: 1, 3, 4, 7, 9, 10, 12, 13, 14, 16 and 19.
Solution:
R = 1, h = 19
∴ Range = h – R = 19 – 1 = 18.

Students should go through these JAC Class 9 Maths Notes Chapter 9 Areas of Parallelograms and Triangles will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 9 Areas of Parallelograms and Triangles

Polygonal Region

Polygon region can be expressed as the union of a finite number of triangular regions in a plane such that if two of these intersect, their intersection is either a point or a line segment. It is the shaded portion including its sides as shown in the figure.

Parallelogram

(a) Base and Altitude of a Parallelogram:
→ Base: Any side of parallelogram can be called its base.
→ Altitude: The perpendicular to the base from the opposite side is called the altitude of the parallelogram corresponding to the given base.
In the given Figure
→ DL is the altitude of ||^gm ABCD corresponding to the base AB.
→ DM is the altitude of ||^gm ABCD, corresponding to the base BC.

Theorem 1.
A diagonal of parallelogram divides it into two triangles of equal area.

Proof:
Given: A parallelogram ABCD whose one of the diagonals is BD.
To prove: ar(ΔABD) = ar(ΔCDB).
Proof: In ΔABD and ΔCDB;
AB = DC [Opp sides of a ||^gm]
AD = BC [Opp. sides of a ||^gm]
BD = BD [Common side]
∴ ΔΑΒD ≅ ΔCDB [By SSS]
ar(ΔABD) = ar (ΔCDB) [Areas of two congruent triangles are equal]
Hence, proved.

Theorem 2.
Parallelograms on the same base or equal base and between the same parallels are equal in area.

Proof:
Given: Two Parallelograms ABCD and ABEF are on the same base AB and between the same parallels AB and FC.
To Prove: ar(||^gm ABCD) = ar(||^gm ABEF)
Proof: In ΔADF and ΔBCE, we have
AD = BC [Opposite sides of a ||^gm]
and AF = BE [Opposite sides of a ||^gm]
AD || BC (Opposite sides of a parallelogram)
⇒ ∠ADF = ∠BCE (Alternate interior angles)
∴ ΔADF ≅ ΔBCE [By AAS]
∴ ar(ΔADF) = ar(ΔBCE) …..(i)
[Congruent triangles have equal area]
∴ ar (||^gm ABCD) = ar(ABED) + ar(ΔBCE)
= ar (ABED) + ar(ΔADF) [Using (1)]
= ar(||^gm ABEF).
Hence, ar(||^gm ABCD) = ar(||^gm ABEF).
Hence, proved.

Theorem 3.
The area of parallelogram is the product of its base and the corresponding altitude.

Proof:
Given: A ||^gm ABCD in which AB is the base and AL is the corresponding height.
To prove: Area (||^gm ABCD) = AB × AL.
Construction: Draw BM ⊥ DC so that rectangle ABML is formed.
Proof: ||^gm ABCD and rectangle ABML are on the same base AB and between the same parallel lines AB and LC.
∴ ar(||^gm ABCD) = ar(rectangle ABML)
= AB × AL
∴ area of a ||^gm = base × height.
Hence, proved.

Area Of A Triangle
Theorem 4.
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Proof:
Given: Two triangles ABC and PCB on the same base BC and between the same parallel lines BC and AP.
To prove: ar(ΔABC) = ar(ΔPBC)
Construction: Through B, draw BD || CA intersecting PA produced in D and through C, draw CQ || BP, intersecting line AP produced in Q.
Proof: We have, BD || CA (By construction) And, BC || DA [Given]
∴ Quad. BCAD is a parallelogram.
Similarly, Quad. BCQP is a parallelogram.
Now, parallelogram BOQP and BCAD are on the same base BC, and between the same parallels.
∴ ar (||^gm BCQP) = ar (||^gm BCAD)….(i)
We know that the diagonals of a parallelogram divides it into two triangles of equal area.
∴ ar(ΔPBC) = \(\frac{1}{2}\)(||^gm BCQP) …..(ii)
And ar (ΔABC) = \(\frac{1}{2}\)(||^gm BCAD)….(iii)
Now, ar (||^gm BCQP) = ar(||^gm BCAD) [From (i)]
\(\frac{1}{2}\)ar(||^gm BCAD) = \(\frac{1}{2}\)ar(||^gm BCQP)
Hence, \(\frac{1}{2}\)ar(ABC) = ar(APBC) (Using (ii) and (iii) Hence, proved.

Theorem 5.
The area of a trapezium is half the product of its height and the sum of the parallel sides.

Proof:
Given: Trapezium ABCD in which AB || DC, AL ⊥ DC, CN ⊥ AB and AL = CN = h (say). AB = a, DC = b.
To prove: ar(trap. ABCD) = \(\frac{1}{2}\)h × (a + b).
Construction: Join AC.
Proof: AC is a diagonal of quad. ABCD.
∴ ar(trap. ABCD) = ar(ΔABC) + ar(ΔACD)
= \(\frac{1}{2}\)h × a+\(\frac{1}{2}\)h × b= \(\frac{1}{2}\)h(a + b).
Hence, proved.

Theorem 6.
Triangles having equal areas and having one side of the triangle equal to corresponding side of the other, have their corresponding altitudes equal.

Proof:
Given: Two triangles ABC and PQR such that
(i) ar(ΔABC) = ar(ΔPQR) and
(ii) AB = PQ.
CN and RT and the altitudes corresponding to AB and PQ respectively of the two triangles.
To prove: CN = RT.
Proof: In ΔABC, CN is the altitude corresponding to the side AB
ar(ΔABC) = \(\frac{1}{2}\)AB × CN ……(i)
Similarly, ar(ΔPQR) = \(\frac{1}{2}\)PQ × RT ……(ii)
Since, ar(ΔABC) = ar(ΔPQR) [Given]
∴ \(\frac{1}{2}\)AB × CN = PQ × RT
Also, AB = PQ [Given]
∴ CN = RT Hence, proved.

Students should go through these JAC Class 9 Maths Notes Chapter 15 Probability will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 15 Probability

Probability:
Theory of probability deals with measurement of uncertainty of the occurrence of some event or incident in terms of percentage or ratio.
→ Sample Space: Set of all possible outcomes.
→ Trial: Trial is an action which results in one of several outcomes.
→ An experiment: An experiment is any kind the of activity such as throwing a die, tossing a coin, drawing a card. The different possibilities which can occur during an experiment. e.g. on throwing a dice, 1 dot, 2 dots, 3 dots, 4 dots, 5 dots, 6 dots can occur.
→ An event: Getting a ‘six’ in a throw of dice, getting a head, in a toss of a coin.
→ A random experiment: is an experiement that can be repeated numerous times under the same conditions.
→ Equally likely outcomes: The outcomes of a sample space are called equally likely if all of them have same chance of occurring.
→ Probability of an event A: Written as P(A) in a random experiment and is defined as –
P(A) = \(\frac{\text { Number of outcomes in favour of A }}{\text { Total number of possible outcomes }}\)

Important Properties:
(i) 0 ≤ P(A) ≤ 1
(ii) P (not happening of A) + P (happening of A) = 1
or, P(\(\bar{A}\)) + P(A) = 1
∴ P(\(\bar{A}\)) = 1 – P(A)
Probability of the happening of A = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}}\)
Probability of not happening of A (failing of A) = \(\frac{\mathrm{n}}{\mathrm{m}+\mathrm{n}}\)
where event A can happen in m ways and fail in n ways. All these ways being equally likely to occur.

Problems of Die:
A die is thrown once. The probability of
→ Getting an even number in the throwing of a die: the total number of outcomes is 6. Let A be the event of getting an even number then there are three even numbers 2, 4, 6.
∴ Number of favourable outcomes = 3.
P(A) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{3}{6}=\frac{1}{2}\)

→ Getting an odd number: Total no. of outcomes = 6,
favourable outcomes = 3 i.e. {1, 3, 5}
∴ P(A) = \(\frac{3}{6}=\frac{1}{2}\)
→ Getting a natural number P(A) = \(\frac{6}{6}\) = 1
→ Getting a number which is multiple of 2 and 3 = \(\frac{1}{6}\)
→ Getting a number ≥3 i.e. {3, 4, 5, 6},
P(A) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
→ Getting a number 5 or 6, P(A) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
→ Getting a number ≤5 i.e. {1, 2, 3, 4, 5},
P(A) = \(\frac{5}{6}\)

Problems Concerning Drawing a Card:

• A pack of 52 cards
• Face cards (King, Queen, Jack)

JAC Board Class 9th Science Important Questions Chapter 11 Work and Energy

Multiple Choice Questions

Question 1.
The gravitational potential energy of an object is due to
(a) it’s mass
(b) its acceleration due to gravity
(c) its height above the earth’s surface
(d) all of the above
Answer:
(d) all of the above

Question 2.
The unit of work is the joule. The other physical quantity that has the same unit is
(a) power
(b) velocity
(c) energy
(d) force
Answer:
(c) energy

Question 3.
If the velocity of a body is doubled, its kinetic energy
(a) gets doubled
(b) becomes half
(c) does not change
(d) becomes 4 times
Answer:
(d) becomes 4 times

Question 4.
A student carries a bag weighing 5 kg from the ground floor to his class on the first floor that is 2 m high. The work done by the boy is
(a) 1J
(b) 10J
(c) 100J
(d) 1000J
Answer:
(c) 100J

Question 5.
How much time will be required to perform 520 J of work at the rate of 20 W?
(a) 24s
(b) 16s
(c) 20s
(d) 26s
Answer:
(d) 26s

Question 6.
A body of mass 2 kg is dropped from a height of lm. Its kinetic energy as it touches the ground is
(a) 19.6N
(b) 19.6J
(c) 9.8m
(d) 9.8J
Answer:
(b) 19.6J

Question 7.
The unit of power is
(a) watt per second
(b) joule
(c) kilojoule
(d) joule per second
Answer:
(d) joule per second

Question 8.
A coolie carries a load of 500 N to a distance of 100 m. The work done by him is
(a) 5 Nm
(b) 50,000 Nm
(c) 0 Nm
(d) 1/5 N m
Answer:
(c) 0 Nm

Question 9.
Power is a measure of the
(a) rate of change of momentum
(b) force which produces motion
(c) change of energy
(d) rate of change of energy
Answer:
(d) rate of change of energy

Question 10.
If the speed of an object is doubled, its kinetic energy is
(a) doubled
(b) quadrupled
(c) halved
(d) tripled
Answer:
(b) quadrupled

Question 11.
Which of the following is not correct?
(a) Energy is the ability of doing work
(b) Work can be expressed as F × s
(c) Unit of power is joule
(d) Power is the amount of work done per unit of time
Answer:
(d) Power is the amount of work done per unit of time

Question 12.
kW h is the unit of
(a) acceleration
(b) work
(c) power
(d) energy
Answer:
(c) power

Question 13.
Considering air resistance negligible, the sum of potential and kinetic energies of a free falling body would be
(a) zero
(b) increasing
(c) decreasing
(d) fixed
Answer:
(d) fixed

Question 14.
Two bodies of masses m] and m2 have equal kinetic energies. If P₁ and P₂ are their respective momenta, the ratio of P₁ to P₂ is
(a) m₁ : m₂
(b) m₂ : m₁
(c) \(\sqrt{\mathrm{m}_{1}} : \sqrt{\mathrm{m}_{2}}\)
(d) \(m_{1}^{2} : m_{2}^{2}\)
Answer:
(c) \(\sqrt{\mathrm{m}_{1}} : \sqrt{\mathrm{m}_{2}}\)

Question 15.
A light and a heavy body have equal momenta Which one has greater kinetic energy?
(a) The lighter body
(b) The heavier body
(c) Both have same KE
(d) None of these
Answer:
(b) The heavier body

Analysing & Evaluating Questions

Question 16.
A car is accelerated on a levelled road and attains a velocity four times of its initial velocity. In this process the potential energy of the car
(a) does not change
(b) becomes twice to that of initial
(c) becomes 4 times that of intial
(d) becomes 16 times that of initial
Answer:
(a) does not change

Question 17.
An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same
(a) acceleration
(b) momenta
(c) potential energy
(d) kinetic energy
Answer:
(a) acceleration

Question 18.
A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be (g = 10 m^-2)
(a) 6 × 10³J
(b) 6J
(c) 0.6J
(d) zero
Answer:
(d) zero

Assertion Reason Questions

Directions: In the following questions, the Assertions and the Reasons have been put forward. Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both the statements are false.
1. Assertion: Work, the product of force and displacement, is a vector quantity.
Reason: Product of two vector quantities is always a vector quantity.
Answer:
(D) Both the statements are false.

2. Assertion: When a book is moved from a table to the top of an almirah, its potential energy increases.
Reason: Higher the height of a body from the ground level, higher is its potential energy.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

3. Assertion: A person carrying a suitcase on his head is not doing any work. Reason: The force applied by the person is acting in the downward direction.
Answer:
(C) The assertion is true but the reason is false.

4. Assertion: The work done by the force of gravity on the moon revolving around the earth is zero.
Reason: The gravitational force and the displacement of moon are at right angles to each other.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

5. Assertion: Work done on an object can be positive, negative or zero.
Reason: Work done is a scalar quantity.
Answer:
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.

Very Short Answer Type Questions

Question 1.
Define work.
Answer:
Work is said to be done when a force applied on a body produces a displacement of the body. It is given by W = F x s where ‘F’ is the force applied and ‘s’ is the displacement caused.

Question 2.
State reason why work is a scalar quantity.
Answer:
Work is the product of force (F) and displacement (s). Since both F and s are vector quantities and the dot product of vector quantities produces a scalar quantity, therefore, work is a scalar quantity.

Question 3.
Name two kinds of potential energies.
Answer:
Gravitational potential energy and elastic potential energy.

Question 4.
If the work done is 20J and displacement is 2m, find the force applied.
Answer:
Given, W = 20J and s = 2m.
W = F × s
20 = F × 2
F= 10 N

Question 5.
Name the energy stored in a rubber band when it is stretched?
Answer:
On stretching a rubber band, potential energy is stored in it.

Question 6.
State the law of conservation of energy.
Answer:
It states that energy can neither be created nor destroyed. It can only change its form.

Question 7.
When a book is lifted from a table, against which force is the work done?
Answer:
Work is done against the force of gravity.

Question 8.
Define the commercial unit of energy.
Answer:
The commercial unit of energy is kW h (kilowatt hour). 1 kW h is the energy used in one hour at the rate of 1000 J per second.
1 kWh = 1 kW × 1 h = 3.6 × 10⁶j

Question 9.
A light and a heavy body have equal kinetic energies. Which one is moving faster?
Answer:
The lighter body is moving faster because for the same kinetic energy, velocity is inversely proportional to the mass.
Analysing & Evaluating Question uestions

Question 10.
Two objects of same mass are placed at positions A and B as shown in the figure. Both of them are raised to the position C. In which case, the potential energy is more?
Answer:
The object at A will gain more potential energy than the object at B. But the final potential energy of both A and B will be equal when raised to position C.

Question 11.
A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of the itwo kinetic energies?
Answer:
K.E. °c (velocity)²
When the velocity is tripled, K.E. increases by a factor of 9 ( = 3²)
Thus, the ratio of the two kinetic energies is 1 : 9.

Question 12.
Can any object have momentum even if its mechanical energy is zero? Explain.
Answer:
No. Zero mechanical energy means no potential energy and no kinetic energy. Zero kinetic energy means, zero velocity. As a result, momentum is also zero (as P = mv).

Short Answer Type Questions

Question 1.
Calculate the work done by a man in rotating a wheel of an amusement slide in a fair 40 times in 1 minute?
Answer:
The man is rotating the wheel of an amusement slide by just standing at a place. This concludes that the wheel is not undergoing any displacement. Since displacement is zero, therefore, work done is zero.

Question 2.
Define positive work done and negative work done.
Answer:

1. Positive work (done: Work done is positive when the displacement occurs in the direction of force.
2. Negative work done: Work done is negative when the displacement occurs opposite to the direction of force.

Question 3.
In which of the following cases, work is said to be done?
1. When we push a table and the table is displaced.
2. When a person holds a book in his hand and keeps it stationary.
3. When a wire is twisted.

Answer:

1. When we push a table and the force applied by us is large enough to move it from its original position, then work is said to be done.
2. When a person holds a book in his hand and keeps it stationary, there occurs no movement of the book. In this case, though a force is constantly being applied, there is no displacement and hence work done is zero.
3. When a wire is twisted, the shape of the wire changes which concludes that work is done as there occurred changes in the configuration of the wire.

Question 4.
What will be the nature of work done when the force acting on a body retards its motion? Justify your answer by quoting examples.
Answer:
When force retards the motion of a body, the motion is stopped, i.e., a force opposite to the direction of the motion is applied. Thus, a negative work is done by the force.
For example:

1. In tug of war, the work done by the losing team is negative.
2. When a ball is thrown up in the air, the gravitational force acting downwards upon the ball does negative work on the ball.

Question 5.
What is gravitational potential energy?
Answer:
The gravitational potential energy of an object at a point above the ground
is defined as the work done in raising an object from the ground to that point against gravity.
G_PE = mgh

Question 6.
Differentiate between potential energy and kinetic energy.
Answer:

Question 7.
State two situations where energy is supplied but no work is done.
Answer:
(a) A person pushing a heavy rock is using all the energy but if the rock does not move, no work is done, (b) A person standing with heavy load on his head is spending energy in doing this, but no work is done.

Question 8.
How are work and energy related to each other?
Answer:
An object having a capability to do work is said to possess energy. The object which does work loses energy and the object on which work is done, gains energy. The unit of both energy and work is joule.

Question 9.
On what factor does the gravitational potential energy depend?
Answer:
Gravitational potential energy depends on the height of object from the ground level zero level we choose. Example: A ball tossed from the second floor of a building will attain a height, say h, from its roof, but from the first floor its height will be h ‘where h > h’.
Hence, the potential energy of the ball on the first floor level is less as compared to that on the second floor.

Question 10.
Write the form of energy possessed by the body in the following situations:
(a) A coconut falling from tree
(b) An object raised to a certain height
(c) Blowing wind
(d) A child driving a bicycle on the road
Answer:
(a) Kinetic energy + Potential energy
(b) Potential energy
(c) Kinetic energy
(d) Kinetic energy

Question 11.
What is energy? Give the unit of energy. Name the different forms of energy.
Answer:
Energy of a body is defined as its capacity or ability to do work. When a body is capable of doing more work, it is said to possess more energy.
The SI unit of energy is joule (J).
Energy has many forms: potential energy, kinetic energy, heat energy, chemical energy, electrical energy, light energy, solar energy, etc.

Question 12.
Derive an equation for kinetic energy of an object?
Answer:
The kinetic energy of a body can be determined by calculating the amount of work required to set the body into motion with the velocity ‘v’ from its state of rest. Suppose,
m = mass of the body
u = 0 = initial velocity of the body
F = force applied on the body
a = acceleration produced in the body in
the direction of force
v = final velocity of the body
s = distance covered by the body
As v² – u² = 2as
= v² – 0² = 2as
a= \(\frac{v^{2}}{2 s}\)
As the force and displacement are in the same direction, the work done on the body is
W = Fs = mas = m \( \frac{\mathrm{v}^{2}}{2 \mathrm{~s}}\)s = \(\frac{1}{2}\) mv²
This work done appears as the kinetic energy of the body.
∴ KE = \(\frac{1}{2}\) mv²

Question 13.
Derive an equation for potential energy?
Answer:
Let the work done on the object against gravity be W.
Work done, W = force × displacement
Work done, W = mg × h
Work done, W = mgh
Since work done on the object is equal to mgh, an energy equal to mgh units is gained by the object. This is the potential energy (E_p) of the object.
E_p = mgh

Question 14.
An electric heater of 1000 W is used for 2 hours a day. What is the cost of using it for a month of 28 days, if 1 unit costs ₹ 3.00?
Answer:
Here, P = 1000W = lkW
Total time, t = 2 × 28 hours = 56 hours
Total energy consumed = P × t
= 1 kW × 56 h = 56 kW h
Cost of 1 kWh = ₹ 3.00
Cost of 56 kWh = 3 × 56 = ₹ 168.

Analysing & Evaluating Questions

Question 15.
Two identical pointed objects made from iron and wood are allowed to fall on a heap of sand from the same height. The iron object penetrates more in sand than the wooden object. Which of the objects has more potential energy?
Answer:
Of the two identical objects, the one made from iron will have greater mass. So when it falls from a height, it will possess greater kinetic energy as compared to the wooden object. As a result,
(a) the iron object will penetrate more in sand.
(b) the iron object will have more potential energy.

Question 16.
The weight of a person on a planet A is about half that on the earth. He can jump upto 0.4 m height on the surface of the earth. How high can he jump on the planet A?
Answer:
The weight of a person on planet A is about half that on the earth. This means, the acceleration due to gravity of the planet A is half that of the earth. So, Height of the jump on the surface of planet A = \(\frac{0.4 \mathrm{~m}}{1 / 2}\) = 0.8 m

Long Answer Type Questions

Question 1.
How is work done measured when a body moves in a direction inclined to the direction of the applied force?
Answer:
In the figure, a force F pulls a block making angle 0 with the horizontal surface. Under this force, suppose the block moves from position A to B after covering a distance ‘s’.

Let, F₁ = Component of force in the direction of displacement ‘s’
Then, \(\frac{F_{1}}{F}\) = cos θ or F₁ = F cos θ
Work done = Component of force in the direction of displacement × displacement
W = F₁ × s = F cos θ x s
W = Fs cos θ
Special cases:
(a) When θ = 0°,
cos θ = 1 and W = Fs
Thus, work done is maximum when the displacement of the body is along the direction of the force.

(b) When θ = 90°,
cos θ = 0 and W = 0
Thus, work done is zero when the displacement of the body is perpendicular to the direction of force.

(c) When θ = 180°,
cos θ = -1 and W = – Fs
Thus, work done is negative when displacement is opposite to the direction of force.

(d) When s = 0, W = 0 Thus, work done on a stationary body is zero.

Question 2.
What is meant by potential energy of a body? Give some examples.
Answer:
The energy possessed by a body by the virtue of its position, shape or configuration is called its potential energy.

1. Examples of P.E. due to position:
   • Water stored in a dam at a height has potential energy.
   • A stone lying on the roof of the building has potential energy.
2. Examples of P.E. due to shape:
   • In a toy car, the wound spring possesses potential energy. As the spring is released, its potential energy changes into kinetic energy which moves the toy car.
   • Energy possessed by a stretched rubber band is potential energy.
3. Example of P.E. due to configuration:
   • A stretched bow possesses potential energy. As soon as it is released, it shoots the arrow in the forward direction with a large velocity.
   • The potential energy of the stretched bow gets converted into the kinetic energy of the arrow.
     

Question 3.
A 100 W electric bulb is lighted for 2 hours every day and five 40 W tubes are lighted for 4 hours every day. Calculate:
(a) the energy consumed for 60 days and
(b) the cost of electricity consumed at the rate of ₹3 per kW h.
Answer:
(a) Energy consumed by a 100 W bulb each day = 100 W × 2 h
= 200 Wh = \(\frac{200}{1000}\)= 0.2 kW h
Energy consumed by five 40 W tubes each day = 5 × 40 W × 4h
= 800 Wh = \(\frac{800}{1000}\) = 0.8 kW h
Total energy consumed each day = 0.2 + 0.8 = 1.0 kWh
Total energy consumed in 60 days = 1.0 x 60 = 60 kWh

(b) Cost of 1 kW h = ₹ 3
Cost of 60 kW h = 3 × 60 = 180

Question 4.
Answer the following:
(a) List any three situations in your daily life where you can say that work has been done.
(b) How much work is done in increasing the velocity of a car from 15 km/h to 30 km/h if the mass of the car is 1000 kg?
Answer:
(a) Three situations where work is done are:

1. Pushing a pebble lying on the ground. The pebble moves through some distance. Here, we apply a force and the pebble gets displaced. So, we have done work on the pebble.
2. We apply a force to lift a book through a height. The book rises up. We have done work in moving up the book.
3. A bullock is pulling a cart and the cart moves. There is a force on the cart and the cart has moved. The bullock has done work on the cart.

(b) According to the question,
u = 15 km/h = 4 m/s
v = 30 km/h = 8 m/s
Mass = 1000 kg
W = ?
W = K.E. = \(\frac{1}{2}\) m(v² – u²)
\(\frac{1}{2}\) × 1000 × ((8)² – (4)²)
= \(\frac{1}{2}\) × 1000 × (64 – 16)
= \(\frac{1}{2}\) × 1000 × (675) = 24,000J
Hence, work done is 24,000J.

Analysing & Evaluating Questions

Question 5.
A boy is moving on a straight road against a frictional force of 5N. After traveling a distance of 1.5 km he forgot the correct path at a roundabout of radius 100 m. However, he moves on that circular path for one and half cycle and then he moves forward up to 2.0 km. Calculate the work done by him.
Answer:
Work done by the boy while moving on a straight road
W = F × s
W = 5N × 1.5 km
= 5 kg m s^-2 × 1500 m = 7500J
Work done during moving around circular path
= 5N × (2 × 100 m) = 1000J
Work done during moving further by
2.0 km = 5N × (2 × 1000 m) = 10,000J
Total work done by the boy = 7500J + 1000J + 10,000J
= 18500J

Activity 1

• Take a heavy ball. Drop it on a thick bed of sand. A wet bed of sand would be better. Drop the ball on the sand bed from a height of about 25 cm and observe the results.
• Repeat this activity from heights of 50 cm, lm and 1.5 m.
• Ensure that all the depressions are distinctly visible.
• Mark the depressions to indicate the height from which the ball was dropped.
• Compare their depths.

Observations

• The ball that falls from the height of 1.5 m creates the deepest depression.
• The ball that falls from the height of 50 cm creates the shallowest depression.
• Larger the height from which the ball is dropped, larger is the kinetic energy gained by the ball on reaching the ground and more is its capability of doing work.

Activity 2

1. Set up the apparatus as shown in the figure.
2. Place a wooden block of known mass in front of the trolley at a convenient fixed distance.
3. Place a known mass on the pan so that the trolley starts moving.
4. The trolley moves forward and hits the wooden block.
5. Fix a stop on the table in such a manner that the trolley stops after hitting the block. The block gets displaced.
6. Note down the displacement of the block.
7. Repeat this activity by increasing the mass on the pan. Observe, in which case is the displacement more.
   

Observations

1. The force of gravity pulls the mass in the pan in the downward direction. This force gets transferred to the trolley through the string.
2. The trolley moves and hits the block with a force.
3. The larger the mass in the pan, the larger is the force with which the trolley hits the block.
4. Consequently, larger will be the displacement and larger will be the work done. The moving trolley possess energy and hence does work on the block.

Activity 3

• Take a slinky as shown below.
• Ask a friend to hold one of its ends. You hold the other end and move away from your friend. Now, you release the slinky and observe.
  

Observations

• When released, the slinky regains its original length. The slinky has acquired potential energy due to the work done on it during stretching. On releasing, potential energy is converted into kinetic energy.
• The slinky will also acquire energy when it is compressed.

Value Based Questions

Question 1.
Apoorva saw few planter pots kept on the balcony sill of fourth floor in her building. She makes an effort and keeps all the planter pots down the sill.
1. What type of energy is present in the pot kept on the balcony sill of fourth floor?
2. If the pot falls from the fourth floor, what type of energy will be seen in the falling pot?
3. What value of Apoorva is reflected in the above act?
Answer:
1. The pot on the sill possesses potential energy.
2. The falling pot possesses both the kinetic energy and the potential energy.
3. Apporva showed the value of moral responsibility and awareness.

Question 2.
Siddharth saw a lady labourer who carried bricks on her head from one point of the construction site to the other end which was some 500 m away. He prepares a trolley for the labourer to carry the bricks to make her work easier.
1. In carrying the bricks from point A to point B on the head by the lady labourer in the construction site, is any work done by the labourer?
2. By pulling the trolley of bricks from point A to point B, is any work done?
3. What value of Siddharth is seen in the above act?
Answer:
1. In carrying the bricks from point A to point B on head by the lady, no work is said to be done.
2. By pulling the trolley of bricks, work is said to be done.
3. Siddharth showed kindness, general awareness and sympathy.

JAC Class 9 Science Important Questions