Class 9

Haryana State Board HBSE 9th Class Maths Notes Chapter 6 Lines and Angles Notes.

Haryana Board 9th Class Maths Notes Chapter 6 Lines and Angles

Introduction
In previous chapter, we have already studied axioms, postulate and some terms allied to geometry. In this chapter, we shall study the angles and their properties and also the properties of the angles formed when a line intersects two or more parallel lines at distinct points.

An architect cannot draw the plan of the multistoried building without the knowledge of the properties of parallel lines, intersecting lines and angles.

In science, to study the refraction property of light when a ray enters from one medium to other medium, we use the properties of parallel lines and intersecting lines. We have to represent forces by direct line segments to study the net effect of the forces on the body.

Key Words
→ Architect: One who designs building and supervises their construction.

→ Congruent: Having the same size and shape.

→ Adjacent: Lying near to.

→ Intersect: Two geometrical figures or curves are said to intersect if they have at least one point in common.

→ Converse: The converse of an implication p ⇒ q is the implication q ⇒ p. If an implication is true, then its converse may or may not be true.

→ Concurrent: A number of lines are said to be concurrent, if there is a point through which they all pass.

→ Incident Ray: The incident ray is the ray of light that strikes the surface before reflection, transmission or absorption.

→ Reflected Ray: A ray extending outward from a point.

→ Angle of Incidence: The angle between the direction of an approaching emission and the normal to the surface upon which it is incident.

→ Angle of Reflection: The angle between the direction of propagation of a reflected emission and the normal to the reflecting surface.

1. Basic Terms and Definitions : (a) Angle: An angle is formed when two rays or two line segments have a common end point. The two rays which form an angle are called the arms and the point at which they meet is called the vertex of the angle.

The adjacent figure represent an angle ABC or ∠ABC or ∠CBA or simply B. AB and BC are the arms of the angle and their common point B is the vertex.

∠ABC divides the plane containing it into two parts:
(i) Interior of an angle
(ii) Exterior of an angle.
(i) Interior of an angle: The interior of an angle ABC is the set of all points in its plane, which lie on the same side of BC as A and also on the same side of AB as C. In the adjoining figure, P is a point in the interior of an angle ABC.

(ii) Exterior of an angle : The exterior of an ∠ABC is the set of all points in its plane, which do not lie on the angle or in its interior.
In the adjoining figure, the point Q is exterior of the ∠ABC.
Note: Point R lies on the angle ∠ABC.

(b) Measure of an angle: The amount of turning of a ray from the initial position of BC to BA is called the measure of ∠ABC which is written as m∠ABC. The unit of angle measure is a standard angle, called a degree.

If the measure of an angle ABC is x degree, then we write
m∠ABC = x°.
Note: 1° = 60′ (i.e., 60 minutes),
1′ = 60″ (60 seconds)

(c) Congruent or equal angles: If the measures of the angles are same then they said to be equal or congruent.
For examples:
If m∠ABC = m∠PQR,
then ∠ABC ≅ ∠PQR
or if ∠ABC ≅ ∠DEF,
then m∠ABC = m∠DEF.

(d) Types of angles: (i) Right angle: The angle which measures equal to 90°, is called a right angle. In the figure 6.4, ∠PQR is a right angle.

(ii) Acute angle: The angle which measures less than one right angle (i.e., less than 90°) is called an acute angle.
In the figure 6.5, ∠PQR is an acute angle.

(iii) Obtuse angle: The angle which measures more than 90° and less than 180° is called an obtuse angle. In the figure 6.6, ∠PQR is an obtuse angle.

(iv) Straight angle: The angle which measures equal to two right angles (i.e., equal to 180°) is called a straight angle.
In the figure 6.7, ∠PQR is a straight angle.

(v) Reflex angle: The angle which measures more than 180° and less than 360° is called a reflex angle.
In the figure 6.8, ∠PQR is a reflex angle.

(vi) Complete angle: The angle which measures is 4 right angles (i.e., 360°) is called a complete angle.
In the figure 6.9, ∠QPR is a complete angle.

(vii) Complementary angles: If the sum of two angles is 90°, they are called complementary angles eg., if ∠a + ∠b= 90°, then ∠PQS and ∠RQS are complementary angles.

(viii)Supplementary angles: If the sum of two angles is 180°, they are called. supplimentary angles eg., if ∠x + ∠y = 180°, then ∠PQS and ∠RQS are supplementary angles.

(e) Bisetor of an angle: A ray is called the bisector of an angle if it divides the angle into two equal parts.

In the figure 6.12, ray QS is the bisector of ∠PQR, then
∠PQS = ∠RQS.

2. Pairs of Angles: (i) Adjacent angles: Two angles are called adjacent angles, if:
(i) they have the same vertex.
(ii) they have a common arm.
(iii) other arms of these angles are on opposite sides of the common arm.

In the figure 6.13, ∠AOC and ∠BOC have the common vertex O and have a common arm OC and other arms OA and OB lie on the opposite sides of the common arm OC. Therefore, ∠AOC and ∠BOC are adjacent angles.

(ii) Linear pair angles: Two adjacent angles are said to form a linear pair of angles, if their non-common arms are two opposite rays.

In the figure 6.14, ∠AOC and ∠BOC are two adjacent angles and OA and OB are two opposite rays. Therefore, ∠AOC and ∠BOC form a linear pair of angles.

Axiom 6.1: If a ray stands on a line, then the sum of two adjacent angles so formed is 180°.
Recall that when the sum of two adjacent angle is 180°, then they are called a linear pair of angles.

In the figure 6.15, ray OB stands on the line AC. It forms adjacent angles ∠AOB and ∠COB.
∵ ∠AOB + ∠COB = 180° and OA and OC are opposite rays. Thus, two angles form a linear pair of angles. The above axiom can be stated in the reverse ways as below:
Axiom 6.2: If the sum of two adjacent angles is 180°, then the non-common arms of the angles form a line.

In the figure 6.16, ∠AOB and ∠COB are adjacent angles, such that
∠AOB + ∠COB = 180°
⇒ ∠AOC = 180°,
[∵ ∠AOC = ∠AOB + ∠COB]
⇒ ∠AOC is a straight line.
Hence, AOC is a straight line.
The two axioms above together is called the linear pair axiom.

(iii) Vertically opposite angles: Two angles are called a pair of vertically opposite angles, if their arms form two pairs of opposite rays.

In the figure 6.17, two lines AB and CD intersect at O, then two pairs of vertically opposite angles are formed.
(i) ∠AOC and ∠BOD
(ii) ∠AOD and ∠BOC.

Theorem 6.18.
If two lines intersect each other, then the vertically opposite angles are equal.
Given: Two lines AB and CD intersect at a point O.
To prove : (i) ∠AOC = ∠BOD
(ii) ∠AOD = ∠BOC.
Proof. Since ray OA stands on line CD.
∴∠AOC + ∠AOD = 180°,
(Linear pair axiom) …….(i)
Ray OD stands on line AB.
∴ ∠AOD + ∠BOD = 180°,
(Linear pair axiom) …….(ii)
From (i) and (ii), we have
∠AOC + ∠AOD = ∠AOD + ∠BOD
⇒ ∠AOC = ∠BOD
Similarly, we can prove that
∠AOD = ∠BOC. Hence proved

Parallel Lines and a Transversal: A line which intersects two or more given lines at distinct points is called a transversal of the given lines.

In the figure 6.32, AB and CD are two lines and a transversal EF intersects them at P and Q respectively. The three lines determine eight angles, four angles at P namely ∠1, ∠2, ∠3, ∠4 and remaining four angles at Q namely ∠5, ∠6, ∠7 and ∠8.

In the figure, ∠2, ∠3, ∠5 and ∠8 are called interior angles and ∠1, ∠4, ∠6 and ∠7 are called exterior angles. We classify these eight angles in the following groups:
(a) Corresponding angles: (i) ∠1 and ∠5, (ii) ∠4 and ∠8, (iii) ∠2 and ∠6, (iv) ∠3 and ∠7.
(b) Alternate interior angles: (i) ∠2 and ∠8, (ii) ∠3 and ∠5.
(c) Alternate Exterior angles: (i) ∠1 and ∠7, (ii) ∠4 and ∠6.
(d) Interior angles on the same side of the transversal: (i) ∠2 and ∠5, (ii) ∠3 and ∠8.
Interior angles on the same side of the transversal are also referred to as consecutive interior angles or allied angles or co-interior angles.

Some Important Relations
If AB || CD and transversal XY cuts them (see in the figure 6.33) then

(i) Corresponding angles are equal i.e., ∠1 = ∠5, ∠4 = ∠8, ∠2 = ∠6 and ∠3 = ∠7.
(ii) Alternate interior angles are equal i.e., ∠2 = ∠8 and ∠3 = ∠5.
(iii) Alternate exterior angles are equal i.e., ∠1 = ∠7 and ∠4 = ∠6.
(iv) Co-interior angles are supplementary i.e., ∠2 + ∠5 = 180° and ∠3 + ∠8 = 180°.
From this we conclude the following axioms:
Axiom 6.3: If a transversal intersects two parallel lines, then each pair of corresponding angles are equal. It is called corresponding angles axiom also.
Axiom 6.4: If a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines are parallel to each other. It is called converse of corresponding angles axiom.

Theorem 6.2:
If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.
Given: A transversal line XY cuts the parallel lines AB and CD at P and Q respectively.

To prove : ∠2 = ∠8 and ∠3 = ∠5.
Proof :
∠4 = ∠2 ……(i)
(vertically opposite angles)
∠4 = ∠8 ……(ii)
(corresponding angles axiom)
From (i) and (ii), we get
∠2 = ∠8
Again, ∠1 = ∠3 ……(iii)
(vertically opposite angles)
∠1 = ∠5 ……(iv)
(corresponding angles axiom)
From (iii) and (iv), we get
∠3 = ∠5
Hence, ∠2 = ∠8 and ∠3 = ∠5.
Proved

Theorem 6.3:
(Converse of theorem 6.2) : If a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel.

Given: A transversal line XY intersects two lines AB and CD at P and Q respectively such that ∠2 and ∠8 are a pair of alternate interior angles and ∠2 = ∠8.
To prove : AB || CD.
Proof :
∠2 = ∠8, (given) ……(i)
∠2 = ∠4 …….(ii)
(vertically opposite angles)
From (i) and (ii), we get
∠4 = ∠8
Thus, a pair of corresponding angles 24 and 28 are equal. Therefore, by converse of corresponding angle axiom, we have
AB || CD.
Hence proved

Theorem 6.4:
If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.

Given: A transversal line XY intersects two parallel lines AB at P and CD at Q making two pairs of consecutive interior angles ∠2, ∠5, ∠3 and ∠8.
To prove :
∠2 + ∠5 = 180° and ∠3 + ∠8 = 180°.
Proof: Ray PB stands on line XY.
∴ ∠3 + ∠4 = 180
(Linear pair axiom)
But ∠8 = ∠4
(corresponding angles axiom)
∴ ∠3 + ∠8 = 180° ……(i)
Now, ray QC stands on line XY.
∴ ∠5 + ∠6 = 180°,
(Linear pair axiom)
But ∠6 = ∠2,
(corresponding angles axiom)
∴ ∠5 + ∠2 = 180°
⇒ ∠2 + ∠5 = 180° ……(ii)
Hence, ∠2 + ∠5 = 180°
and ∠3 + ∠8 = 180°.
Proved.

Theorem 6.5.
(Converse of theorem 6.4): If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel.

Given: A transversal line XY intersects two lines AB at P and CD at Q such that ∠2 and ∠5 are pair of consecutive interior angles and ∠2 + ∠5 = 180°.
To prove : AB || CD.
Proof: Since, Ray PA stands on line XY.
∴ ∠1 + ∠2 = 180°
(Linear pair axiom) …..(i)
∠2 + ∠5 = 180° (Given) …..(ii)
From (i) and (ii), we get
∠1 + ∠2 = ∠2 + ∠5
⇒ ∠1 = ∠5
Thus, a pair of corresponding angles ∠1 and ∠5 are equal.
Therefore, by converse of corresponding angles axiom, we have
AB || CD. Hence proved.

Lines Parallel to Same Line:
Theorem 6.6:
Lines which are parallel to the same line are parallel to each other.

Given: l || m and n || m.
To prove : l || m
Construction: Draw a transversal lines t for the lines l, m and n.
Proof: Since, l || m (Given)
∠1 = ∠2 …….(i)
(Corresponding angles axiom)
Since, n || m (Given)
∠3 = ∠2 …….(ii)
(Corresponding angles axiom)
From (i) and (ii), we get
∠1 = ∠3
But these are corresponding angles.
⇒ l || n,
[By converse of corresponding angles]
Hence proved

1. Triangle: A plane figure bounded by three straight lines, is called a triangle. It is usually denoted by the Greek letter Δ (delta).
The figure, given along side, shows a triangle ABC (ΔABC) bounded by three sides AB, BC and CA. Triangle ABC has three vertices, namely A, B and C and it has three angles, namely ∠A, ∠B and ∠C.
Note: If side BC of ΔABC is produced upto any point D, the angle ∠ACD so formed, is called exterior angle at C, where as the angles A and B are called its interior opposite angles.

2. Kinds of Triangles: (a) With regard to their sides:
(i) Equilateral triangle: If all the sides of a triangle are equal, it is called an equilateral triangle.
In the given figure, ΔABC is an equilateral triangle because AB = BC = CA. Also, all the angles of an equilateral triangle are equal to each other and so each angle = 60°.

(ii) Isosceles triangle: If at least two sides of a triangle are equal, it is called an isosceles triangle. In the given figure, ΔABC is an isosceles triangle, because AB = AC.

In an isosceles triangle, two angles of opposite equal side are also equal.

(iii) Scalene triangle: If all the sides of a triangle are unequal, it is called an scalene triangle.
The figure below shows a scalene triangle because AB ≠ BC ≠ AC.

In a scalene triangle, all its angles are also unequal.

(b) With regard to their angles: (i) Acute angled triangle: When each angle of a triangle is acute (i.e., less than 90°), it is called an acute angled triangle.
The given figure 6.60, shows an acute angled triangle, because each angle of the triangle is less than 90°.

(ii) Right angled triangle: When one angle of a triangle is right angle (ie., equal to 90°), it is called right angled triangle, the side opposite to the right angle is called its hypotenuse and the remaining two sides are called its legs.

The given figure 6.61, shows a right angled triangle because ∠B = 90°. In triangle ABC, AC is hypotenuse and AB and BC are its legs. Each of the other two angles of the right triangle is acute.

(iii) Obtuse angled triangle: When one angle of a triangle is obtuse (i.e., greater than 90° but less than 180°), it is called an obtuse angled triangle. The given figure 6-62, shows an obtuse angled triangle, because B is obtuse.

In an obtuse angled triangle, each of the other two angles is acute.

3. Some Important Terms Related to the Traingle: (i) Median : The median of a triangle, corresponding to any side, is the line joining the mid point of that side with the opposite vertex. e.g.,

AL is the median corresponding to side BC

CN is the median corresponding to side AB

A triangle has three medians and all three medians are always concurrent i.e., they intersect each other at one point only.

(ii) Centroid: The point of intersection of all the three medians of a triangle is called its centroid. In the given tigure 6-66, three medians AD, BE and CF intersect at the point G. Therefore, G is the centroid of the triangle ABC.

(iii) Altitude: An altitude of a triangle, corresponding to any side, is the length of perpendicular drawn from the opposite vertex to that side. e.g.,

AL is the altitude corresponding to side BC

BM is the altitude corresponding to side AC

CN is the altitude co-rresponding to side AB A triangle has three altitudes and all three altitudes are always concurrent i.e., they intersect each other at one point only.

(iv) Ortho centre: The point of intersection of all the three altitudes of a triangle is called ortho centre.
In the given figure 6.69, O is the ortho centre of triangle ΔABC.

4. Angle sum property of a triangle: In previous classes, we have studied that sum of the angles of a triangle is 180°. We can prove this statement using the theorem related to parallel lines.

Theorem 6.7:
The sum of the angles of a triangle is 180°.

Given: A triangle ΔABC and ∠1 + ∠2 + ∠3 are its angles
To prove : ∠1 + ∠2 + ∠3 = 180°.
Construction: Through A, draw a line PQ parallel to BC.
Proof: PQ || BC and AB is the transversal.
∠2 = ∠4 ……(i)
(Alternate interior angles)
Again, PQ || BC and AC is the transversal.
∴ ∠1 = ∠5, ……(ii)
(Alternate interior angles)
Adding (i) and (ii), we get
∠2 + ∠1 = ∠4 + ∠5
⇒ ∠1 + ∠2 + ∠3 = ∠4 + ∠3 + ∠5,.
(Adding ∠3 on both sides)
But, ∠4 + ∠3 + ∠5 = 180°,
(Linear pair axiom)
∴ ∠1 + ∠2 + ∠3 = 180°
Hence, the sum of three angles of a triangle is 180°.
Hence Proved

Theorem 6.8:
If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

Given: A ΔABC whose side BC has been produced to D such that ∠4 is its exterior angle.
To prove : ∠4 = ∠2 + ∠3.
Proof: We know that,
Sum of the angles of a triangle is 180°,
(By theorem 6.7)
⇒ ∠1 + ∠2 + ∠3 = 180°…(i)
∠1 + ∠4 = 180°…(ii)
(Linear pair axiom)
From (i) and (ii), we get
∠1 + ∠4 = ∠1 + ∠2 + ∠3
⇒ ∠4 = ∠2 + ∠3. Hence proved

Corollary: An exterior angle of a triangle is greater than either of its interior opposite angles.

Proof: We know that an exterior angle of a triangle is equal to the sum of the interior opposite angles (By theorem 6.8)
∴ ∠4 = ∠2 + ∠3
∴ ∠4 > ∠2 and ∠4 > ∠3
Hence, an exterior angle of a triangle is greater than each interior opposite angle.

Haryana State Board HBSE 9th Class Maths Notes Chapter 7 Triangles Notes.

Haryana Board 9th Class Maths Notes Chapter 7 Triangles

Introduction
In previous classes, we have learnt about triangles and their various properties. Recall that, a closed figure formed by three line segments, is called a triangle. It is usually denoted by the greek letter Δ (delta).

The figure 7.1 shows a triangle ABC denoted as a ΔABC. A triangle has three sides namely AB, BC and CA, three vertices namely A, B and C and has three angles namely ∠A, ∠B and ∠C.

In this chapter we shall learn about the congruence of triangles, criteria for congruence of triangles, some more properties of triangles and inequalities in a triangle.

Key Words
→ Equalities: The logical relation expressing identity or sameness and denoted by the sign “=”.

→ Axiom: A statement whose truth is either to be taken as self evident or to be assumed.

→ Inequalities: Refers to a statement that a quantity is either greater than or less than but not equal to them.

→ Perimeter: The perimeter of a two dimensional region is the length of its boundary.

→ Concentrate: On one point, to pay attention upon.

→ Slides: A gliding seat or surface for sliding.

→ Swing: To move with to and fro motion.

→ Hexagon: It is a polygon bounded by six straight line. (six-sided figure).

Basic Concepts
1. (a) Definition of Congruence: If two geometrical figures coincide exactly, by placing one over the other, the figures are said to be congruent to each other.
(i) Two line segments are congruent if and only if their lengths are equal. (See fig. 7.2).

Two line segments AB and CD are congruent, if AB = CD.

(ii) Two angles are congruent if and only if their measures are equal. (See fig. 7-3)

Two angles ABC and PQR are congruent if m∠ABC = m∠PQR.

(iii) Two circles are congruent if and only if lengths of their radii are equal. (See fig. 7.4)

Two circles C and C2 are congruent, if r1 = r2.

(iv) Two squares are congruent if and only if they have equal sides in length. (See fig. 7.5)

Two squares ABCD and EFGH are congruent, if AB = BC = CD = DA = EF = FG = GH = HE.

(b) Congruence of Triangles: The three sides and three angles of a triangle are called its six parts or six elements. Two triangles are said to be congruent to each other, if on placing one over the other, they exactly coincide.

Let ΔABC is placed over ΔDEF, such that vertex A falls on vertex D and side AB falls on side DE, then if the two triangles coincide with each other in a such a way that B falls on E, C falls on F; side BC coincides with side EF and side AC coincides with side DF, then the two triangles are congruent to each other. (See fig. 7.6)

The symbol used for congruence is “≅”.
ΔABC is congruent to ΔDEF is written as:
ΔABC ≅ ΔDEF.
In case of congruent ΔABC and ΔDEF as given above, the sides of the two triangles, which coincide with each other are called corresponding sides. Thus the sides AB, BC and CA of AABC are corresponding to the side DE, EF and FD of ADEF respectively. In the same way, the angles of the two traingles which coincide with each other, are called corresponding angles. Thus three pairs of corresponding angles are ∠A and ∠D, ∠B and ∠E and ∠C and ∠F.

It follows from the above discussion that if ΔABC coincide ΔDEF exactly such that the vertices of ΔABC fall on the vertices of ΔDEF, in the order
A ↔ D, B ↔ E, C ↔ F.

Then we have the following six equalities:
(i) AB = DE, BC = EF and CA = FD i.e. corresponding sides are equal (congruent).
(ii) ∠A = ∠D, ∠B = ∠E and ∠C = ∠F i.e. corresponding angles are equal (congruent). Hence, following is the general condition for congruence of two triangles:
Two triangles are congruent if and only if there exists a correspondence between their sides and vertices such that the corresponding sides and the corresponding angles of two triangles are equal.

If ΔABC is congruent to ΔDEF and the correspondence ABC ↔ DEF makes the six pairs of corresponding parts of the two triangles congruent, then we write ΔABC ≅ ΔDEF.
But it will not be correct to write ΔCBA ≅ ΔDEF because corresponding parts of ΔCBA and ΔDEF will not be equal i.e., congruent.
Note: The corresponding parts of congruent triangles (abbreviated as CPCT).

(c) Congruence relation in the set of all triangles: From the definition of congruence triangles and notation above, we obtain the following results:
(i) ΔABC ≅ ΔABC
[Congruence relation is reflexive]
(ii) If ΔABC ≅ ΔDEF,
then ΔDEF ≅ ΔABC,
[Congruence relation is symmetric]
(iii) If ΔABC ≅ ΔDEF
and ΔDEF ≅ ΔKLM,
then ΔABC ≅ ΔKLM,
[Congruence relation is transitive]

2. Criteria for Congruence of Triangles:
Axiom 7.1 (SAS congruence rule): Two triangles are congruent if two sides and included angle of one triangle are equal to the two sides and the included angle of the other triangle.
Given: ΔABC and ΔDEF in which AB = DE, AC = DF and ∠A = ∠D. (See fig. 7.7).

To prove : ΔABC ≅ ΔDEF.
Proof: Place ΔABC over ΔDEF such that vertex A falls on vertex D and side AB falls on side DE. Since side AB side DE. Therefore vertex B falls on vertex E. Since ∠A = ∠D, therefore AC will fall on DF.
But AC = DF and A falls on D.
∴ C will fall on F.
Thus AC will coincide with DF.
Now B falls on E and C falls on F.
∴ BC will coincide with EF.
∴ ΔABC coincide with ΔDEF.
Hence, by definition of congruence ΔABC ≅ ΔDEF.
Note: (1) This axiom refers two sides and the angle between them so it is known as (side-angle-side) or the “SAS congruence axiom” or the “SAS” criteria

(2) SAS axiom is not applicable when two sides and one angle ‘not included between these sides’ of one triangle are equal to two sides and one angle of the other triangle.

Theorem 7.1 (ASA congruence rule):
Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle.
Given: Two triangles ABC and DEF, such that
∠B = ∠E, ∠C = ∠F and BC = EF.
To prove : ΔABC ≅ ΔDEF.
Proof: For proving the congruence of the two triangles there are three cases arise:
Case I: Let AB = DE. (See figure 7.8)

In this case, we have
AB = DE, (Assumed)
∠B = ∠E, (Given)
and BC = EF, (Given)
So, ΔABC ≅ ΔDEF, (By SAS axiom)

Case II: Let if possible AB > DE, so we can take a point P on AB such that PB = DE. Now consider ΔPBC and ΔDEF.

In ΔPBC and ΔDEF, we have
PB = DE, (By construction)
∠B = ∠E, (Given)
and BC = EF, (Given)
So, ΔPBC ≅ ΔDEF,(By SAS axiom)
⇒ ∠PCB = ∠DFE, (CPCT)
But ∠ACB = ∠DFE, (Given)
∴ ∠ACB = ∠PCB
This is possible only when P coincide with A, therefore AB must be equal to DE.
Thus, in ΔABC and ΔDEF, we have
AB = DE, (As proved above)
∠B = ∠E, (Given)
and BC = EF, (Given)
So, ΔABC ≅ ΔDEF, (By SAS axiom)

Case III: If AB < DE, we can choose a point M on DE such that ME = AB and repeating the arguments as given in Case (II), we can conclude that AB = DE and so we can prove that

ΔABC ≅ ΔDEF
Hence, in all three cases, we have
ΔABC ≅ ΔDEF. Proved

Corollary (AAS congruence rule):
If any two angles and non-included side of one triangle are equal to the corresponding angles and side of another triangle, then two triangles are congruent.
Given: Two triangles ABC and DEF, such that
∠A = ∠D, ∠B = ∠E
and BC = EF.
To prove : ΔABC ≅ ΔDEF
Proof: Since, we know that sum of the interior angles of a triangle is 180°.

∴ ∠A + ∠B + ∠C = ∠D + ∠E + ∠F = 180°…(i)
But ∠A = ∠D and ∠B = ∠E, (given) …….(ii)
∠A + ∠B + ∠C = ∠A + ∠B + ∠F,
[From (i) and (ii)]
⇒ ∠C = ∠F
In ΔABC and ΔDEF, we have
∠B = ∠E, (Given)
∠C = ∠F, (As proved above)
and BC = EF, (Given)
So, ΔABC ≅ ΔDEF
(By ASA congruence rule)

Some Properties of a Triangle: In chapter 6 we have studied the triangle and their kinds. Recall that, if a triangle has two sides equal, then it is called an isosceles triangle. Now we shall study some results related to isosceles triangles.

Theorem 7.2:
Angles opposite to equal
Given: In ΔABC, AB = AC.
To prove: ∠B = ∠C. sides of an isosceles triangle are equal.

Construction: Draw AD, the bisector of ∠A which meets BC on D.
Proof: In ΔABD and ΔACD, we have
AB = AC, (Given)
∠BAD = ∠CAD, (By construction)
and AD = AD, (Common)
ΔABD ≅ ΔACD,
(By SAS congruence rule)
⇒ ∠B = ∠C. Hence proved.

Theorem 7.3:
[Converse of Theorem 7.2]: The sides opposite to equal angles of a triangle are equal.
Given: In ΔABC,
∠B = ∠C.
To prove: AB = AC.

Construction: Draw AD perpendicular to BC. It meets BC on D.
Proof: In ΔABD and ΔACD, we have
∠ABD = ∠ACD, (Given)
∠ADB = ∠ADC,
(By construction each = 90°)
and AD = AD, (Common)
∴ ΔABD ≅ ΔACD, (By AAS congruence rule)
⇒ AB = AC (CPCT)
Hence proved

Some More Criteria for Congruence of Triangles :
Theorem 7.4 (SSS congruence rule): If three sides of one triangle are equal to the three sides of another triangle, then two triangles are congruent.
Given: ΔABC and ΔPQR such that AB = PQ, BC = QR and AC = PR.
To prove : ΔABC = ΔPQR.

Construction: Suppose BC is the longest side of ΔABC. Draw QS and RS such that ∠SQR = ∠ABC and ∠SRQ = ∠ACB. Join P to S.
Proof: In ΔABC and ΔSQR, we have
BC = QR, (Given)
∠ABC = ∠SQR, (By construction)
and ∠ACB = ∠SRQ, (By construction)
∴ ΔABC ≅ ΔRQS,
(By ASA congruence rule)
⇒ ∠BAC = ∠QSR, (CPCT) …(i)
AB = QS, (CPCT)
and AC = SR, (CPCT)
Now AB = PQ (given)
and AC = PR (given) …(ii)
AB = QS
and AC = SR (as proved above)…(iii)
From (ii) and (iii), we get
PQ = QS and PR = SR
⇒ ∠QPS = ∠QSP…(iv)
and ∠RPS ∠RSP…(v)
Adding (iv) and (v), we get
∠QPS + ∠RPS = ∠QSP + ∠RSP
⇒ ∠QPR = ∠QSR
and ∠BAC = ∠QSR, [From (i)]
∴ ∠BAC = ∠QPR …(vi)
Now in ΔABC and ΔPQR, we have
AB = PQ, (Given)
∠BAC = ∠QPR, [From (vi)]
and AC = PR, (Given)
∴ ΔABC = ΔPQR (By SAS congruence rule).
Hence proved

Theorem 7.5.
[RHS congruence rule]: If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.
Given: In ΔABC and ΔDEF,
∠B = ∠E = 90°, AC = DF and BC = EF.
To prove: ΔABC ≅ ΔDEF.

Construction: Produce DE to G so that GE = AB. Join G to F.
Proof: In ΔABC and ΔGEF, we have
AB = GE, (by construction)
∠ABC = ∠GEF, [Each = 90°]
and BC = EF, (Given)
∴ ΔABC ≅ ΔGEF, (By SAS congruence rule)
⇒ ∠A = ∠G, (CPCT) …(i)
AC = GF, (CPCT) …(ii)
Now, AC = DF, (Given)
AC = GF, [from (ii)]
∴ DF = GF
⇒ ∠D = ∠G …(iii)
(Angles opposite to equal sides are equal)
From (i) and (iii), we get
∠A = ∠D…(iv)
In ΔABC and ΔDEF, we have
∠B = ∠E, (Each = 90°)
∠A = ∠D, [from (iv)]
∴ ∠C = ∠F …(v)
[third angles of the triangles]
Now in ΔABC and ΔDEF, we have
BC = EF, (Given)
∠C = ∠F, [From (v)]
and AC = DF, (Given)
∴ ΔABC = ΔDEF, (by SAS congruence rule)
Hence proved
Note: RHS stands for Right angle-Hypotenuse-Side.

Inequalities in a Triangle: In this section, we shall learn about some inequality relations among sides and angles of a triangle. Consider scalene triangle ΔABC such that AB = 7 cm, BC = 6 cm and AC = 5 cm. Now measure the three angles A, B and C with the help of protector, we observe that:

(i) AB > AC and ∠C > ∠B
i.e., the longer side has greater angle opposite to it.
(ii) AC < BC and ∠B < ∠A.
i.e., shorter side has smaller angle opposite to it.
(iii) AB > BC and ∠C > ∠A
i.e., angle opposite the larger side is greater.
Below we give some important results of inequalities in a triangle.
It is stated in the form of a theorem.

Theorem 7.6:
If two sides of a triangle are unequal, the angle opposite to the longer side is larger or (greater). (See fig. 7.62)

Given: A triangle ABC in which AB > AC.
To prove: ∠ACB > ∠ABC.
Construction: From AB, cut AD = AC.
Join C and D.
Proof: In ΔACD, we have
AC = AD, (By construction)
⇒ ∠ACD = ∠ADC …..(1)
(Angles opposite to equal sides are equal)
In ΔBDC, we have
Exterior ∠ADC > ∠B
(exterior angle of a triangle is always greater than each its opposite interior angle)
⇒ ∠ACD > ∠B, [From (i)]
⇒ ∠ACB > ∠B, [∵ ∠ACB > ∠ABC]
⇒ ∠ACB > ∠ABC

Theorem 7.7:
In any triangle, the side opposite to the larger (greater) angle is longer.
Given: A triangle ABC in which ∠ABC > ∠ACB.
To prove: AC > AB.

Proof: There are following possibilities.
(i) AC = AB
(ii) AC < AB
(iii) AC > AB
There is only one possibilities must be true.
(i) If AC = AB
⇒ ∠ABC = ∠ACB
[Angles opposite to equal sides are equal]
It is impossible, for it is given that
⇒ ∠ABC > ∠ACB
(ii) If AC < AB
⇒ AB>AC
⇒ ∠ACB > ∠ABC
This is also impossible, being contrary to hypothesis.
(iii) Since AC is neither less than nor equal to AB.
∴ AC must > AB.
Hence AC > AB. Proved.

Theorem 7.8:
The sum of any two sides of a triangle is greater than the third side. (See fig. 7.64).

Given: A triangle ΔABC.
To prove : (i) AB + AC > BC
(ii) AB + BC > AC
(iii) AC+ BC > AB.
Construction: Produce BA to D such that AD = AC. Join C to D

Proof: In ΔACD, we have
AD = AC (By construction)
⇒ ∠ADC = ∠ACD …..(i)
∠BCD > ∠ACD,
[∵ ∠BCD = ∠ACB + ∠ACD]
⇒ ∠BCD > ∠ADC, [from (i), ∠ACD = ∠ADC]
⇒ ∠BCD > ∠BDC
⇒ BD > BC (By theorem 7·7)
⇒ AB + AD > BC
⇒ AB + AC > BC (∵ AD = AC)
Similarly,
AB + BC > AC and AC + BC > AB.
Hence proved

Corollary 1.
The difference between the lengths of any two sides of a triangle is always less than the third side.

Haryana State Board HBSE 9th Class Maths Notes Chapter 3 Coordinate Geometry Notes.

Haryana Board 9th Class Maths Notes Chapter 3 Coordinate Geometry

Introduction
In previous class, we have studied how to locate a point on a real number line. We also know how to describe the position of a point on the line.
In this chapter, we shall study to locate the coordinates of a point in a plane. We shall also learn about the plotting the points in the plane (Cartesian plane).

In the seventeenth century, French mathematician Rene Descartes developed the idea of describing the position of a point in a plane. His method was a development of the older idea of latitude and longitude. In honour of Descartes, the system used for describing the position of a point in a plane is also known as the cartesian system.

Key Words
→ Coordinates: A set of numbers which locate a point. In general, two numbers are needed to locate a point in a plane and is written as (x, y).

→ Quadrant : (\(\frac{1}{4}\))^th part of a plane divided by coordinate axes is known as a quadrant.

→ Abscissa: The first coordinate x, of a pair (x, y) of cartesian coordinates in the plane.

→ Ordinate: The second coordinate y, of a pair (x, y) of cartesian coordinates in the plane.

Basic Concepts
Cartesian System: We have studied the number line. On the number line, distance from a fixed point is marked in equal units positively in one direction and negatively in the other. The point from which the distances are marked is called the origin. A cartesian (or coordinate) plane consists of two mutually perpendicular number lines intersecting each other at their zeroes.

The adjoining figure shows a cartesian plane consisting of two mutually perpendicular number lines XOX’ and YOY’ intersecting each other at O.
1. The horizontal number line XOX’ is called the x-axis.
2. The vertical number line YOY’ is called the y-axis.
3. The point of intersection ‘O’ is called the origin which is zero for both the axes.

The system consisting of the x-axis, the y-axis and the origin is also called cartesian coordinate system. The x-axis and y-axis together are called coordinate axes.
(a) Coordinates of points in a plane: The position of each point in a coordinate plane is determined by means of an ordered pair (a pair of numbers) with reference of the coordinates axes; as stated below:
(i) The distance of the point along x-axis from the origin O is called x-coordinate or abscissa of the point.
(ii) The distance of the point along y-axis from the origin O is called y-coordinate or ordinate of the point.
Thus, coordinates of the point = Position of the point reference to coordinates axes = (abscissa, ordinate)
If the abscissa of a point is x and its ordinate is y, then its coordiantes = (x, y).

(b) Quadrants: The coordinate plane is divided by axes into four parts which are known as quadrants. Each point is located either in one of the quadrants or on one of the axes.
Starting from OX in the anticlockwise direction :
(i) XOY is called the first quadrant.
(ii) YOX’ is called the second quadrant.
(iii) X’OY’ is called the third quadrant.
(iv) Y’OX is called the fourth quadrant which are shown in the adjoining figure.

(c) Rules for signs of coordinates:
(i) In the first quadrant, the abscissa and ordinate both are positive.
(ii) In the second quadrant, the abscissa is negative and ordinate is positive.
(iii) In the third quadrant, the abscissa and ordinate both are negative.
(iv) In the fourth quadrant, the abscissa is positive and ordinate is negative.
Using the convention of signs, we have the signs of the coordinates in quadrants as given below :

(d) Coordinates of a point on the x-axis: The distance of every point on the x-axis from x-axis is 0 unit. So, its ordinate is 0.
So, the coordinates of a point on the x-axis is of the form (x, 0).
e.g., (5, 0), (3, 0), (-5, 0) etc.

(e) Coordinates of a point on the y-axis: The distance of every point on the y-axis from the y-axis is 0 unit. So, its abscissa is zero.
So, the coordinates of a point on the y-axis is of the form (0, y).
e.g., (0,5), (0, 2), (0, -7) etc.

(f) The coordinates of the origin are (0, 0).

Plotting of a point in the plane if its coordinates are given: In this section, we will learn about to locate a point in the cartesian plane if co-rdinates of the points are given. This process is known as the plotting of the point with given coordinates. In this process, we use a graph paper as the coordinates plane. We draw x-axis and y-axis on it which intersect each other at origin O. Any point P in the xy-plane can be located by using an ordered pair (x, y) of real numbers. Let x denote the signed distance of P from the y-axis (signed in the sence that, if x > 0, then P is to the right of the y-axis and if x < 0, then P is to the left of the y-axis); and let y denote the signed distance of P from the x-axis. The ordered pair (x, y) is called the coordinates of P, that gives us enough information to locate the point P in the cartesian plane.

If (a, b) are the coordinates of a point P, then ‘a’ is called the x-coordinate or abscissa of P and ‘b’ is called the y-coordinate or ordinate of P as shown in fig 3.11.

For example, to locate the point (-3, 5), go 3 units along x-axis to the left of O(origin) and then go up straight 5 units and to locate the point (5, -3), go 5 units along the x-axis to the right of O and then go down 3 units. We plot these points by placing a dot at this location. See figure 3.12 in which the points(-4, -7) and (2, 2) are also located.

Any point on the x-axis has coordinates of the form (x, 0) and any point on the y-axis has coordinates of the form (0, y).

Haryana State Board HBSE 9th Class Maths Notes Chapter 1 Number Systems Notes.

Haryana Board 9th Class Maths Notes Chapter 1 Number Systems

Introduction
In our previous classes, we have learnt about various types of numbers such as natural numbers, whole numbers, integers, rational numbers etc. and their representation on the number line. In this chapter, we shall study rational numbers, irrational numbers and their representation on number line and laws of exponents for real numbers. Let us review the various types of numbers.
(a) Natural Numbers: The counting numbers 1, 2, 3, 4, 5, … are called natural numbers. The smallest natural number is 1 but there is no largest as it goes up to infinity. The set of natural numbers is denoted by N.
N = {1, 2, 3, 4, 5, ………}.
(b) Whole Numbers: In the set of natural numbers, if we include the number 0, the resulting set is known as the set of whole numbers. The set of whole numbers is denoted by W.
W = {0, 1, 2, 3, 4,……}.
(c) Integers: Natural numbers along with zero and their negative are called integers. The set of integers is denoted by Z.
Z = {…….., -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ………}.
The set of integers contains negative numbers, zero and positive numbers. Z comes from the German word “Zahlen’ which means “to count”.

In order to represent integers on the number line we draw a straight line and mark origin (O) in the middle of the line. The numbers may be represented by dots along a straight line as shown in below figure. Positive numbers are represented on the right hand side of the origin and negative numbers are represented on the left hand side of the origin.

(d) Rational Numbers: A rational number is a number which can be expressed in the form of \(\frac{p}{q}\), where p and q are integers and q ≠ 0.
Thus, Q = {\(\frac{p}{q}\) : q ≠ 0, p, q ∈ Z}

A rational number may be positive, negative or zero. The rational number is positive if p and q have like signs and negative if p and q have unlike signs.
For example: 5 = \(\frac{5}{1}\), -11 = \(\frac{-11}{1}, \frac{-2}{3}, \frac{4}{7}, \frac{32}{96}, \frac{3}{-5}\)
In above examples 5, \(\frac{4}{7}\), \(\frac{32}{96}\), are positive rational numbers and -11, \(\frac{-2}{3}\), \(\frac{3}{-5}\) are negative rational numbers.
For representing rational numbers on the number line, we draw a line and mark origin (O) in the middle of line. Positive numbers are represented on the right hand side of the origin and negative numbers are represented on the left hand side of the origin. If we mark a point P on the line to the right of O to represent 1, then OP = 1 unit and mark a point Q on the line to the left of O to represent -1, then OQ = -1 unit.

For representing the rational numbers \(\frac{1}{3}\), \(\frac{2}{3}\) on the number line, we divide the segment OP into three equal parts such that OA = AB = BP = \(\frac{1}{3}\). So, A and B represent \(\frac{1}{3}\) and \(\frac{2}{3}\) respectively as shown in the figure 1.2.

For representing rational numbers \(\frac{1}{2}\) on the number line, we divide the segment OQ into four equal parts such that OR = RS = ST = TQ = \(\frac{1}{4}\). So R, S and T represent \(\frac{-1}{4}, \frac{-2}{4}\) and \(\frac{-3}{4}\) respectively.

(e) To find out rational numbers between two rationals: 1. If x and y are two rational numbers such that x < y.
Then a rational number lying between x and y is \(\left(\frac{x+y}{2}\right)\).

2. If a and b are two rational numbers, such that a < b.
Then a rational numbers lying between a and b are

Key Words
→ Numbers: The natural numbers or whole numbers are known as numbers. e.g., 0, 1, 2, 3,…, etc.

→ Even number: A number which is divisible by 2 is called an even number. e.g., 2, 4, 6, 8,…, etc.

→ Odd number: A number which cannot divisible by 2 is called an odd number. e.g., 1, 3, 5, 7,…, etc.

→ Prime number: A natural number which is divisible by 1 and itself, is called a prime number. e.g., 2, 3, 5, 7,…, etc.

→ Composite number: A whole number that can be divided evenly by numbers other than 1 or itself is called composite numbers e.g., 4, 6, 8, 9 …, etc.

→ Coprime numbers: Two numbers are said to be coprime, if their H.C.F. is 1. eg., (4, 5), (8, 9), (15, 16), etc.

→ Number line: Refers to a straight horizontal line on which each point represents a real number.

→ Terminating decimal: A number which divides completly in decimal or whole number is called terminating decimal.
e.g., \(\frac{3}{5}\) = 0.6, \(\frac{20}{4}\) = 5, etc.

→ Non terminating decimal: A number which cannot divide completly in decimal or whole number i.e., process of division will never end, is called non terminating decimal. eg., \(\frac{1}{11}\) = 0.090909…, \(\frac{1}{7}\) = 0.142857…, etc.

→ Recurring (or repeating) decimal: A non terminating decimal, in which a digit or set of digits repeats continually, is called a recurring or repeating decimal.
e.g., \(\frac{4}{9}\) = 0.444….., \(\frac{2}{11}\) = 0.181818…, \(\frac{1}{3}\) = 0.076923076923… etc.

→ Exponent or Index or Power: Suppose that ‘a’ is a real number, when the product a × a × a × a × a is written as a⁵, the number 5 is called the exponent of a.

Basic Concepts
1. Irrational Numbers: In previous section, we observe that some numbers are not rational. These numbers can not write in the form of \(\frac{p}{q}\), where p and q are intergers and q ≠ 0.
For example: 0.141441444…., 0.232332333…..
In 400 B.C., the famous Mathematician Pythagoras was the first to discover the number which were not rational. These numbers are called irrational numbers.
Let us define these numbers as an irrational is a non-terminating and non-recurring decimal, that is, it cannot be written in the form \(\frac{p}{q}\), where p and q are integers and q ≠ 0.

Examples of Irrational Numbers:
(i) 0.2020020002……, 0.313311333111……., 0.535535553… are irrational numbers.
(ii) \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\), \(\sqrt{6}\), \(\sqrt{7}\), \(\sqrt{10}\), \(\sqrt{11}\), \(\sqrt{12}\), \(\sqrt{15}\), \(\sqrt{16}\) etc. all are irrational numbers.
(iii) \(\sqrt[3]{2}, \sqrt[3]{3}, \sqrt[3]{4}, \sqrt[3]{5}, \sqrt[3]{6}, \sqrt[3]{7}, \sqrt[3]{9}, \sqrt[3]{10}\) etc. all are irrational numbers.
(iv) π is irrational, while \(\frac{22}{7}\) is rational.

2. Properties of Irrational Numbers:
A. (a) Sum of a rational number and an irrational number is an irrational number.
(b) Difference of a rational number and an irrational number is an irrational number.
(c) Product of a rational number and an irrational number is an irrational number.
(d) Quotient of a rational number and an irrational number is an irrational number.
For example: (5 + \(\sqrt{2}\)), (5 – \(\sqrt{3}\)), 3\(\sqrt{2}\) and \(\frac{7}{\sqrt{3}}\) all are irrational numbers.

B. (a) It is not necessary that sum of two irrational numbers may be an irrational number.
For example:(4 + \(\sqrt{5}\)) and (3 – \(\sqrt{5}\)) both are irrational number. But (4 + \(\sqrt{5}\))+(3 – \(\sqrt{5}\)) = 7, which is rational number.
(b) It is not necessary that difference of two irrational numbers may be an irrational number.
For example: (4 + \(\sqrt{3}\)) and (3 + \(\sqrt{3}\)) both are irrational numbers. But (4 + \(\sqrt{3}\)) – (3 + \(\sqrt{3}\)) = 1, which is rational number.

(c) It is not necessary that product of two irrational numbers may be an irrational number.
For example: \(\sqrt{2}\) is an irrational number. But \(\sqrt{2}\) × \(\sqrt{2}\) = 2, which is rational number.
(d) It is not necessary that quotient of two irrational numbers may be an irrational number.
For example: 6\(\sqrt{2}\) and 3\(\sqrt{2}\) both are irrational numbers. But \(\frac{6 \sqrt{2}}{3 \sqrt{2}}\) = 2, which is rational number.

3. Representation of Irrational Numbers on the Number Line: We show the representation of irrational numbers on the number line by help of some examples given below:

(a) Real Numbers: Rational numbers and irrational numbers form the collection of all real numbers.
Every real number is either rational or irrational number.
For example: -3\(\frac{1}{2}\), \(\sqrt{3}\), π, 1.596. etc.
(i) Rational Number: A number whose decimal expension is either terminating or non-terminating recurring is called a rational number.
For example: \(\frac{3}{4}\), 2.1, \(\frac{1}{3}\) etc.
(ii) Irrational Number: A number whose decimal expension is non-terminating non recurring is called an irrational number.
For example: \(\sqrt{2}\), \(\sqrt{3}\), 2010010001…., π, etc.

(b) Decimal Expansions of Rational Numbers: A rational number can be expressed as a decimal. For example, if we divide 3 by 5, we will get 0.6, a terminating decimal.

Similarly, if we divide 2 by 8, we will get 0.25, a terminating decimal.

Similarly,
\(\frac{3}{4}\) = 0.75
\(\frac{7}{16}\) = 0.4375
\(\frac{-2}{5}\) = -0.4 etc.
are rational numbers expressed as decimals. Now let us divide 3 by 11

We observe that the continued process of division by 11, in the quotient digits 2, 7 will repreat from the stage marked (→) onwards. Thus
\(\frac{3}{11}\) = 0.2727…..
= \(0 . \overline{27}\)
The process of division, in this case the decimal expressions are known as non-terminating repeating (recurring) decimals.
Similarly,

A repeating decimal such as 0.6666… is often written as \(0.\overline{6}\), where over bar indicates the number that repeats. Hence,
\(\frac{2}{3}\) = \(0.\overline{6}\)
So, we say that every terminating or repeating decimal represents a rational number.

Representing Real Numbers on the Number Line: In the previous section, we have learn about the decimal expansions of real numbers. This help us to represent it on the number line. In this section, we will learn how to visualise the position of real numbers in decimal form.

Suppose we want to visualise the representation of 2.665 on the number line. We observe that 2.665 is located between 2 and 3. So let us look closely at the portion of the number line 2 and 3. We divide it into 10 equal parts and mark each point of division as shown in figure 1.10. Then the first mark to the right of 2 will represent 2.1, the second 2.2 and so on. To see this clearly we may take a magnifying glass and look at the portion between 2 and 3. Now, we observe that 2.665 lies between 2.6 and 2.7. To get more accurate visualization of the representation, we divide this portion of real line into 10 equal parts. The first mark will represent 2.61, the second 2.62 and so on.

Again, 2.665 lies between 2.66 and 2.67, so let us focus on this portion of the number line. We magnify the portion of line between 2.66 and 2.67 and divide it into 10 equal parts. The first mark represents 2.661, the next one represents 2.662 and so on. So 2.665 is the 5th mark in these subdivisions. We call this process of visualisation of representation of numbers on the number line through a magnifying glass, as the process of successive magnification.

1. Operations on Real Numbers: In the previous classes, we have learnt about that rational numbers satisfied the associative, commutative and distributive properties for addition and multiplication. The sum, difference, product and quotient of two rational numbers will be a rational number. Irrational numbers also satisfied the commutative, associative and distributive properties for addition and multiplication. However the sum, difference, quotients and products of irrational numbers are not always irrational.

The facts related to operations on real numbers are given below:

• The sum or difference of a rational number and an irrational number is an irrational number.
• The product or quotient of a non-zero rational number with an irrational number is an irrational number.
• If we, add, subtract, multiply or divide two irrational numbers, the result may be a rational or irrational number.

2. Existence of Square Root of a given Positive Real Number:
To find \(\sqrt{x}\) geometrically, where x is any positive real number.
Construction: Draw a line AB of length x units extend AB to point C such that BC = 1 unit. Find the midpoint of AC and marked that point O. Draw a semicircle with centre O and radius OA. Draw a line perpendicular to AC passing through B and intersecting the semicircle at D, then BD = \(\sqrt{x}\).

Proof:
We have AB = x units, BC = 1 unit
⇒ AC = x + 1 units
OA = OD = OC = \(\frac{x+1}{2}\),
(∵ O is the mid-point of AC)
Now,
OB = x – \(\frac{(x+1)}{2}\)
⇒ OB = \(\frac{2 x-x-1}{2}=\frac{x-1}{2}\)
In right ΔOBD,
OD² = OB² + BD²,
(By Pythagoras theorem)
⇒ BD² = OD² – OB²
⇒ BD² = \(\left(\frac{x+1}{2}\right)^2-\left(\frac{x-1}{2}\right)^2\)
⇒ BD² = \(\frac{x^2+1+2 x}{4}-\frac{x^2+1-2 x}{4}\)
⇒ BD² = \(\frac{1}{4}\)[x² + 1 + 2x – x² – 1 + 2x]
⇒ BD² = \(\frac{1}{4}\) × 4x
⇒ BD² = x
⇒ BD = \(\sqrt{x}\)

3. Rationalisation of Irrational Number:
The process of multiplying a given irrational number by its rationalising factor to get a rational number as product is called rationalisation of the given irrational number.
When the product of two irrational numbers is a rational number, each is called the Rationalising Factors (R.F.) of the other.

For Example:
(i) Since 5\(\sqrt{2}\) × 3\(\sqrt{2}\) = 15 × 2 = 30; which is a rational number. Therefore, 5\(\sqrt{2}\) and 3\(\sqrt{2}\) are rationalising factors of each other.
(ii) (\(\sqrt{3}\) + \(\sqrt{2}\)) (\(\sqrt{3}\) – \(\sqrt{2}\)) = (\(\sqrt{3}\))² – (\(\sqrt{2}\))²
= 3 – 2 = 1
(\(\sqrt{3}\) + \(\sqrt{2}\)) and (\(\sqrt{3}\) – \(\sqrt{2}\)) are the
rationalising factors of each other.
(iii) (3 + \(\sqrt{5}\)) (3 – \(\sqrt{5}\)) = 3² – (\(\sqrt{5}\))²
= 9 – 5 = 4
(3 + \(\sqrt{5}\)) and (3 – \(\sqrt{5}\)) are rationalising factors of each other.

In the same way:
(i) R.F. of \(\sqrt{5}\) – 2 is \(\sqrt{5}\) + 2.
(ii) R.F. of 4 + 3\(\sqrt{5}\) is 4 – 3\(\sqrt{5}\).

Laws of Exponents for Real Numbers:
Let a be real number and m, n be rational numbers, then we have

Note: Here a is called base and m, n are called exponents.

Haryana State Board HBSE 9th Class Maths Notes Chapter 8 Quadrilaterals Notes.

Haryana Board 9th Class Maths Notes Chapter 8 Quadrilaterals

Introduction
In previous chapters 6 and 7, we have studied about triangles and their properties. We know that figure obtained by joining three non-collinear points in pairs, is a triangle. If we have four points in a plane such that no three of them are collinear.

If we joined these points, we obtain a figure with four sides (see figure 8.1).

Key Words
→ Intercept: To cut off.

→ Internal bisector: Name sometimes given to the bisector of the interior angle of a triangle (or polygon).

→ External bisector: The bisector of the exterior angle of a triangle (or polygon) is sometimes called the external bisector of the angle of the triangle (or polygon).

Basic Concepts
1. Quadrilateral: A closed figure which bounded by four line segments is called quadrilateral. Figure 8.2 represents a quadrilateral with vertices A, B, C and D. It is denoted by the symbol quad. ABCD or ☐ABCD.

Four straight lines AB, BC, CD and DA are called the sides of the quadrilateral ABCD and ∠A, ∠B, ∠C and ∠D are the four angles of it.
The straight lines joining the opposite vertices of a quadrilateral are called its diagonals. Thus, a quadrilateral ABCD has two diagonals; namely AC and BD.

In a quadrilateral, we have the following definitions:
(a) Consecutive or adjacent sides: Two sides with common vertex are called the adjacent sides. In fig. 8.2 AB and BC, BC and CD, CD and DA and DA and AB are four adjacent sides of the quadrilateral ABCD.
(b) Opposite sides: Two sides of a quadrilateral having no common end point are called its opposite sides.
In fig. 8.2, AB and CD, AD and BC are two pairs of opposite sides of quad. ABCD.
(c) Consecutive angles: Two angles of a quadrilateral having a common side are called its consecutive angles.
In fig. 8.2, ∠A and ∠B, ∠B and ∠C, ∠C and ∠D and ∠D and ∠A are four pairs of consecutive angles of ☐ABCD.
(d) Opposite angles: Two angles of a quadrilateral having no common side are called its opposite angles.
In figure 8.2, ∠A and ∠C, ∠B and ∠D are two pairs of opposite angles of ☐ABCD.

2. Angle Sum Property of a Quadrilateral:
Let us now recall the angle sum property of a quadrilateral. The sum of the angles of a quadrilateral is 360°. This can be verified by drawing a diagonal and dividing the quadrilateral into two triangles.

Let ABCD be a quadrilateral and AC be a diagonal (see figure 8.3).
In ΔABC, we have
∠1 + ∠B + ∠2 = 180°…(1)
I. Sum of interior angles of a triangle 180°]
In ΔADC, we have
∠3 + ∠D + ∠4 = 180°…(ii)
Adding (i) and (ii), we get
∠1 + ∠3 + ∠B + ∠2 + ∠4 + ∠D = 180° + 180°
⇒ ∠A + ∠B + ∠C + ∠D = 360°
[∵ ∠1 + ∠3 = ∠A and ∠2 + ∠4 = ∠C]
Hence, sum of angles of a quadrilateral is 360°.

3. Types of Quadrilaterals:
(i) Trapezium: A quadrilateral in which one pair of opposite sides is parallel but the other pair of opposite sides is non-parallel is called a trapezium.
The figure 8.4, represents a trapezium ABCD.

In which AB || CD and AD and BC are non-parallel sides. If the non-parallel sides AD and BC of the trapezium are equal, it is called an isosceles trapezium. In this case, ∠D = ∠C and ∠A = ∠B.

(ii) Parallelogram: A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram.

The figure 8.5, represents a parallelogram ABCD, since AD || BC and AB || CD.

(iii) Rectangle: A parallelogram in which each angle is a right angle is called a rectangle.

The figure 8.6, represents a rectangle ABCD in which ∠A = ∠B = ∠C = ∠D = 90°.

(iv) Square: A parallelogram in which all sides are equal and each angle is equal to 90° is called a square.

The figure 8.7, represents a square ABCD in which AB = BC = CD = DA and ∠A = ∠B = ∠C = ∠D = 90°.

(v) Rhombus: A parallelogram in which all sides are equal is called a rhombus.

The figure 8.8, represents a rhombus. ABCD in which AB = BC = CD = DA.

(vi) Kite: A quadrilateral in which both pairs of adjacent sides are equal is called a kite.

The figure 8.9, represents a kite ABCD in which AD = DC and AB = BC.

Note:
(i) Every parallelogram is a trapezium but every trapezium is not a parallelogram.
(ii) Every square is a rectangle and also a rhombus but every rectangle or rhombus is not a square.
(iii) A kite is not a parallelogram.
(iv) Every square, rectangle and rhombus are parallelograms.

4. Properties of a Parallelogram:
Theorem 8.1:
A diagonal of a parallelogram divides it into two congruent triangles.
Given: ABCD is a parallelogram and AC its diagonal.
To prove :
ΔABC ≅ ΔCDA.

Proof: ABCD and AC cuts them
∠BAC = ∠DCA …..(i)
(Alternate interior angles)
and BC || AD and AC cuts them.
∠BCA = ∠DAC (Alternate interior angles)
Now, in ΔABC and ΔCDA, we have
∠BAC = ∠DCA, [From (i)]
AC = AC, (Common)
and ∠BCA = ∠DAC, [From (ii)]
ΔABC ≅ ΔCDA [By ASA congruence rule]
Hence proved

Theorem 8.2:
In a parallelogram, opposite sides are equal.
Given: ABCD is a parallelogram.
To prove: AB = CD and AD = BC.
Construction: Join AC.

Proof: ABDC and AC cuts them.
∠BAC = ∠DCA …(i) (Alternate interior angles)
Again AD || BC and AC cuts them.
∠BCA = ∠DAC …(ii)
(Alternate interior angles).
Now, in AABC and ACDA, we have
∠BAC = ∠DCA, [From (i)]
AC = AC, [Common]
and ∠BCA = ∠DAC, [From (ii)]
ΔABC ≅ ΔCDA (By ASA congruence rule)
AB = CD and AD = BC, (CPCT)
Hence proved

Theorem 8.3:
If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.
Given: A quadrilateral ABCD in which AB = CD and BC = DA.
To prove: Quadrilateral ABCD is a parallelogram.

Construction: Join AC.
Proof: In ΔABC and ΔCDA, we have
AB = CD, (Given)
BC = AD, (Given)
and AC = AC, (Common)
ΔABC ≅ ΔCDA (By SSS congruence rule)
⇒ ∠BAC = ∠DCA, (CPCT)
⇒ AB || CD …..(i)
and ∠BCA = ∠DAC, (CPCT)
AD || BC …..(ii)
From (i) and (ii), we get
AB || CD and AD || BC
Hence, ABCD is a parallelogram. Hence Proved

Theorem 8.4:
In a parallelogram, opposite angles are equal.
Given: ABCD is a parallelogram.
To prove: ∠A = ∠C and ∠B = ∠D.
Proof : BC || AD and AB cuts them.

⇒ ∠A + ∠B = 180° …(1)
[Sum of co-interior angles is 180°]
Again, AB || CD and AD intersects them.
⇒ ∠A + ∠D = 180° …..(ii)
[Sum of co-interior angles is 180°]
From (i) and (ii), we get
∠A + ∠B = ∠A + ∠D
⇒ ∠B = ∠D
Similarly, ∠A = ∠C.
Hence proved

Theorem 8.5:
If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
Given: A quadrilateral ABCD in which ∠A = ∠C and ∠B = ∠D.
To prove: Quad. ABCD is a parallelogram.

Proof: We have
∠A = ∠C…(1)
and ∠B = ∠D…(ii)
Adding (i) and (ii), we have
∠A + ∠B = ∠C + ∠D…(iii)
In a quadrilateral ABCD, we have
∠A + ∠B + ∠C + ∠D = 360°,
[∵ Sum of angles of a quadrilateral is 360°]
⇒ ∠A + ∠B + ∠A + ∠B = 360°, [Using (iii)]
⇒ 2(∠A + ∠B) = 360°
⇒ ∠A + ∠B = \(\frac{360^{\circ}}{2}\) = 180°…(iv)
Now, line segments BC and AD cut by AB,
such that
∠A + ∠B = 180°
⇒ BC || AD
Again, ∠A + ∠B = 180°.
⇒ ∠C + ∠B = 180° [From (i) ∠A = ∠C]
Now, line segments AB and CD cut by BC,
∠B + ∠C = 180°
⇒ AB || CD
Thus, AB || CD and BC || AD.
Hence, ABCD is a parallelogram. Proved

Theorem 8.6:
The diagonals of a parallelogram bisect each other.
Given: A parallelogram ABCD such that its diagonals AC and BD intersect at O.
To prove: OA = OC and OB = OD.
Proof: Since ABCD is a parallelogram.

Therefore AB || CD and BC || AD.
Now, AB || CD and AC intersects.
∠BAC = ∠DCA, (Alternate interior angles)
∠BAO = ∠DCO …(1)
Again, AB || CD and BD intersects them.
∠ABD = ∠CDB, (Alternate interior angles)
⇒ ∠ABO = ∠CDO…(ii)
Now, in ΔAOB and ΔCOD, we have
∠BAO = ∠DCO, [From (i)]
AB = CD, (Opposite sides of parallelogram)
and ∠ABO = ∠CDO, [From (ii)]
∴ ΔAOB ≅ ΔCOD, (By ASA congruence rule)
⇒ OA = OC and OB = OD. Hence proved

Theorem 8.7:
If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
Given: ABCD is a quadrilateral in which the diagonals AC and BD bisect each other at O i.e., AO = OC and BO = OD.
To prove : Quadrilateral ABCD is a parallelogram.

Proof: In ΔAOB and ΔCOD, we have
AO = OC, (Given)
∠AOB = ∠COD, (vertically opposite angles)
and BO = OD, (Given)
∴ ΔAOB ≅ ΔCOD, (by SAS congruence rule)
⇒ ∠BAO = ∠DCO, (CPCT)
⇒ ∠BAC = ∠DCA
But these are alternate interior angles.
∴ AB || CD
and similarly, AD || BC
Hence, quad. ABCD is a parallelogram. Hence Proved

5. Another Conditions for a Quadrilateral to be a parallelogram:
In this chapter, we have studied many properties and verified that if in a quadrilateral any one of those properties is satisfied, then it will be a parallelogram. Now, we study a required condition for a quadrilateral to be a parallelogram. It is stated in the form of a theorem as given below.

Theorem 8.8:
A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.
Given: A quadrilateral ABCD in which AB = CD and AB || CD
To prove: Quadrilateral ABCD is a parallelogram.
Construction : Join AC.

Proof: We have
AB || CD and AC intersects them.
⇒ ∠BAC = ∠DCA …(i)
(Alternate interior angles)
In ΔABC and ∠CDA, we have
AB = CD, (Given)
∠BAC = ∠DCA, [From (i)]
and AC = AC, (Common)
∴ ΔABC ≅ ΔCDA (By SAS congruence rule)
⇒ ∠BCA = ∠DAC, (CPCT)
But these are alternate interios angles.
∴ AD || BC
and AB || CD, (Given)
Hence, quad. ABCD is a parallelogram.
Hence Proved

The Mid Point Theorem: In this section we shall discuss some more properties of a triangle using properties of a parallelogram which is related to the mid points of sides of a triangle.

Theorem 8.9:
The line segment joining the mid points of two sides of a triangle is parallel to the third side and equal to half of it.

Given: D and E are the mid points of the sides AB and AC of the triangle ABC.
To prove : (i) DE || BC, (ii) DE = \(\frac{1}{2}\)BC.
Construction: Produce DE to F, such that EF = DE and join CF.
Proof: In ΔAED and ΔCEF, we have
AE = CE, (∵ E is the mid point of AC)
∠AED = ∠CEF, (Vertically opposite angles)
and DE = EF, (By construction)
∴ ΔAED ≅ ΔCEF, (By SAS congruence rule)
⇒ ∠1 = ∠2, (CPCT)
But these are alternate interior angles.
AB || FC ⇒ BD || FC …(i)
and AD = FC (CPCT)
But AD = BD
(∵ D is the mid point of AB)
∴ BD = FC ……(ii)
From (i) and (ii), we get
BD || FC and BD = FC
∴ BCFD is a parallelogram. [By theorem 8.8]
⇒ DF || BC,
(∵ Opposite sides of a parallelogram)
⇒ DE || BC
and DF = BC,
(∵ Oppsite sides of a parallelogram)
⇒ 2DE = BC,
[∵ DE = EF ⇒ DF = DE + EF = 2DE]
⇒ DE = \(\frac{1}{2}\)BC
Hence, DE || BC and DE = BC. Proved

Theorem 8.10:
The line drawn through the mid point of one side of a triangle, parallel to another side bisects the third side.

Given: A ΔABC, D is the mid point of AB and line through D parallel to BC intersects AC in E.
To prove: E is the mid point of AC.
Construction: Draw CM || BA intersects line l in F.
Proof: Since, AB || CM (By construction)
∴ BD || CF
and DF || BC (Given)
∴ BDFC is a parallelogram.
⇒ BD = CF,
(Opposite sides of a parallelogram)
But, BD = AD,
(∵ D is the mid point of AB)
∴ AD = CF …..(i)
Now, AB || CM and AC intersects them.
∠1 = ∠2,
(Alternate interior angles) …(ii)
In ΔADE and ΔCFE, we have
∠1 = ∠2 [From (i)]
∠AED = ∠CEF
(Vertically opposite angles)
and AD = CF [From (i)]
∴ ΔADE ≅ ΔCFE
(By AAS congruence rule)
⇒ AE = EC
Hence, E is the mid point of AC. Proved

Haryana State Board HBSE 9th Class Maths Notes Chapter 4 दो चरों वाले रैखिक समीकरण Notes.

Haryana Board 9th Class Maths Notes Chapter 4 दो चरों वाले रैखिक समीकरण

→ वह समीकरण जो ax + by + c = 0 के रूप में व्यक्त की जा सकती हो, जहां, a, b और c वास्तविक संख्याएं हैं। a व b दोनों शून्य नहीं हैं, उसे दो चरों वाला रैखिक समीकरण कहा जाता है।

→ दो चरों वाले रैखिक समीकरण के अपरिमित रूप से अनेक हल होते हैं।

→ दो चरों वाले प्रत्येक रैखिक समीकरण का आलेख एक सरल रेखा होता है।

→ एक समीकरण को हल करते समय निम्नलिखित बातों को ध्यान में रखना होता है। एक रैखिक समीकरण पर तब कोई प्रभाव नहीं पड़ता जबकि :

• समीकरण के दोनों पक्षों में समान संख्या जोड़ी या घटाई जाती है।
• समीकरण के दोनों पक्षों को समान शून्येतर संख्या से गुणा या भाग किया जाता है।

→ x = 0, y-अक्ष का और y = 0, x-अक्ष का समीकरण है।

→ x = a का आलेख y-अक्ष के समांतर एक सरल रेखा होता है।

→ y = a का आलेख x-अक्ष के समांतर एक सरल रेखा होता है।

→ y = mx के प्रकार का समीकरण मूल बिंदु से होकर जाने वाली एक रेखा को निरूपित करता है ।

→ दो चरों वाले रैखिक समीकरण के आलेख पर स्थित प्रत्येक बिंदु रैखिक समीकरण का एक हल होता है। विलोमतः रैखिक समीकरण का प्रत्येक हल रैखिक समीकरण के आलेख पर स्थित एक बिंदु होता है।

Haryana State Board HBSE 9th Class Maths Notes Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Notes.

Haryana Board 9th Class Maths Notes Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन

यदि लंबाई = l, चौड़ाई = b, ऊंचाई = h, त्रिज्या = r, भुजा = a, तिर्यक ऊंचाई = l से लिखे जाएं तो-

→ घनाभ का पृष्ठीय क्षेत्रफल = 2 (lb + bh + hl)

→ घन का पृष्ठीय क्षेत्रफल = 6a² [अर्थात् 6 (भुजा)2]

→ बेलन का वक्र पृष्ठीय क्षेत्रफल = 2πrh

→ बेलन का कुल पृष्ठीय क्षेत्रफल = 2πr(r + h)

→ शंकु का वक्र पृष्ठीय क्षेत्रफल = πrl

→ शंकु का कुल पृष्ठीय क्षेत्रफल = πr(l + r)

→ गोले का पृष्ठीय क्षेत्रफल = 4πr²

→ अर्धगोले का वक्र पृष्ठीय क्षेत्रफल = 2πr²

→ अर्धगोले का कुल पृष्ठीय क्षेत्रफल = 3πr²

→ कमरे के फर्श या छत का क्षेत्रफल = b

→ कमरे की चार दीवारों का क्षेत्रफल = 2(l + b) × h

→ आयत का परिमाप = 2 (l + b)

→ घनाभ का आयतन = l × b × h

→ घन का आयतन = a³ अर्थात् [(भुजा)3]

→ बेलन का आयतन = πr²h

→ शंकु का आयतन = \(\frac{1}{3}\)πr²h

→ गोले का आयतन = \(\frac{4}{3}\)πr³

→ अर्धगोले का आयतन = \(\frac{2}{3}\)πr³

Haryana State Board HBSE 9th Class Maths Notes Chapter 12 हीरोन सूत्र Notes.

Haryana Board 9th Class Maths Notes Chapter 12 हीरोन सूत्र

→ त्रिभुज का क्षेत्रफल = \(\frac{1}{2}\) × आधार × ऊँचाई

→ त्रिभुज का क्षेत्रफल निकालने का हीरोन का सूत्र = \(\sqrt{s(s-a)(s-b)(s-c)}\)
जहाँ s = त्रिभुज का अर्धपरिमाप = \(\frac{a+b+c}{2}\) तथा a, b, c, त्रिभुज की भुजाएँ हैं।

→

→

→ वर्ग का क्षेत्रफल = भुजा × भुजा

→ आयत का क्षेत्रफल = लंबाई × चौड़ाई

→ 1 हेक्टेयर = 10,000m²

→ एक चतुर्भुज जिसकी भुजाएँ तथा एक विकर्ण दिए हों तो उसका क्षेत्रफल उसे दो त्रिभुजों में विभाजित करके और फिर हीरोन के सूत्र का प्रयोग करके ज्ञात किया जा सकता है।

→ किसी समकोण त्रिभुज में- (कर्ण)² = (आधार)² + (लंब)²

→ समांतर चतुर्भुज का क्षेत्रफल- आधार × ऊँचाई

Haryana State Board HBSE 9th Class Maths Notes Chapter 3 निर्देशांक ज्यामिति Notes.

Haryana Board 9th Class Maths Notes Chapter 3 निर्देशांक ज्यामिति

→ एक तल में किसी वस्तु या बिंदु का स्थान निर्धारण करने के लिए हमें दो लांबिक रेखाओं की आवश्यकता होती है, जिनमें से एक क्षैतिज तथा दूसरी ऊर्ध्वाधर होती है ।

→ तल को कार्तीय या निर्देशांक तल कहा जाता है और रेखाओं को निर्देशांक अक्ष कहा जाता है।

→ एक तल पर एक बिंदु की स्थिति का निर्धारण करने में प्रयुक्त पद्धति को दकार्ते के सम्मान में कार्तीय पद्धति कहा जाता है ।

→ क्षैतिज रेखा x अक्ष तथा ऊर्ध्वाधर रेखा y-अक्ष कहलाती है।

→ x-अक्ष व y-अक्ष के प्रतिच्छेद बिंदु को मूलबिंदु (origin) कहा जाता है।

→ y-अक्ष से किसी बिंदु की दूरी को उसका x-निर्देशांक या भुज कहते हैं ।

→ x-अक्ष से किसी बिंदु की दूरी को उसका y-निर्देशांक या कोटि कहते हैं।

→ निर्देशांक तल में एक बिंदु के निर्देशांक लिखते समय पहले x-निर्देशांक लिखते हैं और उसके बाद y-निर्देशांक लिखते हैं। हम निर्देशांकों को कोष्ठक के अंदर लिखते हैं।

→ मूलबिंदु के निर्देशांक (0, 0), x-अक्ष पर किसी बिंदु के निर्देशांक (x, 0) व y-अक्ष पर किसी बिंदु के निर्देशांक (0, y) होते हैं।

→ ध्यान रहे-

(i) यदि बिंदु पहले चतुर्थांश में है, तो बिंदु (+, +) के रूप का होगा, क्योंकि पहला चतुर्थांश धनात्मक x-अक्ष और धनात्मक y-अक्ष से परिबद्ध है।
(ii) यदि बिंदु दूसरे चतुर्थाश में है, तो बिंदु (-, +) के रूप का होगा, क्योंकि दूसरा चतुर्थांश ऋणात्मक x-अक्ष और धनात्मक y-अक्ष से परिबद्ध है।
(iii) यदि बिंदु तीसरे चतुर्थांश में है, तो बिंदु (-, -) के रूप में होगा, क्योंकि तीसरा चतुर्थांश ऋणात्मक x-अक्ष और ऋणात्मक y-अक्ष से परिबद्ध है।
(iv) यदि बिंदु चौथे चतुर्थांश में है तो बिंदु (+, -) के रूप में होगा, क्योंकि चौथा चतुर्थांश धनात्मक x-अक्ष और ऋणात्मक y-अक्ष से परिबद्ध है।

Haryana State Board HBSE 9th Class Maths Notes Chapter 11 रचनाएँ Notes.

Haryana Board 9th Class Maths Notes Chapter 11 रचनाएँ

→ अंशाकित पटरी (Ruler)- जिसके एक ओर सेंटीमीटर तथा मिलीमीटर चिह्नित होते हैं तथा दूसरी ओर इंच और उसके भाग चिह्नित होते हैं ।

→ सेट-स्क्वायर- सेट-स्क्वायर का एक युग्म जिसमें एक के कोण 90°, 60° तथा 30° तथा दूसरे के कोण 90°, 45° तथा 45° होते हैं।

→ डिवाइडर- डिवाइडर जिसकी दोनों भुजाओं में दो नुकीले सिरे होते हैं। इसकी भुजाओं को समायोजित किया जा सकता है।

→ परकार- परकार, जिसमें पेंसिल लगाने का विधान होता है, जिससे कोण बनाए जाते हैं ।

Haryana State Board HBSE 9th Class Maths Notes Chapter 2 बहुपद Notes.

Haryana Board 9th Class Maths Notes Chapter 2 बहुपद

→ बहुपद – बहुपद एक ऐसा व्यंजक होता है, जिसमें एक या एक से अधिक पद होते हैं। सामान्यतः चर x में एक बहुपद को निम्नलिखित प्रकार से लिखा जा सकता है-
p(x) = a₀xⁿ + a₁x^n-1 + ……. + a_n-1x + a_n
जिसमें a₀, a₁, a₂, …………., a_n आदि सभी गुणांक वास्तविक संख्याएँ हैं तथा a_n ≠ 0 और n एक ऋणोतर पूर्णांक है।

→ बहुपद की घात- बहुपद की अधिकतम घात वाले पद के घातांक को उस बहुपद की घात कहते हैं; जैसे x⁴ + x³ – x में बहुपद की घात 4 है।

→ एक घात वाले बहुपद को रैखिक बहुपद कहा जाता है।

→ दो घात वाले बहुपद को द्विघाती या द्विघात बहुपद कहा जाता है।

→ तीन घात वाले बहुपद को त्रिघाती बहुपद कहा जाता है।

→ एक शून्येतर अचर बहुपद की घात शून्य होती है अर्थात अचर बहुपद 0 को शून्य बहुपद कहा जाता है।

→ केवल एक पद वाले बहुपद को एकपदी (monomial), दो पदों वाले बहुपद को द्विपद (binomial) तथा तीन पदों वाले बहुपद को त्रिपद (trinomial) कहा जाता है।

→ भाग के किसी प्रश्न के लिए-
भाज्य = (भाजक × भागफल) + शेषफल

→ p(x) = 3x² + x – 1 को (x + 1) से भाग देने पर जो शेषफल प्राप्त होता है, यह वही होता है जोकि बहुपद (x + 1) के शून्यक अर्थात -1 पर बहुपद p(x) का मान होता है।

→ शेषफल प्रमेय- मान लीजिए p(x) एक से अधिक या एक के बराबर घात वाला एक बहुपद है और मान लीजिए a कोई वास्तविक संख्या है। यदि p(x) को रैखिक बहुपद (x – a) से भाग दिया जाए तो शेषफल p(a) होता है।

→ गुणनखंड प्रमेय- यदि p(x) घात n ≥ 1 वाला एक बहुपद हो और कोई वास्तविक संख्या हो तो-
(i) x – a, p(x) का एक गुणनखंड होता है, यदि p(a) = 0 हो और (ii) p(a) = 0 होता है, यदि (x – a), p(x) का एक गुणनखंड हो ।

→ बीजीय सर्वसमिकाएँ-
(i) (x + y)² = x² + 2xy + y²
(ii) (x – y)² = x² – 2xy + y²
(iii) x² – y² = (x + y)(x – y)
(iv) (x + a)(x + b) = x² + (a + b)x + ab
(v) (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
(vi) (x + y)³ = x³ + y³ + 3xy(x + y)
(vii) (x – y)³ = x³ – y³ – 3xy(x – y)
(viii) x + y + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)