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KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.1.

Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Ex 3.1

Question 1.
In Fig 3.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Answer:

∠AOC = ∠BOD = 40°
∠BOD = 40° (Date)
∠AOC = ∠BOD = 40° because vertifically opposite angles.
∴ ∠AOC + ∠BOE = 70°
40° + ∠BOE = 70°
∴ ∠BOE = 70° – 40° = 30°, and
∠AOD = 180° – ∠BOD = 180° – 40° = 140°
Reflex angle COE = ∠COA + ∠AOD + ∠BOD + ∠BOE
= 40° + 140° + 40° + 30°
∴ Reflex angle, ∠COE = 250°.

Question 2.
In Fig. 3.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.

Answer:
∠XOP + ∠POY = 180°
∵ straight supplementary
∠XOP + 90° =180°
∴ ∠XOP = 180° – 90° = 90°
But, ∠XOP = a + b = 90°
a : b = 2 : 3
2 + 3 = 5 Ratio
Ratio 5 means 90°
\(=\frac{2 \times 90}{5}=2 \times 18=36^{\circ}\)
∴ If a = 36° then, ∠b = 54°.
∠XOM = ∠YON = b = 54° (∵ Vertically opposite angles)
∠XON + ∠YON = 180° (∵ Straight angle)
∴ c + 54° = 180°
c = 180 – 54
∴ c = 126°.

Question 3.
In Fig. 3.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Answer:

Data: In this figure, ∠PQR = ∠PRQ.
To prove: ∠PQS = ∠PRT
Proof: In ∆PQR, ∠Q = ∠R.
∴ This is an isosceles triangle.
Let ∠Q = 70°, then ∠R = 70° .
∠PQS + ∠PQR = 180°
∠PQS + 70° = 180°
∴ ∠PQS =180 – 70
∠PQS = 110° … (i)
Similarly,
∠PRT + ∠PRQ = 180°
∠PRT + 70° = 180°
∴ ∠PRT = 180 – 70
∠PRT =100 … (ii)
from (i) and (ii)
∠PQS = ∠PRT =110°
∴ ∠PQS = ∠PRS proved.

Question 4.
If Fig. 3.16, if x + y = w + z, then prove that AOB is a line.

Answer:
Data: In this fiure, ∠BOC = x°
∠AOC = y°
∠BOD = w°
∠AOD = z° and
x + y = w + z.
To Prove: AOB is a straight line.
Proof: x + y + w + z = 360° (∵ one completre angle)
But, x + y = w + z.
∴ x + y = w + z= \(\frac{360}{2}\) = 180°
∴ x + y = 180°
∴ w + z = 180° proved.
But, ∠AOC and ∠BOC are adjacent angles,
∠AOC + ∠BOC = 180°
x = y = 180°
∴ ∠AOB = 180°.
∴ AOB is a straight line.

Question 5.
In Fig. 3.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that
∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS).

Answer:
Data: POQ is a straight line. Ray OR is perpendicular on straight line PQ. OS Ray is in between Rays OP and OR.
To Prove: ∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS)
Proof : ∠QOR = ∠POR = 90° (Data)
∠POR = ∠POS + ∠ROS
∴ ∠ROS = ∠POR – ∠POS
∠ROS = 90°- ∠POS (i)
Now, QOS = ∠QOR + ∠ROS
∠QOS = 90° + ∠ROS
∴ ∠ROS = ∠QOS – 90° (ii)
By adding equation (i) and (ii)

∴ ∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS)

Question 6.
It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYO and reflex ∠QYP.
Answer:

Data: ∠XYZ = 64° and XY are produced up to P. ∠ZYP is bisected.
To Prove: ∠XYQ = ? Reflex ∠QYP = ?
Proof: YQ bisects ∠ZYP
∴ Let ∠PYQ = ∠ZYQ = x°.
∠XYZ + ∠ZYQ + ∠QYP = 180° (∵ ∠XYP is straight angle)
64 + x + x = 180
∴ 64 + 2x = 180
∴ 2x = 180 – 64
2x = 116
∴ \(x=\frac{116}{2}=58^{\circ}\)
∴ ∠PYQ = ∠ZYQ = 58°

(i) ∴ ∠XYQ = ∠XYZ + ∠ZYQ
= 64 + 58
∠XYQ = 122°

(ii) Reflex ∠QYP = ∠PYX + ∠XYZ + ∠ZYQ
= 180 ° + 64° + 58°
= 180 + 122
∴ Reflex ∠QYP = 302°.

We hope the KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.1, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.3 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.3.

Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.3

Question 1.
Find the remainder when x³ + 3x² + 3x + 1 is divided by
i) x + 1
ii) \(x-\frac{1}{2}\)
iii) x
iv) x + π
v) 5 + 2x
Answer:
i) p(x) = x³ + 3x² + 3x + 1 g(x) = x – 1
Let x – 1 = 0, then
x = 1.
As per Remainder theorem, r(x) = p(x) = p(a)
p(x) = x³ + 3x² + 3x + 1
p(1) = (1)³ + 3(1)² + 3(1) + 1
= 1 + 3(1) + 3(1) + 1
= 1 + 3 + 3 + 1
P(1) = 8
∴ r(x) = p(x) = 8
∴ Remainder is 8.

ii) p(x) = x³ + 3x² + 3x + 1 g(x) = \(x-\frac{1}{2}\)
If \(x-\frac{1}{2}=0\) then \(x=\frac{1}{2}\)
p(x) = x³ + 3x² + 3x + 1

∴ r(x) = p(x) = p(a) = \(\frac{15}{8}\)
∴ Remainder is \(\frac{15}{8}\)

iii) p(x) = x³ + 3x² + 3x + 1
g(x) = x
If x = 0, then
p(x) = x³ + 3x² + 3x + 1
p(0) = (0)³ + 3(0)² + 3(0) + 1
= 0 + 3(0) + 3(0) + 1
= 0 + 0 + 0 + 1
p(0) = 0
Remainder r(x) = 1.

iv) p(x) = x³ + 3x² + 3x + 1
g(x) = x + π
If x + π = 0, then x = -π
p(x) = x³ + 3x² + 3x + 1
p(-π) = (-π)³ + 3(-π)² + 3(-π) + 1
p(-πt) = -π²³ – 3π² – 3π + 1
r(x) = -π³ – 3v² – 3π+1

v) p(x) = x³ + 3x² + 3x + 1
g(x) = 5 + 2x
If 5 + 2x = 0, then 2x = -5
\(x=-\frac{5}{2}\)
p(x) = x³ + 3x² + 3x + 1

Question 2.
Find the remainder when x³ – ax² + 6x – a is divided by x – a.
Answer:
p(x) = x³ – ax² + 6x – a
If g(x) = x – a, then r(x) = ?
Let x – a = 0, then x = a
p(x) = x³ – ax² + 6x – a
∴ p(a) = (a)³ – a(a)² + 6(a) – a
= a³ – a³ + 6a – a
∴ p(a) = 5a
∴ r(x) = p(a) = 5a.

Question 3.
Check whether 7 + 3x is a factor of 3x³ + 7x.
Answer:
p(x) = 3x³ + 7x
Let g(x) = 7 + 3x = 0. then
If 7 + 3x = 0, then 3x = -7
\(x=-\frac{7}{3}\)
If p(x) is divided by p(z), remainder r(x) – 0, then g(x) is a factor.
p(x) = 3x³ + 7x

Here, \(r(x)=-\frac{590}{9}\). This is not equal to Zero.
Hence 7 + 3x is not a factor of p(x).

We hope the KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.3 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.3, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 2 Introduction to Euclid Geometry Exercise 2.2.

Karnataka Board Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.2

Question 1.
How would you rewrite Euclid’s fifth postulate so that it would be easier to understand?
Answer:
Euclid’s fifth postulate states that “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 indefinitely meet on that side on which the sum of angles is less than two right angles.”

For example, the line PQ falls on lines AB and CD such that the sum of the interior angles 1 and 2 is less than 180° on the left side of PQ. Therefore, the lines AB and CD will eventually intersect on the left side of PQ.
In the above figure, ∠APQ + ∠PQC < 180° then AB and CD meet at the side of A and C.

Question 2.
Does Euclid’s fifth postulate imply the existence of parallel lines ? Explain.

Answer:

‘n’ is a straight line which intersects l and m straight lines.
∠1 + ∠4 < 180° and ∠2 + ∠3 > 180°.
If l and m, produced by the side of ∠1 and ∠4 they meet at p.
If ∠1 + ∠4 < 180° and ∠2 + ∠3 > 180°,
then l and m straight lines do not meet even produced, which means they are parallel lines.

We hope the KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 2 Introduction to Euclid Geometry Exercise 2.2, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 2 Introduction to Euclid Geometry Exercise 2.1.

Karnataka Board Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1

Question 1.
Which of the following statements are true and which are false? Give reasons for your answers.
(i) Only one line can pass through a single point.
Answer:
False. A number of lines can pass through a single point.

(ii) There is an infinite number of lines which pass through two distinct points.
Answer:
False. Because there is a unique line that passes through two points.

(iii) A terminated line can be produced indefinitely on both sides.
Answer:
True. A terminated line can be produced indefinitely on both sides.

(iv) If two circles’ are equal, then their radii are equal.
Answer:
True. Because the circumference of equal circles is equal their distance is the same from centre. Hence radii of two circles are equal.

(v) In Fig 2.9, if AB = PQ and PQ = XY, then AB = XY.

Answer:
True. Because E.g., If AB = 4 cm, PQ = 4 cm., then, AB = PQ = 4 cm.
PQ = XY = 4 cm.
∴ AB = PQ = XY
∴ AB = XY.

Question 2.
Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them?
(i) parallel lines
(ii) perpendicular lines
(iii) line segment
(iv) the radius of a circle
(v) square.
Answer:
(i) Parallel Lines :

If endpoints of two straight lines do not meet even if they are produced on both sides, they are called parallel lines.
Before the definition concept of point, a straight line is necessary.
Point: A point is that which has no part.
Straight-line: A straight line is a line which lies evenly with the points on itself.

(ii) Perpendicular Lines:

If Angle between \(\overrightarrow{\mathrm{OA}}\) and \(\overrightarrow{\mathrm{OB}}\) is 90°, then only lines are perpendicular mutually.
Before the definition concept of Ray and Angle is necessary.
Ray: Joining one endpoint and the non-end point is called Ray.

Angle:

Ray \(\overrightarrow{\mathrm{OA}}\) and Ray \(\overrightarrow{\mathrm{OB}}\) revolve and forms an angle.

(iii) Segment:

The segment is formed when two endpoints are joined.
Here we must know the point and straight line.
Point: It has only dimension, no length, breadth, and thickness.
A – Starting point
B – Endpoint
AB – Segment
Straight-line: When two endpoints of a segment is produced on both sides, we get a straight line.

iv) The radius of a Circle :

A circle is a set of points that moves from a fixed point.
That equal distance is the radius of the circle.
‘O’ – Fixed point
OA – radius of the circle.
Necessary terms – Point, Segment.

v) Square :

A Square is a figure in which all sides and angles are equal.
AB = BC = CD = DA = 4 cms.
∠A = ∠B = ∠C = ∠D = 90°.
Necessary terms – Quadrilateral, angle.
Quadrilateral In a plane, if 3 points in 4 points are non-collinear and meet in different points, such a figure is called a quadrilateral.
Angle: ‘O’ is common point between \(\overrightarrow{\mathrm{OA}}\) and \(\overrightarrow{\mathrm{OB}}\) Such figure is called an angle.

Question 3.
Consider two ‘postulates’ given below :
(i) Given any two distinct points A and B, there exists a third point C which is in between A and B.
(ii) There exist at least three points that are not on the same line.
Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid’s postulates? Explain.
Answer:
There are many undefined terms for students.
They are consistent because they deal with two different situations.
(i) When two points A and B are given, C is in between two.
(ii) When two points A and B are given take point C which passes through A and B.
These postulates do not follow from Euclid’s postulates. But they follow Axiom 2.1

Question 4.
If a point C lies between two points A and B such that AC = BC, then prove that \(A C=\frac{1}{2} A B\). Explain by drawing the figure.
Answer:

Data: C lies between A and B such that AC = BC.
To Prove: AC = \(\frac{1}{2}\)AB
Proof: In this figure AC = BC.
Adding AC on both sides,
AC + AC = BC + AC (equals are added to equals)
2AC = BC + AC
2AC = AB (∵ BC + AC = AB)
∴ AC = \(\frac{1}{2}\)AB.

Question 5.
In Question 4, point C is called a mid¬point of line segment AB. Prove that every line segment has one and only one mid-point.
Answer:

Let C and D are two points lies on AB.
C is the mid-point of AB.
AC = BC
Adding AC on both sides,
AC + AC = BC + AC
2AC = AB → (i)
D is the mid-point of AB,
AD = DB
Adding AD on both sides,
AD + AD = DB + AD
2AD = AB → (ii)
From equations (i) and (ii),
2AC = 2 AD.
∴ AC = AD.
Then C and D are not different, they are the same.
∴ “Every segment has one and only one mid-point.”

Question 6.
In Fig. 2.10, if AC = BD, then prove that AB = CD.

Answer:
Data: In this figure AC = BD.
To Prove : AB = CD
Proof: AC = BD (Data) → (i)
But, AC = AB + BC
and BD = BC + CD.
Substituting the value of (i)
AC = BD

∴ AB = CD.

Question 7.
Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’ ? (Note that the question is not about the fifth postulate).
Answer:
Euclid’s 5th Axiom of Euclid states that “The whole is greater than the part.”
This is a universal truth. Because it is not applicable to Mathematics only. It is useful for all.
E.g. 1: ‘a’ is whole, ’b’ and ‘c’ are its parts.
a = b + c.
Now, a > b and a > c.
It means a is greater than b.
a is greater than c.
E.g 2: If the human body is full, fingers are its parts.
∴ “Human body is greater than-his fingers.”

We hope the KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 2 Introduction to Euclid Geometry Exercise 2.1, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.5.

Karnataka Board Class 9 Maths Chapter 1 Number Systems Ex 1.5

Question 1.
Classify the following numbers as rational or irrational.
Answer:
i) \(2-\sqrt{5}\)
ii) \((3+\sqrt{23})-\sqrt{23}\)
iii) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\)
iv) \(\frac{1}{\sqrt{2}}\)
v) 2π
Answer:
i) \(2-\sqrt{5}\) = 2 – 2.2360679……….. = -0.2360679
This is a non-terminating, non-recurring decimal.
∴ This is an irrational number.

ii) \((3+\sqrt{23})-\sqrt{23}\)
= \(3+\sqrt{23}-\sqrt{23}\)
= 3
⇒ \(\frac{3}{1}\). This can be written in the form of \(\frac{p}{q}\).
∴ This is a rationl number.

iii) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}=\frac{2 \sqrt{7}}{\sqrt{7}}=2 \Rightarrow \frac{2}{1}\)
This can be written in the form of \(\frac{p}{q}\)
∴ This is a rational number.

iv) \(\frac{1}{\sqrt{2}}\)
\(\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}=\frac{1.4142}{2}\)
= 0.707106…………
This is a non-terminating, non-recurring decimal.
∴ This is an irrational number.

v) 2π
= 2 × 3.1415…..
= 6.2830…….
This is a non-terminating, non-recurring decimal.
∴ This is an irrational number.

Question 2.
Simplify each of the following expressions:
i) \((3+\sqrt{3})(2+\sqrt{2})\)
ii) \((3+\sqrt{3})(3-\sqrt{3})\)
iii) \((\sqrt{5}+\sqrt{2})^{2}\)
iv) \((\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})\)
Answer:

Question 3.
Recall, 2π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, \(\pi=\frac{c}{d}\). This seems to contradict the fact that n is irrational. How will you resolve this contradiction?
Answer:
There is no contradiction. When we measure a length with a scale or any other device we get the quotient.
Therefore we cannot judge whether c is a rational number of d is an irrational number.
∴ Value of \(\frac{c}{d}\) is irrational number.
∴ The value of π is also an irrational number.

Question 4.
Represent \(\sqrt{9.3}\) on the number line.
Answer:
Construction: Mark the distance 9.5 units from a fixed point O such that OB = 9.3 units. Mark midpoint D of OC. Draw a semicircle with centre D. Draw a line perpendicular to OC passing through E and intersecting the semicircle at E. Draw an arc BE which intersect at F. Now, BF = \(\sqrt{9.3}\).

Question 5.
Rationalise the denominators of the following :

Answer:
i) \(\frac{1}{\sqrt{7}}\)
Denominator’s factor is \(\sqrt{7}\) Mulitplying Numerator and denominator by \(\sqrt{7}\).

ii) \(\frac{1}{\sqrt{7}-\sqrt{6}}\)
Denominator’s factor is \(\sqrt{7}+\sqrt{6}\)
Multiplying numerator and denominator by \(\sqrt{7}+\sqrt{6}\),

iii) \(\frac{1}{\sqrt{5}+\sqrt{2}}\)
Denominator’s factor is \(\sqrt{5}-\sqrt{2}\)
Multiplying numerator and denominator by \(\sqrt{5}-\sqrt{2}\).

iv) \(\frac{1}{\sqrt{7}-2}\)
Denominator’s factor is \(\sqrt{7}+2\)
Multiplying numerator and denominator by \(\sqrt{7}+2\).

We hope the KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.5, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.4 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.4.

Karnataka Board Class 9 Maths Chapter 1 Number Systems Ex 1.4

Question 11.
Visualize 3.765 on the number line, using successive magnification.
Answer:
Visualization of 3.765 on the number line:

We know that 3.765 lies between 3 and 4. So, divide the portion between 3 and 4 into 10 equal parts and look at the portion between 3.7 and 3.8 through a magnifying glass. Then, 3.765 lies between 3.7 and 3.8 Fig. Now, we imagine dividing this again into 10 equal parts. The first mark will represent 3.71, the next 3.72, and so on. To see this clearly, we magnify this as shown in Fig. Then, 3.765 lies between 3.76 and 3.77 Fig. So, let us focus on this portion of the number line Fig and imagine to divide it again into 10 equal pans Fig. Here, we can visualize the 3.761 is the first mark, 3.762 is the second mark, and so on.

Question 2.
Visualise \(4 . \overline{26}\) on the number line, up to 4 decimal places.
Answer:
Visulaisation of \(4 . \overline{26}\) upto 4 decimal places on number line :

We hope the KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.4 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.4, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.3.

Karnataka Board Class 9 Maths Chapter 1 Number Systems Ex 1.3

Question 1.
Write the following in decimal form and say what kind of decimal expansion each has :

Answer:
(i)

Terminating decimal expansion.
(ii)

Non-terminating decimal expansion.
\(\begin{aligned} \therefore \quad \frac{1}{11} &=0.090909 \ldots \\ &=0 . \overline{09} \end{aligned}\)
(iii)

∴ Terminating decimal expansion.
(iv)

\(0 . \overline{23069}\) the bar above the digits indicates the block of digits that repeats.
∴ This is Non-terminating, repeating decimal.
(v)

Here, \(0 . \overline{18}\) is block of digits that repeats.
∴ This is non-terminating, repeating decimal.
(vi)

\(\frac{329}{400}=0.8225\) This is terminating decimal because remainder is zero.

Question 2.
You know that\(\frac{1}{7}=0 . \overline{142857}\). Can you predict what the decimal expansions of \(\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\) doing the long division ? If so, how? (Hint: Study the remainders while finding the value of \(\frac{1}{7}\) carefully.).
Answer:


Here the first digit in decimal continued repeated numbers afterward.

Question 3.
Express the following in the form \(\frac{p}{q}\), where p and q are integers and q ≠ 0. (p, q ϵ Z, q ≠ 0).
(i) \(0 . \overline{6}\)
(ii) \(0 . \overline{47}\)
(iii) \(0 . \overline{001}\)
Answer:
(i) \(0 . \overline{6}\) = 0.6666……….
Let x = 0.6666………….
∴ x = 0.6666………..
Multiplying both sides by 10,
10x = 6.666…..
10x = 6 + 0.666
10x = 6 + x
10x – x = 6
9x = 6
\(x=\frac{6}{9}=\text { form of } \frac{\mathrm{p}}{\mathrm{q}}\)
∴ \(0 . \overline{6}=\frac{2}{3}\)

(ii) \(0 . \overline{47}\) = 0.4777…..
Let x = \( 0 . \overline{47}\) then x = 0.4777……..
Here digit 4 is not repeating but 7 is repeated.
Let x = 0.4777
Multiplying both sides by 10.
10x = 4.3 + x
10x -x = 4.3
9x = 4.3
Multiplying both sides by 10.
90x = 43
∴ \(0.4 \overline{7}=\frac{43}{90}\)

(iii) \(0 . \overline{001}\) x = 0.001001……….
Multiplying both sides by 1000
1000x = 1.001001
1000x = 1 + x
1000x – x = 1
999x = 1
\(x=\frac{1}{999}\)
∴ \(0 . \overline{001}=\frac{1}{999}\)

Question 4.
Express 0.99999…….. in the form \(\frac{p}{q}\) , Are you surprised by your answer ? With your teacher and classmates discuss why the answer makes sense.
Answer:
Express 0.99999 in the form of \(\frac{p}{q}\),
Let x = 0.99999 …
Multiplying by 10.
10x = 9.9999……
10x = 9 + 0.9999…….
10x = 9 + x
10x – x = 9
9x = 9
\(x=\frac{9}{9}, x=1=\frac{1}{1}\)
∴ (p = 1 & q = 1)
0.99999 …… goes on forever, so there is no gap between 1 and 0.99999 ………… hence they are equal.

Question 5.
What can the maximum number of digits be in the repeating block of digits in the decimal expansion of \(\frac{1}{17}\) ? Perform the division to check your answer:
Answer:
Expansion of \(\frac{1}{17}\) in decimal:

\(\therefore \quad \frac{1}{17}=0 . \overline{0588235294117647}\)
Here repeated numbers are 16.

Question 6.
Look at several examples of rational numbers in the form \(\frac{p}{q}\) (q≠0), where p
and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
Answer:
Rational numbers in the form \(\frac{p}{q}\) (q ≠ 0) where P and q are integers with no common factors other than 2 or 5 or both.
OR
Prime factors of q have powers of 2 and prime factors of q have powers of 5 or both.

Question 7.
Write three numbers whose decimal expansions are non-terminating non-recurring.
Answer:
Three numbers whose decimal expansions are non-terminating, non-recurring are
i) 0.123123312333
ii) 0.20200200020000
iii) 0.56566566656666

Question 8.
Find three different irrational numbers between the rational numbers \(\frac{5}{7}\) & \(\frac{9}{11}\) .
Answer:
\(\frac{5}{7}=0 . \overline{714285} \quad \frac{9}{11}=0 . \overline{81}\)
Three irrational numbers between \(0 . \overline{714285}\) and \(0 . \overline{81}\) are :
i) 0.72720720072000………….
ii) 0.7357355735555……………..
iii) 0.760760076000………………

Question 9.
Classify the following numbers as rational or irrational :
(i) \(\sqrt{23}\)
(ii) \(\sqrt{225}\)
(iii) 0.3796
(iv) 7.478478
(v) 1.101001000100001…
Answer:
i) \(\sqrt{23}\) = 4.7958………. This is not terminating or non-recurring decimal.
∴ \(\sqrt{23}\) is an irrational number.
ii) \(\sqrt{23}\) = 15 this is a rational number.
iii) 0.3796 This is rational number, because trminating decimal has expansion.
iv) 0.478478… This is rational number, because decimal expansion is recurring.
v) 0.101001000100001… This is an irrational number because termianting or recruring decimal has no expansion.

We hope the KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.3, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Solutions Chapter 3 Lines and Angles Exercise 3.3.

Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Ex 3.3

Question 1.
In Fig. 3.39, sides QP and RQ of ∆PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Answer:

Arms of the ∆PQR QP and RP are produced to S and T and ∠PQT = 110°, ∠SPR = 135°. ∠PRQ =?
Straight-line PR is on straight line SQ.
∠SPR and ∠RPQ are Adjacent angles.
∴ ∠SPR + ∠RPQ = 180°
135 + ∠RPQ = 180°
∴ ∠RPQ =180 – 135
∠RPQ = 45° (i)
Similarly QP straight line is on straight line TR.
∠TQP and ∠PQR are Adjacent angles.
∴ ∠RQP + ∠PQR = 180°
110 + ∠PQR = 180°
∠PQR = 180 – 110
∴ ∠PQR = 70°
Now, in ∆PQR,
∠QPR + ∠PQR + ∠PRQ = 180°
45 + 70 + ∠PRQ = 180°
115 + ∠PRQ = 180°
∠PRQ = 180 – 115
∴ ∠PRQ = 65°.

Question 2.
In Fig. 3.40, ∠X = 62°. ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ

Answer:
In this figure, ∠X = 62°
∠XYZ = 54°
In ∆XYZ, YO and ZO are angular bisectors of ∠XYZ and ∠XZY.
Then, ∠OZY =?
∠OYZ =?

In ∆XYZ
∠X + ∠Y + ∠Z = 180°
62 + 54 + ∠Z= 180
116 + ∠Z= 180
∠Z= 180- 116
∴ ∠Z = 64°
YO is the angular bisector of ∠Y
∴ ∠OYZ = \(\frac{54}{2}\) = 27°
ZO is the angular bisector of ∠Z
∴ ∠OZY = \(\frac{64}{2}\) = 32°
∴ ∠OZY = 32°
Now, in ∆OYZ,
∠OYZ + ∠OZY + ∠YOZ = 180°
27 + 32 + ∠YOZ = 180
59 + ∠YOZ = 180
∠YOZ = 180 – 59
∴∠YOZ = 121°
∴ ∠OZY = 32°
∠YOZ = 121°

Question 3.
In Fig. 3.41, if AB||DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Answer:
If AB || DE, ∠BAC = 35, ∠CDE = 53 then ∠DCE = ?
AB || DE, AE is the bisector.
∴∠BAC = ∠DEC = 35° (∵ Alternate angles)
∴∠DEC= 35°

Now in ∆CDE,
∠DCE + ∠CDE + ∠CED = 180°
∠DCE + 53 + 35 = 180
∠DCE + 88 = 180
∠DEC = 180 – 88
∴ ∠DCE = 92°.

Question 4.
In Fig. 3.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Answer:
PQ and RS straight lines intersect at T.
If ∠PRT = 40°, ∠RPT = 95°, and ∠TSQ = 75°, then ∠SQT =?

In ∆PRT,
∠RPT + ∠PRT + ∠PTR = 180°
95 + 40 + ∠PTR = 180°
135 + ∠PTR = 180
∠PTR = 180 – 135
∴ ∠PTR = 45°
∠PTR = ∠STQ = 45° (∵ Vertically opposite angles)
In ∆TSQ,
∠STQ + ∠TSQ + ∠SQT =180
45 + 75 + ∠SQT = 180
120 + ∠SQT = 180
∴∠SQT = 180 – 120
∴ ∠SQT = 60°.

Question 5.
In Fig. 3.43, PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°. then find the value of x and y.

Answer:
PQ ⊥ PS, PQ ⊥ SR, ∠SQR = 28°, ∠QRT = 65°, Then x = ?, y = ?

Solution: ∠QRT + ∠QRS = 180° (∵ Linear pairs)
65 + ∠QRS = 180
∠QRS = 180 – 65
∴∠QRS =115°
In ∆SRQ,
∠QRS + ∠SQR + ∠RSQ = 180°
115 + 28 + ∠RSQ = 180
∴∠RSQ =180 – 143
∴∠RSQ = 37
Now, ∠RSQ = ∠PQS
37° = x
∴x = 37
In ∆SPQ,
∠SPQ + ∠PSQ + ∠PQS = 180°
90 + y + 37 = 180
∴y = 180- 127
∴y = 53°.

Question 6.
In Fig. 3.44, the side QR of ∆PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = \(\frac{1}{2}\)∠QPR.

Answer:
Data: Arm QR of ∆PQR is produced upto S. Angular bisectors of ∠PQR and ∠PRS meet at T.
To Prove: ∠QTR = \(\frac{1}{2}\) ∠QPR

Proof: Let ∠PQR = 50°, and ∠PRS = 120°
∠PQT = ∠TQR = 25°
∠PRT = ∠TRS = 60°
QR arm of ∆ PQR is produced upto S.
∴Exterior angle ∠PRS = ∠PQR + ∠QPR
120 = 50 + ∠QPR
∴ ∠QPR = 120 – 50
∠QPR = 70°
∴ ∠PRQ = 60°
Now, in ∆TRQ,
∠TQR + ∠TRQ + ∠QTR = 180°
25 + 120 + ∠QTR = 180°
145 + ∠QTR = 180°
∠QTR = 180 – 145
∴ ∠QTR = 35°
Now, ∠QTR = 35° ∠QPR = 70°
∠QTR = \(\frac{70}{2}\)
∴ ∠QTR = \(\frac{10}{2}\) x ∠QPR.

We hope the KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.3, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.2.

Karnataka Board Class 9 Maths Chapter 1 Number Systems Ex 1.2

Question 1.
State whether the following statements are true or false. Justify your answers.
(i) Every irrational number is a real number.
Answer:
True. Because set of real numbers contain both rational and irrational number.

(ii) Every point on the number line is of the form \(\sqrt{\mathrm{m}}\). where’m’ is a natural number.
Answer:
False. Value of \(\sqrt{\mathrm{m}}\) is not netagive number.

(iii) Every real number is an irrational number.
Answer:
False. Because set of real numbers contain both rational and irrational numbers. But 2 is a rational number but not irrational number.

Question 2.
Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rationed number.
Answer:
Square root of all positive integers is not an irrational number.
E.g. \(\sqrt{\mathrm{4}}\) = 2 Rational number.
\(\sqrt{\mathrm{9}}\) = 3 Rational number.

Question 3.
Show how \(\sqrt{\mathrm{5}}\) can be represented on the number line.
Answer:
\(\sqrt{\mathrm{5}}\) can be represented on number line:

In the Right angled ∆OAB ∠OAB = 90°.
OA = 1 cm, AB = 2 cm., then
As per Pythagoras theorem,
OB² =OA² + AB²
= (1)² + (2)²
= 1 + 4
OB² = 5
∴ OB = \(\sqrt{\mathrm{5}}\)
If we draw semicircles with radius OB with ‘O’ as centre, value of \(\sqrt{\mathrm{5}}\) on number line
\(\sqrt{\mathrm{5}}\) = OM = +2.3
and \(\sqrt{\mathrm{5}}\) = ON = -2.3 (accurately).

Question 4.
Classroom activity (Constructing the ‘square root spiral’): Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion.

Start with a point O and draw a line segment OP₁ of unit length. Draw a line segment P₁P2₂ perpendicular to OP₁ of unit length (see fig.). Now draw a line segment P₂P₃ perpendicular to OP₂. Then draw a line segment P₃P₄ perpendicular to OP₃. Continuing in this manner, you can get the line segment P_n-1P_n by drawing a line segment of unit length perpendicular to OP_n-1. In this manner, you will have created the points P₂, P₃, …………….. p_n, …………… and joined them to create a beautiful spiral depicting \(\sqrt{2} \cdot \sqrt{3}, \sqrt{4}, \dots \dots\)
Answer:
Classroom activity :

i) OA = 1 Unit, AB = 1 Unit, ∠A = 90°,
∴OB² = OA² + AB²
= (1)² + (1)²
= 1 + 1
OB² = 2
∴OB = \(\sqrt{2}\)
Similarly, square root spiral can be continued.

We hope the KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.2, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.1.

Karnataka Board Class 9 Maths Chapter 1 Number Systems Ex 1.1

Question 1.
Is a zero a rational number? Can you write it in the form \(\frac{p}{q}\), where p and q are integers and q ≠ 0 ? (p, q,ϵ Z, q ≠ 0)
Answer:
Zero is a rational number.
This can be written in the form of \(\frac{p}{q}\) because \(\frac{o}{q}\) is a rational number.
E.g. \(\frac{0}{2}=0, \quad \frac{0}{5}=0\). etc.
Zero belongs to set of rational number.

Question 2.
Find six rational numbers between 3 and 4.
Answer:
We can write six rational numbers between 3 and 4 as
\(3=\frac{21}{7} \text { and } 4=\frac{28}{7}\)
∴ rational numbers between \(\frac{21}{7}\) and \(\frac{28}{7}\).

Question 3.
Find five rational numbers between \(\frac{3}{5}\) and \(\frac{4}{5}\)
Answer:
Rational numbers between \(\frac{3}{5}\) and \(\frac{4}{5}\) are

∴ Rational numbers between \(\frac{30}{50}\) and \(\frac{40}{50}\)

Question 4.
State whether the following statements are true or false. Give reasons for your answers :
(i) Every natural number is a whole number.
Answer:
True. Because set of natural numbers belongs to a set of whole numbers.
∴ W = {0, 1, 2, 3 …………………….}

(ii) Every integer is a whole number.
Answer:
False. Because zero belongs to a set of integers. But -2, -3, -1 are not whole numbers.

(iii) Every rational number is a whole number.
Answer:
False. Because \(\frac{1}{2}\) is a rational number but not a whole number.

We hope the KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.1, drop a comment below and we will get back to you at the earliest.