Dattu – Page 18 – NCERT MCQ

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 5
Chapter Name	Triangles
Exercise	Ex 5.2
Number of Questions Solved	8
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at 0. Join A to 0. Show that
(i) OB = OC
(ii) AO bisects ∠A
Solution:
(i) In ∆ ABC, we have
AB = AC (Given)
⇒ ∠B = ∠C
(∵ Angles opposite to equal sides are equal)
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 img 1

Question 2.
In ∆ ABC, AD is the perpendicular bisector of BC (see figure). Show that ∆ ABC is an isosceles triangle in which AB = AC.
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 img 2
Solution:
In ∆ABD and ∆ACD, we have ,
DB = DC
∠ ADB = ∠ ADC (∵ D bisect SC)
and AD = AD (Common)
∴ ∆ ABD ≅ ∆ACD (By SAS congruence axiom)
⇒ AB = AC (By CPCT)
Renee,∆ ABC is an isosceles triangle.

Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure). Show that these altitudes are equal.
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 img 3
Solution:
In ∆ ABE and ∆ ACF, we have
∠ AEB = ∠ AFC (BE ⊥ AC, CF ⊥ AS, each 90°)
∠ A = ∠ A (Common)
and AS = AC (Given)
∴ ∆ABE ≅ ∆ACF (By AAS congruence axiom)
⇒ BE = CF (By CPCT)

Question 4. .
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that
(i) ∆ABE = ∆ACF
(ii) AB = AC i.e., ABC is an isosceles triangle.
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 img 4
Solution:
In ∆ABE and ∆ACF, we have
∠ AEB = ∠ AFC (Each 90°)
∠ BAE = ∠ CAF (Common)
and BE =CF (Given)
∴ ∆ABE ≅ ∆ACF (By AAS congruence axiom)
∴ AB = AC
So, ∆ABC is isosceles.

Question 5.
ABC and DBC are isosceles triangles on the same base BC (see figure). Show that ∠ ABD = ∠ACD.
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 img 5
Solution:
In ∆ABC, we have
AB=AC (∵ AABC is an isosceles triangle)
∴ ∠ ABC = ∠ ACB …(i)
(∵ Angles opposite to equal sides are equal)
In ∆ DBC, we have
BD = CD (∵ ADBC is an isosceles triangle)
∴ ∠ DBC = ∠ DCB …(ii)
(∵ Angles opposite to equal sides are equal)
On adding Eqs. (i) and (ii), we have .
∠ ABC + ∠ DBC = ∠ ACB + ∠ DCS
⇒ ∠ ABD = ∠ ACD

Question 6.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠ BCD is a right angle.
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 img 6
Solution:

Question 7.
ABC is a right angled triangle in which ∠ A = 90° and AB = AC, find ∠ B and ∠ C.
Solution:
We have, ∠A = 90° (Given)
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 img 8

AB = AC (Given)
⇒ ∠B = ∠C
(∵ Angles opposite to equal sides are equal)
Now, we have
∠A+∠B+∠C = 180° (By ∆ property)
90° + ∠B+ ∠B = 180°
⇒ 2 ∠B = 90°
⇒ ∠B = 45°
∴ ∠B = ∠C = 45°

Question 8.
Show that the angles of an equilateral triangle are 60° each.
Solution:
Let ∆ ABC be an equilateral triangle, such that
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 img 9
AB = BC = CA (By property)
Now, AB = AC
⇒ ∠B = ∠C …..(i)
(∵Angles opposite to equal sides are equal)
Similarly, CB = CA
⇒∠A = ∠B …(ii)
(∵ Angles opposite to equal sides are equal)
From Eqs. (i) and (ii), we have
∠A=∠B=∠C …(iii)
Now, ∠A+ ∠B+ ∠C = 180° (By ∆ property)
∠A + ∠A + ∠A = 180° [From Eq. (iii)]
3 ∠A = 180°
∠A = 60°
Hence, ∠ A = ∠B = ∠C = 60°

We hope the NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 4
Chapter Name	Lines and Angles
Exercise	Ex 4.3
Number of Questions Solved	6
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3

Question 1.
In figure, sides QP and RQ of APQR are produced to points S and T, respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 1
Solution:

Question 2.
In figure, ∠X – 62°, ∠XYZ = 54°, if YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 4
Solution:
In ∆XYZ,
∵ ∠X+ ∠Y+ ∠Z = 180°
(Sum of all angles of triangle is equal to
∴ 62° + ∠Y + ∠Z = 180° [YZX = 62° (Given)]
⇒ ∠Y + ∠Z = 118° .
⇒ \(\frac { 1 }{ 2 }\)∠Y + \(\frac { 1 }{ 2 }\) ∠Z = \(\frac { 1 }{ 2 }\) x 118°
⇒ ∠OYZ + ∠OZY = 59°
(∵ YO and ZO are the bisectors of ∠XYZ and ∠XZY)
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 5

Question 3.
In figure, if AB || DE, ∠BAC = 35° and ∠CDE = 539 , find ∠DCE.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 7
Solution:
We have AB || DE
⇒ ∠AED= ∠ BAE (Alternate interior angles)
Now, ∠ BAE = ∠ BAC
⇒ ∠ BAE = 35° [ ∵ ∠ BAC = 35° (Given) ]
∴ ∠ AED = 35°
In ∆DCE,
∵ ∠DCE + ∠CED+ ∠EDC= 180° ( ∵Sum of all angles of triangle is equal to 180°)
⇒ ∠ DCE + 35°+ 53° = 180° ( ∵∠ AED = ∠ CED = 35°)
⇒ ∠ DCE = 180° – (35° + 53°)
⇒ ∠ DCE = 92°

Question 4.
In figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠TSQ = 75°, find ∠ SQT.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 8
Solution:
∵∠PTS = ∠RPT + ∠PRT (Exterior angle = Sum of interior opposite angles)
∠ PTS = 95° + 40° [∵ ∠PPT = 95° (Given)]
⇒ ∠ PTS = 135° [and ∠PRT = 40°]
Also, ∠ TSQ + ∠ SQT = ∠ PTS (Exterior angle = Sum of interior opposite angles)
⇒ 75°+ ∠ SQT = 135°
⇒ ∠ SQT = 60° [∵ ∠ TSQ = 75° (Given)]

Question 5.
In figure, if PQ ⊥ PS, PQ||SR, ∠SQR = 2S° and ∠QRT = 65°, then find the values of x and y.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 9
Solution:
Here, PQ || SR
⇒ ∠ PQR = ∠ OPT (Alternate interior angles)
⇒ x + 28° = 65° ⇒ x = 37°
Now, in right angled ASPQ, we have ∠P = 90°
∴ ∠P + x + y = 180° (∵ Sum of all angles of a triangle is equal to 180°)
⇒ 90°+ 37°+ y= 180°
⇒ 127°+ y=180°
⇒ y = 53°

Question 6.
In figure, the side QR of A PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 10
Solution:
In ∆ PQP,
∵ ∠QPR + ∠PQR = ∠PRS …(i)
(∵ Sum of interior opposite angles = Exterior angle)
Now, in ∆ TOR,
∵ ∠QTR + ∠TQR = ∠TRS …..(ii)

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

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 4
Chapter Name	Lines and Angles
Exercise	Ex 4.2
Number of Questions Solved	6
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2

Question 1.
In figure, find the values of x and y and then show that AB || CD.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 1
Solution:
∵ x + 50° = 180° (Linear pair)
⇒ x = 130°
∴ y = 130° (Vertically opposite angle)
Here, ∠x = ∠COD = 130°
These are corresponding angles for lines AB and CD.
Hence, AB || CD

Question 2.
In figure, if AB || CD, CD || EF and y: z = 3:7, find x.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 2
Solution:
Given

⇒ Let y = 3k, z = 7k
x = ∠CHG (Corresponding angles)…(i)
∠CHG = z (Alternate angles)…(ii)
From Eqs. (i) and (ii), we get
x = z …….(iii)
Now, x+y = 180°
(Internal angles on the same side of the transversal)
⇒ z+y = 180° [From Eq. (iii)]
7k + 3k = 180°
⇒ 10k = 180°
⇒ k = 18
∴ y = 3 x 18° = 54°
and z = 7x 18°= 126°
∴ x = z
Now, x + y = 180°
(Internal angles on the same side of the transversal)
⇒ z + y = 180° [From Eq. (iii)]
7k + 3k = 180°
⇒ 10k = 180°
⇒ k = 18
∴ y = 3 x 18° = 54°
and z = 7x 18°= 126°
x = z
⇒ x = 126°

Question 3.
In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 4
Solution:
∵ ∠AGE = ∠GED (Alternate interior angles)
But ∠GED = 126°
⇒ ∠AGE = 126° ….(i)
∴ ∠GEF + ∠FED= 126°
⇒ ∠GEF + 90° =126° (∵ EF ⊥ CD)
⇒ ∠GEF = 36°
Also, ∠AGE + ∠FGE = 180° (Linear pair axiom)
⇒ 126° + ∠FGE =180°
⇒ ∠FGE = 54°

Question 4.
In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 5
Solution:
Drawing a tine parallel to ST through R.

Question 5.
In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 8
Solution:
We have, AB || CD
⇒ ∠APQ = ∠PQR (Alternate interior angles)
⇒ 50° = x
⇒ x = 50°
Now, ∠PQR + ∠QPR = 127°
(Exterior angle is equal to sum of interior opposite angles of a triangle)
⇒ 50°+ ∠QPR = 127°
⇒ y = 77°.

Question 6.
In figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 9
Solution:

We hope the NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid’s Geometry Ex 3.2

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid’s Geometry Ex 3.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid’s Geometry Ex 3.2.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 3
Chapter Name	Introduction to Euclid’s Geometry
Exercise	Ex 3.2
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid’s Geometry Ex 3.2

Question 1.
How would you rewrite Euclid’s fifth postulate so that it would be easier to understand?
Solution:
Two distinct intersecting lines cannot be parallel to the same line.

Question 2.
Does Euclid’s fifth postulate imply the existence of parallel lines? Explain.
Solution:
Yes.
According to Euclid’s fifth postulate when line x falls on line y and z such that ∠1+ ∠2< 180°. Then, line y and line z on producing further will meet in the side of ∠1 arid ∠2 which is less than 180°.
NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid's Geometry Ex 3.2 img 1
We find that the lines which are not according to Euclid’s fifth postulate. i.e., ∠1 + ∠2 = 180°, do not intersect.

We hope the NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid’s Geometry Ex 3.2 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid’s Geometry Ex 3.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 2
Chapter Name	Polynomials
Exercise	Ex 2.4
Number of Questions Solved	5
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4

Question 1.
Determine which of the following polynomials has (x +1) a factor.
(i) x³+x²+x +1
(ii) x⁴ + x³ + x² + x + 1
(iii) x⁴ + 3x³ + 3x² + x + 1
(iv) x³ – x² – (2 +√2 )x + √2
Solution:
The zero of x + 1 is -1.
(i) Let p (x) = x³ + x² + x + 1
Then, p (-1) = (-1)³ + (-1)² + (-1) + 1 .
= -1 + 1 – 1 + 1
⇒ p (- 1) = 0
So, by the Factor theorem (x+ 1) is a factor of x³ + x² + x + 1.
(ii) Let p (x) = x⁴ + x³ + x² + x + 1
Then, P(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1)+1
= 1 – 1 + 1 – 1 + 1
⇒ P (-1) = 1
So, by the Factor theorem (x + 1) is not a factor of x⁴ + x³ + x² + x+ 1.
(iii) Let p (x) = x⁴ + 3x³ + 3x² + x + 1 .
Then, p (-1)= (-1)⁴ + 3 (-1)³ + 3 (-1)² + (- 1)+ 1
= 1- 3 + 3 – 1 + 1
⇒ p (-1) = 1
So, by the Factor theorem (x + 1) is not a factor of x⁴ + 3x³ + 3x² + x+ 1.
(iv) Let p (x) = x³ – x² – (2 + √2) x + √2
Then, p (- 1) =(- 1)3- (-1)2 – (2 + √2)(-1) + √2
= -1 – 1+ 2 +√2+√2
= 2√2
So, by the Factor theorem (x + 1) is not a factor of
x³ – x² – (2 + √2) x + √2.

Question 2.
Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases
(i) p (x)= 2x³ + x² – 2x – 1, g (x) = x + 1
(ii) p(x)= x³ + 3x² + 3x + X g (x) = x + 2
(iii) p (x) = x³ – 4x² + x + 6, g (x) = x – 3
Solution:
(i) The zero of g (x) = x + 1 is x= -1.
Then, p (-1) = 2 (-1)³+ (-1)² – 2 (-1)-1 [∵ p(x) = 2x³ + x² – 2x -1]
= -2 + 1 + 2 – 1
⇒ P (- 1)= 0
Hence, g (x) is a factor of p (x).

(ii) The zero of g (x) = x + 2 is – 2.
Then, p (- 2) = (- 2)³ + 3 (- 2)² +3 (- 2) + 1 [∵ p(x) = x³ + 3x² + 3x + 1]
= – 8 + 12 – 6 + 1
⇒ p(-2) = -1
Hence, g (x) is not a factor of p (x).

(iii) The zero of g (x) = x – 3 is 3.
Then, p (3) = 3³ – 4 (3)²+3 + 6 [∵ p(x) = x³-4x² + x+6]
= 27 – 36+ 3 +6
⇒ p(3) = 0
Hence, g (x) is a factor of p (x).

Question 3.
Find the value of k, if x – 1 is a factor of p (x) in each of the following cases
(i) p (x) = x² + x + k
(ii) p (x) = 2x² + kx + √2
(iii) p (x) = kx² – √2 x + 1
(iv) p (x) = kx² – 3x + k
Solution:
The zero of x – 1 is 1.
(i) (x – 1) is a factor of p (x),then p(1)= 0 (By Factor theorem)
⇒ 1² + 1 + k = 0 [∵ p(x) = x² + x + k]
⇒ 2 + k =0
⇒ k = -2
(ii) ∵ (x -1) is a factor of p (x), then p (1) = 0 (By Factor theorem)
⇒ 2(1)² + k(1)+√2= 0 [∵p(x) = 2x² + kx+ -√2]
⇒ 2 + k + √2 = 0
⇒ k = – (2 + √2)
(iii) ∵ (x-1) is a factor of p (x), then p (1) = 0 (By Factor theorem)
⇒ k (1)² – √2 + 1 = 0 [∵p(x) = kx² – √2x + 1]
⇒ k = (√2 – 1)
(iv) ∵ (x-1) is a factor of p (x), then p (1) = 0 (By Factor theorem)
⇒ k(1)² – 3 + k = 0 [∵p(x) = kx² – 3x + k]
⇒ 2k-3 = 0
⇒ k = \(\frac { 3 }{ 2 }\)

Question 4.
Factorise
(i) 12x² – 7x +1
(ii) 2x² + 7x + 3
(iii) 6x² + 5x – 6
(iv) 3x² – x – 4
Solution:
(i) 12x² – 7x + 1 = 12x² – 4x- 3x + 1 (Splitting middle term)
= 4x (3x – -0 -1 (3x-1)
= (3x -1) (4x -1)
(ii)2x² + 7x + 3 = 2x² + 6x + x + 3 (Splitting middle term)
= 2x (x + 3) +1 (x + 3) = (x + 3) (2x+ 1)
(iii) 6x² + 5x – 6= 6x² + 9x- 4x- 6 (Splitting middle term)
= 3x(2x+3)-2(2x+3)=(2x+3)(3x-2)
(iv) 3x² – x- 4= 3x²-4x+3x-4 (Splitting middle term)
= x (3x – 4) + 1 (3x – 4)= (3x- 4) (x + 1)

Question 5.
Factorise
(i) x³ – 2x² – x + 2
(ii) x³ – 3x² – 9x – 5
(iii) x³ + 13x² + 32x + 20
(iv) 2y³ + y² – 2y – 1
Solution:
(i) Let p (x) = x³ – 2x² – x+ 2, constant term of p (x) is 2.
Factors of 2 are ± 1 and ± 2.
Now, p (1) = 1³ – 2 (1)² – 1 + 2
=1- 2 – 1 + 2
p(1) = 0
By trial, we find that p (1) = 0, so (x – 1) is a factor of p (x).
So, x³ – 2x² – x+ 2
= x³ – x² – x² + x – 2x + 2
= x² ( x -1)- x (x – 1)-2 (x – 1)
= (x – 1)(x² – x – 2)
= (x – 1)(x² – 2x+x-2)
= (x – 1) [x (x – 2) + 1 (x – 2)]
= (x – 1) (x – 2)(x + 1)

(ii) Let p(x) = x³ – 3x² – 9x – 5
By trial, we find that p(5) = (5)³ – 3(5)² – 9(5) – 5
=125 – 75 – 45 – 5 = 0
So, (x – 5) is a factor of p(x).
So, x³– 3x²– 9x – 5
= x³-5x² + 2x²-10x+x-5
= x²(x – 5)+2x(x – 5)+1(x – 5)
= (x – 5) (x² + 2x + 1)
= (x – 5) (x² + x + x + 1)
= (x – 5) [x (x + 1)+ 1 (x+ 1)]
= (x – 5) (x + 1) (x + 1)
= (x – 5)(x+1)²
(iii) Let p (x) = x³ + 13x² + 32x + 20
By trial, we find that p (-1) = (-1)³ + 13(-1)² + 32 (-1) + 20
= -1+13 – 32 + 20 = -33 + 33 = 0
So (x + 1) is a factor of p (x).
So, x³ + 13x² + 32x + 20
= x³+ x² + 12x² + 12x+ 20x+ 20
=x²(x+ 1) + 12x(x+ 1)+ 20 (x+ 1)
= (x+1)(x²+12x+20)
= (x+ 1) (x²+ 10x + 2x+ 20)
= (x+1)[x(x+10)+2(x+10)]
= (x+ 1) (x+ 10) (x + 2)

(iv) Let p (y) = 2y³ + y² – 2y -1
By trial we find that p(1) = 2 (1)³ + (1)² – 2(1) – 1 = 2 + 1 – 2 -1 = 0
So (y -1) is a factor of p (y).
So, 2y³ + y² – 2y -1
= 2y³ – 2y²+ 3y² – 3y + y – 1
= 2y²(y – 1) + 3y(y – 1)+1(y – 1)
= (y – 1) (2y² + 3y + 1)
= (y – 1)(2y² + 2y +y+1)
= (y – 1 [2y (y + 1) + 1 (y + 1)]
= (y – 1)(y+1)(2y+1)

We hope the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4, help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 2
Chapter Name	Polynomials
Exercise	Ex 2.3
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3

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
Solution:
Let p (x) = x³ + 3x² + 3x + 1
(i) The zero of x+ 1 is – 1. (∵ x+ 1 = 0= x = -1)

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3 img 1

Question 2.
Find the remainder when x³-ax² +6x-ais divided by x – a.
Solution:
The zero of x – a is a. (∵ x – a = 0 =» x= a)
Let p (x) = x³ – ax² + 6x – a
So, p (a) = a³ – a (a)² + 6a – a = a³ – a³ + 5a
⇒ p (a) = 5a
∴ Required remainder = 5a (By Remainder theorem)

Question 3.
Check whether 7 + 3x is a factor of 3x³+7x.Solution:
Let f(x) = 3x³+7x
NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3 img 2

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NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 2
Chapter Name	Polynomials
Exercise	Ex 2.2
Number of Questions Solved	4
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

Question 1.
Find the value of the polynomial 5x -4x² + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2
Solution:
Let p (x) = 5x – 4x²+ 3
(i) The value of p (x) = 5x – 4x²+ 3 at x= 0 is
p(0) = 5 x 0 – 4 x 0²+3
⇒ P (0) = 3
(ii) The value of p (x) = 5x – 4x² + 3 at x = -1 is
p(-1) = 5(-D-4(-1)² + 3 = – 5 -4 + 3
⇒ P(-1) = -6
(iii) The value of p (x) = 5x- 4x² + 3 at x = 2 is
p (2) = 5 (2)- 4 (2)² + 3= 10- 16+ 3
⇒ P (2) = – 3

Question 2.
Find p (0), p (1) and p (2) for each of the following polynomials.
(i) p(y) = y² – y +1
(ii) p (t) = 2 +1 + 2t² -t³(iii) P (x) = x³
(iv) p (x) = (x-1) (x+1)
Solution:
(i) p (y) = y² -y +1
∴ p(0) = 0²-0+1
⇒ p(0) = 1
p(1) = 1²-1+ 1
⇒ p(1) = 1
and p (2) = 2² – 2 + 1 =4-2+1
⇒ P (2) = 3
(ii) p (t) = 2 + t + 2t² -1³p(0) = 2+ 0+ 2 x 0²– 0³⇒ P (0) = 2
p (1) = 2 + 1 + 2 x 1² – 1³⇒ p (1) = 3 + 2 – 1
⇒ p(1) = 4
and p (2) = 2 + 2 + 2 x 2² – 2³=4+8-8
⇒ P (2) = 4
(iii) P(x) = x³⇒ p (0) = 0³ ⇒ p (0) = 0 ⇒ p (1) = 1³⇒ P (1) = 1
and p (2) = 2³ ⇒ p (2) = 8
(iv) p(x) = (x-1)(x+ 1)
p(0) = (0-1)(0+1)
⇒ P (0) = – 1
p (1) = (1 – 1) (1 + 1)
⇒ P (1) = 0
and p (2) = (2-1) (2+1)
⇒ P (2) = 3

Question 3.
Verify whether the following are zeroes of the polynomial, indicated against them.
(i)p(x) = 3x + 1,x = –\(\frac { 1 }{ 3 }\)
(ii)p (x) = 5x – π, x = \(\frac { 4 }{ 5 }\)
(iii) p (x) = x² – 1, x = x – 1
(iv) p (x) = (x + 1) (x – 2), x = – 1,2
(v) p (x) = x², x = 0
(vi) p (x) = lx + m, x = – \(\frac { m }{ l }\)
(vii) P (x) = 3x² – 1, x = – \(\frac { 1 }{ \sqrt { 3 } }\),\(\frac { 2 }{ \sqrt { 3 } }\)
(viii) p (x) = 2x + 1, x = \(\frac { 1 }{ 2 }\)
Solution:
NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1 img 1

Question 4.
Find the zero of the polynomial in each of the following cases
(i) p(x)=x+5
(ii) p (x) = x – 5
(iii) p (x) = 2x + 5
(iv) p (x) = 3x – 2
(v) p (x) = 3x
(vi) p (x)= ax, a≠0
(vii) p (x) = cx + d, c≠ 0 where c and d are real numbers.
Solution:
(i) We have, p (x)= x+ 5
Now, p (x) = 0
⇒ x+ 5 = 0
⇒ x = -5
∴ – 5 is a zero of the polynomial p (x).
(ii) We have, p (x) = x – 5
Now, p (x) = 0
⇒ x – 5 = 0
⇒ x = 5
∴ 5 is a zero of the polynomial p (x).
(iii) We have, p (x) = 2x + 5
Now, P (x)= 0
⇒ 2x+ 5= 0
⇒ x = –\(\frac { 5 }{ 2 }\)
∴ –\(\frac { 5 }{ 2 }\) is a zero of the polynomial p (x).
(iv) We have, p (x)= 3x- 2
Now p(x) = 0
⇒ 3x- 2 = 0
⇒ x= \(\frac { 2 }{ 3 }\)
∴ \(\frac { 2 }{ 3 }\) is a zero of the polynomial p (x).
(v) We have, p (x) = 3x
Now, p (x)= 0
⇒ 3x=0
⇒ x =0
∴ 0 is a zero of the polynomial p (x).
(vi) We have, p (x)= ax, a ≠ 0
Now, p (x)= 0 ⇒ ax= 0
⇒ x= 0
∴ 0 is.a zero of the polynomial p (x).
(vii) We have, p (x) = cx + d,c ≠ 0
Now, p (x) = 0
⇒ cx + d = 0
x = – \(\frac { d }{ c }\)
∴ – \(\frac { d }{ c }\) is a zero of the polynomial p (x).

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NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 1
Chapter Name	Number Systems
Exercise	Ex 1.6
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6

Question 1.
Find:

Solution:

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6 img 2

Question 2.
Find:

Solution:
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6 img 4

Question 3.
Simplify:
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6 img 5
Solution:

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6 img 6

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NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 1
Chapter Name	Number Systems
Exercise	Ex 1.5
Number of Questions Solved	5
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5

Question 1.
Classify the following numbers as rational or irrational.
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5 img 1
Solution:
(i) Irrational ∵ 2 is a rational number and √5 is an irrational number.
∴ 2.√5 is an irrational number.
(∵The difference of a rational number and an irrational number is irrational)
(ii) 3 + \( \sqrt{23} \) – \( \sqrt{23} \) = 3 (rational)
(iii) \(\frac { 2\sqrt { 7 } }{ 7\sqrt { 7 } }\) (rational)
(iv) \(\frac { 1 }{ \sqrt { 2 } }\)(irrational) ∵ 1 ≠ 0 is a rational number and \( \sqrt{2} \)≠ 0 is an irrational number.
∴ \(\frac { 1 }{ \sqrt { 2 } }\) is an irrational number. 42
(∵ The quotient of a non-zero rational number with an irrational number is irrational).
(v) 2π (irrational) ∵ 2 is a rational number and π is an irrational number.
∴ 2x is an irrational number. ( ∵The product of a non-zero rational number with an irrational number is an irrational)

Question 2.
Simplify each of the following expressions
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5 img 2
Solution:

Question 3.
Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is π = \(\frac { c }{ d }\). This seems to contradict the fact that n is irrational. How will you resolve this contradiction?
Solution:
Actually \(\frac { c }{ d }\) = \(\frac { 22 }{ 7 }\),which is an approximate value of π.

Question 4.
Represent \( \sqrt{9.3} \) on the number line.
Solution:
Firstly we draw AB = 9.3 units. Now, from S, mark a distance of 1 unit. Let this point be C. Let O be the mid-point of AC. Now, draw aemi – circle with centre O and radius OA. Let us draw a line perpendicular to AC passing through point B and intersecting the semi-circle at point D.
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5 img 4
∴ The distance BD = \( \sqrt{9.3} \)
Draw an arc with centre B and radius BD, which intersects the number line at point E, then the point E represents \( \sqrt{9.3} \) .

Question 5.
Rationalise the denominator of the following
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5 img 5
Solution:

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NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.4

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.4 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.4.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 1
Chapter Name	Number Systems
Exercise	Ex 1.4
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.4

Question 1.
Visualise 3.765 on the number line, using successive magnification.
Solution:
We know that, 3.765 lies between 3 and 4. So, let us divide the part of the number line between 3 and 4 into 10 equal parts and look at the portion between 3.7 and 3.8 through a magnifying glass. Now 3.765 lies between 3.7 and 3.8 [Fig. (i)]. Now, we imagine to divide this again into ten equal parts. The first mark will represent 3.71, the next 3.72 and soon. To see this clearly,we magnify this as shown in [Fig. (ii)].

Again 3.765 lies between 3.76 and 3.77 [Fig. (ii)]. So, let us focus on this portion of the number line [Fig. (iii)] and imagine to divide it again into ten equal parts [Fig. (iii)]. Here, we can visualise that 3.761 is the first mark and 3.765 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.

So, we get seen that it is possible by sufficient successive magnifications of visualise the position (or representation) of a real number with a terminating decimal expansion on the number line.
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.4 img 1

Question 2.
Visualise 4.\(\bar { 26 }\) on the number line, upto 4 decimal places.
Solution:
We adopt process by successive magnification and successively decreasethe lengths of the portion of the number line in which 4.\(\bar { 26 }\) is located. Since 4.\(\bar { 26 }\) is located between 4 and 5 and is divided into 10 equal parts [Fig. (i)]. In further, we locate 4.26between 4.2 and 4.3 [Fig. (ii)].

To get more accurate visualisation of the representation, we divide this portion into 10 equal parts and use a magnifying glass to visualise that 4.\(\bar { 26 }\) lies between 4.26 and 4.27. To visualise 4.\(\bar { 26 }\) more clearly we divide again between 4.26 and 4.27 into 10 equal parts and visualise the repsentation of 4.\(\bar { 26 }\) between 4.262 and 4.263 [Fig. (iii)].

Now, for a much better visualisation between 4.262 and 4.263 is agin divided into 10 equal parts [Fig. (iv)]. Notice that 4.\(\bar { 26 }\) is located closer to 4.263 then to 4.262 at 4.2627.
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.4 img 2

Remark: We can adopt the process endlessly in this manner and simultaneously imagining the decrease in the length of the number line in which 4.\(\bar { 26 }\) is located.

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NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 1
Chapter Name	Number Systems
Exercise	Ex 1.3
Number of Questions Solved	9
Category	NCERT Solutions

NCERT Solutions for 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

Solution:
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 img 2

Question 2.
You know that \(\frac { 1 }{ 7 }\) = \(\bar { 0.142857 }\). Can you predict what the decimal expansions of \(\frac { 2 }{ 7 }\) , \(\frac { 13 }{ 7 }\) , \(\frac { 4 }{ 7 }\) , \(\frac { 5 }{ 7 }\) , \(\frac { 6 }{ 7 }\) are , without actually doing the long division? If so, how?
Solution:
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 img 5

Question 3.
Express the following in the form \(\frac { p }{ q }\)where p and q are integers and q ≠ 0.
(i) 0.\(\bar { 6 }\)
(ii) 0.4\(\bar { 7 }\)
(iii) 0.\(\overline { 001 }\)
Solution:
(i)Let x= 0.\(\bar { 6 }\) = 0.666… ….(i)
Multiplying Eq. (i) by 10, we get
10x = 6.666.. ….(ii)
On subtracting Eq. (ii) from Eq. (i), we get
(10x- x)=(6.666…) – (0.666…)
9x = 6
x= 6/9
⇒ x=2/3

(ii) Let x = 0.4\(\bar { 7 }\) = 0.4777… …(iii)
Multiplying Eq. (iii) by 10. we get
10x = 4.777… . …(iv)
Multiptying Eq. (iv) by 10, we get
100x = 47.777 ….. (v)
On subtracting Eq. (v) from Eq. (iv), we get
(100 x – 10x)=(47.777….)-(4.777…)
90x =43
⇒ x = \(\frac { 43 }{ 90 }\)

(iii) Let x = 0.\(\overline { 001 }\)= 0.001001001… …(vI)
Multiplying Eq. (vi) by (1000), we get
1000x = 1.001001001… .. .(vii)
On subtracting Eq. (vii) by Eq. (vi), we get
(1000x—x)=(1.001001001….) – (0.001001001……)
999x = 1
⇒ x = \(\frac { 1 }{ 999 }\)

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.
Solution:
Let x = 0.99999… ………..(i)
Multiplying Eq. (i) by 10, we get
10x = 9.99999… …(ii)
On subtracting Eq. (ii) by Eq. (i), we get
(10 x – x) = (9.99999..) – (0.99999…)
9x = 9
⇒ x = \(\frac { 9 }{ 9 }\)
x = 1

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.
Solution:
The maximum number of digits in the repeating block of digits in the decimal expansion of \(\frac { 1 }{ 17 }\) is 17-1 = 16 we have,
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 img 6
Thus,\(\frac { 1 }{ 17 }\) = 0.\(\overline { 0588235294117647….., }\)a block of 16-digits is repeated.

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 that 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
Solution:
Consider many rational numbers in the form \(\frac { p }{ q }\) (q≠ 0). where p and q are integers with no common factors other that 1 and having terminating decimal representations.
Let the various such rational numbers be \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 4 }\), \(\frac { 5 }{ 8 }\), \(\frac { 36 }{ 25 }\), \(\frac { 7 }{ 125 }\), \(\frac { 19 }{ 20 }\), \(\frac { 29 }{ 16 }\) etc.
In all cases, we think of the natural number which when multiplied by their respective denominators gives 10 or a power of 10.
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 img 7
From the above, we find that the decimal expansion of above numbers are terminating. Along with we see that the denominator of above numbers are in the form 2m x 5n, where m and n are natural numbers. So, the decimal representation of rational numbers can be represented as a terminating decimal.

Question 7.
Write three numbers whose decimal expansions are non-terminating non-recurring.
Solution:
0.74074007400074000074…
0.6650665006650006650000…
0.70700700070000…

Question 8.
Find three different irrational numbers between the rational numbers \(\frac { 5 }{ 7 }\) and \(\frac { 9 }{ 11 }\) .
Solution:
To find irrational numbers, firstly we shall divide 5 by 7 and 9 by 11,
so,
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 img 8

Question 9.
Classify the following numbers as rational or irrational
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 img 9
Solution:
(i) \( \sqrt{23} \) (irrational ∵ it is not a perfect square.)
(ii) \( \sqrt{225} \) = 15 (rational) (whole number.)
(iii) 0.3796 = rational (terminating.)
(iv) 7.478478… =7.\(\bar { 478 }\) = rational (non-terminating repeating.)
(v) 1.101001000100001… = irrational (non-terminating non-repeating.)

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