NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 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 4.1.

Board CBSE
Textbook NCERT
Class Class 9
Subject Maths
Chapter Chapter 4
Chapter Name Lines and Angles
Exercise Ex 4.1
Number of Questions Solved 6
Category NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1

Question 1.
In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 1
Solution:
Here, ∠AOC and ∠BOD are vertically opposite angles.
∴ ∠AOC = ∠BOD
⇒ ∠AOC = 40° [∵ ∠BOD = 40°(Given)] …(i)
We have, ∠AOC + ∠ BOE = 70° (Given)
40°+ ∠BOE = 70° [From Eq. (i)]
⇒ ∠BOE = 30°
Also, ∠AOC + ∠COE + ∠BOE = 180° (Linear pair axiom)
⇒ 40° + ∠COE + 30° = 180°
⇒ ∠COE = 110°
Now, ∠COE + reflex ∠COE = 360° (Angles at a point)
110°+reflex ∠COE = 360°
⇒ Reflex ∠COE = 250°

Question 2.
In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 2
Solution:
We have, ∠POY = 90°
⇒ ∠POY + ∠POX = 180° (Linear pair axiom)
⇒ ∠POX = 90°
⇒ a+b = 90°
Also, a : b = 2 : 3 (Given)
⇒ Let a = 2k,b = 3k
Now, from Eq. (j), we get
2k + 3k = 90°
⇒ 5k = 90°
⇒ k = 18°
∴ a = 2 x 18°=36°
and b=3 x 18°=54°
Now, ∠MOX + ∠XON = 1800 (Linear pair axiom)
b+ c = 180°
⇒ 540 + c= 180°
⇒ c = 126°

Question 3.
In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 3
Solution:
∵ ∠PQS+ ∠PQR = 180° (Linear pair axiom) ,..(i)
and ∠PRT + ∠PRQ = 180° (Linear pair axiom).. .(ii)
From Eqs. (i) and (ii), we get
∠PQS + ∠PQR =∠PRT + ∠PRQ
∠PQS + ∠PRQ =∠PRT + ∠PRQ
[Given, ∠PQR = ∠PRQ]
⇒ ∠PQS = ∠PRT

Question 4.
In figure, if x + y = w + z, then prove that AOB is a line.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 4
Solution:
∵ x+ y+w+ z = 360° (Angle at a point)
x + y = w + z (Given)…(i)
∴ x+ y+ x+ y = 360° [From Eq. (i)]
2(x + y) = 360°
⇒ x + y = 180° (Linear pair axiom)
Hence, AOB is a straight line.

Question 5.
In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 5
Solution:
We have,
∠POR = ∠ROQ = 90° (∵ Given that, OR is perpendicular to PQ)
∴ ∠POS + ∠ROS = 90°
⇒ ∠ROS = 90° – ∠POS
On adding ∠ROS both sides, we get
2 ∠ROS = 90° – ∠POS + ∠ROS
⇒ 2 ∠ROS = (90° + ∠ROS) – ∠POS
⇒ 2∠ROS = ∠QOS – ∠POS (∵ ∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS)
⇒ ∠ROS = \(\frac { 1 }{ 2 }\) (∠QOS – ∠POS)
Hence proved.

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 ∠XYQ and reflex ∠QYP.
Solution:
Here, YQ bisects ∠ZYP.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 6
Hence, ∠ZYQ = ∠QYP = \(\frac { 1 }{ 2 }\) ∠ZYP ……..(i)
Given, ∠XYZ = 64° ….(ii)
∵ ∠XYZ + ∠ZYQ + ZQYP = 180° (Linear pair axiom)
⇒ 64° + ∠ZYQ + ∠ZYQ = 180° [From Eqs. (i) and (ii)]
⇒ 2 ∠ZYQ = 180° – 64°
⇒ ∠ZYQ = \(\frac { 1 }{ 2 }\) x 116°
⇒ ∠ZYQ = 58°
∴ ∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58° = 122°
Now, ∠QYP + reflex ∠QYP = 360°
58° + reflex ∠QYP = 360°
⇒ reflex ∠ QYP = 302°

We hope the NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.1

NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.1.

Board CBSE
Textbook NCERT
Class Class 9
Subject Maths
Chapter Chapter 8
Chapter Name Linear Equations in Two Variables
Exercise Ex 8.1
Number of Questions Solved 2
Category NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.1

Question 1.
The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
(Take the cost of a notebook to be ₹ x and that of a pen to be ₹ y)
Solution:
Let the cost of a notebook = ₹ x
and the cost of a pen = ₹ y
According to question,
Cost of a notebook = 2(cost of a pen)
∴ x = 2y
⇒ x- 2y = 0

Question 2.
Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case.
(i) 2x + 3y = 9.\(\overline { 35 }\)
(ii) x – \(\frac { y }{ 5 }\) – 10 = 0
(iii) – 2x + 3y = 6
(iv) x = 3y
(v) 2x = – 5y
(vi) 3x + 2 = 0
(vii) y – 2 = 0
(viii) 5 = 2x
Solution:
(i) 2x + 3y = 9.\(\overline { 35 }\)
⇒ 2x + 3y – 9..\(\overline { 35 }\) = 0
On comparing with ax+by+c = 0.
Then, the values of a = 2, b = 3 and c = 9..\(\overline { 35 }\)

(ii) x- \(\frac { y }{ 5 }\) -10= 0
On comparing with ax + by + c = 0, then the values of
a = 1, b = \(\frac { -1 }{ 5 }\) andc = – 10

(iii) – 2x+ 3y = 6
⇒ -2x + 3y – 6=0
On comparing with ax + by + c = 0, then the values of a = – 2, b = 3 and c = – 6

(iv) x = 3y ⇒ x-3y+0=0
On comparing with ax + by + c = 0, then the values of a = 1, b = – 3 and c = 0.

(v) 2x = -5y
⇒ 2x + 5y + 0=0
On comparing with ax + by + c = 0, then the values of a = 2, b = 5 and c = 0.

(vi) 3x + 2 = 0
⇒ 3x + 0y + 2 = 0
On comparing with ax + by + c = 0, then the values of a = 3, b = 0 and , c= 2.

(vii) y – 2 = 0
⇒0x+y-2 = 0
On comparing with ax + by + c = 0, then the values of a = 0, b = 1 and c = – 2.

(viii) 5 = 2x ⇒ 2x + 0y – 5 = 0
On comparing with ax + by + c = 0, then the values of a = 2, b = 0 and c = – 5.

We hope the NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.1 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5.

Board CBSE
Textbook NCERT
Class Class 9
Subject Maths
Chapter Chapter 2
Chapter Name Polynomials
Exercise Ex 2.5
Number of Questions Solved 16
Category NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5

Question 1.
Use suitable identities to find the following products
(i) (x + 4)(x + 10)

(ii) (x+8) (x -10)
(iii) (3x + 4) (3x – 5)
(iv) (y2+ \(\frac { 3 }{ 2 }\)) (y2– \(\frac { 3 }{ 2 }\))
(v) (3 – 2x) (3 + 2x)
Solution:
(i) (x+ 4) (x + 10)
Using identity (iv), i.e., (x+ a) (x+ b) = x2 + (a + b) x+ ab.
We have, (x+4) (x + 10) = x2+(4 + 10) x + (4x 10) (∵ a = 4, b = 10)
= x2 + 14x+40

(ii)
(x+ 8) (x -10)
Using identity (iv), i.e., (x + a) (x + b) = x2 + (a + b) x + ab
We have, (x + 8) (x – 10) = x2 + [8 + (-10)] x + (8) (- 10)(∵a = 8, b = -10)
= x2 – 2x – 80

(iii)
(3x + 4) (3x – 5)
Using identity Eq. (iv), i.e.,
(x + a) (x + b) = x2 + (a + b) x + ab
We have, (3x + 4) (3x – 5) = (3x)2 + (4 – 5) x + (4) (- 5) (∵a = 4, b = -5)
= 9x2 – x – 20

Question 2.
Evaluate the following products without multiplying directly
(i) 103 x 107
(ii) 95 x 96
(iii) 104 x 96
Solution:
(i) 103 x 107 = (100 + 3) (100 + 7)
= 100 x 100+ (3 + 7) (100)+ (3 x 7) [Using identity (iv)]
= 10000+ 1000+21 = 11021
(ii) 95 x 96 = (100-5) (100-4)
= 100 x 100 + [(- 5) + (- 4)] 100 + (- 5 x – 4) [Using identity (iv)]
= 10000 – 900 + 20 = 9120
(iii) 104 x 96 = (100 + 4) (100 – 4)
= (100)2-42 [Using identity (iii)]
= 10000-16 = 9984

Question 3.
Factorise the following using appropriate identities
(i) 9x2+6xy+ y2
(ii) 4y2-4y + 1
(iii) x2 –\(\frac { { y }^{ 2 } }{ 100 }\)
Solution:
NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 img 1

Question 4.
Expand each of the following, using suitable identity
(i) (x+2y+ 4z)2

(ii) (2x – y + z)2
(iii) (- 2x + 3y + 2z)2

(iv) (3a -7b – c)z
(v) (- 2x + 5y – 3z)2

(vi) ( \(\frac { 1 }{ 4 }\)a –\(\frac { 1 }{ 4 }\)b + 1) 2
Solution:
(i) (x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 + 2 (x) (2y) + 2 (2y) (4z) + 2(4z) (x) [Using identity (v)]
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8 zx
(ii) (2x – y + z)2 = (2x)2 + (- y)2 + z2 + 2 (2x) (- y)+ 2 (- y) (z) + 2 (z) (2x) [Using identity (v)]
= 4x2 + y2 + z2 – 4xy – 2yz + 4zx
(iii) (- 2x + 3y + 2z)2 = (- 2x)2 + (3y)2 + (2z)2 + 2 (- 2x) (3y)+ 2 (3y) (2z) + 2 (2z) (- 2x) [Using identity (v)]
= 4x2 + 9y2 + 4z2 – 12xy + 12yz – 8zx
(iv) (3a -7b- c)2 = (3a)2 + (- 7b)2 + (- c)2 + 2 (3a) (- 7b) + 2 (- 7b) (- c) + 2 (- c) (3a) [Using identity (v)]
= 9a2 + 49b2 + c2 – 42ab + 14bc – 6ac
(v) (- 2x + 5y- 3z)2 = (- 2x)2 + (5y)2 + (- 3z)2 + 2 (- 2x) (5y) + 2 (5y) (- 3z) + 2 (- 3z) (- 2x) [Using identity (v)]
= 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx
NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 img 2

Question 5.
Factorise
(i) 4 x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
(ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz
Solution:
(i) 4x2 + 9y2 + 16z2 +12xy-24yz-16xz
= (2x)2 + (3y)2 + (- 4z)2 + 2 (2x) (3y) + 2 (3y) (- 4z) + 2 (- 4z) (2x)
= (2x + 3y – 4z)2
(ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz
= (- √2x)2 + (y)2 + (2 √2z)2 + 2(- √2x) (y)+ 2 (y) (2√2z) + 2 (2√2z) (- √2x)
= (- √2x + y + 2 √2z)2

Question 6.
Write the following cubes in expanded form
NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 img 3
Solution:
(i) (2x + 1)3 = (2x)3 + 13 + 3 (2x) (1) (2x + 1) [Using identity (x + y)3 = x3 + y3 + 3xy (x + y)]
= 8x3 + 1 + 6x (2x + 1)
= 8x3 + 1 + 12x2 + 6x= 8x3 + 12x2 + 6x + 1
(ii) (2a – 3b)3 = (2a)3 – (3b)3 -3(2a)(3b)(2a-3b) [Using identity (x-y)3=x3-y3 -3xy(x-y)]
= 8a3-27b3-18ab(2a-3b)
= 8a3 – 27 b3 – 36a2b + 54ab2
=8a3 – 36a2b + 54ab2 – 27 b3
NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 img 4
NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 img 5

Question 7.
Evaluate the following using suitable identities
(i) (99)3
(ii) (102)3
(iii) (998)3
Solution:
(i) (99)3 =(100-1)3 = 1003 -(1)3 – 3x 100x 1(100-1) [Using identity (x-y)3 =x3-y3-3xy (x-y)]
=1000000-1-300(100-1)
=1000000-1-30000+300
=970299
(ii) (102)3 = (100+ 2)3 = 1003 + 23 + 3x 100x 2 (100+ 2 [Using identity (x + y)3 = x3 + y3 + 3xy (x + y)]
=1000000 + 8 + 600(100+ 2)
=1000000 + 8 + 60000+ 1200=1061208
(iii) (998)3 = (1000-2)3 =10003 – 23 – 3x 1000x 2(1000-2) [Using identity (x-y)3=x3-y3-3xy (x-y)]
=1000000000-8-6000(1000-2)
= 1000000000- 8 – 6000000+12000
=994011992

Question 8.
Factorise each of the following
(i) 8a3 +b3 + 12a2b+6ab2
(ii) 8a3 -b3-12a2b+6ab2
(iii) 27-125a3 -135a+225a2
(iv) 64a3 -27b3 -144a2b + 108ab2

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 img 6
Solution:
(i) 8a3 +b3 +12a2b+6ab2 = (2a)3 + b3 + 3x 2axb (2a+ b)
=(2a+b)3 [Using identity (vi)]
(ii) 8a3 -b3 -12a2b+ 6ab2 = (2a)3 + (-b)3 + 3x2ax (-b)[(2a)+ (-b)]
= (2a)3 – (b)3 – 3 x 2ax b (2a-b) [Using identity (vii)]
=(2 a-b)3
(iii) 27-125a3-135a + 225a2 =(3)3 + (-5a)3 + 3x3x(-5a)[(3)+(-5a)]
= (3)3 -(5a)3– 3x3x5a(3-5a) [Using identity (vii)]
= (3=5 a)3
(iv) 64a3 – 27b3 -144a2b +108ab2 =(4a)3 + (-3b)3 + 3x4ax(-3b)[4a+ (-3b)]
=(4a)3-(3b)3-3x4ax3b(4a-3b)
=(4a-3b)3 [Using identity (vii)]
NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 img 7

Question 9.
Verify
(i) x3 + y3 = (x + y)-(x2 – xy + y2)
(ii) x3 – y3 = (x – y) (x2 + xy + y2)
Solution:
(i) We know that,
(x + y)3 = x3 + y3 + 3xy (x + y)
⇒ x3 + y3 = (x + y)3 – 3xy (x + y)
= (x + y) [(x + y)2 – 3xy]
= (x + y) [x2 + y2 + 2xy – 3xy]
= (x + y) [x2 + y2 – xy]
= RHS
Hence proved,
(ii) We .know that,
(x – y)3 = x3 – y3 – 3xy (x – y) x3 – y3
= (x – y)3 + 3xy (x – y)
= (x – y) [(x – y)2 + 3xy]
= (x – y) [x2 + y2 + 2xy + 3xy]
= (x – y) [x2 + y2 + xy]
= RHS
Hence proved.

Question 10.
Factorise each of the following
(i) 27y3 + 125z
(ii) 64rh3 – 343n [Hint See question 9]
Solution:
(i) 27y3 + 125z3 = (3y)3 + (5z)3
= (3y + 5z) [(3y)2 – (3y) (5z) + (5z)2]
= (3y + 5z) (9y2 – 15yz + 25z2)
(ii) 64 m3 – 343 n3 = (4m)3 – (7n)3
= (4m – 7n) [(4m)2 + (4m) (7n) + (7n)2]
= (4m – 7 n) [16m2 + 28mn + 49n2]

Question 11.
Factorise 27x3 +y3 +z3 -9xyz.
Solution:
27x3 + y3 + z3-9xyz=(3x)3 + y3 + z3 -3x 3xx yx z
=(3x+y+z)[(3x)2 + y2 + z2– 3xy – yz -z(3x)][Using identity (viii)]
=(3x+ y + z) (9x2 + y2 + z2 -3xy -yz -3zx)

Question 12.
Verify that
x3 +y3 +z3 -3xyz = \(\frac { 1 }{ 2 }\) (x + y+z)[(x-y)2 +(y-z)2 +(z-x)2]
Solution:
x3 + y3 + z3-3xyz = (x+y+z)[x2 + y2 + z2-xy-yz-zx]
= \(\frac { 1 }{ 2 }\)(x+ y+ z)[2x2 + 2y2 + 2z2 – 2xy- 2yz – 2zx]
= \(\frac { 1 }{ 2 }\)(x+y+z)[x2 + x2 + y2 + y2 + z2 + z2 – 2xy – 2yz – 2zx]
= \(\frac { 1 }{ 2 }\)(x+y+ z)[x2 + y2 – 2xy+ y2 + z2 – 2yz+ z2 + x2 -2zx]
= \(\frac { 1 }{ 2 }\)(x+y+z)[(x-y)2 + (y-z)2 + (z-x)2]

Question 13.
If x + y + z = 0, show that x3 + y3 + z3 = 3 xyz.
Solution:
We know that,
x3 + y3 + z3 – 3 xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx [Using identity (viii)]
= 0(x2 + y2 + z2 – xy – yz – zx) (∵ x + y + z = 0 given)
= 0
⇒ x3 + y3 + z3 = 3 xyz
Hence proved.

Question 14.
Without actually calculating the cubes, find the value of each of the following
(i) (- 12)3 + (7)3 + (5)3
(ii) (28)3 + (- 15)3 + (- 13)
3
Solution:
We know that, x3 + y3 + z3 -3xyz
=(x+y+z)(x2 + y2 + z2 – xy – yz – zx)
If x+y+z=0, then x3 + y3 + z3-3xyz=0
or x3 + y3 + z3 = 3xyz
(i) We have to find the value of (-12)3 + (7)3 + (5)3
Here, -12+7+5=0
So, (-12)3 + (7)3 + (5)3= 3x(-12)(7)(5)
=-1260
(ii) We have to find the value of (28)3 + (-15)3 + (-13)3.
Here, 28+(-15)+(-13)=28-15-13 =0
So, (28)3 + (-15)3 + (-13)3
= 3x (28) (-15) (-13)
= 16380

Question 15.
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given
(i) Area 25a2 – 35a + 12
(ii) Area 35y2 + 13y – 12
Solution:
(i) We have, area of rectangle
= 25a2 – 35a+12
= 25a2 – 20a – 15a+12
= 5a(5a – 4)-3(5a – 4)=(a-4)(5a – 3)
Possible expression for length =5a-3
and breadth =5a-4
(ii) We have, Area of rectangle = 35y2+13y – 12=35y2 -15y + 28y – 12
=5y(7y – 3)+4(7y – 3) = (7y – 3)(5y+4)
Possible expression on for length =7y-3
and breadth =5y+4

Question 16.
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume 3x2 – 12x
(ii) Volume 12ky2 + 8ky – 20k
Solution:
(i) We have, volume of cuboid =3x2-12x=3x(x – 4)
One possible expressions for the dimensions of the cuboid is 3, x and x-4.
(ii) We have, volume of cuboid =12ky2 + 8ky-20k
=12ky2 + 20ky-12ky-20k
=4ky (3y+5) – 4k(3y+5)
=(3y+5)(4ky – 4k)
=(3y+5)4k(y – 1)
One possible expressions for the dimensions of the cuboid is 4k, 3y+5 and y – 1.

We hope the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5, help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5, 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 3.1

NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid’s Geometry Ex 3.1 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 3.1.

Board CBSE
Textbook NCERT
Class Class 9
Subject Maths
Chapter Chapter 3
Chapter Name Introduction to Euclid’s Geometry
Exercise Ex 3.1
Number of Questions Solved 7
Category NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid’s Geometry Ex 3.1

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.
(ii) There are an infinite number of lines which pass through two distinct points.
(iii) A terminated line can be produced indefinitely on both the sides.
(iv) If two circles are equal, then their radii are equal.
(v) In figure, if AB – PQ and PQ = XY, then AB = XY.
NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid's Geometry Ex 3.1 img 1
Solution:
(i) False. In a single point, infinite number of lines can pass through it.
NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid's Geometry Ex 3.1 img 2
(ii) False. For two distinct points only one straight line is passing.
NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid's Geometry Ex 3.1 img 3
(iii) True.

NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid's Geometry Ex 3.1 img 4
(iv) True. [∵ Radii of congruent (equal) circles are always equal]
(v) True. AB = PQ …..(i)
PQ = XY
⇒ XY = PQ …(ii)
From Eqs. (i) and (ii), we get 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) Radius of a circle
(v) Square
Solution:
(i) Parallel lines Two lines in a plane are said to be parallel, if they have no point in common.
NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid's Geometry Ex 3.1 img 5
In figure, x and y are said to be parallel because they have no point in . common and we write, x\(\parallel\)
y.
Here, the term point is undefined.
(ii) Perpendicular lines Two lines in a plane are said to be perpendicular, if they intersect each other at one right angle.
NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid's Geometry Ex 3.1 img 6
In figure, P and Q are said to be perpendicular lines because they ; intersect each other at 90° and we write Q ⊥ P.
Here, the term one right angle is undefined.
(iii) Line segment The definite length between two points is called the line segment.
NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid's Geometry Ex 3.1 img 7
In figure, the definite length between A and B is line represented by \(\overline { AB }\).
Here, the term definite length is undefined.
(iv) Radius of a circle The distance from the centre to a point on the circle is called the radius of the circle.
NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid's Geometry Ex 3.1 img 8
In the adjoining figure OA is the radius.
Here, the term, point and centre is undefined.
(v) Square A square is a rectangle having same length and breadth.
Here, the terms length, breadth and rectangle are undefined.

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 atleast 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.
Solution:
There are so many undefined words which should be knowledge. They are consistent because they deal with two different situations that is
(i) if two points A and B are given, then there exists a third point C which is in between A and B.
(ii) if two points A and B are given, then we can take a point C which don’t lie on the line passes through the point A and B.
These postulates don’t follow Euclid’s postulates. However, they follow axiom Euclid’s postulate 1 stated as through two distinct points, there is a unique line that passes through them.

Question 4.
If a point C lies between two points A and B such that AC = BC, then prove that AC = \(\frac { 1 }{ 2 }\) AB, explain by drawing the figure.
Solution:
Line AB is drawn by joining A and B using between a point C is taken. AC is taken and put on line AB.
NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid's Geometry Ex 3.1 img 9
i.e., AC covers the line segment AB in two overlaps.
∴ AB = AC + AC
AB = 2 AC
⇒ 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.
Solution:
Here, C is the mid-point of line segment AB, such that AC = BC
NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid's Geometry Ex 3.1 img 10
Let there are two mid-points C and C’ of AB.
⇒ AC = \(\frac { 1 }{ 2 }\) AB
and AC’ = \(\frac { 1 }{ 2 }\) AB
⇒ AC = AC’
which is only possible when C and C’ coincide.
⇒ Point C’ lies on C.
Hence, every line segment has one and only one mid-point.

Question 6.
In figure, if AC = BD, then prove that AB = CD.
NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid's Geometry Ex 3.1 img 11
Solution:
We have
AC = BC
⇒ AC – BC = BD – BC (∵ Equals are subtracted from equals)
⇒ 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.)
Solution:
According to axiom 5, we have The whole is greater than a part, which is a universal truth.
NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid's Geometry Ex 3.1 img 12
Let a line segment PQ = 8 cm. Consider a point R in its interior, such that PR = 5 cm
Clearly, PR is a part of the line segment PQ and ft lies in its interior.
⇒ PR is smaller than PQ.
Hence, the whole is greater than its part.

We hope the NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid’s Geometry Ex 3.1 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 3 Introduction to Euclid’s Geometry Ex 3.1, drop a comment below and we will get back to you at the earliest.

Class 9 Social Science Notes | NCERT Class 9 SST Notes

NCERT Class 9 social science Notes PDF free download provided gives you an overview of the respective chapter and prepared in a manner that every concept is covered as per the syllabus guidelines. NCERT Notes for Class 9 social science SST Geography, History and Civics Standard are very effective for students to have last minute quick revision. Download the required study material from the resources available here and score higher grades in your exams.

Social Science Class 9 Notes CBSE

History Class 9 Notes

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We believe the information shared regarding the NCERT Notes for Class 9 social science & Study Material has aided in your preparation. If you need any further assistance do leave us a comment and we will get back to you at the earliest possible. Stay tuned to our site for more information on Class 9th SST social science Notes, Study Materials, and other preparation related stuff.

Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers

Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers

Here we are providing Polynomials Class 9 Extra Questions Maths Chapter 2 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Polynomials with Answers Solutions

Extra Questions for Class 9 Maths Chapter 2 Polynomials with Solutions Answers

Polynomials Class 9 Extra Questions Very Short Answer Type

Polynomials Class 9 Extra Questions Question 1.
Factorise : 125x3 – 64y3
Solution:
125x3– 6443 = (5x)3 – (4y)3
By using a3 – b3 = (a – b) (a2 + ab + b2), we obtain
125x3– 64y3 = (5x – 4y) (25x2 + 20xy + 16y2)

Class 9 Polynomials Extra Questions Question 2.
Find the value of (x + y)2 + (x – y)2.
Solution:
(x + y)2 + (x – y)2 = x2 + y2 + 2xy + x2 + y2 – 2xy
= 2x2 + 2y2 = 21x2 + y2)

Square polynomial root calculator. grade 10 math linear algebra substitution word problems.

Polynomial Class 9 Extra Questions Question 3.
If p(x)= x2 – 2√2x+1, then find the value of p(2√2)
Solution:
Put x = 2√2 in p(x), we obtain
p(2√2) = (2√2)2 – 2√2(2√2) + 1 = (2√2)2 – (2√2)2 + 1 = 1

Polynomials Extra Questions Class 9 Question 4.
Find the value of m, if x + 4 is a factor of the polynomial x2 + 3x + m.
Solution:
Let p(x) = x2 + 3x + m
Since (x + 4) or (x – (-4)} is a factor of p(x).
∴ p(-4) = 0
⇒ (-4)2 + 3(-4) + m = 0
⇒ 16 – 12 + m = 0
⇒ m = -4

Class 9 Maths Chapter 2 Extra Questions Question 5.
Find the remainder when x3+ x2 + x + 1 is divided by x – \(\frac{1}{2}\) using remainder theorem.
Solution:
Let p(x) = x3+ x2 + x + 1 and q(x) = x – \(\frac{1}{2}\)
Here, p(x) is divided by q(x)
∴ By using remainder theorem, we have
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 1

Extra Questions For Class 9 Maths Chapter 2 Polynomials With Answers Question 6.
Find the common factor in the quadratic polynomials x2 + 8x + 15 and x2 + 3x – 10.
Solution:
x2 + 8x + 15 = x2 + 5x + 3x + 15 = (x + 3) (x + 5)
x2 + 3x – 10 = x2 + 5x – 2x – 10 = (x – 2) (x + 5)
Clearly, the common factor is x + 5.

Polynomials Class 9 Extra Questions Short Answer Type 1

Extra Questions On Polynomials Class 9 Question 1.
Expand :
(i) (y – √3)2
(ii) (x – 2y – 3z)2
Solution: (i)
(y – √3)2 = y2 -2 × y × √3 + (√3)2 = y2 – 2√3 y + 3 (x – 2y – 3z)2
= x2 + 1 – 2y)2 + (-3z)2 + 2 × x × (-2y) + 2 × (-2y) × (-3z) + 2 × (-3z) × x
= x2 + 4y2 + 9z2 – 4xy + 12yz – 6zx

Extra Questions Of Polynomials Class 9 Question 2.
If x + = \(\frac{1}{x}\) = 7, then find the value of x3 + \(\frac{1}{x^{3}}\)
Solution:
We have x + \(\frac{1}{x}\) = 7
Cubing both sides, we have
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 2

Questions On Polynomials Class 9 Question 3.
Show that p – 1 is a factor of p10 + p8 + p6 – p4 – p2 – 1.
Solution:
Let f(p) = p10 + p8 + p6 – p4 – p2 – 1
Put p = 1, we obtain
f(1) = 110 + 18 + 16 – 14 – 12 – 1
= 1 + 1 + 1 – 1 – 1 – 1 = 0
Hence, p – 1 is a factor of p10 + p8 + p6 – p4 – p2 – 1.

Class 9 Maths Polynomials Extra Questions Question 4.
If 3x + 2y = 12 and xy = 6, find the value of 27x3 + 8y3
Solution:
We have 3x + 2y = 12
On cubing both sides, we have
⇒ (3x + 2y)3 = 123
⇒ (3x)3 +(2y)3 + 3 × 3x × 2y(3x + 2y) = √728
⇒ 27x3+ 8y3 + 18xy(3x + 2y) = √728
⇒ 27x3+ 8y3 + 18 × 6 × 12 = √728
⇒ 27x3+ 8y3 + 1296 = √728
⇒ 27x3+ 8y3 = √728 – 1296
⇒ 27x3+ 8y3 = 432

Polynomials Class 9 Questions With Solutions Question 5.
Factorise : 4x2 + 9y2 + 16z22 + 12xy – 24 yz – 16xz.
Solution:
4x2 + 9y2 + 16z22 + 12xy – 24yz – 16xz
= (2x)2 + (3y)2 + (-4z)2 + 2(2x)(3y) + 2(3y)(= 42) + 2(- 42)(2x)
By using a2 + b2 + 2ab + 2bc + 2ca = (a + b + c)2, we obtain
= (2x + 3y – 4z)2 = (2x + 3y – 4z) (2x + 3y – 4z)

Class 9 Maths Ch 2 Extra Questions Question 6.
Factorise : 1 – 2ab – (a2 + b2).
Solution:
1 – 2ab – (a2 + b2) = 1 – (a2 + b2 + 2ab)
= 12 – (a + b)2
= (1 + a + b) (1 – a – b) [∵ x2 – y2 = (x + y)(x – y)]

Polynomials Class 9 Extra Questions Short Answer Type 2

Polynomials Class 9 Extra Questions With Answers Pdf Question 1.
Factorise :
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 3
Solution:
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 4

Important Questions Of Polynomials For Class 9 Question 2.
Factorise 64a3 – 27b3 – 144a2b + 108ab2.
Solution:
64a2 – 27b2 – 144a2b + 108ab2
= (4a)3 – (3b)3 – 36ab(4a – 3b)
= (40)2 – (3b)3 – 3 × 4a × 3b (4a – 3b)
= (4a – 3b)3 [∵ (x – y)3 = x3 – y3 – 3xy(x – y)]
= (40 – 3b) (4a – 3b) (4a – 3b)

Polynomial Extra Questions Class 9 Question 3.
What are the possible expressions for the dimensions of a cuboid whose volume is given below ?
Volume = 12ky2 + 8ky – 20k.
Solution:
We have, volume = 12ky2 + 8ky – 20k
= 4k(3y2 + 2y – 5) = 4k(3y2 + 5y – 3y – 5)
= 4k[y(3y + 5) – 1(3y + 5)] = 4k(3y + 5) (y – 1)
∴Possible expressions for the dimensions of cuboid are 4k units, (3y + 5) units and (y – 1) units.

Polynomials Class 9 Extra Questions With Solutions Question 4.
If p(x) = x3 + 3x2 – 2x + 4, then find the value of p(2) + p(-2) – P(0).
Solution:
Here, p(x) = x3+ 3x2 – 2x + 4
Now, p(2) = 23 + 3(2)2 – 2(2) + 4
= 8 + 12 – 4 + 4 = 20
p(-2) = (-2)3 + 3(-2)2 – 21 – 2) + 4
= 8 + 12 + 4 + 4 = 12
and p(0) = 0 + 0 – 0 + 4 = 4
∴ p(2) + p(-2) – p(0) = 20 + 12 – 4 = 28.

Class 9 Chapter 2 Maths Extra Questions Question 5.
If one zero of the polynomial x2 – √3x + 40 is 5, which is the other zero ?
Solution:
Let p(x) = x2 – √3x + 40
= x2 – 5x – 8x + 40 = x(x – 5) – 8(x – 5) = (x – 5) (x – 8)
Now, for zeroes of given polynomial, put p(x) = 0
∴ (x – 5) (x – 8) = 0
⇒ x = 5 or x = 8
Hence, other zero is 8.

Polynomials Class 9 Important Questions Question 6.
Simplify:
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 5
Solution:
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 6

Extra Questions For Class 9 Maths Polynomials Question 7.
If one zero of the polynomial x2 – √3x + 40 is 5, which is the other zero ?
Solution:
Let
p(x) = x2 – √3x + 40
= x2 – 5x – 8x + 40 = x(x – 5) – 8(x – 5) = (x – 5) (x – 8)
Now, for zeroes of given polynomial, put p(x) = 0
∴ (x – 5) (x – 8) = 0
x = 5 or x = 8
⇒ Hence, other zero is 8.

Polynomials Class 9 Extra Questions Long Answer Type

Ch 2 Maths Class 9 Extra Questions Question 1.
Prove that (a + b + c)3 – a3 – b3 – c3 = 3(a + b) (b + c) (c + a).
Solution:
L.H.S. = (a + b + c)3 – a3 – b3 – c3
= {(a + b + c)3 – 3} – {b3 + c3}
= (a + b + c – a) {(a + b + c)2 + a2 + a(a + b + c)} – (b + c) (b2 + c2 – bc)
= (b + c) {a2 + b2 + 2 + 2ab + 2bc + 2ca + a2 + a2 + ab + ac – b2 – a2 + bc)
= (b + c) (3a2 + 3ab + 3bc + 3ca}
= 3(b + c) {a2 + ab + bc + ca}
= 31b + c) {{a2 + ca) + (ab + bc)}
= 3(b + c) {a(a + c) + b(a + c)}
= 3(b + c)(a + c) (a + b)
= 3(a + b)(b + c) (c + a) = R.H.S.

Class 9 Maths Polynomials Extra Questions With Solutions Question 2.
Factorise : (m + 2n)2 x2 – 22x (m + 2n) + 72.
Solution:
Let m + 2n = a
∴ (m + 2n)2 x2 – 22x (m + 2n) + 72 = a2x2 – 22ax + 72
= a2x2 – 18ax – 4ax + 72
= ax(ax – 18) – 4(ax – 18)
= (ax – 4) (ax – 18)
= {(m + 2n)x – 4)} {(m + 2n)x – 18)}
= (mx + 2nx – 4) (mx + 2nx – 18).

Question 3.
If x – 3 is a factor of x2 – 6x + 12, then find the value of k. Also, find the other factor of the – polynomial for this value of k.
Solution:
Here, x – 3 is a factor of x2 – kx + 12
∴ By factor theorem, putting x = 3, we have remainder 0.
⇒ (3)2 – k(3) + 12 = 0
⇒ 9 – 3k + 12 = 0
⇒ 3k = 21
⇒ k = 7
Now, x2 – 7x + 12 = x2 – 3x – 4x + 12
= x(x – 3) – 4(x – 3)
= (x – 3) (x – 4)
Hence, the value of k is 7 and other factor is x – 4.

Question 4.
Find a and b so that the polynomial x3– 10x2 + ax + b is exactly divisible by the polynomials (x – 1) and (x – 2).
Solution:
Let p(x) = x3– 10x2 + ax + b
Since p(x) is exactly divisible by the polynomials (x – 1) and (x – 2).
∴ By putting x = 1, we obtain
(1)3 – 10(1)2 + a(1) + b = 0
⇒ a + b = 9
And by putting x = 2, we obtain
(2)3 – 10(2)2 + a(2) + b = 0
8 – 40 + 2a + b = 0
⇒ 2a + b = 32
Subtracting (i) from (ii), we have
a = 23
From (i), we have 23 + b = 9 = b = -14
Hence, the values of a and b are a = 23 and b = -14

Question 5.
Factorise : x2 – 6x2 + 11x – 6.
Solution:
Let p(x) = x2 – 6x2 + 11x – 6
Here, constant term of p(x) is -6 and factors of -6 are ± 1, ± 2, ± 3 and ± 6
By putting x = 1, we have
p(1) = (1)3 – 6(1)2 + 11(1) – 6 = 1 – 6 + 11 -6 = 0
∴ (x – 1) is a factor of p(x)
By putting x = 2, we have
p(2) = (2)3 – 6(2)2 + 11(2) – 6 = 8 – 24 + 22 – 6 = 0
∴ (x – 2) is a factor of p(x)
By putting x = 3, we have
p(3) = (3)3 – 6(3)2 + 11(3) – 6 = 27 – 54 + 33 – 6 = 0
∴ (x – 3) is a factor of p(x) Since p(x) is a polynomial of degree 3, so it cannot have more than three linear factors.
∴ x3 – 6x2 + 11x – 6 = k (x – 1) (x – 2) (x – 3)
By putting x = 0, we obtain
0 – 0 + 0 – 6 = k (-1) (-2) (3)
-6 = -6k
k = 1
Hence, x3 – 6x2 + 11x – 6 = (x – 1) (x – 2)(x – 3).

Question 6.
Show that \(\frac{1}{3}\)and \(\frac{4}{3}\) are zeroes of the polynomial 9x3 – 6x2 – 11x + 4. Also, find the third zero of the polynomial.
Solution:
Let p(x) = 9x3– 6x2 – 11x + 4
Put x = \(\frac{1}{3}\), we have
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 7
Thus, x = \(\frac{4}{3}\) is another zero of the polynomial p(x). Since x = \(\frac{1}{3}\) and x = \(\frac{4}{3}\) are the zeroes of p(x), therefore, \(\left(x-\frac{1}{3}\right)\) \(\left(x-\frac{4}{3}\right)\) (3x – 1) (3x – 4) or 9x2 – 15x + 4 exactly divides p(x).
⇒ 9x3 – 6x2 – 11x + 4 = (9x2 – 15x + 4) (x + 1)
Hence, x = -1 is its third zero.

Question 7.
Factorise : 6x2 – 5x2 – √3x + 12
Solution:
Let p(x) = 6x3– 5x2 – √3x + 12
Here, constant term of p(x) is 12 and factors of 12 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 12.
By putting x = 1, we have
p(1) = 6(1)3 – 5(1)2 – √3(1) + 12 = 6 – 5 – √3 + 12 = 0
∴ (x – 1) is a factor of p(x).
Now, by long division, we have
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 8
Thus,
p(x) = (x – 1) (6x2 + x – 12)
p(x) = (x – 1) (6x2 + 9x – 8x – 12)
p(x) = (x – 1) {3x (2x + 3) – 4(2x + 3)}
p(x) = (x – 1) (3x – 4) (2x + 3).

Polynomials Class 9 Extra Questions HOTS

Question 1.
What must be added to polynomial f(x) = x4 + 2x2 – 2x2 + x – 1 so that resulting polynomial is exactly divisible by x2 + 2x – 3?
Solution:
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 9
Here, remainder = -x + 2
To make remainder = 0, we must add -(remainder) in the polynomial
i.e., -(-x + 2) i.e., x – 2
Hence, x4 + 2x3 – 2x2 + x – 1 + (x – 2)
Here, polynomial = x4 + 2x3– 2x2 + 2x – 3 and required addition is (x – 2).

Question 2.
If x = 2 – √3, y = √3 – √7 and 2 = √7 – √4, find the value of x’ + 43 + 2?.
Solution:
Here, x + y + z = 2 – √3+ √3 – √7+√7 – 2 = 0
x3+ √3 + x3= 3(x)(y)(z)
= 3(2 – √3)(√3 – √7)(√7 – 2)
= 3(2√3 – 2√7 – 3 + √21)(√7 – 2)
= 3(2√21 – 14 – 3√7 + 7√3 – 4√3 + 4√7 + 6 – 2√21)
= 3(3√3 + √7 – 8)

Question 3.
If (x – a) is a factor of the polynomials x2 + px – q and x2 + rx – t, then prove that a = \(\frac{t-q}{r-p}\)
Solution:
Let f(x) = x + px -q and g(x) = x2 + x – t
Since x-a is factor of both f(x) and g(x)
⇒ f(a) = g(a) = 0
Now, here f(a) = a2 + pa – q and
g(a) = a2 + ra- t
⇒ a2 + pa – q = a + ra – t (considering f(a) = g(a)]
⇒ pa – q = ra – t
⇒ ra – pa = t – q
⇒ a(r – p) = t – q
a = \(\frac{t-q}{r-p}\)

Polynomials Class 9 Extra Questions Value Based (VBQs)

Question 1.
If a teacher divides a material of volume 27x3 + 54x2 + 36x + 8 cubic units among three students. Is it possible to find the quantity of material ? Can you name the shape of the figure teacher obtained ? Which value is depicted by the teacher ?
Solution:
We know that, √olume = Length × Breadth × Height
Now, 27x3+ 54x2 + 36x + 8
= (3x)3 + 3(3x)2(2) + 3(3x)(2)2 + (2)3
= (3x + 2)2 = (3x + 2) (3x + 2) (3x + 2)
Thus, volume = (3x + 2) (3x + 2) (3x + 2)
Yes, it is possible to find the quantity of material. (3x + 2) units.
Cube.
Apply knowledge and use of example for clarity of subject, student friendly.

Question 2.
In a camp organised by the students of class-9 to donate amount collected to flood victims of Kashmir. At the time of payment of a juice glass at one stall of juice, stall holder asked the students to pay the remainder of x3+ 3x2 + 3x + 1 divided by \(\left(x-\frac{1}{2}\right)\) What is the price of the juice at the stall ? Which value is depicted by class-9 students by organising such camps ?
Solution:
Let
p(x) = x2 + 3x2 + 3x + 1 and
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 10
By long division method, we have
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 11
Remainder = \(\frac{27}{8}\) or 3 \(\frac{3}{8}\)
Thus, price of the juice glass is ₹ 3 \(\frac{3}{8}\)
Caring, kindness, social welfare and helping in development of the needy.

Extra Questions for Class 9 English Beehive, Moments, Literature Reader

Online Education Extra Questions for Class 9 English Beehive, Moments, Literature Reader

In Online Education Extra Questions for Class 9 English: Here we are providing NCERT Extra Questions for Class 9 English Beehive, Moments, English Literature Reader. Students can get Class 9 English NCERT Solutions, Chapter Wise CBSE Class 9 English Important Questions and Answers were designed by subject expert teachers. https://ncertmcq.com/ncert-solutions-for-class-9-english/

You can refer to NCERT Solutions for Class 9 English Beehive, Moments, Literature Reader to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

Class 9 English Important questions with answers for all lessons are provided here on this page. These extra questions are very beneficial for your exam preparation. As it enhances your understanding level, knowledge about the concept, speed, accuracy & time management skills. So, practicing some extra important questions prescribed by NCERT for Class 9 English is mandatory to attain good scores in the final board examination.

Online Education Extra Questions for Class 9 English Beehive, Moments, Literature Reader Important Questions

By accessing the following pdf links you can easily view & downlaod extra questions for class 9 English with solutions that will be useful to ace up your exam preparation.

Extra Questions for Class 9 English Literature Reader

  1. How I Taught My Grandmother to Read Extra Questions
  2. A Dog Named Duke Extra Questions
  3. The Man Who Knew Too Much Extra Questions
  4. Keeping it from Harold Extra Questions
  5. Best Seller Extra Questions
  6. The Brook Extra Questions
  7. The Road Not Taken Extra Questions
  8. The Solitary Reaper Extra Questions
  9. Lord Ullin’s Daughter Extra Questions
  10. The Seven Ages Extra Questions
  11. Oh, I Wish I’d Looked After Me Teeth Extra Questions
  12. Song of the Rain Extra Questions
  13. Villa for Sale Extra Questions
  14. The Bishop’s Candlesticks Extra Questions
  15. Gulliver’s Travels Questions and Answers
  16. Three Men in a Boat Question and Answers

Extra Questions for Class 9 English Beehive Prose

  1. The Fun They Had Extra Questions
  2. The Sound of Music Extra Questions
  3. The Little Girl Extra Questions
  4. A Truly Beautiful Mind Extra Questions
  5. The Snake and the Mirror Extra Questions
  6. My Childhood Extra Questions
  7. Packing Extra Questions
  8. Reach for the Top Extra Questions
  9. The Bond of Love Extra Questions
  10. Kathmandu Extra Questions
  11. If I Were You Extra Questions

Extra Questions for Class 9 English Beehive Poem

  1. The Road Not Taken Extra Questions
  2. Wind Extra Questions
  3. Rain on the Roof Extra Questions
  4. The Lake Isle of Innisfree Extra Questions
  5. A Legend of the Northland Extra Questions
  6. No Men are Foreign Extra Questions
  7. The Duck and the Kangaroo Extra Questions
  8. On Killing a Tree Extra Questions
  9. The Snake Trying Extra Questions
  10. A Slumber did my Spirit Seal Extra Questions

Extra Questions for Class 9 English Moments

  1. The Lost Child Extra Questions
  2. The Adventures of Toto Extra Questions
  3. Iswaran the Storyteller Extra Questions
  4. In the Kingdom of Fools Extra Questions
  5. The Happy Prince Extra Questions
  6. Weathering the Storm in Ersama Extra Questions
  7. The Last Leaf Extra Questions
  8. A House is not a Home Extra Questions
  9. The Accidental Tourist Extra Questions
  10. The Beggar Extra Questions

If you have any doubts or questions regarding NCERT Extra Questions for Class 9 English Beehive, Moments, English Literature Reader, you can reach out to us in the comment section below and we will get back to you as soon as possible.

If I Were You Extra Questions and Answers Class 9 English Beehive

If I Were You Extra Questions and Answers Class 9 English Beehive

Here we are providing If I Were You Extra Questions and Answers Class 9 English Beehive, Extra Questions for Class 9 English was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-english/

If I Were You Extra Questions and Answers Class 9 English Beehive

If I Were You Extra Questions and Answers Short Answer Type

If I Were You Extra Questions Question 1.
Briefly describe Gerrard’s appearance.
Answer:
Gerrard is a man of medium height and wears horn-rimmed glasses. When the play opens, he is dressed in a lounge suit and a great coat. He talks in a cultured voice and his demeanour is confident.

If I Were You Class 9 Extra Questions Question 2.
Who was the Intruder in Gerrard’s house? Why did he break into his house?
Answer:
The Intruder, who broke into Gerrard’s house, was a criminal. He had murdered a cop and was being chased by the police. He broke into Gerrard’s house with the intention of murdering him and taking on his identity to evade the police.

If I Were You Class 9 Extra Questions And Answers Question 3.
How did Gerrard behave on seeing a gun-toting stranger in his cottage?
Answer:
Gerrard kept his cool and remained absolutely unruffled when he saw the gun-totting stranger in his cottage. There was neither any panic nor any ring of tension in his voice. He remained his normal self and talked to him casually.

If I Were You Extra Question Answer Question 4.
Why does the Intruder intend to kill Gerrard?
Answer:
The Intruder is a criminal who is being chased by the police for having murdered a cop. As per his plan, the Intruder intends to kill Gerrard in order to take on his identity and escape capture by the police. In this way, he can lead a peaceful life without living in constant fear of arrest and punishment.

Extra Questions Of If I Were You Question 5.
Why does the Intruder not kill Gerrard immediately?
Answer:
The Intruder does not kill Gerrard immediately because he first wants to get all the necessary information from him. Without this information, his plan to take on Gerrard’s identity will not succeed.

Class 9 If I Were You Extra Questions Question 6.
What impression do you form of the Intruder as he comes in? Give examples to illustrate.
Answer:
The Intruder is similar in build to Gerrard enters from the right silently – revolver in hand. He is flashily dressed in an overcoat and a soft hat. He seems to be a dangerous person as he is carrying a pistol and threatens Gerrard. He claims to have killed a cop. He is mean, heartless and crafty, for he plans to kill Gerrard and assume his identity in order to escape the police. He is over-confident because he claims that Gerrard is no match for him.

If I Were You Important Questions Question 7.
“You’ll soon stop being smart.” Why did Intruder think that Gerrard was being smart?
Answer:
The Intruder thought that Gerrard was being ‘smart’ or clever and facetious because he did not show any fear at the sight of an armed man enter his house and threaten him. To the contrary, he was giving the Intruder smart answers to his questions.

Extra Questions On If I Were You Question 8.
How did the Intruder threaten Gerrard?
Answer:
The Intruder threatened Gerrard by saying that he would soon stop being smart. He would make Gerrard crawl.

Class 9 English If I Were You Extra Questions Question 9.
“I want to know a few things, see.” What sort of information did the Intruder want from Gerrard?
Answer:
The Intruder wanted personal details from Gerrard like whether he lived alone, what his Christian name was, whether he had a car and whether people visited him. All this information was necessary for the execution of his plan to dodge the police by killing Gerrard and taking on his identity.

If I Were You Extra Questions And Answers Question 10.
Why did the Intruder say, “They can’t hang me twice?”
Answer:
The Intruder said this because he was already wanted for having murdered a cop. If he managed to kill Gerrard, as per his plan, and was later arrested, it would not matter as the punishment for this murder, too, would be a hanging. They could not hang him for the two murders twice.

Class 9 English Beehive Chapter 11 Extra Questions Question 11.
How did Gerrard convince the Intruder that he was also wanted by the police?
Answer:
When Gerrard came to know about the Intruder’s plan, he kept his cool. He cooked a story to outwit him. He said he is also a murderer who was in hiding and that is why he was a mystery man who never met anyone including the tradespeople. So, if the Intruder took on his identify, he would not gain anything. He would anyway be accused of murder.

Extra Questions Of Chapter If I Were You Question 12.
Why has the Intruder chosen Gerrard as the man whose identify he wants to take?
Answer:
Gerrard is of the same height and build as the Intruder. There is some similarity in their Appearance. Moreover, Gerard was something of a mystery man who lived alone in the house and very few people visited him. He phoned in his orders, did not meet any tradespeople, and had irregular hours and habits, going away suddenly and coming back just the same. So, the Intruder thought if he killed Gerrard and took on his identity, he would not get caught. He would gain his freedom and would be free to go places. He could live without the fear of cops.

Extra Questions From If I Were You Question 13.
What did Gerrard tell the Intruder about his childhood and his present life? Was he telling the truth? Why/Why not?
Answer:
Gerrard told the Intruder that as a child, he was stolen by the gypsies and now in his thirties he was all alone in life. He was not telling the truth; he was just being funny as he wished to make it clear that he was not afraid of a gun-totting criminal. In fact, Gerrard had already started concocting stories about himself.

If I Were You Extra Question Answers Question 14.
What made Gerrard ask the Intruder, “Are you an American”?
Answer:
Gerrard asked the Intruder if he were an American as he had told him to ‘Put those paws up! ’ and had called him a ‘wise guy’. ‘Paws’ for hands and ‘guy’ for a man are colloquial American expressions. Hence, the usage of these words by the Intruder made Gerrard ask him this question.

Question 15.
The Intruder announced, “I’m going to kill you”. Was Gerrard nervous? How would you describe Gerrard’s reactions?
Answer:
Confident of his presence of mind and his ability to keep his cool in a difficult situation, Gerrard remained unruffled on being threatened by the Intruder. He remained so calm and nonchalant that the Intruder was irritated. His sense of humour also enraged the Intruder. Thus, Gerrard reacted in a calm and composed way.

Question 16.
Where did Gerrard live? Why was it a suitable place for the Intruder’s plan?
Answer:
Gerrard lived in a lonely cottage in a secluded place in the wilds of Essex. With hardly any population around, it was easy for one to commit a crime without getting detected. In addition, the place was visited by only a few people. Therefore, it was suitable for the Intruder to carry out his plan successfully over here.

Question 17.
The Intruder calls himself ‘a poor hunted rat’. Why does he do so?
Answer:
The Intruder describes himself as ‘a poor hunted rat’ because he is being chased by the police for having killed a cop and he has to keep dodging them to escape punishment. He feels he is like a rat being chased by a cat.

Question 18.
Why has the criminal been called an Intruder all through the play
Answer:
An Intruder is a who enters a place without permission in order to commit a crime. The Intruder is called an Intruder throughout the play as he has forced his way into Gerrard’s cottage, with criminal intent. He has come to murder Gerrard and to steal his identity.

Question 19.
Bring out the contrast between the Intruder and Gerrard.
Answer:
The Intruder is flashy, coarse, crude, boastful and an irritable egoist. He is overconfident and thinks that he is very smart. On the other hand, Gerrard is pleasant, cool-headed, cultured, witty and very intelligent. Despite . all this he is a modest and humble person. Although Gerrard does not brag, he proves to be much smarter and more intelligent than the Intruder and is able to outwit him.

Question 20.
Why did very few people come to Gerrard’s house? Who were the few people who visited him?
Answer:
A playwright, Gerrard needed his solitude and congenial surroundings. So, he lived all alone in a secluded place. His theatrical performances made his schedule irregular and he would go away suddenly and come back just the same. Moreover, very few people came to his house. He was visited only by his regular suppliers like the baker, the greengrocer and the milkman, but he did not meet them.

Question 21.
Gerrard tells the Intruder “A mystery I propose to explain.” What is the mystery that he proposes to explain?
Answer:
The Intruder has just told Gerrard that the people in Aylesbury refer to him as a “mystery man”. Gerrard proposes to explain the mystery about his mysterious life, his sudden comings and goings, his irregular routine, and his refusal to see the tradesmen. Gerrard has already concocted a story attributing his strange behaviour to his being a criminal wanted in many, cases of crime.

Question 22.
“This is your big surprise”. Who says these words in the play? What and where? What is the surprise?
Answer:
Gerrard says these words when the Intruder asks him to clarify how he could still be hanged after assuming Vincent Charles Gerrard’s identity. This is a surprise for the Intruder who never suspected Gerrard of being a criminal. According to his information, Gerrard seemed to be the perfect person who could be easily eliminated and then impersonated. However, Gerrard has just turned the tables on him by claiming to be a criminal on the run.

Question 23.
Why and how did Gerrard persuade the Intruder to get into the cupboard?
Answer:
Gerrard concocted a story about his own criminal background. He gave the Intruder the impression that the police were looking for him and he expected a telephone call from a friend, posted as a lookout, informing him of the police’s arrival. So when the telephone rang, he hurried the Intruder into the cupboard and told him that it was connected to the garage which was an escape route.

Question 24.
How does Gerrard propose to use the Intruder’s episode?
Answer:
Being a playwright, Gerrard is amused at being able to turn the tables on the Intruder. He finds the episode of outwitting a criminal by a clever but an innocent man amusing enough to use it as a plot for his next play.

Question 25.
Gerrard describes this encounter with the Intruder as ‘an amusing spot of bother’? What light does this attitude reflect on Gerrard?
Answer:
Any other person in Gerrard’s place would have been paralysed with fear when faced with a gun-totting stranger who has entered one’s house to kill him and steal his identity. But Gerrard finds it ‘a amusing spot of bother’ as his cool and unflappable approach makes him handle the situation comfortably and outwit the Intruder with ease.

Question 26.
What is Gerrard’s profession? Quote the parts of the play that support your answer.
Answer:
Gerrard is associated with theatre as a writer, producer and director. He also supplies props and make-up materials to other theatre agencies. The following facts reveal his profession clearly.

  • He tells the Intruder that his actions are ‘melodramatic’ but not ‘very original’.
  • He welcomes the Intruder as a ‘sympathetic audience’.
  • He comments on the Intruder’s ‘inflection of voice’.
  • He tells someone over the phone that he cannot deliver the props in time.
  • He also tells that person that he had ‘an amusing spot of bother’ which he might put into his next play.

Question 27.
Why was Gerrard’s schedule so irregular?
Answer:
Having a theatrical background, Gerrard devoted time to writing, producing and directing the plays. He also supplied other theatre companies with props and make-up items. Therefore, his schedule was irregular as it had to suit the requirements at the theatres.

Question 28.
Gerrard said, ‘You have been so modest’. Was Gerrard being ironical or truthful?
Answer:
Gerrard’s remark ‘you have been so modest’ was ironical. The Intruder had been boasting of his intelligence and smartness. Hence, Gerrard ironically commented on his modesty and asked him to say something about himself.

Question 29.
What information has the Intruder gathered about Gerrard?
Answer:
The Intruder only knows Gerrard by his last name. He has learnt that, he is the owner of the house in the wilds of Essex. Also, he is a kind of mystery man, who keeps to himself and does not meet anyone. Not many people know about him or visit him.

Question 30.
How did Gerrard fool the Intruder with his false story?
Answer:
Gerrard told the Intruder that he, too, was a criminal on the run. One of his recent crimes had gone wrong and one of his men had been caught. The things which should had been burnt had been found. So, due to that he expected trouble that night. That’s why, he wanted to clear off at the earliest.

Question 31.
How did Gerard lock the Intruder in the cupboard?
Answer:
Having convinced the Intruder that he himself was being wanted by cops and that police could any time break into his home, Gerrard advised the Intruder to escape in his car. When he saw that the Intruder was ready to come with him in his car, Gerrard opened the door of his cupboard and, as the Intruder stepped into the cupboard thinking it was an exit door, Gerrard gave him a push and locked the door from outside.

Question 32.
What precautions did Gerrard take while calling the police?
Answer:
Gerrard was smarter than the Intruder. Having locked the Intruder inside the cupboard, he knocked the revolver out of his hand. To make sure that the Intruder would not break out of the cupboard, Gerrard went to the phone, where he stood with the gun pointed at the cupboard door.

If I Were You Extra Questions and Answers Long Answer Type

Question 1.
Why was Gerrard packing a bag at the beginning of the play? How did it help him to outwit and trap the Intruder?
Answer:
Gerrard was packing a bag in the beginning of the play as he had to deliver some props to some theatrical company for rehearsal. When the Intruder broke into his cottage and threatened to kill him and steal his identity, Gerrard did not lose his cool. He spontaneously concocted a story that he himself was a criminal and was trying to dodge the police. This story was supported by the aura of mystery that surrounded him, his reclusive lifestyle, the bag he had been packing, the disguise outfit, false moustaches etc.

All this misled the Intruder into believing that Gerrard was speaking the truth. He did not doubt Gerrard any longer and unsuspectingly got ready to escape along with him. When Gerrard indicates the door that leads straight to the garage, the Intruder walks into a trap. Hence, the bag played an important role in convincing the Intruder that Gerrard, too, was a criminal like him and was preparing to flee when he broke into his cottage.

Question 2.
Bring out Gerrard’s intelligence, presence of mind and sense of humour. How did these traits help him outwit the Intruder?
Answer:
An intelligent and level-headed person, Gerrard did not show even the slightest of nervousness at the sight of the gun-totting criminal enter his house and threaten to kill him. He knew that his wit and presence of mind would not only help him to manage the crisis but would also contribute towards unnerving the Intruder, and getting the better of him. Keeping the atmosphere light and lively with his sense of humour and funny remarks, Gerrard surprised the Intruder, who had expected him to be afraid.

Once he found out the Intruder was wanted for murder and had been on the run, and thus living in fear, he instantly cooked up a story about his own criminal background. Convincing the Intruder that police would arrive any minute to nab him, he impressed upon the Intruder that they would have to escape immediately. Cleverly, he made him peep into the cupboard saying that it was an escape route.

The moment the Intruder leaned forward to inspect it, Gerrard pushed him into the cupboard and knocked the revolver out of his hand. He closed and locked the door. Thus, his intelligence, sense of humour, and presence of mind turned the tables on the Intruder.

Question 3.
Why did the Intruder find Gerrard’s cooked up story of criminal background convincing?
Answer:
In the beginning, the Intruder suspected every move made by Gerrard. He snubbed him when Gerrard tried to begin a conversation regarding the Intruder’s identity and curtly told him to answer only what was asked. However, he was gullible enough to unsuspiciously walk into Gerrard’s trap because the latter did not lose his cool, and employed his presence of mind to cook up the story that he, too, was wanted by the police.

Actually, Gerrard supported his claim of being a criminal by showing the Intruder his bag, and his disguise outfits and false moustaches etc. The Intruder did not know about the theatrical background of Gerrard and he found his story convincing. He believed that Gerrard was actually trying to evade the police. Hence, the unsuspecting Intruder walked into Gerrard’s trap. This indicates that although he claimed to be the smartest person around, he was in fact not very intelligent. He was outwitted by a smart Gerrard who foiled his plan.

If I Were You Extra Questions and Answers Reference to Context

Read the extracts given below and answer the questions that follow.

Question 1.
Why, this is a surprise, Mr— er—

(a) Who speaks these words and to whom?
Answer:
Vincent Gerrard speaks these words to the Intruder.

(b) Where are they at the time?
Answer:
They are in Gerrard’s cottage, in his sitting room, at the time. The Intruder, who is carrying a revolver has just entered Gerrard’s cottage.

(c) What is the speaker’s tone at the time?
Answer:
The speaker is speaking in a very pleasant tone.

(d) What does this tell you about the speaker?
Answer:
The speaker is a level-headed person. He should have been afraid of the Intruder who was holding a gun, but he was talking in a normal, pleasant manner.

Question 2.
I’m glad you ’re pleased to see me. I don’t think you ’ll be pleased for long. Put those paws up!

(a) Who is speaking these lines and to whom? Where is the conversation taking place?
Answer:
The Intruder is speaking to Gerrard. The conversation is taking place in Gerrard’s lonely cottage situated in the wilds of Essex.

(b) Why is ‘the speaker’ so sure that ‘his listener’ won’t be pleased for long?
Answer:
The speaker is sure that his listener, Gerrard, will not be pleased for long because the speaker plans to kill him and steal his identity.

(c) What does ‘paws’ mean here? Why does the Intruder use the expression?
Answer:
‘Paws’ here stands for ‘hands’. The Intruder wants to convey to Gerrard that he is an American gangster.

(d) Why is the speaker asking the listener ‘to put those paws up’?
Answer:
The Intruder asks Gerrard to put his ‘paws up’ to threaten and intimidate him. He wants to ensure that Gerrard is not able to use his hands for self-defence.

Question 3.
Thanks a lot. You ’ll soon stop being smart. I’ll make you crawl. I want td know a few things, see.

(a) Who is the speaker? Why is he thanking the listener?
Answer:
The Intruder is the speaker here. He is thanking the listener, Gerrard, as the latter had helped him while he was fumbling for a word and Gerrard had suggested the word ‘nonchalant’.

(b) Why does the speaker think that the listener is trying to be smart?
Answer:
The Intruder feels that Gerrard is trying to be smart because instead of displaying any signs of fear, he acts casual and helps the Intruder complete his sentence when the former fails to find the right word.

(c) Why does the speaker expect the listener to soon stop being smart?
Answer:
The speaker feels that Gerrard will be frightened out of his wits the moment he discloses his intention of killing him and will then forget all the witty retorts that he had been making till then.

(d) What does the speaker mean by ‘I’ll make you crawl’?
Answer:
The speaker means that he would bring the listener down on his knees and make him beg for mercy.

Question 4.
At last a sympathetic audience!

(a) Who speaks these words? To whom?
Answer:
Gerrard, the protagonist of the play, speaks these words. He is speaking to the Intruder.

(b) Why does he say it?
Answer:
He wants to throw the Intruder off course by showing him that he does not feel threatened by his presence.

(c) Is he sarcastic or serious?
Answer:
He is certainly sarcastic because he knows that the Intruder wants to gather information about him only to misuse it and he plans to give incorrect information.

(d) Why does the listener wish to know the story of the speaker’s life?
Answer:
The listener is a criminal who resembles Gerrard and wishes to impersonate him. So he wants to know more about him.

Question 5.
I’m sorry. I thought you were telling me, not asking me. A question of inflection; your voice is unfamiliar.

(a) Who is the speaker and who does he speak to?
Answer:
The speaker is Gerrard. He is speaking to the Intruder.

(b) What had the listener asked the speaker?
Answer:
The listener had asked the speaker if he lived in the cottage all by himself.

(c) What does ‘inflection’ mean here? What logic does the speaker give for misinterpreting the inflection of his voice?
Answer:
‘Inflection’ here means ‘tone of voice’. Gerrard says that since the Intruder’s voice was unfamiliar, he couldn’t know whether he was asking a question or telling something.

(d) What do these lines tell us about the speaker?
Answer:
These lines show that the speaker is a very cool-headed man who can think of many ways to elude a question.

Question 6.
That, ’s a lie. You ’re not dealing with a fool. I’m as smart as you and smarter, and I know you run a car. Better be careful, wise guy!

(a) Who is the speaker? Which Tie’ is he talking about?
Answer:
The Intruder is the speaker here. He is talking about the Tie’ that Gerrard told him about not running a car.

(b) Why did the speaker think he was smarter than the listener?
Answer:
The Intruder considered himself smarter because to succeed in his plan of taking on Gerrard’s identity, he had already gathered as much information about Gerrard as he could from the local people.

(c) Why did he warn the listener to be careful?
Answer:
The Intruder wanted to make it clear that Gerrard could not be fool him by telling a lie.

(d) What does the extract reveal about the Intruder?
Answer:
The extract reveals that the Intruder is over-confident about his abilities and that he also under-estimates Gerrard, who is not afraid of him.

Question 7.
You seem to have taken a considerable amount of trouble. Since you know so much about me, won’t you say something about yourself? You have been so modest.

(a) Who speaks these words and to whom?
Answer:
Gerrard speaks these words to the Intruder.

(b) What is his tone when he speaks these words?
Answer:
He is being sarcastic at the time.

(c) Why does he want to know more about the Intruder?
Answer:
He wants to find out more about the Intruder to see if he can get the better of him. He also wants to keep him talking till he receives his telephone call.

(d) What light does this throw on the speaker’s character?
Answer:
The speaker is a quick-thinking person, who does not give way to fear but is looking for a way out of the situation he finds himself in.

Question 8.
I could tell you plenty. You think you ’re smart, but I’m the top of the class round here. I’ve got brains and I use them. That’s how I’ve got where have.

(a) Who speaks these words to whom and in what context?
Answer:
These words are spoken by the Intruder to Gerrard. He utters these words when Gerrard asks him to tell him something about himself

(b) Why does the speaker say “I could tell you plenty”?
Answer:
The Intruder says so because he is over-confident and thinks that he is smart enough to get the better of Gerrard.

(c) What does he mean by ‘the top of the class round here’?
Answer:
The Intruder means to say that no one else is as smart as he is and thatbGerrard, too, is no match for him.

(d) What is his tone at the moment?
Answer:
There is a ring of pride in his words and his ego makes him over-estimate himself and his abilities.

Question 9.
My speciality’s jewel robbery. Your car will do me a treat. It’s certainly a dandy bus.

(a) What does the speaker do? Why does he call it his ‘speciality’?
Answer:
The speaker is a criminal who robs jewellery. He calls it his speciality because robbing jewellery was a pursuit or skill to which he had devoted much time and effort and in which he was an expert.

(b) What does he call ‘a dandy bus’? What does he mean?
Answer:
He calls Gerrard’s car a dandy bus. He means to say that it is a splendid or outstanding car. It will be useful for him and will suit his purpose very well.

(c) What do his words tell you about the speaker?
Answer:
The speaker has made his plans carefully and has found out information about the listener.

(d) What does the speaker intend to do?
Answer:
The speaker uses his brains by planning and committing crimes without getting caught by the police. He now intends to kill Gerrard and assume his identity to escape the law further.

Question 10.
I’m not taking it for fun. I’ve been hunted long enough. I’m wanted for murder already, and they can’t hang me twice.

(a) What ‘step’ is the speaker talking about taking? Why is he taking it?
Answer:
The speaker is talking about taking the ‘step’ of murdering Gerrard. He claims that he is not taking the step for fun but because of his need to escape the police.

(b) By whom has the speaker been hunted? Why?
Answer:
The speaker has been hunted by the police because he is a criminal. He killed a policeman when something went wrong with the job that he did in the town quite a while ago, but since then he is dodging the police.

(c) Why does he say “they can’t hang me twice”?
Answer:
The Intruder has just told Gerrard that he had murdered one man, and that he would not shy away from murdering him too. This is because the police could not hang him twice for two murders.

(d) What light do these lines reflect on the speaker’s state of mind?
The lines reveal that the Intruder does not have any conscience to prick him. He is in a desperate situation now as he fears the punishment he is likely to get if captured.

Question 11.
I’ve got freedom to gain. As for myself I’m a poor hunted rat. As Vincent Charles Gerrard I’m free to go places and do nothing. I can eat well and sleep and without having to be ready to beat it at the sight of a cop.

(a) Why is the speaker a ‘hunted rat’?
Answer:
The Intruder is being chased by the police for having killed a policeman. The fear of being arrested by the police keeps him on the run and he feels that his condition is as miserable as that of a rat being chased.

(b) Why has he chosen to take on Gerrard’s identity?
Answer:
He has chosen to take on Gerrard’s identity because the have a similar height and build and because Gerrard, being a loner, does not meet many people who may catch him out.

(c) Why does the speaker have to run at the sight of a cop?
Answer:
Having killed a cop, the Intruder lives in constant fear of being caught by the police. So, he has to run at the sight of a cop in order to avoid being caught.

(d) What advantage will the speaker have once he impersonates Gerrard?
Answer:
As Gerrard the Intruder will be able to dodge the police. This way he will be able to live in peace and without any fear of the cops.

Question 12.
It brought me to Aylesbury. That’s where I saw you in the car. Two other people saw you and started to talk.
I listened. It looks like you ’re a bit queer — kind of a mystery man.

(a) What is ‘it’? Where did it bring him?
Answer:
‘It’ here refers to the speaker’s dodging the police. While escaping the police he reached Aylesbury.

(b) What did the speaker overhear about the listener? From whom?
Answer:
He overheard two men discuss Gerrard. They referred to him as being strange and a mystery man about whom nothing much was known.

(c) What made the two men conclude that the listener was a mystery man?
Answer:
The two men concluded Gerrard was a mystery man because they did not know much about him. He kept to himself and ordered his supplies on the phone. He did not meet even the tradespeople who delivered the orders. He sometimes went away suddenly and came back just the same.

(d) How did this suit the Intruder’s purpose?
Answer:
This suited the Intruder’s purpose as no one knew Gerrard well enough to recognise him if the Intruder took on his identity. Also, the Intruder would be able to come and go suddenly as Gerrard did.

Question 13.
Don’t be a fool. If you shoot, you ’ll hang for sure. If not as yourself then as Vincent Charles Gerrard.

(a) Why did the speaker say that the listener will be hanged?
Answer:
The speaker said that even if he shot him and took on the speaker’s identity, the listener would be hanged as Gerrard because he was wanted by the police.

(b) What surprise did the speaker give to the listener?
Answer:
Gerrard surprised the Intruder by telling him that he was also a criminal and wanted for murder.

(c) What proof does the speaker give the listener about his being a criminal?
Answer:
Gerrard told the Intruder that he did not meet any trades people and was a bit of a mystery man here today and gone tomorrow because he was a criminal on the run.

(d) What do you think was the speaker’s tone as he spoke to the listener?
Answer:
The speaker’s tone was serious and confidential. The listener was taken in by the speaker.

Question 14.
This is your big surprise. I said you wouldn’t kill me and I was right. Why do you think I am here today and gone tomorrow, never see tradespeople? You say my habits would suit you. You are a crook. Do you think I am a Sunday-school teacher?

(a) What was the big surprise given by the speaker?
Answer:
Gerrard told that the Intruder that he too lived under the threat of being arrested as he too was involved in crime. The Intruder was naturally surprised at this revelation since he was not aware about this aspect of his victim.

(b) What was the speaker right about? Why was he right?
Answer:
Gerrard, the speaker here, was right about the statement that he had made earlier that the Intruder wouldn’t kill him. He was right because the Intruder intended to kill an ordinary person and impersonate him to evade the police. But Gerrard turned out to be a criminal like him. So, killing and impersonating a criminal would not serve the Intruder’s purpose,

(c) Explain the phrase Sunday school teacher? What does the speaker imply by his words?
Answer:
A Sunday school teacher is not just an instructor but is also the responsibility for the spiritual welfare of the students. As such, the Sunday school teacher is an important member of the church and one of high moral standing. By saying he is not a Sunday school teacher, the speaker implies he is a crook.

(d) What light does it throw on the character of the speaker and the listener?
Answer:
The speaker is a quick-thinking cool-headed person, who has retained his presence of mind and lays a trap for the Intruder. He is able to convince the listener. On the other hand, the listener is a gullible person and is taken in by Gerrard’s words.

Question 15.
“I said it with bullets and got away ”.

(a) Who says this?
Answer:
Gerrard, the protagonist of the play ‘If I Were You’, says this.

(b) What does it mean?
Answer:
Gerrard means that he committed a murder with a gun for his escape because things had gone wrong.

(c) Is it the truth? What is the speaker’s reason for saying this?
Answer:
No, it is not the truth. The speaker has concocted a story to befool the Intruder. He shows himself to be a wanted criminal on run from the police so that the Intruder should give up his plan of killing him and taking up his identity.

(d) How was he in imminent danger from the police?
Answer:
One of his men had been caught by the police with some documents.

Question 16.
I have got a man posted on the main road. He’ll ring up if he sees the police, but I don’t want to leave… (telephone bell rings,) Come on! They ’re after us. Through here straight to the garage.

(a) Whose call had Gerrard been expecting?
Answer:
Gerrard had told the person he was speaking to in the beginning to tell someone to call him at once. So, he had been expecting that call.

(b) Whose call had told the Intruder he was expecting?
Answer:
He told the Intruder he was expecting trouble, and had posted a man on the look out who would tell him if the police were coming.

(c) What did he show the Intruder to convince him that he was going to run away?
Answer:
He showed him the packed bag and disguise outfit; false moustaches and what not to show he was ready . to run away.

(d) What is his tone like as he says these words?
Answer:
He says these words in a tone of urgency.

Question 17.
For God’s sake clear that muddled head of yours and let’s go. Come with me in the car. I can use you. If you find it’s a frame, you’ve got me in a car, and you still have your gun.

(a) What does the speaker call the listener’s head “muddled”?
Answer:
The Intruder, who has come to Gerrard’s house to kill him and steal his identity, is told by Gerrard that he, too, is a criminal on the run. The Intruder is thus looking confused.

(b) Where does the speaker invite the other person?
Answer:
The speaker Gerrard is inviting the other person to accompany him in the car and help him escape

(c) What assurance does he give the listener?
Answer:
He tells the Intruder that he has the gun so he can over-power him whenever he feels he has been trapped.

(d) What is in the speaker’s mind?
Answer:
The speaker wants to get the Intruder into a trap where he can hand him over to the police.

NCERT Solutions for Class 9 English Moments Chapter 2 The Adventures of Toto

NCERT Solutions for Class 9 English Moments Chapter 2 The Adventures of Toto are part of NCERT Solutions for Class 9 English. Here we have given NCERT Solutions for Class 9 English Moments Chapter 2 The Adventures of Toto.

Board CBSE
Textbook NCERT
Class Class 9
Subject English Moments
Chapter Chapter 2
Chapter Name The Adventures of Toto
Category NCERT Solutions

NCERT Solutions for Class 9 English Moments Chapter 2 The Adventures of Toto

TEXTUAL EXERCISES
(Page 11)

Think About It

Question 1.
How does Toto come to Grandfather’s private zoo ? (CBSE)
Answer:
Grandfather buys Toto from a tonga-driver for the sum of five rupees. The tonga- driver used to keep the little red monkey. It was tied to a feeding-trough. The monkey looked out of place there. So Grandfather bought him.

Question 2.
“Toto was a pretty monkey”. In what sense is Toto pretty ? (CBSE)
Answer:
Grandfather finds Toto bodily pretty. Toto has bright sparkling eyes. There is mischief beneath his deep-set eyes. He shows his pearly white teeth in a smile. The Anglo- Indian ladies are afraid of his smile. He has quick fingers. His tail serves as a third hand.

Question 3.
Why does Grandfather take Toto to Saharanpur and how ? Why does the ticket collector insist on calling Toto a dog ?
Answer:
Toto does not allow other animals to sleep. So Grandfather took him to Saharanpur. The ticket collector classifies Toto as a dog. He does so because Toto is a four-legged animal.

Question 4.
How does Toto take a bath ? Where has he learnt to do this ? How does Toto almost boil himself alive ?
Answer:
Toto takes a bath like a man. He tests the temperature of the water with his hand. He applies soap and bathes. He has learnt this after seeing human beings bathing. Toto jumps into a kettle. It has water for heating. The water begins to heat up. He tries to come out. He jumps up and down. Grandmother sees him. She saves him from being burnt alive.

Question 5.
Why does the author say, “Toto was not the sort of pet we could keep for long ?’’
Answer:
Toto creates one problem after the other. He doesn’t let other animals in peace. He breaks plates after plates. He tears up clothes, curtains and wallpaper. The family is poor. So it can’t bear this loss. Therefore, Toto is sold back to the tonga-driver for three rupees.

We hope the NCERT Solutions for Class 9 English Moments Chapter 2 The Adventures of Toto help you. If you have any query regarding NCERT Solutions for Class 9 English Moments Chapter 2 The Adventures of Toto, drop a comment below and we will get back to you at the earliest.

Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers

Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers

Here we are providing Triangles Class 9 Extra Questions Maths Chapter 7 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Triangles with Answers Solutions

Extra Questions for Class 9 Maths Chapter 7 Triangles with Solutions Answers

Triangles Class 9 Extra Questions Very Short Answer Type

Triangles Class 9 Extra Questions Question 1.
Find the measure of each exterior angle of an equilateral triangle.
Solution:
We know that each interior angle of an equilateral triangle is 60°.
∴ Each exterior angle = 180° – 60° = 120°

Class 9 Triangles Extra Questions Question 2.
If in ∆ABC, ∠A = ∠B + ∠C, then write the shape of the given triangle.
Solution:
Here, ∠A = ∠B + ∠C
And in ∆ABC, by angle sum property, we have
∠A + ∠B + C = 180°
⇒ ∠A + ∠A = 180°
⇒ 2∠A = 180°
⇒ ∠A = 90°
Hence, the given triangle is a right triangle.

Triangles Extra Questions Class 9 Question 3.
In ∆PQR, PQ = QR and ∠R = 50°, then find the measure of ∠Q.
Solution:
Here, in ∆PQR, PQ = QR
⇒ ∠R = ∠P = 50° (given)
Now, ∠P + ∠Q  + ∠R = 180°
50° + ∠Q + 50° = 180°
⇒ ∠Q = 180° – 50° – 50°
= 80°

Class 9 Maths Chapter 7 Extra Questions With Solutions Question 4.
If ∆SKY ≅ ∆MON by SSS congruence rule, then write three equalities of corresponding angles.
Solution:
Since ∆SKY ≅ ∆MON by SSS congruence rule, then three equalities of corresponding angles
are ∠S = ∠M, ∠K = ∠O and ∠Y = ∠N.

Triangles Class 9 Extra Questions With Solutions Pdf Question 5.
Is ∆ABC possible, if AB = 6 cm, BC = 4 cm and AC = 1.5 cm ?
Solution:
Since 4 + 1.5 = 5.5 ≠ 6
Thus, triangle is not possible.

Triangle Class 9 Extra Questions Question 6.
In ∆MNO, if ∠N = 90°, then write the longest side.
Solution:
We know that, side opposite to the largest angle is longest.
∴ Longest side = MO.

Extra Questions On Triangles Class 9 Question 7.
In ∆ABC, if AB = AC and ∠B = 70°, find ∠A.
Solution:
Here, in ∆ABC AB = AC ∠C = ∠B [∠s opp. to equal sides of a ∆)
Now, ∠A + ∠B + ∠C = 180°
⇒ ∠A + 70° + 70° = 180° [∵ ∠B = 70°]
⇒ ∠A = 180° – 70° – 70° = 40°

Class 9 Maths Triangles Extra Questions Question 8.
In ∆ABC, if AD is a median, then show that AB + AC > 2AD.
Solution:
Triangles Class 9 Extra Questions
Produce AD to E, such that AD = DE.
In ∆ADB and ∆EDC, we have
BD = CD, AD = DE and ∠1 = ∠2
∆ADB ≅ ∆EDC
AB = CE
Now, in ∆AEC, we have
AC + CE > AE
AC + AB > AD + DE
AB + AC > 2AD [∵ AD = DE]

Triangles Class 9 Extra Questions Short Answer Type 1

Extra Questions On Congruence Of Triangles Class 9 Question 1.
In the given figure, AD = BC and BD = AC, prove that ∠DAB = ∠CBA.
Solution:
Class 9 Triangles Extra Questions
In ∆DAB and ∆CBA, we have
AD = BC [given]
BD = AC [given]
AB = AB [common]
∴ ∆DAB ≅ ∆CBA [by SSS congruence axiom]
Thus, ∠DAB =∠CBA [c.p.c.t.]

Extra Questions Of Triangles Class 9 Question 2.
In the given figure, ∆ABD and ABCD are isosceles triangles on the same base BD. Prove that ∠ABC = ∠ADC.
Solution:
Triangles Extra Questions Class 9
In ∆ABD, we have
AB = AD (given)
∠ABD = ∠ADB [angles opposite to equal sides are equal] …(i)
In ∆BCD, we have
CB = CD
⇒ ∠CBD = ∠CDB [angles opposite to equal sides are equal] … (ii)
Adding (i) and (ii), we have
∠ABD + ∠CBD = ∠ADB + ∠CDB
⇒ ∠ABC = ∠ADC

Extra Questions For Class 9 Maths Triangles Question 3.
In the given figure, if ∠1 = ∠2 and ∠3 = ∠4, then prove that BC = CD.
Solution:
Class 9 Maths Chapter 7 Extra Questions With Solutions
In ∆ABC and ACDA, we have
∠1 = ∠2 (given)
AC = AC [common]
∠3 = ∠4 [given]
So, by using ASA congruence axiom
∆ABC ≅ ∆CDA
Since corresponding parts of congruent triangles are equal
∴ BC = CD

Class 9 Maths Chapter 7 Extra Questions Question 4.
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
Triangles Class 9 Extra Questions With Solutions Pdf
Solution:
Here, ∠B < ∠A
⇒ AO < BO …..(i)
and ∠C < ∠D
⇒ OD < CO …..(ii)
[∴ side opposite to greater angle is longer]
Adding (i) and (ii), we obtain
AO + OD < BO + CO
AD < BC

Triangles Class 9 Important Questions With Solutions Question 5.
In the given figure, AC > AB and D is a point on AC such that AB = AD. Show that BC > CD.
Solution:
Triangle Class 9 Extra Questions
Here, in ∆ABD, AB = AD
∠ABD = ∠ADB
[∠s opp. to equal sides of a ∆]
In ∆BAD
ext. ∠BDC = ∠BAD + ∠ABD
⇒ ∠BDC > ∠ABD ….(ii)
Also, in ∆BDC .
ext. ∠ADB > ∠CBD …(iii)
From (ii) and (iii), we have
∠BDC > CD [∵ sides opp. to greater angle is larger]

Class 9 Maths Ch 7 Extra Questions Question 6.
In a triangle ABC, D is the mid-point of side AC such that BD = \(\frac{1}{2}\) AC. Show that ∠ABC is a right angle.
Solution:
Extra Questions On Triangles Class 9
Here, in ∆ABC, D is the mid-point of AC.
⇒ AD = CD = \(\frac{1}{2}\)AC …(i)
Also, BD = \(\frac{1}{2}\)AC… (ii) [given]
From (i) and (ii), we obtain
AD = BD and CD = BD
⇒ ∠2 = ∠4 and ∠1 = ∠3 …..(iii)
In ∆ABC, we have
∠ABC + ∠ACB + ∠CAB = 180°
⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ ∠1 + ∠2 + ∠1 + ∠2 = 180° [using (iii)]
⇒ 2(∠1 + ∠2) = 180°
⇒ ∠1 + ∠2 = 90°
Hence, ∠ABC = 90°

Triangles Class 9 Extra Questions Short Answer Type 2

Chapter 7 Maths Class 9 Extra Questions Question 1.
ABC is an isosceles triangle with AB = AC. P and Q are points on AB and AC respectively such that AP = AQ. Prove that CP = BQ.
Solution:
Class 9 Maths Triangles Extra Questions
In ∆ABQ and ∆ACP, we have
AB = AC (given)
∠BAQ = ∠CAP [common]
AQ = AP (given)
∴ By SAS congruence criteria, we have
∆ABQ ≅ ∆ACP
CP = BQ

Questions On Triangles For Class 9 Question 2.
In the given figure, ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC, AD is extended to intersect BC at P. Show that : (i) ∆ABD ≅ ∆ACD (ii) ∆ABP ≅ ∆ACP
Extra Questions Of Triangles Class 9
Solution:
(i) In ∆ABD and ∆ACD
AB = AC [given]
BD = CD [given]
AD = AD [common)]
∴ By SSS congruence axiom, we have
∆ABD ≅ ∆ACD
(ii) In ∆ABP and ∆ACP
AB = AC [given]
∠BAP = ∠CAP [c.p.cit. as ∆ABD ≅ ∆ACD]
AP = AP [common]
∴ By SAS congruence axiom, we have
∆ABP ≅ ∆ACP

Ch 7 Maths Class 9 Extra Questions Question 3.
In the given figure, it is given that AE = AD and BD = CE. Prove that ∆AEB ≅ ∆ADC.
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 10
Solution:
We have AE = AD … (i)
and CE = BD … (ii)
On adding (i) and (ii),
we have AE + CE = AD + BD
⇒ AC = AB
Now, in ∆AEB and ∆ADC,
we have AE = AD [given]
AB = AC [proved above]
∠A = ∠A [common]
∴ By SAS congruence axiom, we have
∆AEB = ∆ADC

Triangles Class 9 Extra Questions With Solutions Question 4.
In the given figure, in ∆ABC, ∠B = 30°, ∠C = 65° and the bisector of ∠A meets BC in X. Arrange AX, BX and CX in ascending order of magnitude.
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 11
Solution:
Here, AX bisects ∠BAC.
∴ ∠BAX = ∠CAX = x (say)
Now, ∠A + ∠B + C = 180° [angle sum property of a triangle]
⇒ 2x + 30° + 65° = 180°
⇒ 2x + 95 = 180°
⇒ 2x = 180° – 95°
⇒ 2x = 85°
⇒ x = \(\frac{85^{\circ}}{2}\) = 42.59
In ∆ABX, we have x > 30°
BAX > ∠ABX
⇒ BX > AX (side opp. to larger angle is greater)
⇒ AX < BX
Also, in ∆ACX, we have 65° > x
⇒ ∠ACX > ∠CAX
⇒ AX > CX [side opp. to larger angle is greater]
⇒ CX > AX … (ii)
Hence, from (i) and (ii), we have
CX < AX < BX

Question 5.
In figure, ‘S’ is any point on the side QR of APQR. Prove that PQ + QR + RP > 2PS.
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 12
Solution:
In ∆PQS, we have
PQ + QS > PS …(i)
[∵ sum of any two sides of a triangle is greater than the third side]
In ∆PRS, we have
RP + RS > PS …(ii)
Adding (i) and (ii), we have
PQ + (QS + RS) + RP > 2PS
Hence, PQ + QR + RP > 2PS. [∵ QS + RS = QR]

Question 6.
If two isosceles triangles have a common base, prove that the line joining their vertices bisects them at right angles.
Solution:
Here, two triangles ABC and BDC having the common
base BC, such that AB = AC and DB = DC.
Now, in ∆ABD and ∆ACD
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 13
AB = AC [given]
BD = CD [given]
AD = AD [common]
∴ ΔABD ≅ ΔΑCD [by SSS congruence axiom]
⇒ ∠1 = ∠2 [c.p.c.t.]
Again, in ∆ABE and ∆ACE, we have
AB = AC [given]
∠1 = ∠2 [proved above]
AE = AE [common]
∆ABE = ∆ACE [by SAS congruence axiom]
BE = CE [c.p.c.t.]
and ∠3 = ∠4 [c.p.c.t.]
But ∠3 + ∠4 = 180° [a linear pair]
⇒ ∠3 = ∠4 = 90°
Hence, AD bisects BC at right angles.

Triangles Class 9 Extra Questions Long Answer Type

Question 1.
In the given figure, AP and DP are bisectors of two adjacent angles A and D of quadrilateral ABCD. Prove that 2 ∠APD = ∠B + 2C.
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 14
Solution:
Here, AP and DP are angle bisectors of ∠A and ∠D
∴ ∠DAP = \(\frac{1}{2}\)∠DAB and ∠ADP = \(\frac{1}{2}\)∠ADC ……(i)
In ∆APD, ∠APD + ∠DAP + ∠ADP = 180°
⇒ ∠APD + \(\frac{1}{2}\) ∠DAB + \(\frac{1}{2}\)∠ADC = 180°
⇒ ∠APD = 180° – \(\frac{1}{2}\)(∠DAB + ∠ADC)
⇒ 2∠APD = 360° – (∠DAB + ∠ADC) ……(ii)
Also, ∠A + ∠B + C + ∠D = 360°
∠B + 2C = 360° – (∠A + ∠D)
∠B + C = 360° – (∠DAB + ∠ADC) ……(iii)
From (ii) and (iii), we obtain
2∠APD = ∠B + ∠C

Question 2.
In figure, ABCD is a square and EF is parallel to diagonal BD and EM = FM. Prove that
(i) DF = BE (i) AM bisects ∠BAD.
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 15
Solution:
(i) EF || BD = ∠1 = ∠2 and ∠3 = ∠4 [corresponding ∠s]
Also, ∠2 = ∠4
⇒ ∠1 = ∠3
⇒ CE = CF (sides opp. to equals ∠s of a ∆]
∴ DF = BE [∵ BC – CE = CD – CF)
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 16
(ii) In ∆ADF and ∆ABE, we have
AD = AB [sides of a square]
DF = BE [proved above]
∠D = ∠B = 90°
⇒ ∆ADF ≅ ∆ABE [by SAS congruence axiom]
⇒ AF = AE and ∠5 = ∠6 … (i) [c.p.c.t.]
In ∆AMF and ∆AME
AF = AE [proved above]
AM = AM [common]
FM = EM (given)
∴ ∆AMF ≅ ∆AME [by SSS congruence axiom]
∴ ∠7 = ∠8 …(ii) [c.p.c.t.]
Adding (i) and (ii), we have
∠5 + ∠7 = ∠6 + ∠8
∠DAM = ∠BAM
∴ AM bisects ∠BAD.

Question 3.
In right triangle ABC, right-angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see fig.). Show that : (i) ∆AMC ≅ ∆BMD (ii) ∠DBC = 90° (ii) ∆DBC ≅ ∆ACB (iv) CM = \(\frac{1}{2}\)AB
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 17
Solution:
Given : ∆ACB in which 4C = 90° and M is the mid-point of AB.
To Prove :
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC = 90°
(iii) ∆DBC ≅ ∆ACB
(iv) CM = \(\frac{1}{2}\)AB
Proof : Consider ∆AMC and ∆BMD,
we have AM = BM [given]
CM = DM [by construction]
∠AMC = ∠BMD [vertically opposite angles]
∴ ∆AMC ≅ ∆BMD [by SAS congruence axiom]
⇒ AC = DB …(i) [by c.p.c.t.]
and ∠1 = ∠2 [by c.p.c.t.]
But ∠1 and ∠2 are alternate angles.
⇒ BD || CA
Now, BD || CA and BC is transversal.
∴ ∠ACB + ∠CBD = 180°
⇒ 90° + CBD = 180°
⇒ ∠CBD = 90°
In ∆DBC and ∆ACB,
we have CB = BC [common]
DB = AC [using (i)]
∠CBD = ∠BCA
∴ ∆DBC ≅ ∆ACB
⇒ DC = AB
⇒ \(\frac{1}{2}\)AB = \(\frac{1}{2}\)DC
⇒ \(\frac{1}{2}\)AB = CM or CM = \(\frac{1}{2}\)AB (∵ CM = \(\frac{1}{2}\)DC)

Question 4.
In figure, ABC is an isosceles triangle with AB = AC. D is a point in the interior of ∆ABC such that ∠BCD = ∠CBD. Prove that AD bisects ∠BAC of ∆ABC.
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 18
Solution:
In ∆BDC, we have ∠DBC = ∠DCB (given).
⇒ CD = BD (sides opp. to equal ∠s of ∆DBC)
Now, in ∆ABD and ∆ACD,
we have AB = AC [given]
BD = CD [proved above]
AD = AD [common]
∴ By using SSS congruence axiom, we obtain
∆ABD ≅ ∆ACD
⇒ ∠BAD = ∠CAD [c.p.ç.t.]
Hence, AD bisects ∠BAC of ∆ABC.

Question 5.
Prove that two triangles are congruent if any two angles and the included side of one triangle is equal to any two angles and the included side of the other triangle.
Solution:
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 19
Given : Two As ABC and DEF in which
∠B = ∠E,
∠C = ∠F and BC = EF
To Prove : ∆ABC = ∆DEF
Proof : We have three possibilities
Case I. If AB = DE,
we have AB = DE,
∠B = ∠E and BC = EF.
So, by SAS congruence axiom, we have ∆ABC ≅ ∆DEF
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 20
Case II. If AB < ED, then take a point Mon ED
such that EM = AB.
Join MF.
Now, in ∆ABC and ∆MEF,
we have
AB = ME, ∠B = ∠E and BC = EF.
So, by SAS congruence axiom,
we have ΔΑΒC ≅ ΔΜEF
⇒ ∠ACB = ∠MFE
But ∠ACB = ∠DFE
∴ ∠MFE = ∠DFE
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 21
Which is possible only when FM coincides with B FD i.e., M coincides with D.
Thus, AB = DE
∴ In ∆ABC and ∆DEF, we have
AB = DE,
∠B = ∠E and BC = EF
So, by SAS congruence axiom, we have
∆ABC ≅ ∆DEF
Case III. When AB > ED
Take a point M on ED produced
such that EM = AB.
Join MF
Proceeding as in Case II, we can prove that
∆ABC = ∆DEF
Hence, in all cases, we have
∆ABC = ∆DEF.

Question 6.
In the given figure, side QR is produced to the point S. If the bisectors of ∠PQR and ∠PRS meet at T,
prove that ∠QTR = \(\frac{1}{2}\) ∠QPR.
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 22
Solution:
Here, QT is angle bisector of ∠PQR
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 23

Triangles Class 9 Extra Questions HOTS

Question 1.
Show that the difference of any two sides of a triangle is less than the third side.
Solution:
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 24
Consider a triangle ABC
To Prove :
(i) AC – AB < BC
(ii) BC – AC < AB
(iii) BC – AB < AC
Construction : Take a point D on AC
such that AD = AB.
Join BD.
Proof : In ∆ABD, we have ∠3 > ∠1 …(i)
[∵ exterior ∠ is greater than each of interior opposite angle in a ∆]
Similarly, in ∆BCD, we have
∠2 > ∠4 …..(ii) [∵ ext. ∠ is greater then interior opp. angle in a ∆]
In ∆ABD, we have
AD = AB [by construction]
∠1 = ∠2 …(iii) [angles opp. to equal sides are equal in a triangle]
From (i), (ii) and (iii), we have
⇒ ∠3 > ∠4 =
⇒ BC > CD
⇒ CD < BC
AC – AD < BC
AC – AB < BC [∵ AD = AB]
Hence, AC – AB < BC
Similarly, we can prove
BC – AC < AB
and BC – AB < AC

Question 2.
In the figure, O is the interior point of ∆ABC. BO meets AC at D. Show that OB + OC < AB + AC.
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 25
Solution:
In ∆ABD, AB + AD > BD …(i)
∵ The sum of any two sides of a triangle is greater than the third side. Also, we have
BD = BO + OD
AB + AD > BO + OD ….(ii)
Similarly, in ∆COD, we have
OD + DC > OC … (iii)
On adding (ii) and (iii), we have
AB + AD + OD + DC > BO + OD + OC
⇒ AB + AD + DC > BO + OC
⇒ AB + AC > OB + OC
or OB + OC < AB + AC
Hence, proved.

Triangles Class 9 Extra Questions Value Based (VBQs)

Question 1.
A campaign is started by volunteers of mathematical club to boost school and its surrounding under Swachh Bharat Abhiyan. They made their own logo for this campaign. What values are acquired by mathematical club ?
If it is given that ∆ABC ≅ ∆ECD, BC = AE.
Prove that ∆ABC ≅ ∆CEA.
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 26
Solution:
Here, it is given that
∆ABC ≅ ∆ECD
AB = CE [c.p.c.t.]
BC = CD [c.p.c.t.]
AC = ED [c.p.c.t.]
Now, in ∆ABC and ∆CEA
BC = AE [given]
AB = EC [proved above]
AC = AC [common]
∴ By using SSS congruence axiom, we have
∆ABC ≅ ∆CEA
Value : Cleanliness and social concerning.

Question 2.
Rajiv, a good student and actively involved in applying knowledge A of mathematics in daily life. He asked his classmate Rahul to make triangle as shown by choosing one of the vertex as common. Rahul tried but not correctly. After sometime Rajiv hinted Rahul about congruency of triangle. Now, Rahul fixed vertex C as common vertex and locate point D, E such that AC = CD and BC = CE. Was the triangle made by Rahul is congruent ? Write the condition satisfying congruence.
What value is depicted by Rajiv’s action?
Triangles Class 9 Extra Questions Maths Chapter 7 with Solutions Answers 27
Solution:
In ∆ABC and ∆DEC, we have
AC = DC [by construction]
BC = EC [by construction]
∠ACB = ∠ECD [vert. opp. ∠s]
By using SAS congruence axiom, we have
∆ABC ≅ ∆DEC
Value : Cooperative learning, use of concept and friendly nature.

Class 9 Geography Chapter 1 Extra Questions and Answers India-Size and Location

Class 9 Geography Chapter 1 Extra Questions and Answers India-Size and Location

Check the below Online Education NCERT MCQ Questions for Class 9 Geography Chapter 1 Extra Questions and Answers India-Size and Location Pdf free download. MCQ Questions for Class 9 Geography with Answers were prepared based on the latest exam pattern. We have Provided India Size and Location Class 9 Geography MCQs Questions with Answers to help students understand the concept very well. https://ncertmcq.com/extra-questions-for-class-9-social-science/

Online Education for India-Size and Location Class 9 Extra Questions Geography Chapter 1

Class 9 Geography Chapter 1 Extra Questions And Answers Question 1.
Name the island group of India lying in the Bay of Bengal.
Answer:
The Andaman and Nicobar Islands.

Geography Class 9 Chapter 1 Extra Questions Question 2.
Which island group of India lies to its south-west ?
Answer:
Lakshdweep.

India Size And Location Class 9 Extra Questions Question 3.
Which canal shortened the distance between India and Europe ?
Answer:
Suez canal.

Extra Questions For Class 9 Geography Chapter 1 Question 4.
Which ocean will have to be crossed by a ship going from Singapore to Mogadishu. •
Answer:
Indian ocean.

Important Questions For Class 9 Social Science Geography Chapter  Question 5.
Point out the latitudinal extent of India.
Answer:
The latitudinal extent of India extends between 8° 4′ N and 37° 6′ latitudes touching Jammu and Kashmir in the North and Tamil Nadu in the south.

Class 9 Geography Chapter 1 Extra Questions and Answers India-Size and Location World

Chapter 1 Geography Class 9 Extra Questions Question 6.
Name the southernmost point of the Indian Union.
Answer:
The Indira Point.

Geography Chapter 1 Class 9 Extra Questions Question 7.
Where does the Indira point lie?
Answer:
In Andaman and Nicobar.

Class 9 Geo Ch 1 Extra Questions Question 8.
When was the Suez Canal opened?
Answer:
In 1869.

Class 9 Geography Chapter 1 Extra Questions Question 9.
What is the importance of the Suez Canal for India?
Answer:
Suez Canal has reduced the distance between India and Europe by 7000 km.

Ncert Solutions For Class 9 Geography Chapter 1 Extra Questions Question 10.
Where is Cape of Good Hope?
Answer:
In South Africa.

Class 9th Geography Chapter 1 Extra Questions Question 11.
Write the two routes by which India can reach North America, South America and Europe.
Answer:
These routes are:

  1. The Cape of Good Hope
  2. The Suez Canal.

Cbse Class 9 Geography Chapter 1 Extra Questions Question 12.
Name the countries which are- larger than India in area.
Answer:

  • Russia
  • Canada
  • China
  • United States, of America
  • Brazil
  • Australia.

Class 9 Chapter 1 Geography Extra Questions Question 13.
Which tropic divides India into two arts?
Answer:
Tropic of Cancer.

Learn Insta Class 9 Geography Chapter 1 Question 14.
What is the shape of the areas to the south of the Tropic of Cancer?
Answer:
Triangular.

Class 9 Geography Chapter 1 Extra Questions and Answers India-Size and Location World

Class 9 Geography Ch 1 Extra Questions Question 15.
What is the extension of the mainland of India?
Answer:
The mainland of India extends between 8° 4′ N and 37° 6′ N latitudes and 68° 7′ E and 97° 25’E longitude.

Class 9 Geography Chapter 1 Short Questions And Answers Question 16.
In which hemisphere does India lie?
Answer:
In northern hemisphere.

Extra Questions Of Geography Class 9 Chapter 1 Question 17.
If the globe is divided vertically into eastern and western hemispheres, in which hemisphere will India lie?
Answer:
Eastern.

Ncert Class 9 Geography Chapter 1 Extra Questions Question 18.
What is the length of the Indian coastline?
Answer:
About 7500 km.

India Size And Location Class 9 Short Questions Answers Question 19.
How many Indian States are touched by the Tropic of Cancer?
Answer:
9.

India Size And Location Class 9 Important Questions Question 20.
Name the two extreme, states of India touched by the Tropic of Cancer.
Answer:

  1. Gujarat,
  2. Mizoram.

Class 9 Geography Chapter 1 Extra Questions and Answers India-Size and Location World

Question 21.
State the seven largest countries of the world. ‘
Answer:
The seven largest countries, along with their sizes, are as follows:
Russia.: 17.07 million sq km;
China: 9.59 million sq km;
Canada: 9.21 million sq km;
USA: 9.07 million sq km;
Brazil: 8.51 million sq km;
Australia: 7.68 million sq km;
India: 3.28 million sq km.
This can be explained through a graph as under.

Question 22.
Explain with examples India’s link I with other countries.
Answer:
The exchange of ideas and commodities dates back to the ancient times. India’s link with West Asia, East Asia, Central and South Asia are a noteworthy feature. While Buddhism travelled from India to – Tibet, China and as far as Japan and Korean peninsula, on the other hand, the Mongols, the Turks, the Arabs and the Iranians contributed richly to the country’s architectural heritage. The ideas of the Upanishads, the Indian numeral and the decimal systems could also diffuse to many parts of the world due to these contacts.

Question 23.
Why the sun’s rays are more direct on the Nicobar Islands than on Jammu and Kashmir?
Answer:
It is a well-known fact that the sun always shines vertically on the Equator. Because of Bus fact a place the never to the Equator, the hotter it is Nicobar Islands are nearer to the Equator as compared to Jammu and Kashmir. Hence sun’s rays are more direct on the Nicobar Islands than on Jammu and Kashmir.

Question 24.
In which parts of India would be the sun’s rays more oblique?
Answer:
The sun’s rays would be more oblique on the northern frontiers of India. This is the reason why these areas have lesser amount of heat.

Class 9 Geography Chapter 1 Extra Questions and Answers India-Size and Location World

Question 25.
Point out the latitudinal extends of India. What are its implications?
Answer:
The latitudinal extent of India is about 3,200 km. It is the 1/2 of the total circumference of the earth. This fact has following implications:

  • The southern parts of India get more heat from the sun than the northern parts.
  • The difference between the duration of the day and night is lesser in southern parts of- India in comparison with the northern parts.
  • As a matter of fact, the difference between the length of the day and night in the southernmost part of India is only 45 minutes while in the northern parts this difference is about 5 hours.

Question 26.
Point out the value of the Prime Meridian of India.
Answer:
To adopt one timing for the whole country, India has accepted 82° 30′ E longitude as the Standard Meridian of India. In fact India lies to the east of the Prime Meridian between 68° 7′ to 97° 25′ East Longitude. The local time at this Meridian has been accepted as the Indian Standard Time through India. It helped in avoiding a lot of confessions which, otherwise would have been created due to the difference among local timings. In fact, it has brought harmony in timings of the whole country.

Question 27.
Point out the reason behind the two hours time difference between the two eastern and western extremities of India.
Answer:
The earth takes about 4 minutes to rotate through 1° of longitude. The sun rises earlier in the east than in the west. The longitudinal extent of India is 30 degrees. Because of these facts, the difference in time in the easternmost part and the most western part of India would be 30 × 4=120 minutes=2 hours. This is the reason that the sun rises two hours earlier in the easternmost part of India.

Class 9 Geography Chapter 1 Extra Questions and Answers India-Size and Location World

Question 28.
What is the importance of the partially enclosed character of the land of India?
Answer:
The partially enclosed character of the land of India has strengthened the uniqueness of the country by assimilating new cultural elements coming from outside and yet fostering unity and homogeneity in the Indian society.

Question 29.
Answer the following questions after studying the map:
(i) Any two countries located in the East of India.
(ii) Any three countries located in the North of India.
(iii) Two neighbouring countries of India in the west.
Answer:
(i) Myanmar, Thailand.
(ii) Nepal, Bhutan, China.
(iii) Pakistan and Afghanistan.

Question 30.
Answer the following questions after studying the map:
(i) The nearest southern neighbour of India.
(ii) Any three southern states of India.
(iii) Any three northern states of India.
Answer:
(i) Sri Lanka
(ii) Karnataka, Kerala, Tamil Nadu,
(iii) Jammu-Kashmir, Punjab, Himachal Pradesh.

Class 9 Geography Chapter 1 Extra Questions and Answers India-Size and Location World

Question 31.
Give reasons:
(i) While the sun has already risen in Arunachal Pradesh, it is still dark in Gujarat.
(ii) Ahmedabad will see the sun overhead twice in a year,
(iii) Why has Meghalaya been named as such?
(iv) India is often referred to as a subcontinent.
Answer:
(i) Sun rises in Gujarat two hours later than in Arunachal Pradesh.
(ii) The Tropic of cancer which runs almost halfway through India divides it into two separate climatic zones. The areas which are situated beyond the Tropic of Cancer or are lying to the north of the Tropic of Cancer, never have mid-day Sun overhead.

On the other hand, the places which are within the tropics or the places which are lying to the north of the Tropic of Cancer experience the mid-day sun when the sun is Overhead, at Tropic of Cancer. Ahmedabad and Kolkata both fall within the tropics hence are able to see the noon sun exactly overhead twice a year.

(iii) Because Meghalaya literally means the abode of clouds.
(iv) It is rightly referred to as a subcontinent because it is separated from the rest of the continent by natural features such as mountains and river. Furthermore, it has its own specific climatic characteristics and distinct cultural identity.

Question 32.
Why 82° 30′ E has been selected as the Standard Meridian of India?
Answer:
From Gujarat to Arunachal Pradesh, there is a time lag of two hours. Hence along the Standard Meridian of India (82° 30′ E) passing through Mirzapur in Uttar Pradesh is taken as the standard time of the whole of India.

Question 33.
What is the longitudinal extent of our country? State its significance.
Or
What do you mean by the longitudinal extent of India? Point out the implications of the longitudinal extent of India.
Answer:
Longitudinal extent of our country shows the geographical conditions of our country. Our country India lies Between 68° 7′ E to 97° 25’E longitudinal extent, Hence its longitudinal extent is about 30°. The longitudinal extent of India is quite vast.

Its east-west extent is quite wider which is very much significant The importance and implications of the Longitudinal extent of India can be pointed out in the following manner:
1. Because of the vast east-west extent of India it is quite dosed to East Asian Countries on the eastern side and to the West Asian countries like Afghanistan, Iran, Iraq, and Arabian countries on the Western side. This situation presides a vast seep for trade and commerce. It also helps India in having close relations between East Asian Countries and the West Asian Countries.

2. Because of the Vast east-west longitudinal extent India is quite near to Japan, Australia and other East African countries, as well as to many of the European countries. America lies equidistant to India whether. from the eastern or western side.

Class 9 Geography Chapter 1 Extra Questions and Answers India-Size and Location World

Question 34.
India’s strategic location on the head of the Indian Ocean has helped her in establishing land and maritime contacts with the outside world in the ancient and medieval times. Explain.
Answer:
India’s contacts with the world, in fact, have continued through the ages. In fact, the exchange of ideas and commodities dates back to the ancient times.

This fact is supported by the following facts:
1. The ideas of Upanishads and the Ramayana as well as the stories of the Panchatantra have reached many parts of the world.

2. The Indian numerals and the decimal system could reach many parts of the world from India because of this contact.

3. The spices, muslin and other merchandise were taken from India to different countries since the ancient time.

4. India was also influenced by the other countries and cultures. For example, the influence of Greek sculpture, and the architectural styles of dome and minarets from West Asia can be seen in different parts of our country.

5. The traders from India established links with Egyp, Sudan, Somalia, Kenya and Tanzania.

6. Several Indian merchants were even able to establish new kingdoms far away from their country.

Objective Type Questions

1. Fill in the following blanks with correct words:

(i) The highest peak of India is the ……………………. .
Answer:
Kanchenjunga

(ii) The Tropic of ………………………………… divides India into two almost equal parts.
Answer:
Cancer

(iii) India belongs to the ……………………. Hemisphere.
Answer:
Eastern

(iv) India is the ……………………. largest country in the world.
Answer:
Seventh

Class 9 Geography Chapter 1 Extra Questions and Answers India-Size and Location World

(v) The southern tip of the Indian mainland misses the ……………………. by only a few degrees.
Answer:
Equator

(vi) The southern part of the coast is called ……………………. .
Answer:
Koromandal.

2. Match the following two lists.

(i) Ganga: The largest country of the world.
(ii) Russia: River associated with religion.
(iii) Everest: Hill station.
(iv) Shimla: River of southern peninsula.
(v) Tapi The highest peak of the world.
Answer:
(i) Ganga: The river associated with religion.
(ii) Russia: The largest country of the world.
(iii) Everest. The highest peak of the world.
(iv) Shimla : Hill station.
(v) Tapi: River of southern peninsula.

3. Put (✓) before each correct sentence:

(i) The Tropic of cancer exists in 23° 30′ N.
Answer:
(✓)

(ii) The Tropic of Cancer divides the country into almost two equal parts.
Answer:
(✗)

(iii) The areas to the south of the Tropic of cancer is round in shape.
Answer:
(✗)

Class 9 Geography Chapter 1 Extra Questions and Answers India-Size and Location World

(iv) The southernmost point of the India is Kanyakumari.
Answer:
(✗)

(v) Lakshadweep islands are comparatively more scattered.
Answer:
(✗)

(vi) The Tropic of cancer does not pass through Chhattisgarh.
Answer:
(✗)

(vii) The Equator passes through Kerala.
Answer:
(✗)

(viii) Guwahati is thie capital of Meghalaya.
Answer:
(✗)

4. Choose the right answer from the four alternatives given below:

(i) India lies in the following hemisphere:
(a) northern
(b) eastern
(c) southern
(d) western:
Answer:
(a) northern

(ii) The Bay of Bengal is located on the following side of India:
(a) south-west
(b) south-east
(c) north-west
(d) north-east.
Answer:
(b) south-east

Class 9 Geography Chapter 1 Extra Questions and Answers India-Size and Location World

(iii) The Indira point was submerged in seawater during Tsunami in the following year:
(a) 2001
(b) 2002
(c) 2003
(d) 2004
Answer:
(d) 2004

(iv) The following is, territory wise, larger country than India:
(a) France
(b) Bangladesh
(c) Brazil
(d) Germany.
Answer:
(c) Brazil

(v) The Standard Meridian of India is:
(a) 97°25 E
(b) 68°7′ E
(c) 82°30′ E
(d) 83°20′ E.
Answer:
(c) 82°30′ E

(vi) The total number of the Union Territories, in India, are:
(a) 6
(b) 7
(c) 5
(d) 4
Answer:
(b) 7

(vii) The Meridian in India passes through:
(a) Mirzapur
(b) Kanpur
(c) Jaipur
(d) Rajpur.
Answer:
(a) Mirzapur.

Extra Questions for Class 9 Social Science