CBSE Class 9 – Page 10 – MCQ Questions

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4

January 5, 2022August 17, 2021 by Dattu

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 5
Chapter Name	Triangles
Exercise	Ex 5.4
Number of Questions Solved	6
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4

Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:
Let ABC be a right angled triangle, such that ∠ ABC = 90°
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 img 1
We know that,
∠ABC + ∠BCA + ∠CAB = 180° (By A property)
⇒ 90° + ∠BCA + ∠CAB = 180°
⇒ ∠BCA + ∠CAB = 90°
From above, we have ∠ BCA and ∠ CAB are acute angles.
⇒ ∠BCA < 90°
and ∠CAB < 90°
⇒ ∠BCA < ∠ABC
and ∠CAB < ∠ABC
⇒ AB < AC and BC < AC (∵ Side opposite to greater angle is longer)
Hence, the hypotenuse (AC) is the longest side.

Question 2.
In figure, sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
Solution:
We know that,
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 img 2
∠ACB + ∠QCB = 180° (Linear pair)…(i)
and ∠ABC+ ∠PBC = 180° (Linear pair)…(ii)
From Eqs. (i) and (ii), we have
∠ABC + ∠PBC = ∠ACB + ∠QCB ….(iii)
But ∠PBC < ∠QCB (Given)…(iv) From Eqs. (iii) and (iv), we have ∠ ABC > ∠ ACB
⇒ AC > AB
(∵ Side opposite to greater angle is longer)

Question 3.
In figure, ∠B <∠ A and ∠C <∠ D. Show that AD < BC.
Solution:
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 img 3

Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠ A> ∠C and ∠B > ∠D.
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 img 4
Solution:
Given ABCD is a quadrilateral. AB is the smallest side and CD is the longest side.
To prove ∠A > ∠C and ∠B > ∠D
Construction Join A to C and B to D.
Proof In ∆ABC, we have AB is the smallest side.
∴ AB < BC
⇒ ∠5 < ∠1
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 img 5
(∵ Angle opposite to longer side is greater) …(i)
In ∆ADC, we have CD is the largest side.

Question 5.
In figure, PR > PQ and PS bisect ∠QPR. Prove that
∠PSR >∠PSQ.
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 img 7
Solution:
In ∆ PQR, we have
PR > PQ (Given)
⇒ ∠ PQR > ∠ PRQ …(i)
(∵ Angle opposite to longer side is greater)

Question 6.
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Solution:
Given x is a line and A is a point not lying on x. AB ⊥ x, Cis any point on x other than B.

To prove AB∠C
⇒ AC> AB
(∵ the Side opposite to greater angle is longer)
⇒ AB < AC
Hence, the perpendicular line segment is the shortest.
We hope the NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1

January 5, 2022August 17, 2021 by Dattu

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 5
Chapter Name	Triangles
Exercise	Ex 5.1
Number of Questions Solved	8
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1

Exercise 5.1

Question 1.
In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see figure). Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD?
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 img 1
Solution:
In ∆ABC and ∆ABD, we have
AC = AD (Given)
∠ CAB = ∠ DAB (∵ AB bisects ∠A)
and AB = AB (Common)
∴ ∆ ABC ≅ ∆ABD (By SAS congruence axiom)
∴ BC = BD (By CPCT)

Question 2.
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see figure). Prove that
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 img 2
(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠ BAC
Solution:
In ∆ ABC and ∆ BAC, we have
AD = BC (Given)
∠DAB = ∠CBA (Given)
and AB = AB (Common)
∴ ∆ ABD ≅ ∆BAC (By SAS congruence axiom)
Hence, BD = AC (By CPCT)
and ∠ABD= ∠BAC (By CPCT)

Question 3.
AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 img 3
Solution:
In ∆AOD and ∆BOC, we have
∠AOD = ∠BOC
∵ AB and CD intersects at O.
∴ Which are vertically opposite angle
∵ ∠DAO = ∠CBO = 90°
and AD = BC (Given)
∴ ∆AOD ≅ ∆BOC (By SAS congruence axiom)
⇒ O is the mid-point of AB.
Hence, CD bisects AB.

Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ∆ABC = ∆CDA.
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 img 4
Solution:
From figure, we have
∠ 1 = ∠ 2 (Vertically opposite angles).. .(i)
∠ 1 = ∠ 6 (Corresponding angles)…(ii)
∠ 6 = ∠ 4 (Corresponding angles) …(iii)
From Eqs. (i) (ii) and (iii), we have
∠ 1 = ∠ 4
and ∠ 2 = ∠ 4 …..(iv)
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 img 5
In ∆ABC and ∆CDA, we have
∠ 4 = ∠ 2 [From Eq. (iv]
∠5 = ∠ 3 (Alternate interior angles)
and AC = AC (Common)
∴ ∆ABC ≅ ∆ CDA (By AAS congruence axiom)

Question 5.
Line l is the bisector of an ∠ A and ∠ B is any point on l. BP and BQ are perpendiculars from B to the arms of LA (see figure). Show that
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms ot ∠A.
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 img 6
Solution:
In ∆ APB and ∆AQB, we have
∠ APB = ∠ AQB = 90°
∠ PAB = ∠ QAB (∵ AB bisects ∠ PAQ)
and AB = AB (Common)
∴ ∆ APB ≅ ∆ AQB (By AAS congruence axiom)
⇒ BP = BQ (By CPCT)
⇒ B is equidistant from the arms of ∠ A.

Question 6.
In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 img 7
Solution:
In ∆ABC and ∆ADE, we have
AB = AD (Given)
∠ BAD = ∠ EAC (Given)…(i)
On adding ∠ DAC on both sides in Eq. (i)
⇒ ∠ BAD + ∠ DAC = ∠ EAC + ∠ DAC
⇒ ∠ BAC = ∠ DAE
and AC = AE (Given)
∴ ∆ABC ≅ ∆ADE (By AAS congruence axiom)
⇒ BC = DE (ByCPCT)

Question 7.
AS is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB. (see figure). Show that
(i) ∆DAP ≅ ∆EBP
(ii) AD = BE
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 img 8
Solution:
We have,
AP = BP [∵ P is the mid-point of AB (Given)]… (i)
∠ EPA = ∠ DPB (Given)…(ii)
∠ BAD = ∠ ABE (Given).. .(iii)
On adding ∠ EPD on both sides in Eq. (ii); we have
⇒ ∠ EPA + ∠ EPD = ∠DPB + ∠ EPD
⇒ ∠ DP A = ∠ EPB …..(iv)
Now, In ∆DAP and ∆EBP, we have
∠ DPA = ∠ EPB [ From Eq(4)]
∠ DAP = ∠ EBP (Given)
and AP = BP [From Eq. (i)]
∴ ∆ DAP ≅ ∆ EBP (By ASA congruence axiom)
Hence, AD = BE (By CPCT)

Question 8.
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 figure). Show that
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC is a right angle
(iii) ∆DBC ≅ ∆ACB
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 img 9
(iv) CM = \(\frac { 1 }{ 2 }\) AB
Solution:
Given ∆ACB in which ∠C = 90° and M is the mid- point of AB.

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 img 10
To prove (i) ∆ AMC = ∆ BMD
(ii) ∠DBC is a right angle
(iii) ∆DBC ≅ ∆ACB
(iv) CM = \(\frac { 1 }{ 2 }\) AB
Construction Produce CM to D, such that CM = MD. Join DB.
Proof In ∆ AMC and ∆ BMD, we have
AM = BM (M is the mid-point of AB)
CM = DM (Given)
and ∠AMC=∠BMD (Vertically opposite angles)
∴ ∆ AMC ≅ ∆ BMD (By SAS congruence axiom)
⇒ AC = DB (By CPCT) …(i)
and ∠1 = ∠2 (By CPCT)
Which are alternate angles
∴ BD || CA
Now, BD || CA and BC is transversal
∴ ∠ ACB + ∠CBD =180°
⇒ 90°+ ∠CBD = 180°
⇒∠CBD =90°
⇒ ∠ DBC = 90° [Which is part (ii)]
In ∆ DBC and ∆ ACB, we have
CB = BC (Common)
DB = AC [Using part (i)]
and ∠ CBD = ∠ BCA (Each 90°)
∴ ∆ DBC ≅ ∆ACB (By SSA congruence axiom) [Which is part (iii)]
⇒ DC=AB (by CPCT)
⇒ \(\frac { 1 }{ 2 }\) DC = \(\frac { 1 }{ 2 }\) AB
⇒ CM = \(\frac { 1 }{ 2 }\) AB (∵ CM = \(\frac { 1 }{ 2 }\) DC) [ Which is part(iv)]

We hope the NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1, 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

January 5, 2022August 17, 2021 by Dattu

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) ²Solution:
(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

Question 7.
Evaluate the following using suitable identities
(i) (99)³
(ii) (102)³
(iii) (998)³Solution:
(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

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)³Here, -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 2 Polynomials Ex 2.4

January 5, 2022August 17, 2021 by Dattu

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

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

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4

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

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

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

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

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

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

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

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

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

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

NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3

January 5, 2022August 17, 2021 by Dattu

NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 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 8.3.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 8
Chapter Name	Linear Equations in Two Variables
Exercise	Ex 8.3
Number of Questions Solved	8
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3

Question 1.
Draw the graph of each of the following linear equations in two variables
(i) x + y = 4
(ii) x – y = 2
(iii) y = 3x
(iv) 3 = 2x + y
Solution:
(i) x + y = 4
To draw the graph, we need atleast two solutions of the
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 1
So, draw the graph by plotting the two points from table and then joining by a line.

(ii) x – y = 2
To draw the graph, we need atleast two solutions of the equation.
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 3
So, draw the graph by plotting the two points from table and then joining by a line.

(iii) y = 3x
To draw the graph, we need atleast two solutions of the equation.
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 5
So, draw the graph by plotting the two points from table and then by

(iv) 3 = 2x + y
To draw the graph, we need atleast two solutions of the equation.
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 7
So, draw the graph by plotting the points from the table and the by joining the same by a line.

Question 2.
Give the equations of two lines passing through (2,14). How many more such lines are there, and why?
Solution:
Here, (2,14) is a solution of a linear equation. One example of such a linear equation is y = 7x, others are x + y- 16.
There are infinitely many lines because there are infinitely many linear equations which are satisfied by the. coordinates of the point (2,14).

Question 3.
If the point (3,4) lies on the graph of the equation 3y – ax – 7, find the value of a.
Solution:
If the point (3, 4) lies on the graph, then it will satisfies the equations.
Hence, 3 (4) – a (3) – 7 = 0
⇒ 12 – 3a – 7 = 0
⇒ 5 – 3a = 0
⇒ 3a = 5
⇒ a = \(\frac { 5 }{ 3 }\)

Question 4.
The taxi fare in a city is as follows :
For the first kilometre, the fare is ₹ 8 and for the subsequent distance it is ₹5 per km. Taking the distance covered as x km and the total fare as ₹y, write a linear equation for this information and draw its graph.
Solution:
Distance covered = x km = 1+ (x – 1) km
Fare for first kilometre = ₹ 8
Fare for next (x-1 ) km = (x – 1) x 5 = 5(x-1)
According to question, total fare = y
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 9
Now, plot the points A (0,3) and B (1, 8) on a graph paper and joining them, to form q line AB.

Question 5.
From the choices given below, choose the equation whose graphs are given in Fig. (a) and Fig. (b).
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 10

For Fig. (a)
(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x

For Fig. (b)
(i) y = x + 2
(ii) y = x – 2
(iii) y = – x + 2
(iv) x + 2y = 6
Solution:
In Fig. (a), we observe that the points (-1,1) and (1,-1) passes through the equation x + y = 0
∵ At (-1,1) x + y = -1 + 1 = 0
and at (1, -1) x + y = 1 -1 = 0
In Fig. (b), we observe that the points (-1, 3), (0, 2) and (2, 0) passes through the equation x + y = 2.
∵ At (-1,3), x + y = -1 + (3) = + 2
At ( 0 , 2 ) , x +y = 0+2 = 2
and at ( 2, 0 ) x+ y = 2 + 0= 2

Question 6.
If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also, read from the graph the work done when the distance travelled by the body is.
(i) 2 units
(ii) 0 unit.
Solution:
Given: that, work done by a body on application of a constant force is directly proportional to the distance travelled by the body.
i.e., (W) work done ∝ distance (s)
⇒ W = F. s
(Where, F = arbitrary constant which take the value 5 units)
∴ W = 5s …(i)
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 11
Now, plot the points O (0,0), A (1,5) and B (2,10) on graph paper and joining them to get a line AB.

(i) From point B (2,10), draw a line parallel to OY to intersect the x-axis at (2,0) and draw a line parallel to x-axis intersect at C (0, 10).
∴ Work done when the distance travelled by the body is 2 units = 10units.
(ii) Clearly y = 0 when x = 0 so the work done when the distance travelled by the body is 0 unit.

Question 7.
Yamini and Fatima, two students of class IX of a school, together contributed ₹100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as ₹ x and ₹ y.) Draw the graph of the same.
Solution:
Let the contributions of Yamini and Fatima together towards the Prime Minister’s Relief Fund to help the earthquake victims are ₹x and ₹ y, respectively.
Then, by given condition,
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 13
Here, we plot the points B (0,100) and A (100, 0) on graph paper and join all these points to form a line AB.

Question 8.
In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius.
F = (\(\frac { 9 }{ 5 }\) ) c + 32
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.
(ii) If the temperature is 30 °C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95°F, what is the temperature in Celsius?
(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Solution:
(i) Given, linear equation in Fahrenheit and Celsius is
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 15

(ii) If temperature is 30°C i.e., C = 30°C
Then, from Eq. (i), we get
F = \(\frac { 9 }{ 5 }\) x 30 + 32 = 9 x 6 + 32 = 54 + 32 = 86
∴ Temperature in Fahrenheit = 86°F

(iii) If temperature is 95°F i.e., F= 95°F
Then, from Eq. (i),we get
5 x 95 – 9C = 160 ⇒ 9C= 475 – 160= 315 ⇒ C= 35°
∴ Temperature in Celsius = 35°C

(iv) If the temperature is 0°C i.e., C= 0
Then, from Eq. (i), we get
F = \(\frac { 9 }{ 5 }\) x 0 + 32 = -32
∴ Temperature in Fahrenheit = 32°F
If the temperature is 0°F i.e., F = 0
Then, from Eq. (i), we get
5 x 0 – 9C = 160
⇒ C = \(\frac { -160 }{ 9 }\) = – 17.8° C (Approx.)
∴ Temperature in Celsius = -17.8° C

(v) Yes, if we take both temperature are equal i.e., C = F.
Now, from Eq. (i), we get
F = \(\frac { 9 }{ 5 }\) F + 32
⇒ 5F = 9F + 160
⇒ -4F = 160
⇒ F = -40°
∴ F = C = – 40°
We hope the NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3, drop a comment below and we will get back to you at the earliest.