## RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2I

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2I.

Other Exercises

Question 1.
Solution:
(i)(3x+2)3

Question 2.
Solution:

Question 3.
Solution:
(i)(95)3 = (100 – 5)3

Hope given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2I are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2F

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2F.

Other Exercises

Factorize :

Question 1.
Solution:
25x2 – 64y2
= (5x)2 – (8y)2
= (5x + 8y) (5x – 8y) Ans.

Question 2.
Solution:
100 – 9x2
(10)2 – (3x)2
= (10 + 3x) (10 – 3x) Ans.

Question 3.
Solution:
5x2 – 7y2
=(√5x)2 +(√7y)2
=(√5x – √7y)(√5x – √7y) Ans

Question 4.
Solution:
(3x + 5y)2 – 4z2
= (3x + 5y)2 – (2z)2
= (3x + 5y + 2z) (3x + 5y – 2z) Ans.

Question 5.
Solution:
150 – 6x2
= 6(25 – x2)
= 6{(5)2 – (x2)}
= 6 (5 + x) (5 – x) Ans.

Question 6.
Solution:
20x2 – 45
= 5 (4x2 – 9)
= 5{(2x)2 – (3)2}
= 5 (2x + 3) (2x – 3) Ans.

Question 7.
Solution:
3x3 – 48x
= 3x (x2 – 16)
= 3x {(x)2 – (4)2}
= 3x (x + 4) (x – 4) Ans.

Question 8.
Solution:
2 – 50x2
= 2(1 – 25x2) = 2 {(1)2 – (5x)2}
= 2 (1 + 5x) (1 – 5x) Ans.

Question 9.
Solution:
27a2 – 48b2
= 3(9a2 – 16b2)
= 3 {(3a)2 – (4b)2}
= 3(3a + 4b) (3a – 4b) Ans.

Question 10.
Solution:
x – 64x3
= x (1 – 64x2)
= x{(1)2 – (8x)2}
= x (1 + 8x) (1 – 8x) Ans.

Question 11.
Solution:
8ab2 – 18a3
= 2a (4b2 – 9a2)
= 2a {(2b)2 – (3a)2}
= 2a (2b + 3a) (2b – 3a) Ans

Question 12.
Solution:
3a3b – 243ab3
= 2ab (a2 -81 b2)
= 3ab {(a)2 – (9b)2}
= 3ab (a + 9b) (a – 9b) Ans.

Question 13.
Solution:
(a + b)3 – a – b
= (a + b)3 – 1 (a + b)
= (a + b) {(a + b)2 – 1}
= (a + b) {(a + b)2 – (1)2}
= (a + b) (a + b + 1) (a + b – 1) Ans.

Question 14.
Solution:
108 a2 – 3(b – c)2
= 3 {36a2 – (b – c)2}
= 3 {(6a)2 – (b – c)2}
= 3 {(6a + (b – c)} {6a – (b – c)}
= 3 (6a + b – c) (6a – b + c) Ans.

Question 15.
Solution:
x3 – 5x2 – x + 5
= x2(x – 5) -1(x – 5)
= (x – 5) (x2 – 1)
= (x – 5) {(x)2 – (1)2}
= (x – 5) (x + 1) (x – 1) Ans.

Question 16.
Solution:
a2 + 2ab + b2 – 9c2
= (a + b)2 – (3c)2
= {a2 + b2 + 2ab – (a + b)2}
= (a + b + 3c) (a + b – 3c) Ans.

Question 17.
Solution:
9 – a2 + 2ab – b2
= 9 – (a2 – 2ab + b2)
= (3)2 -(a- b)2
{ ∴a2 + b2 – 2ab = (a – b)2}
= (3 + a – b) (3 – a + b) Ans.

Question 18.
Solution:
a– b2 – 4ac + 4c2
= a2 – 4ac + 4c2 – b2
= (a)2 – 2 x a x 2c + (2c)2 – (b)2
= (a – 2c)2 – (b)2
= (a – 2c + b) (a – 2c – b)
= (a + b – 2c) (a – b – 2c) Ans.

Question 19.
Solution:
9a2 + 3a – 8B – 64b2
= 9a2 – 64b2 + 3a – 8b
= (3a)2 – (8b)2 + 1(3a – 8b)
= (3a + 8b) (3a – 8b) + 1 (3a – 8b)
= (3a – 8b) (3a + 8b + 1) Ans.

Question 20.
Solution:
x2 – y2 + 6y _ 9
= (x)2 – (y2 – 6y + 9)
= (x)2 – {(y)2 – 2 x y x 3 + (3)2}
= (x)2 – (y – 3)2
= (x + y – 3) (x – y + 3) Ans.

Question 21.
Solution:
4x2 – 9y2 – 2x – 3y
= (2x)2 – (3y)2 – 1(2x + 3y)
= (2x + 3y) (2x – 3y) – 1( 2x + 3y)
= (2x + 3y) (2x – 3y – 1) Ans.

Question 22.
Solution:
x4 – 1
= (x2)2 – (1)2
= (x2 + 1) (x2 – 1)
= (x2 + 1) {(x)2 – (1)2}
= (x2 + 1) (x + 1) (x – 1) Ans.

Question 23.
Solution:
a – b – a2+ b2
= 1 (a – b) – (a2 – b2)
= 1 (a – b) – (a + b) (a – b)
= (a – b) (1 – a – b) Ans.

Question 24.
Solution:
x4 – 625
= (x2)2 – (25)2
= (x2 + 25) (x2 – 25)
= (x2 + 25) {(x)2 – (5)2}
= (x2 + 25) (x + 5) (x – 5) Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2F are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A.

Question 1.
Solution:
In ∆ ABC, ∠A = 70° and AB = AC (given)
∴ ∠C = ∠B
(Angles opposite to equal sides)
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)

=> 70° + ∠B + ∠B = 180°
(∴ ∠B = ∠C)
=> 2∠B = 180°- 70° = 110°
∠B = $$\frac { { 110 }^{ o } }{ 2 }$$ = 55° and
Hence ∠B = 55°,∠C = 55° Ans.

Question 2.
Solution:
In ∆ ABC, ∠A= 100°
It is an isosceles triangle

∴AB = AC
∠B = ∠C
(Angles opposite to equal sides)
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
=> 100° + ∠B + ∠B = 180°
(∴ ∠B = ∠C)
=> 2∠B + 100° = 180°
=> 2∠B = 180°- 100° = 80°
∠B = $$\frac { { 80 }^{ o } }{ 2 }$$ = 40°
and ∠C = 40°
∴ Base angles are 40°, 40° Ans.

Question 3.
Solution:
In ∆ ABC,
AB = AC
∴∠C = ∠B
(Angles opposite to equal sides)

But ∠B = 65°
∴ ∠ C = 65°
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
=> ∠A + 65° + 65° = 180°
=> ∠ A + 130° = 180°
=> ∠ A = 180° – 130°
=> ∠ A = 50°
Hence ∠ A = 50° and ∠ C = 65° Ans.

Question 4.
Solution:
In ∆ ABC
AB = AC
∴ ∠C = ∠B
(Angles opposite to equal sides)

But ∠A = 2(∠B + ∠C)
=> ∠B + ∠C = $$\frac { 1 }{ 2 }$$ ∠A 2
But ∠A + ∠B + ∠C = 180°
(sum of angles of a triangle)
=> ∠A+ $$\frac { 1 }{ 2 }$$ ∠A = 180°
=> $$\frac { 3 }{ 2 }$$ ∠A = 180°
=> ∠A = 180° x $$\frac { 2 }{ 3 }$$ = 120°
and ∠B + ∠C = $$\frac { 1 }{ 2 }$$ ∠A = $$\frac { 1 }{ 2 }$$ x 120°
= 60°
∴ ∠ B = ∠ C
∴ ∠ B = ∠ C = $$\frac { { 60 }^{ o } }{ 2 }$$ = 30°
Hence ∠ A = 120°, ∠B = 30°, ∠C = 30° Ans.

Question 5.
Solution:
In ∆ ABC,
AB = BC and ∠B = 90°
∴ AB = BC
∴ ∠ C = ∠ A
(Angles opposite to equal sides)

Now ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
=> ∠ A + 90° + ∠ A = 180°
(∴ ∠C = ∠A)
=> 2∠A + 90° – 180°
=> 2∠ A = 180° – 90° = 90°
∠ A = $$\frac { { 90 }^{ o } }{ 2 }$$ = 45°
∴ ∠ C = 45° (∴ ∠ C = ∠ A)
Hence, each of the equal angles is 45° Ans.

Question 6.
Solution:
Given : ∆ ABC is an isosceles triangle in which AB = AC

Base BC is produced to both sides upto D and E respectively forming exterior angles
∠ ABD and ∠ ACE
To Prove : ∠ABD = ∠ACE
Proof : In ∆ ABC
∴ AB = AC (given)
∴ ∠C = ∠B
(Angles opposite to equal sides)
=> ∠ ABC = ∠ACB
But ∠ ABC + ∠ABD = 180° (Linear pair)
Similarly ∠ACB + ∠ACE = 180°
∠ ABC + ∠ABD = ∠ACB + ∠ACE
But ∠ ABC = ∠ ACB (proved)
∠ABD = ∠ACE
Hence proved.

Question 7.
Solution:
∆ ABC is an equilateral triangle
∴ AB = BC = CA
and ∠A = ∠B = ∠C = 60°

The sides of the ∆ ABC are produced in order to D,E and F forming exterior angles
∠ACP, ∠BAE and ∠CBF
∆ACD + ∠ACB = 180°
=> ∠ACD + 60° = 180°
=> ∠ACD = 180° – 60°
=> ∠ACD = 120°
Similarly, ∠BAE + ∠BAC = 180°
=> ∠BAE + 60° = 180°
=> ∠BAE = 180° – 60° = 120°
and ∠BAF + ∠ABC = 180°
=> ∠BAF + 60° = 180°
=> ∠BAF = 180° – 60° = 120°
Hence each exterior angle of an equilateral triangle is 120°.

Question 8.
Solution:
Given : In the figure,
O is mid-point of AB and CD.
i.e. OA = OB and OC = OD
To Prove : AC = BD and AC || BD.
Proof : In ∆ OAC and ∆ OBD,
OA = OB {given}
OC = OD {given}
∠AOC = ∠BOD
(Vertically opposite angles)
∴∆ OAC ≅ ∆ OBD (S.A.S. axiom)
∴AC = BD (c.p.c.t.)
and ∠ C = ∠ D
But these are alt. angles
∴AC || BD
Hence proved.

Question 9.
Solution:
Given : In the figure,
PA ⊥ AB, QB ⊥ AB and
PA = QB, PQ intersects AB at O.

To Prove : O is mid-point of AB and PQ.
Proof : In ∆ AOP and ∆ BOQ,
∠ A = ∠ B
AP = BQ (given)
∠ AOP = ∠BOQ
(Vertically opposite angles)
∴ ∆AOP ≅ ∆ BOQ (A.A.S. axiom)
∴OA = OB (c.p.c.t)
and OP = OQ (c.p.c.t)
Hence, O is the mid-point of AB as well as PQ

Question 10.
Solution:
Given : Two line segments AB and CD intersect each other at O and OA = OB, OC = OD
AC and BD are joined.
To Prove : AC = BD
Proof : In ∆ AOC and ∆ BOD,
OA = OB {given}
OC = OD
∠AOC = ∠BOD
(vertically opposite angles)
∴ ∆ AOC ≅ ∆ BOD (S.A.S. axiom)
∴ AC = BD (c.p.c.t)
and ∠A = ∠D (c.p.c.t.)
Hence AC ≠ BD
Hence proved.

Question 11.
Solution:
Given : In the given figure,
l || m, m is mid-point of AB
CD is another line segment, which intersects AB at M.
To Prove : M is mid-point of CD
Proof : l || m
∴ ∠ CAM = ∠ MBD (Alternate angles)
Now, in ∆ AMC and ∆ BMD,
AM = MB (Given)
∠ CAM = ∠ MBD (proved)
∠ AMC = ∠BMD
(vertically opposite angles)
∆ AMC ≅ ∆ BMD (ASA axiom)
∴ CM = MD (c.p.c.t.)
Hence, M is mid-point of CD.
Similarly we can prove that M is mid point of any other line whose end points are on l and m.
Hence proved.

Question 12.
Solution:
Given : In ∆ ABC, AB = AC and in ∆ OBC,
OB = OC

To Prove : ∠ABO = ∠ACO
Construction. Join AO.
Proof : In ∆ ABO and ∆ ACO,
AB = AC (Given)
OB = OC (Given)
AO = AO (Common)
∴ ∆ ABO ≅ ∆ ACO (S.S.S. Axiom)
∴ ∠ABO = ∠ACO (c.p.c.t.)
Hence proved.

Question 13.
Solution:
Given : In ∆ ABC, AB = AC
D is a point on AB and a line DE || AB is drawn.
Which meets AC at E
To Prove : AD = AE

Proof : In ∆ ABC
AB = AC (given)
∴ ∠ C = ∠ B (Angles opposite to equal sides)
But DE || BC (given)
∴ ∠ D = ∠ B {corresponding angles}
and ∠ E = ∠ C
But ∠C = ∠B
∴ ∠E = ∠D
∴ AD = AE (Sides opposite to equal angles)
Hence proved.

Question 14.
Solution:
Given : In ∆ ABC,
AB = AC
X and Y are two points on AB and AC respectively such that AX = AY
To Prove : CX = BY
Proof : In ∆ AXC and ∆ AYB
AC = AB (given)
AX = AY (given)
∠ A = ∠ A (common)

∴ ∆ AXC ≅ ∆ AYB (S.A.S. axiom)
∴ CX = BY (c.p.c.t.)
Hence proved.

Question 15.
Solution:
Given : In the figure,
C is mid point of AB
∠DCA = ∠ECB and
∠DBC = ∠EAC
To Prove : DC = EC
Proof : ∠ DCA = ∠ FCB (given)
∠DCA +∠DCE = ∠DCE + ∠ECB
=> ∠ACE = ∠ BCD
Now, in ∆ ACE and ∆ BCD,
AC = BC (C is mid-point of AB)
∠EAC = ∠DBC (given)
∠ACE =∠ BCD (proved)
∴ ∆ ACE ≅ ∆ BCD (ASA axiom)
∴ CE = CD (c.p.c.t.)
=> EC = DC
or DC = EC
Hence proved.

Question 16.
Solution:
Given : In figure,
BA ⊥ AC, DE ⊥ EF .
BA = DE and BF = DC
To Prove : AC = EF
Proof : BF = DC (given)
BF + FC = FC + CD
=> BC = FD.

Now, in right-angled ∆ ABC and ∆ DEF,
Hyp. BC = Hyp. FD (proved)
Side AB = side DE (given)
∴ ∆ ABC ≅ ∆DEF (RHS axiom)
∴ AC = EF (c.p.c.t.)
Hence proved.

Question 17.
Solution:
To prove : AE = CD
Proof: x° + ∠ BDC = 180° (Linear pair)
Similarly y°+ ∠AEB = 180°
∴ x° + ∠BDC = y° + AEB
But x° = y° (given)
∠ BDC = ∠ AEB
Now, in ∆ AEB and ∆ BCD,
AB = CB (given)
∠B = ∠B (common)
∠ AEB = ∠ BDC (proved)
∴ ∆ AEB ≅ ∆ BCD (AAS axiom)
∴ AE = CD. (c.p.c.t.)
Hence proved.

Question 18.
Solution:
Given : In ∆ ABC,
AB = AC.
Bisectors of ∠ B and ∠ C meet AC and AB in D and E respectively

To Prove : BD = CE
Proof : In ∆ ABC,
AB = AC
∠ B = ∠ C (Angles opposite to equal sides)
Now, in ∆ ABD and ∆ ACE,
AB = AC (given)
∠ A = ∠ A (common)
∠ ABD = ∠ACE
(Half of equal angles)
∴ ∆ A ABD ≅ ∆ ACE (ASA axiom)
∴ BD = CE (c.p.c.t)
Hence proved.

Question 19.
Solution:
Given : In ∆ ABC,
To prove : BL = CM.
Proof : In ∆ BLD and ∆ CMD,
BD = DC (D is mid-point of BC)
∠LDB = ∠CDM
(Vertically opposite angles)

∠L = ∠M (each 90°)
∴ ∆ BLD ≅ ∆ CMD (A.A.S. axiom)
Hence, BL = CM (c.p.c.t)
Hence proved.

Question 20.
Solution:
Given : In ∆ ABC, D is mid-point of BC. DL ⊥ AB and DM ⊥ AC and DL = DM
To prove : AB = AC
Proof: In right angled ∆ BLD and ∆ CMD
Hyp. DL = DM (given)
Side BD = DC (D is mid-point of BC)
∴ ∆ BLD ≅ ∆ CMD (R.H.S. axiom)
∴ ∠B = ∠C (c.p.c.t.)
∴ AC = AB
(sides opposite to equal angles)
Hence AB = AC
Hence proved.

Question 21.
Solution:
Given : In ∆ AB = AC and bisectors of ∠B and ∠C meet at a point O. OA is joined.
To Prove : BO = CO and Ray AO is the bisector of ∠ A
Proof : AB = AC (given)
∴ ∠C = ∠B
(Angles opposite to equal sides)
=> $$\frac { 1 }{ 2 }$$ ∠C = $$\frac { 1 }{ 2 }$$ ∠B
=>∠OBC = ∠OCB
∴ in ∆OBC,

OB = OC (Sides opposite to equal angles)
Now in ∆ OAB and ∆ OCA
OB = OC (proved)
OA = OA (common)
AB = AC (given)
∴ ∆ OAB ≅ ∆ OCA (S.S.S. axiom)
∴ ∠OAB = ∠OAC (c.p.c.t.)
Hence OA is the bisector of ∠ A.
Hence proved.

Question 22.
Solution:
Given : ∆ PQR is an equilateral triangle and QRST is a square. PT and PS are joined.
To Prove : (i) PT = PS
(ii) ∠PSR = 15°
Proof : In ∆ PQT and ∆ PRS,
PQ = PR (Sides of equilateral ∆ PQR)
QT = RS (sides of in square PRST) .

∠PQT = ∠PRS
(each angle = 90° + 60° = 150°)
∴ ∆ PQT ≅ ∆ PRS (S.A.S. axiom)
∴ PT = PS (c.p.c.t)
In ∆ PRS, ∠PRS = 60° + 90° = 150°
∠ RPS + ∠ PSR = 180° – 150° = 30°
But ∠RPS = ∠PSR ( ∴PR = RS)
∴∠PSR + ∠PSR = 30°
=> 2∠PSR = 30°
∴ ∠PSR = $$\frac { { 30 }^{ o } }{ 2 }$$ = 15°
Hence proved.

Question 23.
Solution:
Given : In right angle ∆ ABC, ∠B is right angle. BCDE is square on side BC and ACFG is also a square on AC.
To Prove : AD = BF
Proof : ∠ACF = ∠BCD (Each 90°)
∠ ACF + ∠ ACB = ∠ BCD + ∠ ACB
=> ∠ BCF = ∠ ACD
Now in ∆ ACD and ∆ BCF,
AC = CF (sides of a squares)
CD = BC (sides of a square)
∴∠ ACD = ∠ BCF (proved)
∴ ∆ ACD ≅ ∆ BCF (S.A.S. axiom)
Hence proved.

Question 24.
Solution:
Given : ∆ ABC is an isosceles in which AB = AC. and AD is the median which meets BC at D.

To Prove : AD is the bisector of ∠ A
Proof : In ∆ ABD and ∆ ACD,
AB = AC (Given)
BD = CD (D is mid-point of BC)
∴ ∆ ABD ≅ ∆ ACD (S.S.S. axiom)
Hence, AD in the bisector of ∠ A.

Question 25.
Solution:
Given : ABCD is a quadrilateral in which AB || DC. P is mid-point of BC. AP and DC are produced to meet at Q.
To Prove : (i) AB = CQ.
(ii) DQ = DC + AB
Proof:
(i) In ∆ ABP and ∆ CPQ,
BP = PC (P is mid-point of BC)
∠ BAP = ∠ PQC (Alternate angles)
∠ APB = ∠ CPQ
(Vertically opposite angles)
∴ ∆ ABP ≅ ∆ CPQ (A.A.S. axiom)
∴ AB = CQ. (c.p.c.t.)
(ii) Now DQ = DC + CQ
=> DQ = DC + AB ( CQ = AB proved)
Hence proved.

Question 26.
Solution:
Given : In figure,
OA = OB and OP = OQ.
To Prove : (i) PX = QX (ii) AX = BX
Proof : In ∆OAQ and ∆OBP,
OA = OB. (Given)
OQ = OP (Given)
∠O = ∠O (Common)
∴ ∆ OAQ ≅ ∆ OBP (SAS axiom)
∴ ∠ A = ∠ B (c.p.c.t)
Now OA = OB and OP = OQ
Subtracting
OA – OP = OB – OQ
=>PA = QB
Now, in ∆ XPA and ∆ XQB,
PA = QB (Proved)
∠AXP = ∠BXQ
(Vertically opposite angles)
∠ A = ∠ B (Proved)
∴ ∆ XPA ≅ ∆ XQB (A.A.S. axiom)
∴ AX = BX (c.p.c.t.) and PX = QX (c.p.c.t.)
Hence proved.

Question 27.
Solution:
Given : ABCD is a square in which a point P is inside it. Such that PB = PD.
To prove : CPA is a straight line.
Proof : In ∆ APB and ∆ ADP,
AB = AD (Sides of a square)
AP = AP (Common)
PB = PD (Given)
∴ ∆ APB ≅ ∆ ADP (S.S.S. axiom)
∠APD = ∠ APB (c.p.c.t) …(i)
Similarly, in ∆ CBP and ∆ CPD,
CB = CD (Sides of a square)
CP – CP (Common)
PB = PD (Given)
∴ ∆ CBP ≅ ∆ CPD (S.S.S. axiom)
∴ ∠BPC = ∠CPD (c.p.c.t.) …(i)
∴ ∠ APD + ∠ CPD = ∠ APB + ∠ BPC
∠APC = ∠APC
∠APC = 180°
(v sum of angles at a point is 360°)
APC or CPA is a straight line.
Hence proved.

Question 28.
Solution:
Given :∆ ABC is an equilateral triangle PQ || AC and AC is produced to R such that CR = BP
To prove : QR bisects PC
Proof : ∴ PQ || AC
∴ ∠BPQ = ∠BCA
(Corresponding angles)
But ∠ BCA = 60°
(Each angle of the equilateral triangle)
and ∠ ABC or ∠QBP = 60°
∴ ∆ BPQ is also an equilateral triangle.
∴ BP = PQ
But BP = CR (Given)
∴ PQ = CR
Now in ∆ PQM and ∆ RMC,
PQ = CR (proved)
∠QMP = ∠RMC
(vertically opposite angles)
∠ PQM = ∠ MRC (alternate angles)
∴ ∆ PQM ≅ ∆ RMC (AAS axiom)
∴ PM = MC (c.p.c.t.)
Hence, QR bisects PC.
Hence proved

Question 29.
Solution:
AB = AD and BC = DC
AC and BD are joined.
To prove : (i) AC bisects ∠ A and ∠ C
(ii) AC is perpendicular bisector of BD.
Proof : In ∆ ABC and ∆ ADC,
BC = DC (given)
AC = AC (common)
∴ ∆ ABC ≅ ∆ ADC (S.S.S. axiom)
∴ ∠BCA = ∠DCA (c.p.c.t.)
and ∠BCA = ∠DAC (c.p.c.t.)
Hence AC bisects ∠ A and ∠ C

(ii) In ∆ ABO and ∆ ADO,
AO = AO (common)
∠BAO = ∠DAO
(proved that AC bisects ∠ A)
∴ ∆ ABO ≅ ∆ ADO (SAS axiom)
∴ BO = OD and ∠AOB = ∠AOD (c.p.c.t.)
But ∠ AOB + ∠ AOD = 180° (linear pair)
∠AOB = ∠AOD = 90°
Hence AC is perpendicular bisector of BD. Hence proved.

Question 30.
Solution:
Given : In ∆ ABC,
Bisectors of ∠ B and ∠ C meet at I
From I, IP ⊥ BC. IQ ⊥ AC and IR ⊥ AB.
IA is joined.
To prove : (i) IP = IQ = IR
(ii) IA bisects ∠ A.
Proof : (i) In ∆ BIP and ∆ BIR
BI = BI (Common)

and ∠P = ∠R (Each 90°)
∴ ∠ IBP ≅ ∠ IBR
( ∴ IB is bisector of ∠ B)
∴ ∆ BIP = ∆ BIR (A.A.S. axiom)
∴ IP = IR (c.p.c.t) …(i)
Similarly, in ∆ CIP and ∆ CIQ,
CI = CI (Common)
∠P = ∠Q (each = 90°) and ∠ICP = ∠ ICR
( IC is bisector of ∠ C)
∴ ∆CIP ≅ ∆CIQ (A.A.S. axiom)
∴ IP = IQ (c.p.c.t.) …(ii)
From (i) and (ii),
IP = IQ = IR.
(i) In right angled ∆ IRA and ∆ IQA,
Hyp. IA = IA (Common)
side IR = IQ (Proved)
∴ ∆ IRA ≅ ∆ IQA (R.H.S. axiom)
∴ ∠IAR = ∠ IRQ (c.p.c.t.)
Hence IA is the bisector of ∠ A
Hence proved.

Question 31.
Solution:
Given. P is a point in the interior of ∠ AOB
PL ⊥ OA and PM ⊥ OC are drawn and PL = PM.

To prove : OP is the bisector of ∠ AOB
Proof: In right angled ∆ OPL and ∆ OPM,
Hyp. OP OP (Common)
Side PL = PM (Given)
∴ ∆ OPL ≅ ∆ OPM (R.H.S. axiom)
∴ ∠POL = ∠POM (c.p.c.t.)
Hence, OP is the bisector of ∠AOB.
[ Hence proved.

Question 32.
Solution:
Given : ABCD is a square, M is midpoint of AB.

PQ is perpendicular to MC which meets CB produced at Q. CP is joined.
To prove : (i) PA = BQ
(ii) CP = AB + PA
Proof : (i) In ∆ PAM and ∆ QBM,
AM = MB (M is midpoint of AB)
∠AMP = ∠BMQ
(vertically opposite angles)
∠ PAM = ∠ QBM
(Each = 90° angles of a square)
∴ ∆ PAM ≅ ∆ QBM (A.S.A. axiom)
∴ AP = BQ
=> PA = BQ (c.p.c.t.)
and PM = QM (c.p.c.t.)
Now, in ∆ CPM and ∆ CQM,
CM = CM (Common)
PM = QM (Proved)
∠CMP =∠CMQ (Each 90° as PQ ⊥ MC)
∴ ∆ CPM ≅ ∆ CQM (SAS axiom)
∴ CP = CQ (c.p.c.t.)
= CB + BQ
= AB + PA
( CB = AB sides of squares and BQ = PA proved)
Hence, CP = AB + PA.

Question 33.
Solution:
Construction. Let AB be the breadth of the river. Mark any point M on the bank on which B is situated. Let O be the midpoint of BM. From M move along the path MN perpendicular to BM to a point N such that A, O, N are on the same straight line. Then MN is the required breadth of the river.
Proof : In ∆ ABO and ∆ MNO,
BO = OM (const.)
∠ AOB = ∠ MON (vertically opposite angle)
∠B = ∠M (each 90°)
∴ ∆ ABO ≅ ∆ MNO (A.S.A. axiom)
∴ AB = MN. (c.p.c.t.)
Hence, MN is the required breadth of the river.

Question 34.
Solution:
In ∆ ABC,
∠A = 36°, ∠B = 64°
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
=> 36° + 64° + ∠C = 180°
=> 100° + ∠C = 180°
=> ∠C = 180° – 100° = 80°
∴ ∠ C = 80° which is the greatest angle.
∴ The side AB, opposite to it is the longest side.
∴∠ A is the shortest angle
∴ BC is the shortest side.

Question 35.
Solution:
In ∆ ABC,
∴ ∠A = 90°
∴ ∠B + ∠C 180° – 90° = 90°
Hence, ∠ A is the greatest angle of the triangle.
Side BC, opposite to this angle be the longest side.

Question 36.
Solution:
In ∆ ABC,
∠ A = ∠ B = 45°
∴ ∠A + ∠B = 45° + 45° = 90°
and ∠C= 180°-90° = 90°
∴ ∠ C is the greatest angle.
∴ Side AB, opposite to ∠ C will be the longest side of the triangle.

Question 37.
Solution:
Given : In ∆ ABC, side AB is produced to D such that BD = BC.
∠B = 60°, ∠A = 70°
Proof : In ∆ BCD,
Ext. ∠ B = 60°
∴ ∠ CBD = 180° – 60° = 120°
and ∠ BCD + ∠ BDC = Ext. ∠ CBA = 60°
But ∠BCD = ∠ BDC ( BC = BD)
∴ ∠BCD = ∠BDC = $$\frac { { 60 }^{ o } }{ 2 }$$ = 30°.
∴ ∠ ACD = 50° + 30° = 80°
(i) Now in ∆ ACD,
∴ ∠ ACD > ∠ CAD ( 80° > 70°)
(ii) and ACD > ∠D (80° > 30°)
Hence proved.

Question 38.
Solution:
Given : In ∆ ABC, ∠B = 35°and ∠C = 65°
AX is the bisector of ∠ BA C meeting BC in X
∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
=> ∠ A + 35° + 65° = 180°
=> ∠A + 100° = 180°
=> ∠A = 180° – 100° = 80°
∴ AX is the bisector of ∠ BAC
∴ ∠ BAX = ∠ CAX = 40°
In ∆ ABX,
∠BAX = 40° and ∠B = 35°
∴ ∠BAX > ∠B
∴ BX > AX …(i)
and in ∆ AXC,
∠C = 65° and ∠CAX = 40°
∴ ∠C > ∠CAX
∴ AX > XC …(ii)
From (i) and (ii),
BX > AX > XC
or BX > AX > CX

Question 39.
Solution:
Given : In ∆ ABC,
AD is the bisector of ∠ A
To prove : (i) AB > BD and
(ii) AC > DC
Proof : (i) In ∆ ADC,
( AD is bisector of ∠ A)
In ∆ ABD,
AB > BD.
Now in ∆ ACD.
AC > DC.
Hence proved

Question 40.
Solution:
Given : In ∆ ABC, AB = AC
BC is produced to D and AD is joined.
To Prove : AD > AC
Proof : In ∆ ABC,
AB = AC (given)
∴ ∠B = ∠C (Angles opposite to equal sides)
Ext. ∠ ACD > ∠ ABC
∠ACB = ∠ABC
Now, in ∆ ABD,
Hence proved.

Question 41.
Solution:
Given : In ∆ ABC,
AC > AB and AD is the bisector of ∠ A which meets BC in D.
Proof : In ∆ ABC,
AC > AB
∴∠B > ∠C
∴∠ 1 = ∠ 2 (AD is the bisector of ∠ A)
∴ ∠B + ∠2 = ∠C + ∠1

Ext. ∠ADC = ∠B + ∠2
Ext. ∠ADB > ∠C + ∠1
∴ ∠B + ∠2 > ∠C + ∠1 (Proved)

Question 42.
Solution:
Given : In ∆ PQR,
S is any point on QR and PS is joined
To Prove : PQ + QR + RP > 2PS
Proof : In ∆ PQS,
PQ + QS > PS
(Sum of two sides of a triangle is greater than the third side) …(i)
Similarly, in ∆ PRS.
PQ + SR > PS …(ii)
PQ + QS + PR + SR > PS + PS
=> PQ + QS + SR + PR > 2PS
=> PQ + QR + RP > 2PS
Hence proved.

Question 43.
Solution:
Given : O is the centre of the circle and XOY is its diameter.
XZ is the chord of this circle
To Prove : XY > XZ
Const. Join OZ
Proof : OX, OZ and OY are the radii of the circle
∴ OX = OZ = OY
In ∆ XOZ,
OX + OZ > XZ (Sum of two sides of a triangle is greater than its third side)
=> OX + OY > XZ (∴ OZ = OY)
=> OXY > XZ
Hence proved.

Question 44.
Solution:
Given : In ∆ ABC, O is any point within it OA, OB and OC are joined.
To Prove : (i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) OA + OB + OC > $$\frac { 1 }{ 2 }$$
(AB + BC + CA)
Const. Produce BO to meet AC in D.

Proof : (i) In ∆ ABD,
AB + AD > BD …(i)
(Sum of two sides of a triangle is greater than its third side)
=> AB + AD > BO + OD …(i)
Similarly in ∆ ODC
OD + DC > OC. …(ii)
AB + AD + OD + DC > OB + OD + OC
=> AB + AD + DC > OB + OC
AB + AC > OB + OC.
(ii) In (i) we have proved that
AB + AC > OB + OC
Similarly, we can prove that
AC + BC > OC + OA
and BC + AB > OA + OB
AB + AC + AC + BC + B + AB > OB + OC + OC + OA + OA + OB
=> 2(AB + BC + CA) > 2(OA + OB + OC)
=> AB + BC + CA > OA + OB + OC.
(iii) In ∆ AOB,
OA + OB > AB
Similarly, in ∆ BOC
OB + OC > BC
and in ∆ COA
OC + OA > CA
OA + OB + OB + OC + OC + OA > AB + BC + CA
=> 2(OA + OB + OC) > (AB + BC + CA)
=> OA + OB + OC > $$\frac { 1 }{ 2 }$$ (AB + BC + CA)
Hence proved.

Question 45.
Solution:
Sides of ∆ ABC are AB = 3cm, BC = 3.5cm and CA = 6.5cm
We know that sum of any two sides of a triangle is greater than its third side.
Here, AB = 3 cm and BC = 3.5 cm
∴ AB + BC = 3cm + 3.5 cm = 6.5 cm and CA = 6.5 cm
∴ AB + BC = CA
Which is not possible to draw the triangle.
Hence, we cannot draw the triangle with the given data.

Hope given RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2J

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2J.

Other Exercises

Factorize :

Question 1.
Solution:
x3 + 27
= (x)3 + (3)3

Question 2.
Solution:
8x3 + 27y3

Question 3.
Solution:
343 + 125b3

Question 4.
Solution:
(1)3+(4x)3

Question 5.
Solution:
125a3+ $$\frac { 1 }{ 8 }$$

Question 6.
Solution:
216x3+$$\frac { 1 }{ 125 }$$

Question 7.
Solution:
16x4 + 54x

Question 8.
Solution:
7a3 + 56b3
=7(a3+8b3)

Question 9.
Solution:
x5 + x2
=x2(x3+1)

Question 10.
Solution:
a3 + 0.008

Question 11.
Solution:
x6 + y6

Question 12.
Solution:
2a3 + 16b3 – 5a – 10b

Question 13.
Solution:
x3 – 512
=(x)– (8)3

Question 14.
Solution:
64x3 – 343
=(4x)– (7)3

Question 15.
Solution:
1 – 27x3
=(1)3– (3x)3

Question 16.
Solution:
x3 – 125y3

Question 17.
Solution:
8x3 – $$\frac { 1 }{ { 27y }^{ 3 } }$$

Question 18.
Solution:
a3 – 0.064
=(a)– (0.4)3

Question 19.
Solution:
(a + 6)3 – 8
=(a+b)– (2)3

Question 20.
Solution:
x6 – 729
=(x2)– (9)3

Question 21.
Solution:
(a + b)3 – (a – b)3

Question 22.
Solution:
x – 8xy3
=x(1 – 8y3)
=x{(1)– (2y)3}

Question 23.
Solution:
32x4 – 500x

Question 24.
Solution:
3a7b – 81a4 b4

Question 25.
Solution:
$${ a }^{ 3 }-\frac { 1 }{ { a }^{ 3 } } -2a+\frac { 2 }{ a }$$

Question 26.
Solution:
8a3 – b3 – 4ax + 2bx

Question 27.
Solution:
a3 + 3a2b + 3ab2 + b3 – 8

Hope given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2J are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2K

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2K.

Other Exercises

Question 1.
Solution:
125a3 + b3 + 64c3 – 60abc

Question 2.
Solution:
a3 + 8b3 + 64c3 – 24abc

Question 3.
Solution:
1 + b3 + 8c3 – 6bc

Question 4.
Solution:
216 + 27b3 + 8c3 – 108bc

Question 5.
Solution:
27a3 – b3 + 8c3 + 18abc

Question 6.
Solution:
8a3 + 125b3 – 64c3 + 120abc

Question 7.
Solution:
8 – 27b3 – 343c3 – 126bc

Question 8.
Solution:
125 – 8x3 – 27y3 – 90xy

Question 9.
Solution:
2√2a3 + 16√2b3 + c3 – 12abc

Question 10.
Solution:
x3 + y3 – 12xy + 64

Question 11.
Solution:
(a – b)3 + (b – c)3 + (c – a)3

Question 12.
Solution:
(3a – 2b)3 + (26 – 5c)3 + (5c – 3a)3

Question 13.
Solution:
a3 (b – c)3 + b3 (c – a)3 + c3 (a – b)3

Question 14.
Solution:
(5a – 7b)3 + (9c – 5a)3 + (7b – 9c)3

Question 15.
Solution:
(x + y – z) (x2 + y2 + z2 – xy + yz + zx)

Question 16.
Solution:
(x – 2y + 3) (x2 + 4y2 + 2xy -3x + 6y + 9)

Question 17.
Solution:
(x – 2y – z) (x2 + 4y2 + z2 + 2xy + zx- 2yz)

Question 18.
Solution:
x + y + 4 = 0,

Question 19.
Solution:
x = 2y + 6

Hope given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2K are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2E

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2E.

Other Exercises

Factorize:

Question 1.
Solution:
9x2 + 12xy
= 3x (3x + 4y) Ans.

Question 2.
Solution:
18x2y – 24xyz
= 6xy (3x – 4z) Ans.

Question 3.
Solution:
27a3b3 – 45a4b2
= 9a3b2 (3b – 5a) Ans.

Question 4.
Solution:
2a (x + y) – 3b(x + y)
= (x + y) (2a – 3b) Ans.

Question 5.
Solution:
2x (p2 + q2) + 4y (p2 + q2)
= 2(p2 + q2) (x + 2y) Ans

Question 6.
Solution:
x (a – 5) + y (5 – a)
= x (a – 5) -y (a – 5)
= (a – 5) (x – y) Ans.

Question 7.
Solution:
4(a + b) – 6 (a + b)2
= 2(a + b) {2 – 3 (a + b)}
= 2(a + b) (2 – 3a – 3b) Ans.

Question 8.
Solution:
8(3a – 2b)2 – 10 (3a – 2b)
= 2(3a – 2b) {4 (3a – 2b) – 5}
= 2(3a – 2b) (12a – 8b – 5) Ans.

Question 9.
Solution:
x (x + x)3 – 3x2 y (x + y)
= x (x + y) {(x + y)2 – 3xy}
= x (x + y) [x2 + y2 + 2xy – 3xy)
= x (x + y) (x2 + y2 – xy) Ans.

Question 10.
Solution:
x3 + 2x2 + 5x + 10
= x2 (x + 2) + 5 (x + 2)
= (x + 2) (x2 + 5) Ans.

Question 11.
Solution:
x2 + xy – 2xz – 2yz
= x (x + y) -2z (x + y)
= (x + y) (x – 2z) Ans.

Question 12.
Solution:
a3 b – a2b + 5ab – 5b.
= b (a3 – a2 + 5a – 5)
= b {(a2 (a – 1) + 5 (a – 1)}
= b (a – 1) (a2 + 5) Ans.

Question 13.
Solution:
8 – 4a – 2a3 + a4
= 4 (2 – a) – a3 (2 – a)
= (2 – a) (4 – a3) Ans.

Question 14.
Solution:
x3 – 2x2y + 3xy2 – 6y3
= x2 (x – 2y) + 3y2 (x – 2y)
= (x – 2y) (x2 + 3y2) Ans

Question 15.
Solution:
px – 5q + pq – 5x
= px – 5x + pq – 5q
= x(p – 5) + q(p – 5)
= {p – 5) (x + q) Ans.

Question 16.
Solution:
x2 + y – xy – x
= x2 – x – xy + y
= x (x – 1) – y (x – 1)
= (x – 1) (x – y) Ans.

Question 17.
Solution:
(3a – 1)2 – 6a + 2
= (3a – 1)2 – 2 (3a – 1)
= (3a – 1) (3a – 1 – 2)
= (3a – 1) (3a – 3)
= 3(3a – 1) (a – 1) Ans.

Question 18.
Solution:
(2x – 3)2 – 8x + 12
= (2x – 3)2 – 4(2x – 3)
= (2x – 3) (2x – 3 – 4)
= (2x – 3) (2x – 7) Ans.

Question 19.
Solution:
a3 + a – 3a2 – 3
= a3 – 3a2 + a – 3
= a2 (a – 3) + 1 (a – 3)
= (a – 3) (a2 + 1) Ans.

Question 20.
Solution:
3ax – 6ay – 8by + 4bx
= 3ax – 6ay + 4bx – 8by
= 3a (x – 2y) + 4b (x – 2y)
= (x – 2y) (3a + 4b) Ans

Question 21.
Solution:
abx2 + a2x + b2x + ab
= ax (bx + a) + b (bx + a)
= (bx + a) (ax + b) Ans.

Question 22.
Solution:
x3 – x2 + ax + x – a – 1
= x3 – x2 + ax – a + x – 1
= x2 (x – 1) + a (x – 1) + 1 (x – 1)
= (x – 1) (x2 + a + 1) Ans.

Question 23.
Solution:
2x + 4y – 8xy – 1
= 2x – 8xy -1+4y
= 2x (1 – 4y) -1 (1 – 4y)
= (1 – 4y) (2x – 1) Ans.

Question 24.
Solution:
ab (x2 +y2) – xy (a2 + b2)
= abx2 + aby2 – a2xy – b2xy
= abx2 – a2xy – b2xy + aby2
= ax (bx – ay) – by (bx – ay)
= (bx – ay) (ax – by) Ans.

Question 25.
Solution:
a2 + ab (b + 1) + b3
= a2 + ab2 + ab + b3
= a (a + b2) + b (a + b2)
= (a + b2) (a + b) Ans

Question 26.
Solution:
a3 + ab (1 – 2a) – 2b2
= a3 + ab – 2a2b – 2b2
= a3 – 2a2b + ab – 2b2
= a2 (a – 2b) + b (a – 2b)
= (a – 2b) (a2 + b) Ans.

Question 27.
Solution:
2a2 + bc – 2ab – ac
= 2a2 – 2ab – ac + bc
= 2a (a – b) – c (a – b)
= (a – b) (2a – c) Ans.

Question 28.
Solution:
(ax + by)2 + (bx – ay)2
= a2x2 + b2y2 + 2abxy + b2x2 + a2y2 – 2bxy
= a2x2 + b2y2 + b2x2 + a2y2
= a2x2 + b2x2 + a2y2 + b2y2
= x2(a2 + b2) + y2(a2 + b2)
= (a2 + b2) (x2 + y2) Ans.

Question 29.
Solution:
a (a + b – c) – bc
= a2 + ab – ac – bc
= a (a + b) – c (a + b)
(a + b) (a – c) Ans.

Question 30.
Solution:
a(a – 2b – c) + 2bc
= a2 – 2ab – ac +2bc
= a2 – ac – 2ab + 2bc
= a (a – c) – 2b (a – c)
= (a – c) (a – 2b) Ans.

Question 31.
Solution:
a2x2 + (ax2 + 1) x + a
= a2x2 + ax3 + x + a
= a2x2 + ax3 + a + x
= ax2 (a + x) + 1 (a + x)
– (a + x) (ax2 + 1) Ans

Question 32.
Solution:
ab (x2 + 1) + x (a2 + b2)
= abx2 + ab + a2x + b2x
= abx2 + a2x + b2x + ab
= ax (bx + a) + b (bx + a)
= (bx + a) (ax + b) Ans.

Question 33.
Solution:
x2 – (a + b) x + ab
= x2 – ax – bx + ab
= x (x – a) – b (x – a)
= (x – a) (x – b) Ans.

Question 34.
Solution:

Hope given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2E are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4B

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4B.

Other Exercises

Question 1.
Solution:
AOB is a straight line
∠AOC + ∠BOC = 180° (Linear pair)
=> 62° + x= 180°
=> x = 180° – 62°
=> x = 118°
Hence, x = 118° Ans.

Question 2.
Solution:
AOB is straight line
∠AOC + ∠COD + ∠DOB – 180°
=> (3x – 5)° + 55° + (x + 20)° = 180°
=> 3x – 5° + 55° + x + 20° = 180°
=> 4x – 5° + 75° = 180°
=> 4x + 70° = 180°
=> 4x = 180° – 70°
=> 4x = 110°
=> $$x=\frac { { 110 }^{ o } }{ 4 } =27.{ 5 }^{ o }$$
Hence x = 27.5°
and ∠AOC = 3x – 5° = 3 x 27.5° – 5°
= 82.5° – 5° = 77.5°
∠BOD = x + 20° = 27.5° + 20°
= 47.5° Ans.

Question 3.
Solution:
AOB is a straight line
∠AOC + ∠COD + ∠DOB = 180°
{angles on the same side of line AB}
=> (3x + 7)° + (2x – 19)° + x = 180°
=> 3x + 7° + 2x – 19° + x = 180°
=> 6x – 12° – 180°
=> 6x = 180° + 12° = 192°
=> $$x=\frac { { 192 }^{ o } }{ 6 } =32^{ o }$$
Here x = 32°
∠AOC = 3x + 7° = 3 x 32° + 7°
= 96° + 7°= 103°
∠COD = 2x – 19° = 2 x 32° – 19°
= 64 – 19° = 45°
and ∠BOD = x = 32° Ans.

Question 4.
Solution:
In the given figure,
x + y + z = 180°
But x : y : z = 5:4:6
Let ∠XOP = x° – 5a
∠POQ =y° = 4a
and ∠QOY = z = 6a
then 5a + 4a + 6a = x + y + z = 180°
=> 15a = 180°
=> a = $$\frac { { 180 }^{ o } }{ 15 } ={ 12 }^{ o }$$
=> x = 5a = 5 x 12° = 60°
y = 4a = 4 x 12° = 48°
and z = 6x = 6 x 12° = 72° Ans.

Question 5.
Solution:
AOB will be a straight line
If ∠AOC + ∠COB = 180°
If (3x + 20)° + (4x – 36)° = 180°
If 3x + 20 + 4x – 36 = 180°
If 7x – 16 = 180°
If 7x = 180° + 16 = 196°
If $$x=\frac { { 196 }^{ o } }{ 7 } ={ 28 }^{ o }$$
Hence, if x = 28°, then AOB will be a straight line.

Question 6.
Solution:
AB and CD intersect each other at O
AOC = ∠BOD and ∠BOC = ∠AOD (vertically opposite angles)
But ∠AOC = 50°
∠BOD = ∠AOC = 50°
But ∠AOC + ∠BOC = 180° (Linear pair)
=> 50° + ∠BOC = 180°
=> ∠BOC = 180° – 50° = 130°
∠AOD = ∠BOC = 130°
Hence,∠AOD = 30°,∠BOD = 50° and ∠BOC = 130° Ans.

Question 7.
Solution:
In the figure,
AB, CD and EF are coplanar lines intersecting at O.
∠AOF = ∠BOE
∠DOF = ∠COE and ∠BOD = ∠AOC (Vertically opposite angles)
x = y,
z = 50°
t = 90°
But AOF + ∠DOF + ∠BOD = 180° (Angles on the same side of a st. line)
=> x + 50° + 90° = 180°
=> x° + 140° + 180°
=> x = 180° – 140° = 40°
Hence, x = 40°, y = x = 40°, z = 50° and t = 90° Ans.

Question 8.
Solution:
Three coplanar lines AB, CD and EF intersect at a point O
∠AOD = ∠BOC
∠DOF = ∠COE
and ∠AOE = ∠BOF
(Vertically opposite angles)
But ∠AOD = 2x
∠BOC = 2x
and ∠BOF = 3x
∠AOE = 3x
and ∠COE = 5x
∠DOF = 5x
But ∠AOD + ∠DOF + ∠BOF + ∠BOC + ∠COE + ∠AOE = 360° (Angles at a point)
=> 2x + 5x + 3x + 2x + 5x + 3x = 360°
=> 20x = 360° => x = $$\frac { { 360 }^{ o } }{ 20 }$$ = 18°
Hence x = 18°
∠AOD = 2x = 2 x 18° = 36°
∠COE = 5x = 5 x 18° = 90°
and ∠AOE = 3x = 3 x 18° = 54° Ans.

Question 9.
Solution:
AOB is a line and CO stands on it forming ∠AOC and ∠BOC
But ∠AOC : ∠BOC = 5:4
Let ∠AOC = 5x and ∠BOC = 4x
But ∠AOC + ∠BOC = 180° (Linear pair)
=> 5x + 4x = 180° => 9x = 180°
=> x = $$\frac { { 180 }^{ o } }{ 9 }$$ = 20°
∠AOC = 5x = 5 x 20° = 100°
and ∠BOC = 4x = 4 x 20° = 80° Ans.

Question 10.
Solution:
Two lines AB and CD intersect each other at O and
∠AOC = 90°
∠AOC = ∠BOD
(Vertically opposite angles)
∠BOD = 90°
But ∠AOC + ∠BOC = 180° (Linear pair)
=> 90° + ∠BOC – 180°
=> ∠BOC = 180° – 90° = 90°
But ∠AOD = ∠BOC
(Vertically opposite angles)
∠AOD = 90°
Hence each of the remaining angle is 90°.

Question 11.
Solution:
Two lines AB and CD intersect each other at O and

∠BOC + ∠AOD = 280°
∠AOD = ∠BOC
(vertically opposite angles)
∠BOC + ∠BOC = 280°
(∠AOD = ∠BOC)
=> 2 ∠BOC = 280°
=> ∠BOC = $$\frac { { 280 }^{ o } }{ 2 }$$ = 140°
But ∠BOC + ∠AOC = 180° (Linear pair)
=> 140° + ∠AOC = 180°
=> ∠AOC = 180° – 140° = 40°
But ∠BOD = ∠AOC
(vertically opposite angles)
∠BOD = 40°
Hence ∠AOC = 40°, ∠BOC = 140°,
∠BOD = 40°
and ∠AOD = 140° Ans.

Question 12.
Solution:
OC is the bisector of ∠AOB. and OD is the ray opposite to OC.
Now ∠AOC = ∠BOC (OC is bisector of ∠AOB)
But ∠BOC + ∠BOD = 180° (Linear pair)
Similarly, ∠AOD + ∠AOC = 180°
=> ∠BOC + ∠BOD = ∠AOD + ∠AOC
But ∠AOC = ∠BOC (Given)
∠BOD = ∠AOD
=> ∠AOD = ∠BOD
Hence proved.

Question 13.
Solution:
AB is the mirror.
PQ is the incident ray, QR is its reflected ray.
=> ∠BQR = ∠PQA
But ∠BQR + ∠PQR + ∠PQA = 180° (Angles on one side of a straight line)
=> ∠PQA + ∠PQA + 112° = 180°
=> 2∠PQA + 112° = 180°
=> 2∠PQA = 180° – 112° = 68°
∠PQA = $$\frac { { 68 }^{ o } }{ 2 }$$ = 34° Ans.

Question 14.
Solution:
Given. Two lines AB and CD intersect each other at O.
OE is the bisector of ∠BOD and EO is produced to F.
To Prove : OF bisects ∠AOC.
Proof : AB and CD intersect each other at O
∠AOC = ∠BOD
(Vertically opposite angles)
OE is the bisector of ∠BOD
∠1 = ∠2
But ∠1 = ∠3
and ∠2 = ∠4 (Vertically opposite angles)
and ∠1 = ∠2 (proved)
∠3 = ∠4
Hence, OF is the bisector of ∠AOC.
Hence proved.

Question 15.
Solution:
Given ∠AOC and ∠BOC are supplementary angles
OE is the bisector of ∠BOC and OF is the bisector of ∠AOC
To Prove : ∠EOF = 90°
Proof : ∠1 = ∠2
∠3 = ∠4
{OE and OF are the bisectors of ∠BOC and ∠AOC respectively}
But ∠AOC + ∠BOC = 180°
(Linear pair)
=> ∠1 + ∠2 + ∠3 + ∠4 = 180°
=> ∠1 + ∠1 + ∠3 + ∠3 = 180°
=> 2∠1 + 2∠3 = 180°
=> 2(∠1 + ∠3) = 180°
=> ∠1 + ∠3 = $$\frac { { 180 }^{ o } }{ 2 }$$ 90°
=> ∠EOF = 90°
Hence proved.

Hope given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4B are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2C.

Other Exercises

Using remainder theorem, find the remainder when :

Question 1.
Solution:
f(x) = (x3 – 6x2 + 9x + 3)
Let x-1 = 0, then x = 1

Question 2.
Solution:
f(x) = 2x3 – 5x2 + 9x – 8)
Let x-3 = 0, then x = 3

Question 3.
Solution:
f(x) = (3x4 – 6x2 – 8x + 2)
Let x-2 = 0, the x = 2

Question 4.
Solution:
f(x) = (x3 – 7x2 + 6x + 4)
Let x-6 = 0,then x=6

Question 5.
Solution:
f(x)=(x3 – 6x2 + 13x + 60)
Let x+2=0,then x=-2

Question 6.
Solution:
f(x)=(2x4 + 6x3 + 2x2 + x – 8)
Let x+3=0,then x=-3

Question 7.
Solution:
f(x)=(4x3 – 12x2 + 11x – 5)

Question 8.
Solution:
f(x)=(81x4 + 54x3 – 9x2 – 3x + 2)

Question 9.
Solution:
f(x)=(x3 – ax2 + 2x – a)
let x-a=0, then x=a

Question 10.
Solution:
f(x)=(ax3+ 3x2 – 3)
g(x)=(2x3 -5x + a)

Question 11.
Solution:
f(x)=x4 – 2x3 + 3x2 – ax + b
let x-1=0, then x=1

Hope given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2C are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2H

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2H.

Other Exercises

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:
9x2 + 16y2 + 4z2 – 24xy + 16yz – 12xz

Question 5.
Solution:
25x2 + 4y2 + 9z2 – 20xy – 12yz + 30xz

Question 6.
Solution:

Hope given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2H are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4C.

Other Exercises

Question 1.
Solution:
AB || CD and a line t intersects them at E and F forming angles ∠1, ∠ 2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8.

Question 2.
Solution:
AB || CD and a transversal t intersects them at E and F respectively forming angles ∠l, ∠2, ∠3, ∠4, ∠5, ∠ 6, ∠ 7 and ∠ 8

Question 3.
Solution:

To Prove : ∠ADC = ∠ABC
Proof : AB || DC and AD intersects their
(sum of co-interior angles)
∠DAB + ∠ABC = 180° …(ii)
from (i) and (ii),
∠ DAB + ∠ ADC = ∠DAB + ∠ABC

Question 4.
Solution:
(i) In the figure, AB || CD
∠ABE = 35° and ∠EDC = 65°
Draw FEG || AB or CD
∴ AS || FG (const.)

Question 5.
Solution:
In the figure, AB || CD || EF,
∠ ABC = 70° and ∠ CEF = 130°
∴EF || CD (given)

∴∠ CEF +∠1 – 180°
(sum of Co-interior angles)
=> 130° + ∠ 1 = 180°
=> ∠ 1 = 180° – 130° = 50°
Again, AB || CD (given)
∴ ∠ ABC = ∠ BCD (Alternate angles)
=> 70° = ∠ BCD = x + ∠ 1
=> x + 50° – 70°
=> x = 70° – 50° = 20°
Hence x = 20° Ans.

Question 6.
Solution:
In the figure, AB || CD.
∠DCE = 130°
and ∠ CEA – 20°
From E, draw EF || AB or CD.

Question 7.
Solution:
Given. In the given figure, AB || CD.

Question 8.
Solution:
In the figure, AB || CD

Question 9.
Solution:
In the figure, AB || CD, ∠AEF = p

Question 10.
Solution:
In the figure, AB || PQ.
A transversal LM cuts them at E and F
∠ LEB = 75°
∠BEG = 20°
∠ EFG = 25°
∠ EGF = x° and ∠ GFD = y°
∴∠ LEB + ∠ BEF = 180° (Linear pair)

Question 11.
Solution:
In the figure

Question 12.
Solution:
In the figure, AB || CD
∠PEF = 85°, ∠QHC = 115°
∴ ∠ GHF = ∠ QHC
(Vertically opposite angles)
∠ GHF = 115°
∴AB || CD
∴∠ PEF = ∠ EGH
(Corresponding angles)
∴∠EGH = 85°
But ∠ QGH + ∠ EGH = 180°, (Linear pair)
=> ∠QGH + 85° = 180°
=> QGH = 180° – 85° = 95°
In ∆ GHQ,
Ext. ∠GHF = ∠QGH + ∠GQH
=> 115° = 95° + x
=> x = 115° – 95°
Hence, x = 20° Ans.

Question 13.
Solution:
In the figure, AB || CD
∠BAD = 75°, ∠ BCD = 35°
∴AB || CD
∴∠ABC = ∠BCD (Alternate angles)
=> x = 35°
=> 75° = z

Question 14.
Solution:
In the figure, AB || CD
∠APQ = 75°, ∠ PRD = 125°
∴AB || CD.
∠ APQ = ∠ PQR (Alternate angles)
∴75° = y°
=> y° = 75°

Question 15.
Solution:
In the figure, AB || CD and EF || GH
∠APR = 110°, ∠LRF = 60°
∴∠PRQ = ∠LRF
(vertically opposite angles)

Question 16.
Solution:
(i) l is parallel to m
if 3x – 20° = 2x +10°
(Alternate angles are equal)

Question 17.
Solution:
Given. Two lines AB and CD are perpendiculars on EF
To Prove : AB ⊥ CD.

Hope given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4C are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2D.

Other Exercises

Using factor theorem, show that :

Question 1.
Solution:
By factor theorem, x – 2 will be a factor of f(x) = x3 – 8 if f(2) = 0
(∴ x-2 = 0=>x = 2)
Now f(2) = (2)3 – 8 = 8- 8 = 0
Hence (x – 2) is a factor of f(x) Ans.

Question 2.
Solution:
By factor theorem,

Question 3.
Solution:
By factor theorem,
(x – 1) is a factor of f(x)=(2x4 + 9x3 + 6x2– 11x – 6)

Question 4.
Solution:
By factor theorem, (x + 2) will
a factor of f (x) = x4 – x4 + 2

Question 5.
Solution:
By factor theorem, (x + 5) will be a factor of f(x) = 2x3 + 9x2 – 11x – 30 if f(-5) = 0

Question 6.
Solution:
By factor theorem, (2x – 3) is a factor of f(x) = 2x4 + x3 – 8x2 – x + 6

Question 7.
Solution:
By factor theorem, (x – √2 ) will be a factor of f(x) = 7x2 – 4√2x – 6

Question 8.
Solution:
By factor theorem, (x + √2) will be a factor of f(x) = 2√2 x3 + 5x + √2

Question 9.
Solution:
Let f(x) = 2x3 + 9x2 + x + k and x – 1 is a factor of f(x)

Question 10.
Solution:
Let f(x) = 2x3 – 3x2 – 18x + a and x – 4 is its factor

Question 11.
Solution:
Let f(x) – x4 – x3 – 11x2 – x + a
f(x) is divisible by (x + 3)

Question 12.
Solution:
Let f(x) = 2x3 + ax2 + 11x + a + 3
and (2x – 1) is its factor
Let 2x – 1 = 0 then 2x = 1

Question 13.
Solution:
Let f(x) = x3 – 10x2 + ax + b and (x – 1) and (x – 2) are its factors
∴ x – 1 = 0 =>x=1
and x – 2 = 2 =>x=2

Question 14.
Solution:
Let f(x) = x4 + ax3 – 7x2 – 8x + b ,
and (x + 2) and (x + 3) are its factors
∴x + 2 = 0 => x = -2
and x + 3= 0 => x = -3

Question 15.
Solution:
Let f(x) = x3 – 3x2 – 13x + 15
Now x2 + 2x – 3 = x2 + 3x – x – 3

Question 16.
Solution:
Let f(x) = x3 + ax2 + bx + 6 and (x – 2) is its factor
Let x – 2 = 0 then x = 2

Hope given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2D are helpful to complete your math homework.

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