Class 9: Maths Chapter 5 solutions. Complete Class 9 Maths Chapter 5 Notes.
RS Aggarwal Solutions for Class 9 Maths Chapter 5–Congruence of Triangles and Inequalities in a Triangle
RS Aggarwal 9th Maths Chapter 5, Class 9 Maths Chapter 5 solutions
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 = 110o2 = 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 = 80o2 = 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 = 12 ∠A 2
But ∠A + ∠B + ∠C = 180°
(sum of angles of a triangle)
=> ∠A+ 12 ∠A = 180°
=> 32 ∠A = 180°
=> ∠A = 180° x 23 = 120°
and ∠B + ∠C = 12 ∠A = 12 x 120°
= 60°
∴ ∠ B = ∠ C
∴ ∠ B = ∠ C = 60o2 = 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 = 90o2 = 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)
Adding ∠ DCE both sides,
∠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)
Adding FC both sides,
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,
AD is median. BL and CM are perpendiculars on AD and AD is produced to E
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)
=> 12 ∠C = 12 ∠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 = 30o2 = 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.
AD and BF are joined.
To Prove : AD = BF
Proof : ∠ACF = ∠BCD (Each 90°)
Adding ∠ ACB both sides,
∠ 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)
AD = BF (c.p.c.t)
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,
AD = AD (Common)
AB = AC (Given)
BD = CD (D is mid-point of BC)
∴ ∆ ABD ≅ ∆ ACD (S.S.S. axiom)
∠ BAD = ∠ CAD (c.p.c.t.)
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)
Adding (i) and (ii),
∴ ∠ 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:
Given : In quadrilateral ABCD,
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,
AB = AD (given)
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,
AB = AD (given)
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°
To Prove : (i) AD > CD (ii) AD > AC
Proof : In ∆ BCD,
Ext. ∠ B = 60°
∴ ∠ CBD = 180° – 60° = 120°
and ∠ BCD + ∠ BDC = Ext. ∠ CBA = 60°
But ∠BCD = ∠ BDC ( BC = BD)
∴ ∠BCD = ∠BDC = 60o2 = 30°.
∴ ∠ ACD = 50° + 30° = 80°
(i) Now in ∆ ACD,
∴ ∠ ACD > ∠ CAD ( 80° > 70°)
AD > CD .
(ii) and ACD > ∠D (80° > 30°)
∴ AD > AC
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,
Ext. ∠ ADB > ∠ CAD
=>∠ ADB > ∠BAD ,
( AD is bisector of ∠ A)
In ∆ ABD,
AB > BD.
(ii) Again, in ∆ ADB,
Ext. ∠ADC > ∠BAD
=> ∠ADC > ∠CAD
( ∠ CAD = ∠BAD)
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
∴ ∠ABC > ∠ADC
Now, in ∆ ABD,
∴ ∠ABC > ∠ADC or ∠ADB
∴ AD > AC
Hence proved.
Question 41.
Solution:
Given : In ∆ ABC,
AC > AB and AD is the bisector of ∠ A which meets BC in D.
To Prove : ∠ADC > ∠ADB
Proof : In ∆ ABC,
AC > AB
∴∠B > ∠C
∴∠ 1 = ∠ 2 (AD is the bisector of ∠ A)
∴ ∠B + ∠2 = ∠C + ∠1
But in ∆ ADB
Ext. ∠ADC = ∠B + ∠2
and in ∆ ADC,
Ext. ∠ADB > ∠C + ∠1
∴ ∠B + ∠2 > ∠C + ∠1 (Proved)
∴ ∠ ADC > ∠ ADB Hence 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)
Adding (i), and (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 > 12
(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)
Adding (i) and (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
Adding, we get:
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
adding we get:
OA + OB + OB + OC + OC + OA > AB + BC + CA
=> 2(OA + OB + OC) > (AB + BC + CA)
=> OA + OB + OC > 12 (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.
RS Aggarwal Solutions for Class 9 Maths Chapter 5: Download PDF
RS Aggarwal Solutions for Class 9 Maths Chapter 5–Congruence of Triangles and Inequalities in a Triangle
Chapterwise RS Aggarwal Solutions for Class 9 Maths :
- Chapter 1–Real Numbers
- Chapter 2–Polynomials
- Chapter 3–Introduction to Euclid’s Geometry
- Chapter 4–Lines and Triangles
- Chapter 5–Congruence of Triangles and Inequalities in a Triangle
- Chapter 6–Coordinate Geometry
- Chapter 7–Areas
- Chapter 8–Linear Equations in Two Variables
- Chapter 9–Quadrilaterals and Parallelograms
- Chapter 10–Area
- Chapter 11–Circle
- Chapter 12–Geometrical Constructions
- Chapter 13–Volume and Surface Area
- Chapter 14–Statistics
- Chapter 15–Probability
About RS Aggarwal Class 9 Book
Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal's book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.
He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.
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RS Aggarwal is one of the most important reference books for high school grades and is recommended to every high school student. The book covers every single topic in detail. It goes in-depth and covers every single aspect of all the mathematics topics and covers both theory and problem-solving. The book is true of great help for every high school student. Solving a majority of the questions from the book can help a lot in understanding topics in detail and in a manner that is very simple to understand. Hence, as a high school student, you must definitely dwell your hands on RS Aggarwal!
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