Congruence of Triangles – Maharashtra Board Class 9 Solutions for Geometry
AlgebraGeometryScience and TechnologyHindi
Exercise – 3.1
Solution 1:
Congruent sides:
side AB ≅ side MN
side BC ≅ side NP
side AC ≅ side MP
Congruent angles:
∠A ≅ ∠M
∠B ≅ ∠N
∠C ≅ ∠P
Solution 2:
Seg PQ ≅ seg QR
∴∠R ≅ ∠P ….(Isosceles triangle theorem)
m∠P = 70° …(Given)
∴m∠R = 70° ….(1)
Now, the sum of the measures of the angles of a triangle is 180°.
∴ m∠P + m∠Q + m∠R = 180°
∴70° + m∠Q + 70° = 180° …[Given and from (1)]
∴140° + m∠Q = 180°
∴m∠Q = 180° – 140°
∴m∠Q = 40°
The measures of the remaining angles are m∠R = 70° and m∠Q = 40°.
Solution 3:
In ∆PST, seg PS ≅ seg PT. ….(Given)
∴ By Isosceles Triangle Theorem,
∠PTS ≅ ∠PST ….(1)
The sum of the measures of the angles of a triangle is 180°.
∴m∠P + m∠PTS + m∠PST = 180° ….(2)
∴m∠P + 2m∠PTS = 180° ….. [From (1) and (2)] …..(3)
Seg PS ≅ seg PT and seg SQ ≅ seg TR …(Given)
∴ PS = PT and SQ = TR
∴ PS + SQ = PT + TR
∴ PQ = PR ….(P-S-Q and P-T-R)
∴ seg PQ ≅ seg PR.
In ∆PQR, seg PQ ≅ seg PR ….(Proved)
∴ ∠PRQ ≅ ∠PQR ….(4)
m∠P + m∠PQR + m∠PRQ = 180° …..(5)
∴m∠P + 2m∠PQR = 180° ……[From (4) and (5)]….(6)
From (3) and (6), ∠PTS = ∠PRQ
∴ seg ST ∥ seg QR ….(corresponding angles test for parallel lines)
i.e. Side ST ∥ Side QR.
Solution 4:
Proof with justification:
In ∆POR and ∆SOQ,
seg OP ≅ seg OS and seg OR ≅ seg OQ …..(Given)
∠POR ≅ ∠SOQ ….(Vertically opposite angles)
∴ ∆POR ≅ ∆SOQ ….(SAS test)
∴ seg PR ≅ seg SQ ….(c.s.c.t.)
Also, ∠P ≅∠S and ∠R ≅∠Q …..(c.a.c.t.)
For seg PR and seg SQ, PQ is the transversal.
∴ For seg PR and seg SQ to be parallel, the alternate angles, ∠ P and ∠Q, should be congruent.
But these angles are not congruent.
∴ seg PR and seg SQ are not parallel.
Also, taking RS as the transversal, we can show that seg PR and seg SQ are not parallel.
Solution 5:
∠ABP ≅ ∠CBQ …..(Given)
∴ ∠ABP = ∠CBQ
Adding ∠PBC to both the sides
∠ABP + ∠PBC = ∠CBQ + ∠PBC
∠ABC ≅ ∠PBQ ….(1)
seg AB = seg PB and seg BC ≅ seg BQ …..(Given)
∴∆ABC ≅ ∆PBQ …..(SAS test)
∴ seg AC ≅ seg PQ ….(c.s.c.t.)
Solution 6:
Solution with justification:
In ∆ ABC, seg AB ≅ seg AC ……(Given)
∴ by Isosceles Triangle Theorem,
∠ABC ≅ ∠ACB.
i.e. ∠ABC = ∠ACB …(1)
The sum of the measures of the angles of a triangle is 180°
∴m∠A + m∠ABC + m∠ACB = 180°
∴m∠A + m∠ABC + m∠ABC = 180° …..[From (1)]
∴m∠A + 2m∠ABC = 180°
∴40° + 2m∠ABC = 180° …[Given: m∠A = 40°]
∴2m∠ABC = 180° – 40° = 140°
∴m∠ABC = 70°
∴m∠ABC = m∠ACB = 70° …(2)
In ∆ABP, seg AB ≅ seg PB …(Given)
∴∠P ≅ ∠PAB i.e. ∠P = ∠PAB …(3)
By Remote Interior Angle Theorem,
∠ABC = ∠P + ∠PAB
From (2) and (3), 70° = m∠P + m∠P
∴m∠P = m∠PAB = 35° ….(4)
Similarly, we can prove that
m∠Q = m∠QAC = 35° ….(5)
In ∆APQ, m∠PAQ + m∠P + m∠Q = 180°
∴m∠PAQ + 35° + 35° = 180° ….[From (4) and (5)]
∴m∠PAQ = 180° – 70°
∴m∠PAQ = 110° …..(6)
So, m∠P = m∠Q = m∠PAB = m∠QAC = 35°
PC = PB + BC and QB = QC + BC …..(7)
But PB = QC ….(Given) ….(8)
∴ PC = QB ….[From (7) and (8)]
In ∆PAC and ∆QAB,
PC = QB …..(Proved)
AC = AB ….(Given)
∠ACB = ∠ABC …..[From (2)]
i.e. ∠ACP = ∠ABQ
∴∆PAC ≅ ∆QAB …..(SAS test)
Similarly, we can prove that
∆PAB ≅ ∆QAC.
Hence, m∠PAQ = 110°; Also ∆PAC ≅ ∆QAB and ∆PAB ≅ ∆QAC are the congruent triangles.
Exercise – 3.2
Solution 1:
In ∆ACO and ∆DBO,
∠ACO ≅ ∠DBO ….(Given)
seg CO ≅ seg BO …(Given)
∠AOC ≅ ∠DOB …..(Vertically opposite angles)
∴∆ACO ≅ ∆DBO ….(ASA test)
The remaining congruent parts:
∠A ≅ ∠D, seg AC ≅ seg DB, and seg AO ≅ seg DO
Solution 2(i):
∆BDA ≅ ∆BDC
1. Test of congruence: SAA test
2. Correspondence: BDA ↔ BDC
3. Congruent Parts:
∠ABD ≅ ∠CBD
seg AB ≅ seg CB
seg AD ≅ seg CD
Solution 2(ii):
∆PQR ≅ ∆SRQ
1. Test of congruence: SAS test
2. Correspondence: PQR ↔ SRQ
3. Congruent Parts:
seg PR ≅ seg SQ
∠QPR ≅ ∠RSQ
∠PRQ ≅ ∠SQR
Solution 2(iii):
∆MQN ≅ ∆PQN
1. Test of congruence: SAS test
2. Correspondence: MQN ↔ PQN
3. Congruent Parts:
seg MN ≅ seg PN
∠QMN ≅ ∠QPN
∠QNM ≅ ∠QNP
Solution 2(iv):
∆BPA ≅ ∆BPC
1. Test of congruence: ASA test
2. Correspondence: BPA ↔ BPC
3. Congruent Parts:
∠BAP ≅ ∠BCP
seg BA ≅ seg BC
seg PA ≅ seg PC
Solution 2(v):
∆ ABP ≅ ∆CBP
1. Test of congruence: SSS test
2. Correspondence: ABP ↔ CBP
3. Congruent Parts:
∠PBA ≅ ∠PBC
∠BAP ≅ ∠BCP
∠BPA ≅ ∠BPC
Solution 2(vi):
BD + DC = FC + DC
∴BC = FD
∆ ABC ≅ ∆EFD
1. Test of congruence: SAS test
2. Correspondence: ABC ↔ EFD
3. Congruent Parts:
seg AB ≅ seg EF
∠ABC ≅ ∠EFD
∠BAC ≅ ∠FED
Solution 2(vii):
AP = AQ
PB = QC
∴ AP + PB = AQ + QC
∴AB = AC …..(A-P-B; A-Q-C)
∆ ABQ ≅ ∆ACP
1. Test of congruence: SAS test
2. Correspondence: ABQ ↔ ACP
3. Congruent Parts:
seg BQ ≅ seg CP
∠ABQ ≅ ∠ACP
∠AQB ≅ ∠APC
Solution 2(viii):
∆ PSQ ≅ ∆PSR
1. Test of congruence: Hypotenuse-Side theorem
2. Correspondence: PSQ ↔ PSR
3. Congruent Parts:
seg QS ≅ seg RS
∠QPS ≅ ∠RPS
∠PQS ≅ ∠PRS
Solution 2(ix):
∆ APR ≅ ∆BPQ ≅ ∆CQR ≅ ∆QPR
1. Test of congruence: SSS test
2. Correspondence:
APR ↔ BPQ ↔ CQR ↔ QPR or any correspondence
3. Congruent Parts:
All the angles of ∆APR, ∆BPQ, ∆CQR and ∆QPR are 60° each.
Solution 2(x):
∆ PQS ≅ ∆RQS
1. Test of congruence: Hypotenuse – side theorem
2. Correspondence: PQS ↔ RQS
3. Congruent Parts:
seg PS ≅ seg RS
∠PQS ≅ ∠RQS
∠PSQ ≅ ∠RSQ
Solution 3:
i.
1. ∆XYZ and ∆DYZ are drawn as above.
2. Congruent parts seg XZ ≅ seg DY and
∠XZY ≅ ∠ZYD are shown in the figure by identical marks.
3. Test of congruence: SAS test
4. Congruent triangle: ∆ XYZ ≅ ∆DZY.
Correspondence: XYZ ↔ DZY
ii.
1. ∆EFG and ∆LMN are drawn as above.
2. Congruent parts seg FG ≅ seg MN,
∠G ≅ ∠N and ∠F ≅ ∠M are shown in the figure by identical marks.
3. Test of congruence: ASA test
4. Congruent triangle: ∆ EFG ≅ ∆LMN.
Correspondence: EFG ↔ LMN
iii.
1. ∆MNO and ∆CNR are drawn as above.
2. Congruent parts seg MN ≅ seg CN,
seg NO ≅ seg NR and ∠MNO ≅ ∠CNR are shown in the figure by identical marks.
3. Test of congruence: SAS test
4. Congruent triangle: ∆MNO ≅ ∆CNR.
Correspondence: MNO ↔ CNR
iv.
1. ∆BTR and ∆BTP are drawn as above.
1. Congruent parts seg BR ≅ seg BP,
Seg RT ≅ seg PT are shown in the figure by identical marks.
2. Test of congruence: SSS test
3. Congruent triangle: ∆BTR ≅ ∆BTP.
4. Correspondence: BTR ↔ BTP
v. In ∆LMP and ∆LPN,
∠LMP ≅ ∠LNP, ∠LPM ≅ ∠NPL.
Ans. (1) and (2)
1. ∆LMP and ∆LPN are drawn as above.
2. Congruent parts ∠LMP ≅ ∠LNP and ∠LPM ≅ ∠NPL are shown in the figure by identical marks.
3. Test of congruence: SAA test
4. Congruent triangle: ∆LMP ≅ ∆LNP.
Correspondence: LMP ↔ LNP
Solution 4:
Missing information required and sufficient to prove
∆ADC ≅ ∆CBA:
SAS test
seg AD ≅ seg CB
SAA test
∠D ≅ ∠B
ASA test
∠ACD ≅ ∠CAB
Solution 5:
Seg QS is the bisector of ∠PQR ….(Given)
∴∠PQS ≅ ∠RQS …(1)
Seg QS is the bisector of ∠PSR ….(Given)
∴∠PSQ ≅ ∠RSQ …(2)
In ∆PQS and ∆RQS,
∠PQS ≅ ∠RQS …[From (1)]
seg QS ≅ seg QS …(Common side)
∠PSQ ≅ ∠RSQ …..[From (2)]
∴ ∆PQS ≅ ∆RSQ ….(ASA test)
∴∠P ≅ ∠R ….(c.a.c.t.)
Solution 6:
Diagonal MP is the perpendicular bisector of diagonal NQ.
∴ seg NT ≅ seg QT and
m∠NTM = m∠QTM = m∠PTN = m∠PTQ = 90° …..(1)
Now, in ∆MNT and ∆MQT,
seg NT ≅ seg QT …[From (1)]
∠MTN ≅ ∠MTQ …[From (1)]
seg MT ≅ seg MT …(Common side)
∴ ∆MNT ≅∆MQT …(SAS test)
∴ seg MN ≅ seg MQ …(c.s.c.t.)
In ∆PNT and ∆PQT,
seg NT ≅ seg QT ….[From (1)]
∠PTN ≅ ∠PTQ ….[From (1)]
seg PT ≅ seg PT …(Common side)
∴ ∆PNT ≅∆PQT …(SAS test)
∴ seg NP ≅ seg QP …(c.s.c.t.)
In ∆NMP and ∆QMP,
seg MN ≅ seg MQ
seg NP ≅ seg QP
seg MP ≅ seg MP …(Common side)
∴ ∆PNT ≅∆PQT …(SSS test)
Solution 7:
i. In ∆ABC and ∆CDE,
m∠B = m∠D = 90°
hypotenuse AC ≅ hypotenuse CE ….(Given)
seg BC ≅ seg ED ….(Given)
∴ ∆ABC ≅ ∆CDE ….(Hypotenuse- side theorem)
ii. As ∆ABC ≅ ∆CDE
∴∠BAC ≅ ∠DCE ….(c.a.c.t.) …(1)
iii. In ∆ABC, m∠BAC + m∠ACB = 90° … (Acute angles of a right angled triangle)…(2)
From (1) and (2),
m∠DCE + m∠ACB = 90° ….(3)
m∠ACB + m∠ACE + m∠DCE = 180°
∴ m∠ACB + m∠DCE + m∠ACE = 180°
∴ 90° + m∠ACE = 180° ….[From (3)]
∴ m∠ACE = 90°
Solution 8:
i. The pairs of congruent triangles:
a. ∆ABP and ∆ACP
b. ∆ABQ and ∆ACQ
c. ∆BQP and ∆CQP
a.
In ∆ABP and ∆ACP,
seg AB ≅ seg AC …(Given)
∠BAP ≅ ∠CAP ….(AP is the bisector of ∠BAC)
seg AP ≅ seg AP …(Common side)
∴ ∆ABP ≅ ∆ACP ….(SAS test)
∴ seg BP ≅ seg CP …(c.s.c.t.)….(1)
b.
In ∆ABQ and ∆ACQ,
seg AB ≅ seg AC …(Given)
∠BAQ ≅ ∠CAQ ….(AP, i.e. AQ is the bisector of ∠BAC)
seg AQ ≅ seg AQ …(Common side)
∴ ∆ABQ ≅ ∆ACQ ….(SAS test)
∴ seg BQ ≅ seg CQ …(c.s.c.t.)….(2)
c.
In ∆BQP and ∆CQP,
seg BP ≅ seg CP …[ From (1)]
seg BQ ≅ seg CQ …[ From (2)]
seg QP ≅ seg QP …(Common side)
∴ ∆BQP ≅ ∆CQP …(SSS test)
ii ∆BQC is an isosceles triangle. ….[From (2)]
seg AB ≅ seg AC …(Given)
∆ABC is an isosceles triangle.
Exercise – 3.3
Solution 1:
m∠A + m∠B + m∠C = 180° …(Angles of a triangles)
∴40° + 80° + m∠C = 180°
∴m∠C = 180° – 120° ∴m∠C = 60°
The descending order of the measures of the angles is ∠B > ∠C > ∠A
∴ side AC > side AB > side BC ….(Sides opposite to the angles)
The shortest side is side BC and the longest side is side AC.
Solution 2:
AB = 5 cm, BC = 8 cm, AC = 10 cm.
∴ AC > BC > AB
∴∠B > ∠A > ∠C …..(Inequality property of a triangle)
The smallest angle is ∠C and the greatest angle is ∠B.
Their descending order is ∠B > ∠A > ∠C.
Solution 3:
In ∆ABC, AB = 4 cm, AC = 6 cm.
∴ AC > AB.
Angle opposite to the greater side is greater.
∴a > b
b is the exterior angle of ∆ACD.
∴ By exterior angle theorem b > d ….(2)
From (1) and (2) a > b > d ….(3)
Now, c is the exterior angle of ∆ABC.
∴ c > a ….(By exterior angle theorem) …(4)
From (3) and (4) c > a > b > d.
Solution 4:
In ∆ABC, m∠B = 30°, m∠C = 25°,
∴ ∠B > ∠C
∴side AC > side AB
…..(In a triangle, the side opposite to the greater angle is greater) ….(1)
m∠ADB + m∠ADC = 180° …(Angles in a linear pair)
∴m∠ADB + 70° = 180°
∴m∠ADB = 110°
In ∆ABD, m∠ADB = 110° and m∠B = 30°
∴ ∠ADB > ∠B
∴ side AB > side AD …(2)
From (1) and (2),
side AC > side AB > side AD.
Solution 5:
(i) In ∆ABC, side AB ≅ side BC and A-P-C. …(Given)
∴ ∠A ≅ ∠C …(Isosceles Triangle Theorem) ..(1)
∠BPC > A …(Exterior Angle Theorem) …(2)
From (1) and (2), ∠BPC > ∠C
∴ BC > BP
…(Side opposite to greater angle)
i.e. BP < BC …(3)
AB ≅ BC ….(Given)…(4)
From (3) and (4),
BP
∴ BP< congruent sides.
(ii) In ∆ABC, side AB ≅ side BC and A-C-P.
….(Given)
∴ ∠A ≅ ∠BCA ….(Isosceles Triangle Theorem) …(1)
∠BCA > ∠P …(Exterior Angle Theorem)…(2)
From (1) and (2), ∠A > ∠P
∴ BP > BA …(Side opposite to greater angle) ..(3)
side AB ≅ side BC …(Given) …(4)
From (3) and (4),
BP > BA and BP > BC.
∴ BP > congruent sides.
Solution 6:
∠APC > ∠B …(Exterior Angle Theorem) …(1)
∠B ≅ ∠ACB …(Isosceles Triangle Theorem) …(2)
From (1) and (2), ∠APC > ∠ACP.
∴ AC > AP …(Side opposite to greater angle) … (3)
AB = AC ….(Given) …(4)
From (3) and (4), AB > AP
i.e. AP < AB …(5)
∠ACB > ∠AQC …(Exterior Angle Theorem) ..(6)
∴ ∠ABC > ∠AQC …[From (2) and (6)]
i.e. ∠ABQ > ∠AQC
∴ in ∆ABQ, AQ > AB
…(Side opposite to greater angle)
i.e. AB < AQ …(7)
From (5) and (7), AP < AQ.
Solution 7:
Let the length of side PR be x cm.
The sum of the lengths of any two sides of a triangle is greater than the third side.
∴ PQ + QR > PR
∴ 4 + 6 > PR
∴ 10 > PR
∴ 10 > x …(1)
The difference between the lengths of any two sides of a triangle is less than the length of the third side.
∴ QR – PQ < PR
∴ 6 – 4 < PR
∴ 2 < PR i.e. PR > 2
∴ x > 2 …(2)
From (1) and (2), 10 > x > 2
The length of side PR is greater than 2 cm but less than 10 cm.