Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RS Aggarwal Solutions: Congruence
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Page 1
Question 1.
Solution:
(i) ?ABC ?
?EFD, Then
A ? E, B ? F and C ? D
AB = EF, BC = FD and CA = DE
?A = ?E, ?B = ?F and ?C = ?D
(ii) ?CAB ?
?QRP
C ? Q, A ? R and B ? P
CA = QR, AB = RP and BC = PQ
?C = ?Q, ?A = ?R and ?B = ?P
(iii) ?XZY ?
?QPR
X ? Q, Z ? P, Y ? R
XZ = QP, ZY = PR and YX = RQ
?X = ?Q, ?Z = ?P and ?Y = ?R
(iv) ?MPN ?
?SQR
M ? S, P ? Q and N ? R
MP = SQ, PN = QR and NM = RS
?M = ?S, ?P = ?Q and ?N = ?R.
Question 2.
Page 2
Question 1.
Solution:
(i) ?ABC ?
?EFD, Then
A ? E, B ? F and C ? D
AB = EF, BC = FD and CA = DE
?A = ?E, ?B = ?F and ?C = ?D
(ii) ?CAB ?
?QRP
C ? Q, A ? R and B ? P
CA = QR, AB = RP and BC = PQ
?C = ?Q, ?A = ?R and ?B = ?P
(iii) ?XZY ?
?QPR
X ? Q, Z ? P, Y ? R
XZ = QP, ZY = PR and YX = RQ
?X = ?Q, ?Z = ?P and ?Y = ?R
(iv) ?MPN ?
?SQR
M ? S, P ? Q and N ? R
MP = SQ, PN = QR and NM = RS
?M = ?S, ?P = ?Q and ?N = ?R.
Question 2.
RS Aggarwal Solutions Class 7 Chapter 16 - Congruence (Ex 16) Exercise- 16
Page 3
Question 1.
Solution:
(i) ?ABC ?
?EFD, Then
A ? E, B ? F and C ? D
AB = EF, BC = FD and CA = DE
?A = ?E, ?B = ?F and ?C = ?D
(ii) ?CAB ?
?QRP
C ? Q, A ? R and B ? P
CA = QR, AB = RP and BC = PQ
?C = ?Q, ?A = ?R and ?B = ?P
(iii) ?XZY ?
?QPR
X ? Q, Z ? P, Y ? R
XZ = QP, ZY = PR and YX = RQ
?X = ?Q, ?Z = ?P and ?Y = ?R
(iv) ?MPN ?
?SQR
M ? S, P ? Q and N ? R
MP = SQ, PN = QR and NM = RS
?M = ?S, ?P = ?Q and ?N = ?R.
Question 2.
RS Aggarwal Solutions Class 7 Chapter 16 - Congruence (Ex 16) Exercise- 16
RS Aggarwal Solutions Class 7 Chapter 16 - Congruence (Ex 16) Exercise- 16
Solution:
(i) In
fig (i)
In ?ABC and ?DEF
?C = ?E
CA = ED
CB = EF
?ACB ?
?DEF (SAS condition)
(ii) In fig (ii)
In ?RPQ and ?LNM
Side PQ = NM
Hyp. RQ = LM
?RPQ ?
?LNM (RHS condition)
(iii) In ?YXZ and ?TRS
XY = RT
?X = SR and YZ = TS
?YXZ ? ?TRS (SSS condition)
(iv) In ?DEF and ?PNM
?E = ?N
?F = ?M
EF = NM
?DEF ? ?PNM (ASA condition)
(v) In ?ABC and ?ADC
AC = AC (common)
? CAB = ? CAD (each 50°)
? ACB = ? DCA (each 60°)
?ABC ? ?ADC (ASA condition)
Page 4
Question 1.
Solution:
(i) ?ABC ?
?EFD, Then
A ? E, B ? F and C ? D
AB = EF, BC = FD and CA = DE
?A = ?E, ?B = ?F and ?C = ?D
(ii) ?CAB ?
?QRP
C ? Q, A ? R and B ? P
CA = QR, AB = RP and BC = PQ
?C = ?Q, ?A = ?R and ?B = ?P
(iii) ?XZY ?
?QPR
X ? Q, Z ? P, Y ? R
XZ = QP, ZY = PR and YX = RQ
?X = ?Q, ?Z = ?P and ?Y = ?R
(iv) ?MPN ?
?SQR
M ? S, P ? Q and N ? R
MP = SQ, PN = QR and NM = RS
?M = ?S, ?P = ?Q and ?N = ?R.
Question 2.
RS Aggarwal Solutions Class 7 Chapter 16 - Congruence (Ex 16) Exercise- 16
RS Aggarwal Solutions Class 7 Chapter 16 - Congruence (Ex 16) Exercise- 16
Solution:
(i) In
fig (i)
In ?ABC and ?DEF
?C = ?E
CA = ED
CB = EF
?ACB ?
?DEF (SAS condition)
(ii) In fig (ii)
In ?RPQ and ?LNM
Side PQ = NM
Hyp. RQ = LM
?RPQ ?
?LNM (RHS condition)
(iii) In ?YXZ and ?TRS
XY = RT
?X = SR and YZ = TS
?YXZ ? ?TRS (SSS condition)
(iv) In ?DEF and ?PNM
?E = ?N
?F = ?M
EF = NM
?DEF ? ?PNM (ASA condition)
(v) In ?ABC and ?ADC
AC = AC (common)
? CAB = ? CAD (each 50°)
? ACB = ? DCA (each 60°)
?ABC ? ?ADC (ASA condition)
RS Aggarwal Solutions Class 7 Chapter 16 - Congruence (Ex 16) Exercise- 16
Question 3.
Solution:
In fig,
PL ?
OA and PM ?
OB and PL = PM
Now in right ?PLO and ?PMO,
Side PL = PM (given)
Hypotenuse OP = OP (common)
?PLO ?
?PMO (RHS condition)
Yes ?PLO ?
?PMO
Hence proved.
Question 4.
Solution:
In the figure,
AD = BC and AD || BC.
In ?ABC and ?ADC,
AC = AC (common)
BC = AB (given)
?ACB = ?CAD (Alternate angles)
?ABC ? ?ADC (SAS condition)
AB = DC (c.p.c.t)
Hence proved.
Page 5
Question 1.
Solution:
(i) ?ABC ?
?EFD, Then
A ? E, B ? F and C ? D
AB = EF, BC = FD and CA = DE
?A = ?E, ?B = ?F and ?C = ?D
(ii) ?CAB ?
?QRP
C ? Q, A ? R and B ? P
CA = QR, AB = RP and BC = PQ
?C = ?Q, ?A = ?R and ?B = ?P
(iii) ?XZY ?
?QPR
X ? Q, Z ? P, Y ? R
XZ = QP, ZY = PR and YX = RQ
?X = ?Q, ?Z = ?P and ?Y = ?R
(iv) ?MPN ?
?SQR
M ? S, P ? Q and N ? R
MP = SQ, PN = QR and NM = RS
?M = ?S, ?P = ?Q and ?N = ?R.
Question 2.
RS Aggarwal Solutions Class 7 Chapter 16 - Congruence (Ex 16) Exercise- 16
RS Aggarwal Solutions Class 7 Chapter 16 - Congruence (Ex 16) Exercise- 16
Solution:
(i) In
fig (i)
In ?ABC and ?DEF
?C = ?E
CA = ED
CB = EF
?ACB ?
?DEF (SAS condition)
(ii) In fig (ii)
In ?RPQ and ?LNM
Side PQ = NM
Hyp. RQ = LM
?RPQ ?
?LNM (RHS condition)
(iii) In ?YXZ and ?TRS
XY = RT
?X = SR and YZ = TS
?YXZ ? ?TRS (SSS condition)
(iv) In ?DEF and ?PNM
?E = ?N
?F = ?M
EF = NM
?DEF ? ?PNM (ASA condition)
(v) In ?ABC and ?ADC
AC = AC (common)
? CAB = ? CAD (each 50°)
? ACB = ? DCA (each 60°)
?ABC ? ?ADC (ASA condition)
RS Aggarwal Solutions Class 7 Chapter 16 - Congruence (Ex 16) Exercise- 16
Question 3.
Solution:
In fig,
PL ?
OA and PM ?
OB and PL = PM
Now in right ?PLO and ?PMO,
Side PL = PM (given)
Hypotenuse OP = OP (common)
?PLO ?
?PMO (RHS condition)
Yes ?PLO ?
?PMO
Hence proved.
Question 4.
Solution:
In the figure,
AD = BC and AD || BC.
In ?ABC and ?ADC,
AC = AC (common)
BC = AB (given)
?ACB = ?CAD (Alternate angles)
?ABC ? ?ADC (SAS condition)
AB = DC (c.p.c.t)
Hence proved.
RS Aggarwal Solutions Class 7 Chapter 16 - Congruence (Ex 16) Exercise- 16
Solution:
In ?ABD and ?ACD,
AD = AD (common)
AB = AC (given)
BD = CD (given)
?ABD ?
?ADC (SSS condition)
?BAD = ?CAD (c.p.c.t.)
and ?ADB = ?ADC (c.p.c.t.)
But ?ADB + ?ADC = 180° (Linear pair)
?ADB = ?ADC = 90°
Hence proved.
Question 6.
Question 5.
Solution:
given : In ?ABC, AD is the bisector of ?A i.e. ?BAD = ?CAD
AD ? BC.
To prove : ?ABC is an isosceles
Proof : In ?ADB and ?ADC.
AD = AD (common)
? BAD = ? CAD (AD is the bisector of ?A)
? ADB = ? ADC (each = 90°, AD ? BC)
?ADM ? ?ADC (ASA condition)
AB = AC (c.p.c.t)
Hence ?ABC is an isosceles triangle.
Hence proved.
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