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Mathsight Class 7 Solutions Chapter 13 Congruence
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students to solve MathSight Class 7 Mathematics Book, Chapter 13, Congruence. Here students can easily get all the exercise questions solution for Chapter 13, Congruence Exercise 13.1, 13.2, 13.3, 13.4 and 13.5
Congruence Exercise 13.1 Solution :
Question no – (1)
Solution :
(a) and (d) are congruent.
AB ≅ HJ,
DC ≅ JK,
AD ≅ IJ,
BC ≅ HK
Question no – (2)
Solution :
(a) Same measurement
(b) Same length
(c) Same radius
(d) Same side
(e) Same length and breath
Question no – (3)
Solution :
The measure of the other angle will be 120°
Question no – (4)
Solution :
∠SQR = 90°
So, the angle 90° is congruent to ∠SQR
Question no – (5)
Solution :
(a) – True
(b) – True
(c) – True
(d) – True
(e) – False
Question no – (6)
Solution :
∴ Yes, they are congruent.
Question no – (7)
Solution :
Due to same Diameter, two circular tabletops are congruent.
Congruence Exercise 13.2 Solution :
Question no – (1)
Solution :
Required photo,
(a) ⇒ Congruent angle,
⇒ ∠ABC ≅ ∠MNO
∠BAC ≅ ∠NMO
∠ACB ≅ ∠MON
Congruent sides,
⇒ AB ≅ MN
BC ≅ NO
AC ≅ MO
(d) Photo ⇒ congruent angle
⇒ ∠ABC ≅ ∠FED
∠ACB ≅ ∠EFD
∠CAB ≅ ∠FDE
Congruent sides,
⇒ CB ≅ FE
AC ≅ FD
AB ≅ ED
Question no – (2)
Solution :
Required photo :
(a) ∠P ≅ ∠L
(b) PQ2 = LM
(c) ∠Q = ∠M
(d) QR2 ≅ MN
(e) ∠R2 ≅ ∠N
(f) PR2 ≅ LN
Question no – (3)
Solution :
△ABC ≅ △DEF
| Correspondence Side | Correspondence Angles |
| AB2 ≅ ED | ∠A ≅ ∠D |
| BC2 ≅ EF | ∠B ≅ ∠E |
| AC ≅ DF | ∠C ≅ ∠F |
Question no – (4)
Solution :
| Correspondence Side | Correspondence Angles |
| IM ≅ RS | ∠L ≅ ∠R |
| NM2 ≅ TS | ∠N ≅ ∠T |
| IN ≅ RT | ∠M ≅ ∠S |
Question no – (5)
Solution :
3n – 30° = 57°
Or, 3n = 57 + 30° = 87°
Or, n = 87/3 = 29°
∠B ≅ ∠N
Congruence Exercise 13.3 Solution :
Question no – (1)
Solution :
| Common side | Corresponding side | Congruent triangle | |
| A | BD ≅ BD | AD ≅ BC
AB ≅ CD |
△ABD ≅ △BDC |
| B | BC ≅ BC | AB ≅ BD
AC ≅ CD |
△ABC ≅ △BDC |
| C | AC ≅ FD | AB ≅ ED
BC ≅ FE |
△ABC ≅ △DEF |
Question no – (2)
Solution :
(a) Not true
(b) QR ≅ SR
(c) QP ≅ ST
(d) PR ≅ TR
∴ △PQR ≅ △TSR
(c) Not True.
Question no – (3)
Solution :
(a) Due to equal of correspondence part △DGF ≅ △JKH
(b) Due to equal of correspondence part △PTQ ≅ △SRT
Question no – (4)
Solution :
DE = GE, DF = GE
∴ EF = EF (common side)
∴ △DEF ≅ △GEF [Proved]
Question no – (5)
Solution :
AB = AC
BO = OC [∵ O, midpoint of BC]
AO = AO [common side]
∴ △AOB ≅ △AOC [Proved]
Question no – (6)
Solution :
(a) EG = EG [Common side]
ED = EF
DG = FG
∴ △DEG ≅ △FEG
(b) Yes, EG is common side of both triangle.
Question no – (7)
Solution :
(a) SQ = SQ, PS = SR, RQ = PQ
(b) Yes, their correspondence part are equal.
(c) Yes, it is common side.
Question no – (8)
Solution :
(a) Given statement is True.
(b) Given statement is False.
Congruence Exercise 13.4 Solution :
Question no – (1)
Solution :
(a) Correspondence part,
| Sides | Angles |
| EG = EG | ∠GED = ∠GEF |
| EF = ED | ∠EDG = ∠EFG |
| FG = DG |
∴ △EDG ≅ △EFG
(b) Correspondence part,
| Sides | Angles |
| QR = CB | ∠PQR = ∠ABC |
| PR = AC | |
| PQ = AB |
∴ △PQR ≅ △ABC
(c) Correspondence part,
| Sides | Angles |
| BD = BD | ∠ABD = ∠CBD |
| AB = BC | |
| AD = BC |
∴ △ABD ≅ △BCD
Question no – (2)
Solution :
(a) True
(b) False. Because they are not fulfilled the condition of S.A.S
(c) False. Because they are not fulfilled the condition of S.A.S
Question no – (3)
Solution :
(a) △PQS ≅ △QSR
(b) △ACB ≅ △FED
Question no – (4)
Solution :
(a) △ABC ≅ △DFE
Question no – (5)
Solution :
We know that,
2a + 30° = a + 50°
Or, 2a – a = 50 – 30° = 20°
Or, a = 20°
Again, 3b – 10° = b + 20°
Or, 3b – b = 20° + 10° = 30°
Or, 2b = 30°
Or, b = 30/2 = 15°
Question no – (6)
Solution :
O is midpoint of YZ.
∴ X bisect ∠X on YZ
∴ YO = OZ
∠XOY = ∠XOZ
XO = XO (Common side)
∴ △YXO ≅ △ZOX
Question no – (7)
Solution :
∠D ≅ ∠M
DF ≅ MO
EF ≅ NO
DE = MN
∠E = ∠N
∠F = ∠O
∴ △DEF ≅ △MNO ….[Proved]
Question no – (8)
Solution :
QR bisect the angle ∠QO.
PR bisect the angle ∠R
∴ PR = SQ, QR = common side
∴ △PQR = △RQS ….[Proved]
Congruence Exercise 13.5 Solution :
Question no – (1)
Solution :
(a) △DEF ≅ △DFG
(b)△ABC ≅ △DEF
(c) △PRQ ≅ △PQS
Question no – (2)
Solution :
(a) False. It does not fulfilled the condition of SAS sides should be shower here.
(b) True.
(c) False. It does not fulfilled the condition of SAS. Angles and side should be properly show here.
Question no – (3)
Solution :
∠BAC = 90° = ∠BDC
AC = BD
BC = BC [common side]
∴ ∠ABC = ∠DCB
∴ AB = DC
∴ They are satisfied the condition of SAS.
∴ △ABC ≅ △DCB [P]
Question no – (4)
Solution :
PQ = PQ = common side
∴ ∠RQP = ∠SPQ
∠RPQ = ∠SQP = 90°
RP = QS
RQ = PS, ∠R = ∠S
∴ They are satisfied the condition of SAS.
∴ △RPQ ≅ △SPQ [P]
Question no – (5)
Solution :
∠XPY = ∠XOZ = 90°
∴ XP = ∠OZ
XY = XZ
XO = YP
∠Y = ∠X = ∠Z
∴ △XYP ≅ △XZO [P]
Question no – (6)
Solution :
Yes, the information’s are enough.
Question no – (7)
Solution :
No, Because the diaries are not satisfied the condition of SAS. They all are diff height.
Question no – (8)
Solution :
We know that,
12 = 4x
Or, x = 12/4
∴ x = 3 cm.
Previous Chapter Solution :
