# Mathsight Class 7 Solutions Chapter 13

## 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,

BC ≅ HK

Question no – (2)

Solution :

(a) Same measurement

(b) Same length

(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 :

Updated: May 27, 2023 — 8:32 am