In the figure, given two triangles are congruent. So, the corresponding parts are: A ↔ O, R ↔ W, T ↔ N. We can write, ∆RAT ≅ ∆WON [By SAS congruence rule] Class 7 Maths Chapter 7 Exercise 7.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-7/
In the figure, given two triangles are congruent. So, the corresponding parts are:
A ↔ O, R ↔ W, T ↔ N.
We can write, ∆RAT ≅ ∆WON [By SAS congruence rule]
No, because the two triangles with equal corresponding angles need not be congruent. In such a correspondence, one of them can be an enlarged copy of the other. Class 7 Maths Chapter 7 Exercise 7.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
No, because the two triangles with equal corresponding angles need not be congruent. In
such a correspondence, one of them can be an enlarged copy of the other.
(i) PM = QM (i) Given (ii) ∠PMA = ∠QMA (ii) Given (iii) AM = AM (iii) Common (iv) ∆AMP ≅ ∆AMQ (iv) SAS congruence rule Class 7 Maths Chapter 7 Exercise 7.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
(i) PM = QM (i) Given
(ii) ∠PMA = ∠QMA (ii) Given
(iii) AM = AM (iii) Common
(iv) ∆AMP ≅ ∆AMQ (iv) SAS congruence rule
(a) Using SSS criterion, ∆ART ≅ ∆PEN (i) AR = PE (ii) RT = EN (iii) AT = PN (b) Given: ∠T = ∠N Using SAS criterion, ∆ART ≅ ∆PEN (i) RT = EN (ii) PN = AT (c) Given: AT = PN Using ASA criterion, ∆ART ≅ ∆PEN (i) ∠RAT = ∠EPN (ii) ∠RTA = ∠ENP Class 7 Maths Chapter 7 Exercise 7.2 for more answers vist to:Read more
(a) Using SSS criterion, ∆ART ≅ ∆PEN
(i) AR = PE (ii) RT = EN (iii) AT = PN
(b) Given: ∠T = ∠N
Using SAS criterion, ∆ART ≅ ∆PEN
(i) RT = EN (ii) PN = AT
(c) Given: AT = PN
Using ASA criterion, ∆ART ≅ ∆PEN
(i) ∠RAT = ∠EPN (ii) ∠RTA = ∠ENP
(a) By SSS congruence criterion, since it is given that AC = DF, AB = DE, BC = EF The three sides of one triangle are equal to the three corresponding sides of another triangle. Therefore, ∆ABC ≅ ∆DEF (b) By SAS congruence criterion, since it is given that RP = ZX, RQ = ZY and ∠PRQ = ∠XZY The two siRead more
(a) By SSS congruence criterion,
since it is given that AC = DF, AB = DE, BC = EF
The three sides of one triangle are equal to the three corresponding sides of
another triangle.
Therefore, ∆ABC ≅ ∆DEF
(b) By SAS congruence criterion,
since it is given that RP = ZX, RQ = ZY and ∠PRQ = ∠XZY
The two sides and one angle in one of the triangle are equal to the corresponding
sides and the angle of other triangle.
Therefore, ∠PRQ ≅ ∠XyY
(c) By ASA congruence criterion,
since it is given that ∠MLN = ∠FGH, ∠NML = ∠HFG, ML = FG.
The two angles and one side in one of the triangle are equal to the corresponding
angles and side of other triangle.
Therefore, ∆LMN ≅ ∆GFH
(d) By RHS congruence criterion,
since it is given that EB = BD, AE = CB, ∠A = ∠C = 90°
Hypotenuse and one side of a right angled triangle are respectively equal to the
hypotenuse and one side of another right angled triangle.
Therefore, ∆ABE ≅ ∆CDB
In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ∆RAT ≅ ?
In the figure, given two triangles are congruent. So, the corresponding parts are: A ↔ O, R ↔ W, T ↔ N. We can write, ∆RAT ≅ ∆WON [By SAS congruence rule] Class 7 Maths Chapter 7 Exercise 7.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-7/
In the figure, given two triangles are congruent. So, the corresponding parts are:
A ↔ O, R ↔ W, T ↔ N.
We can write, ∆RAT ≅ ∆WON [By SAS congruence rule]
Class 7 Maths Chapter 7 Exercise 7.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-7/
In ∆ABC, ∠A = 30° , ∠B = 40° and ∠C = 110°. In ∆ PQR, ∠P = 30° , ∠Q = 40° and ∠R = 110° . A student says that ∆ABC ≅ ∆PQR by AAA congruence criterion. Is he justified? Why or why not?
No, because the two triangles with equal corresponding angles need not be congruent. In such a correspondence, one of them can be an enlarged copy of the other. Class 7 Maths Chapter 7 Exercise 7.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
No, because the two triangles with equal corresponding angles need not be congruent. In
such a correspondence, one of them can be an enlarged copy of the other.
Class 7 Maths Chapter 7 Exercise 7.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
You have to show that ∆AMP ≅ ∆AMQ. In the following proof, supply the missing reasons: (i) PM = QM (ii) ∠PMA = ∠QMA (iii) AM = AM (iv) ∆AMP ≅ ∆AMQ
(i) PM = QM (i) Given (ii) ∠PMA = ∠QMA (ii) Given (iii) AM = AM (iii) Common (iv) ∆AMP ≅ ∆AMQ (iv) SAS congruence rule Class 7 Maths Chapter 7 Exercise 7.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
(i) PM = QM (i) Given
(ii) ∠PMA = ∠QMA (ii) Given
(iii) AM = AM (iii) Common
(iv) ∆AMP ≅ ∆AMQ (iv) SAS congruence rule
Class 7 Maths Chapter 7 Exercise 7.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
You want to show that ∆ART ≅ ∆PEN: (a) If you have to use SSS criterion, then you need to show: (i) AR = (ii) RT = (iii) AT = (b) If it is given that ∠ T = ∠ N and you are to use SAS criterion, you need to have:(i) RT = and (ii) PN = (c) If it is given that AT = PN and you are to use ASA criterion, you need to have: (i) ? (ii) ?
(a) Using SSS criterion, ∆ART ≅ ∆PEN (i) AR = PE (ii) RT = EN (iii) AT = PN (b) Given: ∠T = ∠N Using SAS criterion, ∆ART ≅ ∆PEN (i) RT = EN (ii) PN = AT (c) Given: AT = PN Using ASA criterion, ∆ART ≅ ∆PEN (i) ∠RAT = ∠EPN (ii) ∠RTA = ∠ENP Class 7 Maths Chapter 7 Exercise 7.2 for more answers vist to:Read more
(a) Using SSS criterion, ∆ART ≅ ∆PEN
(i) AR = PE (ii) RT = EN (iii) AT = PN
(b) Given: ∠T = ∠N
Using SAS criterion, ∆ART ≅ ∆PEN
(i) RT = EN (ii) PN = AT
(c) Given: AT = PN
Using ASA criterion, ∆ART ≅ ∆PEN
(i) ∠RAT = ∠EPN (ii) ∠RTA = ∠ENP
Class 7 Maths Chapter 7 Exercise 7.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
Which congruence criterion do you use in the following? (a) Given: AC = DF, AB = DE, BC = EF So If ∆ ABC ≅ ∆ DEF (b) Given: RP = ZX, RQ = ZY, ∠PRQ = ∠XZY So ∆ PQR ≅ ∆ XYZ (c) Given: ∠MLN = ∠FGH, ∠NML = ∠HFG, ML = FG So ∆LMN ≅ ∆GFH (d) Given: EB = BD, AE = CB, ∠A = ∠C = 90° So ∆ABE ≅ ∆CDB
(a) By SSS congruence criterion, since it is given that AC = DF, AB = DE, BC = EF The three sides of one triangle are equal to the three corresponding sides of another triangle. Therefore, ∆ABC ≅ ∆DEF (b) By SAS congruence criterion, since it is given that RP = ZX, RQ = ZY and ∠PRQ = ∠XZY The two siRead more
(a) By SSS congruence criterion,
since it is given that AC = DF, AB = DE, BC = EF
The three sides of one triangle are equal to the three corresponding sides of
another triangle.
Therefore, ∆ABC ≅ ∆DEF
(b) By SAS congruence criterion,
since it is given that RP = ZX, RQ = ZY and ∠PRQ = ∠XZY
The two sides and one angle in one of the triangle are equal to the corresponding
sides and the angle of other triangle.
Therefore, ∠PRQ ≅ ∠XyY
(c) By ASA congruence criterion,
since it is given that ∠MLN = ∠FGH, ∠NML = ∠HFG, ML = FG.
The two angles and one side in one of the triangle are equal to the corresponding
angles and side of other triangle.
Therefore, ∆LMN ≅ ∆GFH
(d) By RHS congruence criterion,
since it is given that EB = BD, AE = CB, ∠A = ∠C = 90°
Hypotenuse and one side of a right angled triangle are respectively equal to the
hypotenuse and one side of another right angled triangle.
Therefore, ∆ABE ≅ ∆CDB
Class 7 Maths Chapter 7 Exercise 7.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/