Given: ∠A = ∠F, BC = ED, ∠B = ∠E In ∆ABC and ∆FED, ∠B = ∠E = 90 ∠A = ∠F BC = ED Therefore, ∆ABC ≅ ∆FED [By RHS 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/
Given: ∠A = ∠F, BC = ED, ∠B = ∠E
In ∆ABC and ∆FED,
∠B = ∠E = 90
∠A = ∠F
BC = ED
Therefore, ∆ABC ≅ ∆FED [By RHS congruence rule]
∆ABC and ∆PQR are congruent. Then one additional pair is BC = QR. Given: ∠B = ∠Q = 90 ∠C = ∠R BC - QR Therefore, ∆ABC ≅ ∆PQR [By ASA 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/
∆ABC and ∆PQR are congruent. Then one additional pair is BC = QR.
Given: ∠B = ∠Q = 90
∠C = ∠R
BC – QR
Therefore, ∆ABC ≅ ∆PQR [By ASA congruence rule]
Let us draw two triangles PQR and ABC. All angles are equal, two sides are equal except one side. Hence, ∆PQR are not congruent to ∆ABC. Class 7 Maths Chapter 7 Exercise 7.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-7/
Let us draw two triangles PQR and ABC.
All angles are equal, two sides are equal except one side. Hence, ∆PQR are not congruent to ∆ABC.
In a squared sheet, draw ∆ABC and ∆ PQR. When two triangles have equal areas and (i) these triangles are congruent, i.e., ∆ABC ≅ ∆PQR [By SSS congruence rule] Then, their perimeters are same because length of sides of first triangle are equal to the length of sides of another triangle by SSS congrueRead more
In a squared sheet, draw ∆ABC and ∆ PQR.
When two triangles have equal areas and
(i) these triangles are congruent, i.e., ∆ABC ≅ ∆PQR [By SSS congruence rule]
Then, their perimeters are same because length of sides of first triangle are
equal to the length of sides of another triangle by SSS congruence rule.
(ii) But, if the triangles are not congruent, then their perimeters are not same
because lengths of sides of first triangle are not equal to the length of
corresponding sides of another triangle.
In ∆BAT and ∆BAC, given triangles are congruent so the corresponding parts are: B ↔ B, A ↔ A, T ↔ C Thus, ∆BCA ≅ ∆BTA [By SSS congruence rule] In ∆QRS and ∆TPQ, given triangles are congruent so the corresponding parts are: P ↔ R, T ↔ Q, Q ↔ S Thus, ∆QRS ≅ ∆TPQ [By SSS congruence rule] Class 7 MathsRead more
In ∆BAT and ∆BAC, given triangles are congruent so the corresponding parts are:
B ↔ B, A ↔ A, T ↔ C
Thus, ∆BCA ≅ ∆BTA [By SSS congruence rule]
In ∆QRS and ∆TPQ, given triangles are congruent so the corresponding parts are:
P ↔ R, T ↔ Q, Q ↔ S
Thus, ∆QRS ≅ ∆TPQ [By SSS congruence rule]
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
Explain, why ∆ABC ≅ ∆FED.
Given: ∠A = ∠F, BC = ED, ∠B = ∠E In ∆ABC and ∆FED, ∠B = ∠E = 90 ∠A = ∠F BC = ED Therefore, ∆ABC ≅ ∆FED [By RHS 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/
Given: ∠A = ∠F, BC = ED, ∠B = ∠E
In ∆ABC and ∆FED,
∠B = ∠E = 90
∠A = ∠F
BC = ED
Therefore, ∆ABC ≅ ∆FED [By RHS 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/
If ∆ABC and ∆PQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?
∆ABC and ∆PQR are congruent. Then one additional pair is BC = QR. Given: ∠B = ∠Q = 90 ∠C = ∠R BC - QR Therefore, ∆ABC ≅ ∆PQR [By ASA 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/
∆ABC and ∆PQR are congruent. Then one additional pair is BC = QR.
Given: ∠B = ∠Q = 90
∠C = ∠R
BC – QR
Therefore, ∆ABC ≅ ∆PQR [By ASA 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/
Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.
Let us draw two triangles PQR and ABC. All angles are equal, two sides are equal except one side. Hence, ∆PQR are not congruent to ∆ABC. Class 7 Maths Chapter 7 Exercise 7.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-7/
Let us draw two triangles PQR and ABC.
All angles are equal, two sides are equal except one side. Hence, ∆PQR are not congruent to ∆ABC.
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 a squared sheet, draw two triangles of equal area such that: (i) the triangles are congruent. (ii) the triangles are not congruent. What can you say about their perimeters?
In a squared sheet, draw ∆ABC and ∆ PQR. When two triangles have equal areas and (i) these triangles are congruent, i.e., ∆ABC ≅ ∆PQR [By SSS congruence rule] Then, their perimeters are same because length of sides of first triangle are equal to the length of sides of another triangle by SSS congrueRead more
In a squared sheet, draw ∆ABC and ∆ PQR.
When two triangles have equal areas and
(i) these triangles are congruent, i.e., ∆ABC ≅ ∆PQR [By SSS congruence rule]
Then, their perimeters are same because length of sides of first triangle are
equal to the length of sides of another triangle by SSS congruence rule.
(ii) But, if the triangles are not congruent, then their perimeters are not same
because lengths of sides of first triangle are not equal to the length of
corresponding sides of another triangle.
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/
Complete the congruence statement:
In ∆BAT and ∆BAC, given triangles are congruent so the corresponding parts are: B ↔ B, A ↔ A, T ↔ C Thus, ∆BCA ≅ ∆BTA [By SSS congruence rule] In ∆QRS and ∆TPQ, given triangles are congruent so the corresponding parts are: P ↔ R, T ↔ Q, Q ↔ S Thus, ∆QRS ≅ ∆TPQ [By SSS congruence rule] Class 7 MathsRead more
In ∆BAT and ∆BAC, given triangles are congruent so the corresponding parts are:
B ↔ B, A ↔ A, T ↔ C
Thus, ∆BCA ≅ ∆BTA [By SSS congruence rule]
In ∆QRS and ∆TPQ, given triangles are congruent so the corresponding parts are:
P ↔ R, T ↔ Q, Q ↔ S
Thus, ∆QRS ≅ ∆TPQ [By SSS 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 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/