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]
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/