Rs Aggarwal 2020 2021 for Class 7 Maths Chapter 16 - Congruence

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Rs Aggarwal 2020 2021 Solutions for Class 7 Maths Chapter 16 Congruence are provided here with simple step-by-step explanations. These solutions for Congruence are extremely popular among Class 7 students for Maths Congruence Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 7 Maths Chapter 16 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

Page No 199:

Answer:

$We have to state the correspondence between the vertices, sides and angles of the following pairs of congruent triangles . (i) △ A B C ≅ △ E F D Correspondence between vertices : A \leftrightarrow E, B \leftrightarrow F, C \leftrightarrow D Correspondence between sides : A B = E F, B C = F D, C A = D E Correspondence between angles : ∠ A = ∠ E, ∠ B = ∠ F, ∠ C = ∠ D (ii) △ C A B ≅ △ Q R P Correspondence between vertices : C \leftrightarrow Q, A \leftrightarrow R, B \leftrightarrow P Correspondence between sides : C A = Q R, A B = R P, B C = P Q Correspondence between angles : ∠ C = ∠ Q, ∠ A = ∠ R, ∠ B = ∠ P (iii) △ X Z Y ≅ △ Q P R Correspondence between vertices : X \leftrightarrow Q, Z \leftrightarrow P, Y \leftrightarrow R Correspondence between sides : X Z = Q P, Z Y = P R, Y X = R Q Correspondence between angles : ∠ X = ∠ Q, ∠ Z = ∠ P, ∠ Y = ∠ R (iv) △ M P N ≅ △ S Q R Correspondence between vertices : M \leftrightarrow S, P \leftrightarrow Q, N \leftrightarrow R Correspondence between sides : M P = S Q, P N = Q R, N M = R S Correspondence between angles : ∠ M = ∠ S, ∠ P = ∠ Q, ∠ N = ∠ R$

Page No 200:

Answer:

$(i) △ A C B ≅ △ D E F (SAS congruence property) (i i) △ R P Q ≅ △ L N M (RHS congruence property) (i i i) △ Y X Z ≅ △ T R S (SSS congruence property) (i v) △ D E F ≅ △ P N M (ASA congruence property) (v) △ A C B ≅ △ A C D (ASA congruence property)$

Page No 200:

Answer:

$Given : P L ⊥ O A P M ⊥ O B P L = P M To prove : △ P L O ≅ △ P M O Proof : I n △ P L O and △ P M O : ∠ P L O = ∠ P M O (90 ° each) P O = P O (common) P L = P M (given) B y RHS congruence property : △ P L O ≅ △ P M O$

Page No 200:

Answer:

$Given : A D = B C A D ∥ B C We have to show that A B = D C . Proof : A D ∥ B C ∴ ∠ B C A = ∠ D A C (alternate angles) In △ A B C and △ C D A : B C = D A (given) ∠ BCA = ∠ DAC (proved above) AC = AC (common) B y SAS c ongruence property : △ A B C ≅ △ C D A = > A B = C D (corresponding parts of the congruent triangles)$

Page No 200:

Answer:

$Given : A B = A C, B D = D C To prove : △ A D B ≅ △ A D C Proof : (i) I n △ A D B and △ A D C : A B = A C (given) BD = DC (given) DA = DA (common) B y SSS congruence property : △ A D B ≅ △ A D C ∠ A D B = ∠ A D C (corresponding parts of the congruent triangles) . . . (1) ∠ A D B and ∠ A D C are on the straight line . ∴ ∠ A D B + ∠ A D C = 180 ° ∠ A D B + ∠ A D B = 180 ° = > 2 ∠ A D B = 180 ° = > ∠ A D B = 90 ° F rom (1) : ∠ A D B = ∠ A D C = 90 ° (ii) ∠ B A D = ∠ C A D (corresponding parts of the congruent triangles)$

Page No 200:

Answer:

$Given : A D is a bisector of ∠ A . = > ∠ D A B = ∠ D A C . . . (1) A D ⊥ B C = > ∠ B D A = ∠ C D A (90 ° each) To prove : △ A B C is isosceles . Proof : In △ D A B and △ D A C : ∠ B D A = ∠ C D A (90 ° each) D A = D A (common) ∠ DAB = ∠ DAC (from 1) B y ASA congruence property : △ D A B ≅ △ D A C = > A B = A C (corresponding parts of the congruent triangles) Therefore, △ A B C is isosceles .$

Page No 201:

Answer:

$Given : A B = A D C B = C D To prove : △ A B C ≅ △ A D C Proof : In △ A B C and △ A D C : A B = A D (given) BC = DC (given) AC = AC (common) ∴ △ A B C ≅ △ A D C (b y SSS congruence property)$

Page No 201:

Answer:

$Given : P A ⊥ A B Q B ⊥ A B P A = Q B To prove : △ O A P ≅ △ O B Q Find whether O A = O B . Proof : In △ O A P and △ O B Q : ∠ P O A = ∠ Q O B (v ertically opposite angles) ∠ O A P = ∠ O B Q (90 ° each) P A = Q B (g iven) B y A A S congruence property : △ O A P ≅ △ O B Q = > O A = O B (corresponding parts of the congruent triangles)$

Page No 201:

Answer:

$Given : Triangles A B C a n d D C B are right angled at A and D, respectively . A C = D B To prove : △ A B C ≅ △ D C B In △ A B C and △ D C B : ∠ C A B = ∠ B D C (90 ° each) B C = B C (common) A C = D B (given) B y R . H . S . congruence property : △ A B C ≅ △ D C B$

Page No 201:

Answer:

$G i v e n : △ A B C is an isosceles triangle in which A B = A C . E and F are midpoints of A C and A B, respectively . To prove : B E = C F Proof : E and F are midpoints of A C and A B, respectively . = > A F = F B, A E = E C A B = A C = > \frac{1}{2} A B = \frac{1}{2} A C = > F B = E C ∠ A B C = ∠ A C B (angle opposite to equal sides are equal) = > ∠ F B C = ∠ E C B Consider △ B C F and △ C B E : B C = B C (common) ∠ F B C = ∠ E C B (proved above) F B = E C (proved above) B y S A S congruence property : △ B C F ≅ △ C B E B E = C F (corresponding parts of the congruent triangles)$

Page No 201:

Answer:

$Given : A B = A C △ A B C is an isosceles triangle . A P = A Q To prove : B Q = C P Proof : A B = A C (given) A P = A Q (given) A B - A P = A C - A Q = > B P = C Q ∠ A B C = ∠ A C B (angle opposite to the equal sides are equal) = > ∠ P B C = ∠ Q C B I n △ P B C and △ Q C B : P B = Q C (proved above) ∠ P B C = ∠ Q C B (proved above) B C = B C (common) B y S A S congruence property : △ P B C ≅ △ Q C B B Q = C P (corresponding parts of the congruent triangles)$

Page No 201:

Answer:

$Given : A B C is an isosceles triangle . A B = A C B D = C E To prove : B E = C D Proof : A B + B D = A C + C E (A s, A B = A C, B D = C E) = > A D = A E Consider △ A C D and △ A B E : A C = A B (given) ∠ C A D = ∠ B A E (common) A D = A E (proved above) B y S A S congruence property : △ A C D ≅ △ A B E = > C D = B E (corresponding parts of the congruent triangles)$

Page No 202:

Answer:

. $Given : △ A B C is an isosceles triangle . A B = A C B D = C D To prove : A D bisects ∠ A and ∠ D . Proof : Consider △ A B D and △ A C D : A B = A C (given) BD = CD (given) AD = AD (common) B y S S S congruence property : △ A B D ≅ △ A C D = > ∠ B A D = ∠ C A D (by cpct) = > ∠ B D A = ∠ C D A (by cpct)$

Page No 202:

Answer:

No, its not necessary. If the corresponding angles of two triangles are equal, then they may or may not be congruent.
They may have proportional sides as shown in the following figure:

Page No 202:

Answer:

No, two triangles are not congruent if their two corresponding sides and one angle are equal. They will be congruent only if the said angle is the included angle between the sides.

Page No 202:

Answer:

Both triangles have equal area due to the the same product of height and base. But they are not congruent.

Page No 202:

Answer:

(i) the same length

(ii) the same measure

(iii)the same side length

(iv) the same radius

(v) the same length and the same breadth

(vi) equal parts

Page No 202:

Answer:

(i) False
This is because they can be equal only if they have equal sides.

(ii) True
This is because if squares have equal areas, then their sides must be of equal length.

(iii) False
For example, if a triangle and a square have equal area, they cannot be congruent.

(iv) False
For example, an isosceles triangle and an equilateral triangle having equal area cannot be congruent.

(v) False
They can be congruent if two sides and the included angle of a triangle are equal to the corresponding two sides and the included corresponding angle of another triangle.

(vi) True
This is because of the AAS criterion of congruency.

(vii) False
Their sides are not necessarily equal.

(viii) True
This is because of the AAS criterion of congruency.

(ix) False
This is because two right triangles are congruent if the hypotenuse and one side of the first triangle are respectively equal to the hypotenuse and the corresponding side of the second triangle.

(x) True

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