RD Sharma Solutions For Class 7 Maths Exercise 16.4 Chapter 16 Congruence

Our expert team has designed the solutions for RD Sharma Solutions for Class 7 Maths Chapter 16 to help students prepare for their exams at ease.

RD Sharma Solutions for Class 7 is one of the best reference materials for CBSE students. Learners can download the pdf from the links provided below. Experts suggest students practice the solutions many numbers of times to yield good results in their exams. In Exercise 16.4 of Chapter 16 Congruence, we shall discuss the ASA congruence condition for two congruent triangles and their properties.

Download the PDF of RD Sharma Solutions For Class 7 Chapter 16 – Congruence Exercise 16.4

 

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rd sharma class 7 maths solution ch 16 ex 4
rd sharma class 7 maths solution ch 16 ex 4
rd sharma class 7 maths solution ch 16 ex 4
rd sharma class 7 maths solution ch 16 ex 4

 

Access answers to Maths RD Sharma Solutions For Class 7 Chapter 16 – Congruence Exercise 16.4

1. Which of the following pairs of triangle are congruent by ASA condition?


RD Sharma Solutions for Class 7 Maths Chapter 16 Congruence Image 22RD Sharma Solutions for Class 7 Maths Chapter 16 Congruence Image 23

RD Sharma Solutions for Class 7 Maths Chapter 16 Congruence Image 24

RD Sharma Solutions for Class 7 Maths Chapter 16 Congruence Image 25

Solution:

(i) We have,

Since ∠ABO = ∠CDO = 45o and both are alternate angles, AB parallel to DC, ∠BAO = ∠DCO (alternate angle, AB parallel to CD and AC is a transversal line)

∠ABO = ∠CDO = 45o (given in the figure) Also,

AB = DC (Given in the figure)

Therefore, by ASA ΔAOB ≅ ΔDOC

(ii) In ABC,

Now AB =AC (Given)

∠ABD = ∠ACD = 40o (Angles opposite to equal sides)

∠ABD + ∠ACD + ∠BAC = 180o (Angle sum property)

40o + 40o + ∠BAC = 180o

∠BAC =180o – 80o =100o

∠BAD + ∠DAC = ∠BAC

∠BAD = ∠BAC – ∠DAC = 100o – 50o = 50o

∠BAD = ∠CAD = 50°

Therefore, by ASA, ΔABD ≅ ΔADC

(iii) In Δ ABC,

∠A + ∠B + ∠C = 180o (Angle sum property)

∠C = 180o– ∠A – ∠B

∠C = 180o – 30o – 90o = 60o

In PQR,

∠P + ∠Q + ∠R = 180o (Angle sum property)

∠P = 180o – ∠Q – ∠R

∠P = 180o – 60o – 90o = 30o

∠BAC = ∠QPR = 30o

∠BCA = ∠PRQ = 60o and AC = PR (Given)

Therefore, by ASA, ΔABC ≅ ΔPQR

(iv) We have only

BC = QR but none of the angles of ΔABC and ΔPQR are equal.

Therefore, ΔABC and Cong ΔPRQ

2. In fig. 37, AD bisects A and AD and AD ⊥ BC.

(i) Is ΔADB ≅ ΔADC?

(ii) State the three pairs of matching parts you have used in (i)

(iii) Is it true to say that BD = DC?

RD Sharma Solutions for Class 7 Maths Chapter 16 Congruence Image 26

Solution:

(i) Yes, ΔADB≅ΔADC, by ASA criterion of congruency.

(ii) We have used ∠BAD = ∠CAD ∠ADB = ∠ADC = 90o

Since, AD ⊥ BC and AD = DA

(iii) Yes, BD = DC since, ΔADB ≅ ΔADC

 

3. Draw any triangle ABC. Use ASA condition to construct other triangle congruent to it.

Solution:

RD Sharma Solutions for Class 7 Maths Chapter 16 Congruence Image 27

We have drawn

Δ ABC with ∠ABC = 65o and ∠ACB = 70o

We now construct ΔPQR ≅ ΔABC has ∠PQR = 65o and ∠PRQ = 70o

Also we construct ΔPQR such that BC = QR

Therefore by ASA the two triangles are congruent

4. In Δ ABC, it is known that ∠B = C. Imagine you have another copy of Δ ABC

(i) Is ΔABC ≅ ΔACB

(ii) State the three pairs of matching parts you have used to answer (i).

(iii) Is it true to say that AB = AC?

Solution:

RD Sharma Solutions for Class 7 Maths Chapter 16 Congruence Image 28

(i) Yes ΔABC ≅ ΔACB

(ii) We have used ∠ABC = ∠ACB and ∠ACB = ∠ABC again.

Also BC = CB

(iii) Yes it is true to say that AB = AC since ∠ABC = ∠ACB.

5. In Fig. 38, AX bisects ∠BAC as well as ∠BDC. State the three facts needed to ensure that ΔACD ≅ ΔABD

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Solution:

As per the given conditions,

∠CAD = ∠BAD and ∠CDA = ∠BDA (because AX bisects ∠BAC)

AD = DA (common)

Therefore, by ASA, ΔACD ≅ ΔABD

6. In Fig. 39, AO = OB and ∠A = ∠B.

(i) Is ΔAOC ≅ ΔBOD

(ii) State the matching pair you have used, which is not given in the question.

(iii) Is it true to say that ∠ACO = ∠BDO?

RD Sharma Solutions for Class 7 Maths Chapter 16 Congruence Image 30

Solution:

We have

∠OAC = ∠OBD,

AO = OB

Also, ∠AOC = ∠BOD (Opposite angles on same vertex)

Therefore, by ASA ΔAOC ≅ ΔBOD


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