RD Sharma Solutions - Ex-16.4, Congruence, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions for Class 7 Mathematics

Created by: Abhishek Kapoor

Class 7 : RD Sharma Solutions - Ex-16.4, Congruence, Class 7, Math Class 7 Notes | EduRev

The document RD Sharma Solutions - Ex-16.4, Congruence, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
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Question 1:

Which of the following pairs of triangles are congruent by ASA condition?

 

RD Sharma Solutions - Ex-16.4, Congruence, Class 7, Math Class 7 Notes | EduRev           RD Sharma Solutions - Ex-16.4, Congruence, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions - Ex-16.4, Congruence, Class 7, Math Class 7 Notes | EduRev                                 RD Sharma Solutions - Ex-16.4, Congruence, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions - Ex-16.4, Congruence, Class 7, Math Class 7 Notes | EduRev

 

Answer 1:

1)  We have

Since ∠ABO = ∠CDO = 45° and both are alternate angles,  
AB∥DC
∠BAO = ∠DCO  (alternate angle , AB∥CD and AC is a transversal line)
∠ABO = ∠CDO = 45° (given in the figure) 
Also, AB = DC    (Given in the figure)
Therefore, by ASA △AOB ≅△DOC
 

2)  
In △ABC ,
Now AB = AC (Given)
∠ABD = ∠ACD = 40° (Angles opposite to equal sides)
∠ABD +∠ACD+∠BAC=180° (Angle sum property)
40°+40°+∠BAC=180°
∠BAC=180°-80°=100°
∠BAD +∠DAC=∠BAC
∠BAD=∠BAC-∠DAC=100°-50°=50°
∠BAD =∠CAD = 50°
Therefore, by ASA, △ABD ≅△ADC

3)
  InΔABC,
∠A+∠B+∠C=180°(Angle sum property)
∠C=180°-∠A-∠B
∠C=180°-30°-90°=60°
InΔPQR,
∠P+∠Q+∠R=180°(Angle sum property)
∠P=180°-∠Q-∠R
∠P=180°-60°-90°=30°
∠BAC = ∠QPR = 30°
∠BCA=∠PRQ = 60°
and AC = PR (Given)
Therefore, by ASA, △ABC ≅△PQR

4)  
We have only BC =QR  but none of the angles of △ABC AND △PQR are equal.
Therefore, △ABC≆ △PRQ

 

Question 2:

In Fig. 37, AD bisects ∠A and AD ⊥ BC.
 (i) Is ∆ ADB ≅ ∆ ADC?
 (ii) State the three pairs of matching parts you have used in (i).
 (iii) Is it true to say that BDDC?

RD Sharma Solutions - Ex-16.4, Congruence, Class 7, Math Class 7 Notes | EduRev

 

Answer 2:

(i) Yes, △ADB ≅△ADC, by ASA criterion of congruency 
(ii) We have used  ∠BAD =∠CAD
∠ADB=∠ADC = 90° since AD⊥BC
and AD = DA
(iii) Yes, BD= DC since, △ADB ≅△ADC

 

Question 3:

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

Answer 3:

We have drawn
△ABC with ∠ABC = 60°and ∠ACB = 70°
We now construct △PQR≅△ABC
△PQR has ∠PQR =60° and ∠PRQ = 70°
Also we construct △PQR such that BC =QR
Therefore by ASA the two triangles are congruent

RD Sharma Solutions - Ex-16.4, Congruence, Class 7, Math Class 7 Notes | EduRev              RD Sharma Solutions - Ex-16.4, Congruence, Class 7, Math Class 7 Notes | EduRev

 

Question 4:

In ∆ ABC, it is known that ∠B = ∠C. Imagine you have another copy of ∆ ABC
 (i) Is ∆ ABC ≅ ∆ ACB?
 (ii) State the three pairs of matching parts you have used to answer (i).
 (iii) Is it true to say that ABAC?

Answer 4:

RD Sharma Solutions - Ex-16.4, Congruence, Class 7, Math Class 7 Notes | EduRev

(i) Yes△ABC ≅△ACB
(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

 

Question 5:

In Fig. 38, AX bisects ∠BAC as well as ∠BDC. State the three facts needed to ensure that ∆ ABD ≅ ∆ ACD.

RD Sharma Solutions - Ex-16.4, Congruence, Class 7, Math Class 7 Notes | EduRev

Answer 5:

As per the given conditions,∠CAD=∠BAD and ∠CDA=∠BDA (because AX bisects ∠BAC ) AD=DA (common)
Therefore, by ASA, △ACD≅△ABD


Question 6:

In Fig. 39, AOOB and ∠A = ∠B.

RD Sharma Solutions - Ex-16.4, Congruence, Class 7, Math Class 7 Notes | EduRev

(i) Is ∆ AOC ≅ ∆ BOD?
 (ii) State the matching pair you have used, which is not given in the question.
 (iii) Is it true to say that ∠ACO = ∠BDO?

Answer 6:

We have
∠OAC =∠OBD, AO = OBAlso, ∠AOC = ∠BOD (Opposite angles on same vertex)        

 Therefore, by ASA △AOC ≅△BOD

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