RD Sharma solution class 7 chapter 16 Congruence Exercise 16.4

Exercise 16.4

Page-16.19

Question 1:

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

 

Answer 1:

1)  We have

Since ∠ABO = ∠CDO = 45° and both are alternate angles,  ABDC∠BAO = ∠DCO  (alternate angle , ABCD 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)
   In∆ABC,∠A+∠B+∠C=180°(Angle sum property)∠C=180°-∠A-∠B∠C=180°-30°-90°=60°In∆PQR,∠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 BD = DC?


Answer 2:

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

Page-16.20

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 PQRABCPQR has PQR =60° and PRQ = 70°Also we construct PQR such that BC =QRTherefore by ASA the two triangles are congruent

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 AB = AC?

Answer 4:



(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.

Answer 5:

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

Question 6:

In Fig. 39, AO = OB and A = ∠B.


(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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