RD Sharma Solutions - Ex-16.3, 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-16.3, Congruence, Class 7, Math Class 7 Notes | EduRev

The document RD Sharma Solutions - Ex-16.3, Congruence, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
All you need of Class 7 at this link: Class 7

Question 1:

By applying SAS congruence condition, state which of the following pairs (Fig. 28) of triangles are congruent. State the result in symbolic form

RD Sharma Solutions - Ex-16.3, Congruence, Class 7, Math Class 7 Notes | EduRev     RD Sharma Solutions - Ex-16.3, Congruence, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions - Ex-16.3, Congruence, Class 7, Math Class 7 Notes | EduRev        RD Sharma Solutions - Ex-16.3, Congruence, Class 7, Math Class 7 Notes | EduRev

 

Answer 1:

1) We have OA = OC and OB = OD and∠AOB =∠COD which are vertically opposite angles.
Therefore by SAS condition, △AOC ≅ △BOD.

2) We have BD = DC
∠ADB =∠ADC = 90°
and AD = AD
Therefore by SAS condition, △ADB ≅ △ADC.

3)   We have AB = DC
∠∠ABD = ∠CDB and BD = DB
Therefore by SAS condition, △ABD ≅ △CBD.

4)  We have BC = QR
∠ABC =∠PQR = 90°
and AB = PQ
Therefore by SAS condition, △ABC ≅ △PQR.

 

Question 2:

State the condition by which the following pairs of triangles are congruent.

 

RD Sharma Solutions - Ex-16.3, Congruence, Class 7, Math Class 7 Notes | EduRev   RD Sharma Solutions - Ex-16.3, Congruence, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions - Ex-16.3, Congruence, Class 7, Math Class 7 Notes | EduRevRD Sharma Solutions - Ex-16.3, Congruence, Class 7, Math Class 7 Notes | EduRev

 

Answer 2:

1)  AB = AD
BC = CD
and AC = CA
Therefore by SSS condition, △ABC≅△ADC.

2) AC = BD
AD = BC and AB = BA
Therefore by SSS condition, △ABD≅△BAC.

3)  AB = AD
∠BAC=∠DAC∠BAC=∠DAC
and AC = AC
Therefore by SAS condition, △BAC≅△DAC.

4)  AD = BC
∠ DAC = ∠BCA
and AC = CA
Therefore by SAS condition, △ABC≅△ADC.

 

Question 3:

In Fig. 30, line segments AB and CD bisect each other at O. Which of the following statements is true?
  (i) ∆ AOC ≅ ∆ DOB
  (ii) ∆ AOC ≅ ∆ BOD
  (iii) ∆ AOC ≅ ∆ ODB.
  State the three pairs of matching parts, yut have used to arive at the answer.

RD Sharma Solutions - Ex-16.3, Congruence, Class 7, Math Class 7 Notes | EduRev

Answer 3:

We have AO = OB.
And CO = OD

Also ∠AOC = ∠BOD
Therefore by SAS condition, △AOC ≅ △BOD.

Therefore, statement (ii) is true.

 

Question 4:

Line-segments AB and CD bisect each other at OAC and BD are joined forming triangles AOC and BOD. State the three equality relations between the parts of the two triangles, that are given or otherwise known. Are the two triangles congruent? State in symbolic form. Which congruence condition do you use?

Answer 4:

RD Sharma Solutions - Ex-16.3, Congruence, Class 7, Math Class 7 Notes | EduRev

We have AO = OB and CO = OD since AB and CD bisect each other at O.

Also ∠AOC = ∠BOD since they are opposite angles on the same vertex.
Therefore by SAS congruence condition,

△AOC≅△BOD

 

Question 5:

∆ ABC is isosceles with ABAC. Line segment AD bisects ∠A and meets the base BC in D.
  (i) Is ∆ ADB ≅ ∆ ADC?
  (ii) State the three pairs of matching parts used to answer (i).
  (iii) Is it true to say that BDDC?

Answer 5:

RD Sharma Solutions - Ex-16.3, Congruence, Class 7, Math Class 7 Notes | EduRev

(i) We have AB = AC (given)
∠BAD =∠CAD∠BAD =∠CAD (AD bisects ∠BAC)
and AD = AD (common)
Therefore by SAS condition of congruence, △ABD≅△ACD

(ii) We have used AB, AC; ∠BAD =∠CAD;  AD, DA.

(iii) Now△ABD≅△ACD  therefore by c.p.c.t BD = DC.

 

Question 6:

In Fig. 31, ABAD and ∠BAC = ∠DAC.
 (i) State in symbolic form the congruence of two triangles ABC and ADC that is true.
 (ii) Complete each of the following, so as to make it true:
 (a) ∠ABC = ........
 (b) ∠ACD = ........
 (c) Line segment AC bisects ..... and .....

RD Sharma Solutions - Ex-16.3, Congruence, Class 7, Math Class 7 Notes | EduRev

Answer 6:

i) AB = AD (given)
∠BAC =∠DAC  (given)
AC = CA (common)
Therefore by SAS conditionof congruency, △ABC ≅△ADC

ii) ∠ABC =∠ADC (c.p.c.t)
∠ACD =∠ACB (c.p.c.t)

 

Question 7:

In Fig. 32, AB || DC and ABDC.
 (i) Is ∆ ACD ≅ ∆ CAB?
 (ii) State the three pairs of matching parts used to answer (i).
 (iii) Which angle is equal to ∠CAD?
 (iv) Does it follow from (iii) that AD || BC?

RD Sharma Solutions - Ex-16.3, Congruence, Class 7, Math Class 7 Notes | EduRev

Answer 7:

(i) Yes △ACD≅△CAB by SAS condition of congruency.
(ii) We have used AB = DC, AC = CA and ∠DCA=∠BAC.
(iii) ∠CAD =∠ACB  since the two triangles are congruent.
(iv) Yes, this follows from AD∥∥BC as alternate angles are equal.If alternate angles are equal the lines are parallel.

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