RD Sharma Solutions - Ex-16.2, 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-16.2, Congruence, Class 7, Math Class 7 Notes | EduRev

The document RD Sharma Solutions - Ex-16.2, 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:

In the following pairs of triangles (Fig. 12 to 15), the lengths of the sides are indicated along sides. By applying SSS condition, determine which are congruent. State the result in symbolic form.

RD Sharma Solutions - Ex-16.2, Congruence, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions - Ex-16.2, Congruence, Class 7, Math Class 7 Notes | EduRev  RD Sharma Solutions - Ex-16.2, Congruence, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions - Ex-16.2, Congruence, Class 7, Math Class 7 Notes | EduRev      RD Sharma Solutions - Ex-16.2, Congruence, Class 7, Math Class 7 Notes | EduRev

Answer 1:

1)    In ΔABC and ΔDEF∆ABC and ∆DEF
       AB = DE = 4.5 cm (Side)
      BC = EF = 6 cm (Side)
      and AC = DF = 4 cm (Side)

Therefore, by SSS criterion of congruence, △ABC≅△DEF△ABC≅△DEF.

2) 
     In △ACB and △ADB
AC =AD (Side)
BC = BD (Side)
and AB=AB (Side)

Therefore, by SSS criterion of congruence, △ACB≅△ADB


3)   
In △ABD and △FEC, 
AB = FE (Side)
AD = FC  (Side)
BD = CE  (Side)

Therefore, by SSS criterion of congruence, △ABD≅△FEC△ABD≅△FEC.

 

Question 2:

In Fig. 16, ADDC and ABBC.
 (i) Is ∆ ABD ≅ ∆ CBD?
 (ii) State the three parts of matching pairs you have used to answer (i).

RD Sharma Solutions - Ex-16.2, Congruence, Class 7, Math Class 7 Notes | EduRev

Answer 2:

Yes △ABD≅△CBD△ABD≅△CBD by the SSS criterion.
We have used the three conditions in the SSS criterion as follows:
AD = DC
AB = BC
and DB = BD

 

Question 3:

In  Fig. 17, ABDC and BCAD.

RD Sharma Solutions - Ex-16.2, Congruence, Class 7, Math Class 7 Notes | EduRev

 

(i) Is ∆ ABC ≅ ∆ CDA?
 (ii) What congruence condition have you used?
 (iii) You have used some fact, not given in the question, what is that?

 

Answer 3:

We have AB = DC
BC = AD
and AC = AC
Therefore by SSS △ABC≅△CDA.

We have used Side Side Side congruence condition with one side common in both the triangles.

Yes, we have used the fact that AC = CA.

 

Question 4:

If ∆ PQR ≅ ∆ EFD,
 (i) Which side of ∆ PQR equals ED?
 (ii) Which angle of ∆ PQR equals ∠E?

Answer 4:


△PQR ≅ △EDF

1) Therefore PR = ED since the corresponding sides of congruent triangles are equal.

2) ∠QPR = ∠FED since the corresponding angles of congruent triangles are equal.

RD Sharma Solutions - Ex-16.2, Congruence, Class 7, Math Class 7 Notes | EduRev            RD Sharma Solutions - Ex-16.2, Congruence, Class 7, Math Class 7 Notes | EduRev

 

Question 5:

Triangles ABC and PQR are both isosceles with ABAC and PQPR respectively. If also, ABPQ and BCQR, are the two triangles congruent? Which condition do you use?
 If ∠B = 50°, what is the measure of ∠R?

Answer 5:

RD Sharma Solutions - Ex-16.2, Congruence, Class 7, Math Class 7 Notes | EduRevRD Sharma Solutions - Ex-16.2, Congruence, Class 7, Math Class 7 Notes | EduRev


We have AB = AC in isosceles △ABC
and PQ = PR in isosceles △PQR.
Also, we are given that AB = PQ and QR = BC.

Therefore, AC = PR (AB = AC, PQ = PR and AB = PQ)
Hence, △ABC≅△PQR.
Now
∠ABC = ∠PQR (Since triangles are congruent)
However, △PQR is isosceles.
Therefore, ∠PRQ =∠PQR =∠ABC = 50°

 

Question 6:

ABC and DBC are both isosceles triangles on a common base BC such that A and D lie on the same side of BC. Are triangles ADB and ADC congruent? Which condition do you use? If ∠BAC = 40° and∠BDC = 100°; then find ∠ADB.

Answer 6:

RD Sharma Solutions - Ex-16.2, Congruence, Class 7, Math Class 7 Notes | EduRev

YES  △ADB ≅△ADC (By SSS)
AB = AC , DB = DC AND AD= DA
∠BAD=∠CAD  (c.p.c.t)
∠BAD+∠CAD=40°
2∠BAD=40°
∠BAD=40°/2=20°

 

∠ABC+∠BCA+∠BAC=180°  (Angle sum property)
Since ΔABC is an isosceles triangle, 
∠ABC=∠BCA
∠ABC+∠ABC+40°=180°
2∠ABC=180°−40°=140°
∠ABC=140°/2=70°

∠DBC+∠BCD+∠BDC=180°  (Angle sum property)
Since ΔABC is an isosceles triangle, 
∠DBC=∠BCD
∠DBC+∠DBC+100°=180°
2∠DBC=180°−100°=80°
∠DBC=80°/2=40°

In ΔBAD, 
∠ABD+∠BAD+∠ADB=180°(Angle sum property)
30°+20°+∠ADB=180°  (∠ABD=∠ABC-∠DBC)
∠ADB=180°−20°−30°
∠ADB=130°

∠ADB =130°

 

Question 7:

∆ ABC and ∆ ABD are on a common base AB, and ACBD and BCAD as shown in Fig. 18. Which of the following statements is true?
 (i) ∆ ABC ≅ ∆ ABD
 (ii) ∆ ABC ≅ ∆ ADB
 (iii) ∆ ABC ≅ ∆ BAD

 

RD Sharma Solutions - Ex-16.2, Congruence, Class 7, Math Class 7 Notes | EduRev

Answer 7:

In △ABC and △BAD we have,
AC = BD (given)
BC = AD (given)
and AB = BA (common)
Therefore by SSS criterion of congruency, △ABC ≅≅△BAD.
There option (iii) is true.

 

Question 8:

In Fig. 19, ∆ ABC is isosceles with ABACD is the mid-point of base BC.
 (i) Is ∆ ADB ≅ ∆ ADC?
 (ii) State the three pairs of matching parts you use to arrive at your answer.

RD Sharma Solutions - Ex-16.2, Congruence, Class 7, Math Class 7 Notes | EduRev

Answer 8:

We have AB = AC.
Also since D is the midpoint of BC, BD = DC.
And AD = DA.
Therefore by SSS condition, △ABD ≅△ADC△ABD ≅△ADC.

We have used AB, AC : BD, DC and AD, DA.

 

Question 9:

In Fig. 20, ∆ ABC is isosceles with ABAC. State if ∆ ABC ≅ ∆ ACB. If yes, state three relations that you use to arrive at your answer.

RD Sharma Solutions - Ex-16.2, Congruence, Class 7, Math Class 7 Notes | EduRev

Answer 9:

Yes △ABC ≅△ACB  by SSS condition.
Since ABC is an isosceles triangle, AB = AC, BC = CB and AC = AB.

 

Question 10:

Triangles ABC and DBC have side BC common, ABBD and ACCD. Are the two triangles congruent? State in symbolic form. Which congruence condition do you use? Does ∠ABD equal ∠ACD? Why or why not?

Answer 10:

Yes.

In ΔABC and ΔDB
CAB=DB (Given)
AC=DC (Given)
BC=BC (Common) 
By SSS criterion of congruency, ΔABC≅ΔDBC    

No, ∠ABD and∠ACD are not equal
because AB ≠ AC.

RD Sharma Solutions - Ex-16.2, Congruence, Class 7, Math Class 7 Notes | EduRev  

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