**Rd Sharma Solutions Class 7 Chapter 16 Congruence**

Welcome to NCTB Solution. Here with this post we are going to help 7th class students for the Solutions of Rd Sharma Class 7 Mathematics, Chapter 16, Congruence. Here students can easily find Exercise wise solution for chapter 16, Congruence. Students will find proper solutions for Exercise 16.1, 16.2, 16.3, 16.4 and 16.5. Our teachers solved every problem with easily understandable methods so that every students can understand easily. Here all solutions are based on the CBSE latest curriculum.

**Congruence Exercise 16.1 Solution**

**Question no – (1)**

**Solution : **

**Congruent** : The word congruent means same shape and size, that is equal in every respect. Thus if two figure have exactly the same shape and size they are said to be congruent.

**Question no – (2) **

**Solution : **

**(i)** Two line segments are congruent if ‘they are of equal lengths’.

**(ii)** Two angles are congruent if ‘they measures are equal’.

**(iii)** Two squares are congruent if ‘the same side length’.

**(iv)** Two rectangles are congruent if ‘the dimensions are same’.

**(v)** Two circles are congruent if ‘they have Same radius’.

**Question no – (3)**

**Solution : **

Yes, we can say that ∠POR ≅ ∠QOS

**Question no – (4)**

**Solution : **

Angle which is congruent to ∠AOC is,

∠AOC is congruent to ∠DOB

**Question no – (6)**

**Solution : **

∠AOB is congruent to ∠FYQ.

**Question no – (7)**

**Solution : **

**(i)** All squares are congruent – **(False).**

**(ii)** If two squares have equal areas, they are congruent – **(True).**

**(iii)** If two rectangles have equal area, they are congruent – **(False).**

**(iv)** If two triangles are equal in area, they are congruent – **(False).**

**Congruence Exercise 16.2 Solution**

**Question no – (1)**

**Solution : **

**Figure – (i)**

Since in △ABC and △DEF

AB = DE, BC = EF and AC = DF

Thus, △ABC ≅ DEF

**Figure – (ii)**

In triangle △ABC and △ ABD

AD = AC and BC = BD

Hence, △ABC ≅ △ ABD

**Figure – (iii)**

In △ABD and △CEF

AB = EF and AD = CF

Therefore, △ABC ≅ △CEF

**Figure – (iv)**

In △ABO and △COD

AB = CD, AO = OC and BO = OD

So, △ABO ≅ △COD

**Question no – (2)**

**Solution : **

**(i)** Is △ABD ≅ △ CBD?

Since AD = DC

and AB = BC

So, △ ABD ≅ △ BDC.

**(ii)** The three pairs are,

AD = CP, BD = BD and AB = BC

**Question no – (3)**

**Solution :**

**(i)** Is △ ABC ≅ △CDA?

AB = DC

and BC = AD

Therefore, △ABC ≅ △CDA

**(ii)** What congruence condition have you used?

AB = DC,

BC = AD

and AC = AC

**(iii)** You have used some fact, not given in the question, what is that?

= AC = CA

**Question no – (4)**

**Solution :**

**(i)** Which side of △ POQ equals ED

= ED = PR

**(ii)** Which angle of △ POQ equals ∠E

= ∠ED = ∠QPR

**Question no – (5)**

**Solution :**

In △ ABC and △PR

FB = AC and PQ = PR

AB = PQ and BC = QR

Yes, △ ABC ≅ △ PQR

**Question no – (6)**

**Solution :**

AB = BD,

AC = DC

BC = BC

Therefore, △ABC ≅ △DBC

**Question no – (8)**

**Solution :**

As per the question,

△ADB and △ADC

**∴ **AB = AC

AD = AD

and BD = DC

Since, D is the mid-point of BC

Thus, △ ADB ≅ △ ADC

**Question no – (9)**

**Solution :**

In △ABC and △ACB

AB = AC

BC = CB

AC = AB

Therefore, △ABC ≅ △ACB

**Congruence Exercise 16.3 Solution**

**Question no – (1)**

**Solution :**

**Figure – (i)**

△AOB ≅ △DOC

**Figure – (ii)**

△ADC ≅ △APB

**Figure – (iii)**

△ABD ≅ △BDC

**Figure – (iv)**

△ABC ≅ △PQR

**Question no – (2)**

**Solution :**

**Figure – (i)**

AB = AD

BC = CD

AC = AC

**∴** SSS

**Figure – (ii)**

AC = BD

AD = BC

AB = AB

**∴** SSS

**Figure – (iii)**

AD = AB

∠AC = ∠AB

AC = AC

**∴** SAS

**Figure – (iv)**

AC = AC

∠DAC = ∠ACB

AD = BC

**∴** SAS

**Question no – (3)**

**Solution :**

According to the question,

AO = BO

CO = DO

∠AOC = ∠BOD

△AOC ≅ △BOD

Therefore, (ii) △AOC ≅ △BOD

The three pairs of matching parts are,

AO,BO; CO, DO; ∠AOC, ∠BOD

**Question no – (5)**

**Solution :**

In the figure,

AB = AC

AD = AC

and ∠BAD = ∠CAD

Hence, △ABD ≅ △ADC

**Question no – (6)**

**Solution :**

**(i)** So, △ABC ≅ △ADC

**(ii)** (a) ∠ABC = ∠ADC

**(b)** ∠ACD = ∠ACB

**(c)** Line segment AC bisects ∠BAC and ∠BCD

**Question no – (7)**

**Solution :**

**(i)** In triangles △ACD and △CAB

∠BCA = ∠CAD

AC = AC

DC = AB

Thus, △ACD ≅ △CAB

**(ii)** The three pairs of matching parts,

AC, CA; DC, BA; ∠DCA, ∠BAC.

**(iii)** ∠CAD is equal to ∠ACB

**(iv)** Yes since ∠BAC = ∠ACD

**Congruence Exercise 16.4 Solution**

**Question no – (1)**

**Solution : **

**Figure – (i)**

In △ABD and △CDO

∠B = ∠D

∠BOA = ∠COD

AB = DC

So, △ABD ≅ △CDO

**Figure – (ii)**

∠C = 180 – (50 – 90)

= 40

**∴** In triangle ABD and ADC

∠B = ∠C

AD = AC

AB = AC

So, △ABD ≅ △ADC`

**Figure – (iii)**

∠D = 180 – (90 + 60)

= 30

∠C = 180 – 90 – 30

= 60

**∴** ∠A = ∠ D

AC = PR

∠C = ∠ R

**∴** △ABC ≅ △PQR

**Question no – (2)**

**Solution : **

In given triangle ABD and ADC

AD = AD

BD = D

Since D ids midpoint BC

∠BAD = ∠DAC

Therefore, △ADB ≅ △ADC

**(ii)** Matching parts are,

∠ABC, ∠ACB; ∠ACB, ∠ABC; BC, CD

**(ii)** Yes BD = DC since AD ⊥ BC

**Question no – (3)**

**Solution : **

∠A = ∠R

AC = PR

∠C = ∠P

**Question no – (4)**

**Solution :**

**(i)** ∠B = ∠C

**∴** AB = AC

AC = AB

Therefore, △ABC ≅ △ACB

**(ii)** Matching parts are,

∠ABC, ∠ACB; ∠ACB,∠ABC; BC,CD

**(iii)** Yes, it is true.

**Question no – (5)**

**Solution :**

In triangle,

ACD and ABD

∠CAD = ∠BAD

AD = AD

and ∠ACD ≅ △ABD

Therefore, △ACD ≅ △ABD

**Question no – (6)**

**Solution :**

**(i)** Yes, △AOC ≅ △BOD

**(ii)** Matching pair,

AO = OB

∠A = ∠B

∠AOC = ∠BOD

**(iii)** Yes, it is true.

**Congruence Exercise 16.5 Solution**

**Question no – (1)**

**Solution : **

**Figure – (i)**

△ABD ≅ △ABC

**Figure – (ii)**

△ABD ≅ △ACD

**Figure – (iii)**

△AOB ≅ △DOC

**Figure – (iv)**

△ABC ≅ △ADC

**Figure – (v)**

△ABD ≅ △CBD

**Question no – (2)**

**Solution : **

**(i)** Yes, △ABD ≅ △ACD

**(ii)** AB = AC

AD = AD

∠ADB = ∠ADC

**(iii)** Yes, it is true since AD ⊥ BC

**Question no – (3)**

**Solution :**

Yes, △ABD ≅ △ACD

BD is equal to CD

In △ADC ∠B = ∠C

**Question no – (5)**

**Solution :**

**(i)** Yes, △BCD ≅ △CBE

**(ii)** DD = CE

CB = CB

∠D = ∠E

Therefore, △BCD ≅ △CBE

**Next Chapter Solution :**

👉 Chapter 18 👈