Class 7: Maths Chapter 16 solutions. Complete Class 7 Maths Chapter 16 Notes.

Contents

## RS Aggarwal Solutions for Class 7 Maths Chapter 16–Congruence

RS Aggarwal 7th Maths Chapter 16, Class 7 Maths Chapter 16 solutions

**Question 1.****Solution:**

(i) ∆ABC ≅ ∆EFD, Then

A ↔ E, B ↔ F and C ↔ D

AB = EF, BC = FD and CA = DE

∠A = ∠E, ∠B = ∠F and ∠C = ∠D

(ii) ∆CAB ≅ ∆QRP

C ↔ Q, A ↔ R and B ↔ P

CA = QR, AB = RP and BC = PQ

∠C = ∠Q, ∠A = ∠R and ∠B = ∠P

(iii) ∆XZY ≅ ∆QPR

X ↔ Q, Z ↔ P, Y ↔ R

XZ = QP, ZY = PR and YX = RQ

∠X = ∠Q, ∠Z = ∠P and ∠Y = ∠R

(iv) ∆MPN ≅ ∆SQR

M ↔ S, P ↔ Q and N ↔ R

MP = SQ, PN = QR and NM = RS

∠M = ∠S, ∠P = ∠Q and ∠N = ∠R.

**Question 2.****Solution:**

(i) In fig (i)

In ∆ABC and ∆DEF

∠C = ∠E

CA = ED

CB = EF

∆ACB ≅ ∆DEF (SAS condition)

(ii) In fig (ii)

In ∆RPQ and ∆LNM

Side PQ = NM

Hyp. RQ = LM

∆RPQ ≅ ∆LNM (RHS condition)

(iii) In ∆YXZ and ∆TRS

XY = RT

∠X = SR and YZ = TS

∆YXZ ≅ ∆TRS (SSS condition)

(iv) In ∆DEF and ∆PNM

∠E = ∠N

∠F = ∠M

EF = NM

∆DEF ≅ ∆PNM (ASA condition)

(v) In ∆ABC and ∆ADC

AC = AC (common)

∠ CAB = ∠ CAD (each 50°)

∠ ACB = ∠ DCA (each 60°)

∆ABC ≅ ∆ADC (ASA condition)

**Question 3.****Solution:**

In fig,

PL ⊥ OA and PM ⊥ OB and PL = PM

Now in right ∆PLO and ∆PMO,

Side PL = PM (given)

Hypotenuse OP = OP (common)

∆PLO ≅ ∆PMO (RHS condition)

Yes ∆PLO ≅ ∆PMO

Hence proved.

**Question 4.****Solution:**

In the figure,

AD = BC and AD || BC.

In ∆ABC and ∆ADC,

AC = AC (common)

BC = AB (given)

∠ACB = ∠CAD (Alternate angles)

∆ABC ≅ ∆ADC (SAS condition)

AB = DC (c.p.c.t)

Hence proved.

**Question 5.****Solution:**

In ∆ABD and ∆ACD,

AD = AD (common)

AB = AC (given)

BD = CD (given)

∆ABD ≅ ∆ADC (SSS condition)

∠BAD = ∠CAD (c.p.c.t.)

and ∠ADB = ∠ADC (c.p.c.t.)

But ∠ADB + ∠ADC = 180° (Linear pair)

∠ADB = ∠ADC = 90°

Hence proved.

**Question 6.****Solution:**

given : In ∆ABC, AD is the bisector of ∠A i.e. ∠BAD = ∠CAD

AD ⊥ BC.

To prove : ∆ABC is an isosceles

Proof : In ∆ADB and ∆ADC.

AD = AD (common)

∠ BAD = ∠ CAD (AD is the bisector of ∠A)

∠ ADB = ∠ ADC (each = 90°, AD ⊥ BC)

∆ADM ≅ ∆ADC (ASA condition)

AB = AC (c.p.c.t)

Hence ∆ABC is an isosceles triangle.

Hence proved.

**Question 7.****Solution:**

In the figure,

AB = AD, CB = CD

To prove : ∆ABC ≅ ∆ADC

Proof : In ∆ABC and ∆ADC

AC = AC (common)

AB = AD (given)

CB = CD (given)

∆ABC ≅ ∆ADC (SSS condition)

Hence proved.

**Question 8.****Solution:**

Given : In the figure,

PA ⊥ AB, QB ⊥ AB and PA = QB.

To prove : ∆OAP ≅ ∆OBQ,

Is OA = QB ?

Proof : In ∆OAP and ∆OBQ,

∠ A = ∠ B (each 90°)

AP = BQ (given)

∠AOP = ∠BOQ (vertically opposite angles)

∆OAP ≅ ∆OBQ (AAS condition)

OA = OB (c.p.c.t.)

Hence proved.

**Question 9.****Solution:**

Given : In right triangles ABC and DCB right angled at A and D respectively and AC = DB

To prove : ∆ABC ≅ ∆DCB.

Proof: In right angled ∆ABC and ∆DCB,

Hypotenuse BC = BC (common)

side AC = DB (given)

∆ABC ≅ ∆DCB (RHS condition)

Hence proved.

**Question 10.****Solution:**

Given: ∆ABC is an isosceles triangle in which AB = AC.

E and F are the midpoints of AC and AB respectively.

To prove : BE = CF

Proof : In ∆BCF and ∆CBE,

BC = BC (common)

BF = CE (Half of equal sides AB and AC)

∠CBF = ∠BCF (Angles opposite to equal sides)

∆BCF ≅ ∆CBE (SAS condition)

CF = BE (c.p.c.t.)

or BE = CF

Hence proved.

**Question 11.Solution:**

Given : In isosceles ∆ABC,

AB = AC.

P and Q are the points on AB and AC respectively such that AP = AQ.

To prove : BQ = CP

Proof : In ∆ABQ and ∆ACP,

AB = AC (given)

AQ = AP (given)

∠ A = ∠ A (common)

∆ABQ ≅ ∆ACP (SAS condition)

BQ = CP (c.p.c.t.)

Hence proved.

**Question 12.****Solution:**

Given : ∆ABC is an isosceles triangle in which AB = AC.

AB and AC are produced to D and E respectively such that BD = CE.

BE and CD are joined.

To prove : BE = CD.

Proof : AB = AC and BD = CE

Adding we get:

AB + BD = AC + CE

AD = AE

Now, in ∆ACD and ∆ABE

AC = AB (given)

AD = AE (proved)

∠ A = ∠ A (common)

∆ACD ≅ ∆ABE (SSA condition)

CD = BE (c.p.c.t.)

Hence, BE = CD.

**Question 13.****Solution:**

Given : In ∆ABC,

AB = AC.

D is a point such that BD = CD.

AD, BD and CD are joined.

To prove : AD bisects ∠A and ∠D.

Proof : In ∆ABD and ∆CAD,

AD = AD (common)

AB = AC (given)

BD = CD (given)

∆ABD ≅ ∆CAD (SSS condition)

∠BAD = ∠CAD (c.p.c.t.)

and ∠BDA = ∠CDA (c.p.c.t.)

Hence AD is the bisector of ∠A and Z D.

Hence proved.

**Question 14.****Solution:**

Two triangles whose corresponding angles are equal, it is not necessarily that they should be congruent. It is possible if atleast one side must be equal. Below given a pair of triangles whose angles are equal but these are not congruent.

**Question 15.****Solution:**

In two triangles, if two sides and and included angle of the one equal to the corresponding two sides and included angle, then the two triangles are congruent.

If another angle except included angles are equal to each other and two sides are also equal these are not congruent. In the above figures, in ∆ABC and ∆PQR, two corresponding sides and one angle are equal, but these are not congruent.

**Question 16.****Solution:**

In ∆ABC,

Area = 12 x BC x AL = 12 x 5 x 4 = 10 cm²

and in ∆PQR

Area = 12 x QR x PR = 12 x 5 x 4 = 10 cm²

In these triangles

Areas of both triangles are equal but are not congruent to each other

**Question 17.****Solution:**

(i) Two line segments are congruent if they have the same length.

(ii) Two angles are congruent if they have equal measure.

(iii) Two squares are congruent if they have same side length.

(iv) Two circles are congruent if they have equal radius.

(v) Two rectangles are congruent if they have the same length and same breadth.

(vi) Two triangles are congruent if they have all parts equal.

**Question 18.****Solution:**

(i) False : Only those squares are congruent which have the same side.

(ii) True :

(iii) False : It is not necessarily, that those figures which have equal areas, must be congruent.

(iv) False : It is not necessarily that those triangles whose areas are equal, must be congruent.

(v) False : It is not necessarily that such triangles must be congruent.

(vi) True : It two angles and one side of a triangle are equal to the corresponding two angles and one side of the other are equal they are congruent.

(vii) False : Only three angles of one are equal the three angles of is not necessarily that these must be congruent.

(viii) True.

(ix) False : Only hypotenuse and one right angle of the one are equal to the hypotenuse and one right angles of the other, the triangles are not necessarily congruent, one side except them, must be equal.

(x) True : It is the definition of congruency of two triangles.

## RS Aggarwal Solutions for Class 7 Maths Chapter 16: **Download PDF**

RS Aggarwal Solutions for Class 7 Maths Chapter 16–Congruence

**Download PDF**: RS Aggarwal Solutions for Class 7 Maths Chapter 16–Congruence PDF

**Chapterwise RS Aggarwal Solutions for Class 7 Maths :**

- Chapter 1–Integers
- Chapter 2–Fractions
- Chapter 3–Decimals
- Chapter 4–Rational Numbers
- Chapter 5–Exponents
- Chapter 6–Algebraic Expressions
- Chapter 7–Linear Equations in One Variable
- Chapter 8–Ratio and Proportion
- Chapter 9–Unitary Method
- Chapter 10–Percentage
- Chapter 11–Profit and Loss
- Chapter 12–Simple Interest
- Chapter 13–Lines and Angles
- Chapter 14–Properties of Parallel Lines
- Chapter 15–Properties of Triangles
- Chapter 16–Congruence
- Chapter 17–Constructions
- Chapter 18–Reflection and Rotational Symmetry
- Chapter 19–Three-Dimensional Shapes
- Chapter 20–Mensuration
- Chapter 21–Collection and Organisation of Data (Mean, Median and Mode)
- Chapter 22–Bar Graphs
- Chapter 23–Probability

## About RS Aggarwal Class 7 Book

Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal’s book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.

He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.

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