# NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles

NCERT solutions for class 7 maths chapter 7 Congruence of Triangles- Congruence of triangles is one of the important topics of the geometry. In this chapter, there are two exercises and topic wise practice questions. The solutions of NCERT class 7 chapter 7 congruence of triangles give detailed explanations to all these questions. Students can do their homework easily if they have a tool like CBSE NCERT solutions for class 7 maths chapter 7 congruence of triangles in hand. In geometry, two objects or two figures are congruent if they have the same dimension and same shape, or in other words, we can say that two objects or figures are congruent if both are exact copies of one another. The relation of two objects or two figures being congruent is called congruence. NCERT class 7 maths chapter 7 congruence of triangles deal with plane figures or 2D only, although congruence is a general concept applicable to 3D figures also. Congruence of plane figures, congruence among line segments, congruence of angles, congruence of triangles, some criteria for congruence of triangles like SSS congruence of two triangles, SAS congruence of two triangles, ASA congruence of two triangles, RHS congruence of two right-angled triangles are the concepts which are covered in this chapter. Questions on all these concepts are discussed in the NCERT solutions for class 7 maths chapter 7 congruence of triangles. The NCERT solutions are prepared in such a manner that students are able to understand the concepts easily and prepare themselves very well for CBSE final exams to score higher marks. Here you will get solutions to two exercises of this chapter.

Exercise:7.1

Exercise:7.2

## Topics of NCERT Grade 7 Maths Chapter 7 Congruence of Triangles-

7.1 Introduction

7.2 Congruence of Plane Figures

7.3 Congruence among Line Segments

7.4 Congruence of Angles

7.5 Congruence of Triangles

7.6 Criteria for Congruence of Triangles

7.7 Congruence among Right-Angled Triangles

## 1.   When two triangles, say ABC and PQR are given, there are, in all, six possible matchings or correspondences. Two of them are:

$(i) ABC\leftrightarrow PQR$ and $(ii) ABC\leftrightarrow QRP$

Find the other four correspondences by using two cutouts of triangles. Will all these correspondences lead to congruence? Think about it.

the other four correspondences by using two cutouts of triangles are :

$(i) ABC\leftrightarrow RPQ$

$(ii) BCA\leftrightarrow PQR$

$(iii) CAB\leftrightarrow PQR$

$(iv) CBA \leftrightarrow PQR$

## NCERT solutions for class 7 chapter 7 congruence of triangles topic 7.6

i) Since

AB = PQ

BC = QR

CA = PR

So, by SSS congruency rule both triangles are congruent to each other.

$\Delta ABC \cong\Delta PQR$

ii) Since,

ED = MN

DF = NL

FE = LM

So, by SSS congruency rule both triangles are congruent to each other.

$\Delta EDF\cong\Delta MNL$.

iii) Since

AC = PR

BC = QR But

$AB\neq QR$

So the given triangles are not congruent.

iv) Since,

AB = AC

BD = CD

So, By SSS Congruency rule, they both are congruent to each other.

$\Delta ADB\cong\Delta ADC$.

(i) State the three pairs of equal parts in $\Delta ADB$ and $\Delta ADC.$
(ii) Is $\Delta ADB\cong \Delta ADC ?$ Give reasons.
(iii) Is $\angle B = \angle C ?$ Why?

Here in  $\Delta ADB$ and $\Delta ADC.$

i) Three pair of equal parts are:

BD = CD ( as d is the mid point of BC)

AB = AC (given in the question)

ii) Now,

by SSS Congruency rule,

$\Delta ADB\cong \Delta ADC$

iii) As both triangles are congruent to each other we can compare them and say

$\angle B = \angle C$.

$(i) \Delta ABC\cong \Delta ABD$            $(ii) \Delta ABC\cong \Delta BAD$

Given,

$AC=BD$  and

$AD=BC.$.

AB = AB ( common side )

So By SSS congruency rule,

$\Delta ABC\cong \Delta BAD$.

So this statement is meaningfully written as all given criterions are satisfied in this.

## 1.  ABC is an isosceles triangle with $AB=AC$ (Fig 7.17). Take a trace-copy of $\Delta ABC$ and also name it as $\Delta ABC$ (i) State the three pairs of equal parts in $\Delta ABC \; and \: \Delta ACB$. (ii) Is $\Delta ABC \cong \Delta ACB$? Why or why not? (iii) Is $\angle B = \angle C$ ? Why or why not?

Here, in $\Delta ABC \; and \: \Delta ACB$.

i)the three pairs of equal parts in $\Delta ABC \; and \: \Delta ACB$ are

AB = AC

BC = CB

AC = AB

ii)

Hence By SSS Congruency rule, they both are congruent.

$\Delta ABC \cong \Delta ACB$

iii) Yes, $\angle B = \angle C$ because $\Delta ABC \; and \: \Delta ACB$ are congruent and by equating the corresponding parts of the triangles we get,

$\angle B = \angle C$.

Since both the sides  $\overline{DE}$  and $\overline{EF}$ intersects at E,

$\angle E$  is included between the sides $\overline{DE}$  and $\overline{EF}$ of  $\bigtriangleup DEF$.

To prove congruency by SAS rule, we need to equate two corresponding sides and one corresponding angle,

so in   proving $\bigtriangleup PQR\cong \bigtriangleup FED$ we need,

$PQ= FE$

$RP= DF$

And

$\angle P = \angle F$.

Hence the extra information we need is $\angle P = \angle F$.

i) in $\Delta ABC$ and $\Delta DEF$

AB = DE

AC = DF

$\angle A \neq \angle D$

Hence, they are not congruent.

ii) In $\Delta ACB$ and $\Delta RPQ$

AC = RP = 2.5 cm

CB = PQ = 3 cm

$\angle C = \angle P = 35^0$

Hence by SAS congruency rule, they are congruent.

$\Delta ACB \cong\Delta RPQ$.

iii) In $\Delta DFE$ and $\Delta PQR$

DF= PQ = 3.5 cm

FE= QR = 3 cm

$\angle F = \angle Q$

Hence, by SAS congruency rule, they are congruent.

$\Delta DFE\cong\Delta PQR$

iv) In $\Delta QPR$ and $\Delta SRP$

QP = SR = 3.5 cm

PR = RP (Common side)

$\angle QPR = \angle SRP$

Hence, by SAS congruency rule, they are congruent.

$\Delta QPR\cong\Delta SRP$.

The side MN is the side which is included between the angles $M$ and N of $\bigtriangleup M\! N\! P$.

## 2. You want to establish $\bigtriangleup DEF\cong \bigtriangleup MNP$, using the ASA congruence rule. You are given that $\angle D= \angle M$ and$\angle F= \angle P$. What information is needed to establish the congruence? (Draw a rough figure and then try!)

As we know, in ASA congruency two angles and one side is equated to their corresponding parts. So

To Prove $\bigtriangleup DEF\cong \bigtriangleup MNP$

$\angle D= \angle M$

$\angle F= \angle P$

And The side joining these angles is

$\overline {DF}= \overline {MP}$.

So the information that is needed in order to prove congruency is $\overline {DF}= \overline {MP}$.

(i) State the three pairs of equal parts in two triangles $AOC$ and $BOD$.

(ii) Which of the following statements are true?

$(a) \bigtriangleup AOC\cong \bigtriangleup DOB$
$(b) \bigtriangleup AOC\cong \bigtriangleup BOD$

i) The three pairs of equal parts in two triangles $AOC$ and $BOD$ are:

CO = DO (given)

OA = OB (given )

$\angle COA = \angle DOB$ ( As opposite angles are equal when two lines intersect.)

ii) So by SAS congruency rule,

$\Delta COA \cong\Delta DOB$

that is

$\bigtriangleup AOC\cong \bigtriangleup BOD$

Hence, option B is correct.

i) in $\Delta ABC$ and $\Delta FED$

AB = FE = 3.5 cm

$\angle A = \angle F = 40 ^0$

$\angle B = \angle E = 60^0$

So by ASA congruency rule, both triangles are congruent.i.e.

$\Delta ABC \cong \Delta FED$

ii) in $\Delta PQR$ and $\Delta FDE$

$\angle Q= \angle D= 90 ^0$

$\angle R= \angle E = 50^0$

But,

$\overline {EF}\neq\overline {RP}$

So, given triangles are not congruent.

iii) in $\Delta RPQ$ and $\Delta LMN$

RQ = LN = 6 cm

$\angle R = \angle L = 60 ^0$

$\angle Q= \angle N= 30^0$

So by ASA congruency rule, both triangles are congruent.i.e.

$\Delta RPQ\cong \Delta LMN$.

iv) in $\Delta ADB$ and $\Delta BCA$

AB = BA (common side)

$\angle CAB = \angle DBA= 30 ^0$

$\angle D= \angle C=180^0-30^0-30^0-45^0=75^0$

So by ASA congruency rule, both triangles are congruent.i.e.

$\Delta ADB \cong \Delta BCA$

## 4. Given below are measurements of some parts of two triangles. Examine whether the two triangles are congruent or not, by ASA congruence rule. In the case of congruence, write it in symbolic form.

$\bigtriangleup DEF$                                                                                                 $\bigtriangleup PQR$

$(i)\angle D= 60^{o},\angle F= 80^{o},DF= 5cm$                                            $\angle Q= 60^{o},\angle R= 80^{o},QR= 5cm$
$(ii)\angle D= 60^{o},\angle F= 80^{o},DF= 6cm$                                            $\angle Q= 60^{o},\angle R= 80^{o},QP= 6cm$
$(iii)\angle D= 80^{o},\angle F= 30^{o},EF= 5cm$                                           $\angle P= 80^{o},PQ= 5 cm,\angle R= 30^{o}$

i)

Given in $\bigtriangleup DEF$ and  $\bigtriangleup PQR$.

$\\\angle D=\angle Q= 60^{o}\\\angle F=\angle R= 80^{o}\\DF=QR= 5cm$

So, by ASA congruency criterion, they are congruent to each other.i.e.

$\bigtriangleup DEF\cong\Delta QPR$.

ii)

Given in $\bigtriangleup DEF$ and  $\bigtriangleup PQR$.

$\\\angle D=\angle Q= 60^{o}\\\angle F=\angle R= 80^{o}\\DF=QP= 6cm$

For congruency by ASA criterion, we need to be sure of equity of the side which is joining the two angles which are equal to their corresponding parts. Here the side QR is not given which is why we cannot conclude the congruency of both the triangles.

iii)

Given in $\bigtriangleup DEF$ and  $\bigtriangleup PQR$.

$\\\angle D=\angle Q= 60^{o}\\\angle F=\angle R= 80^{o}\\DF=QP= 6cm$

For congruency by ASA criterion, we need to be sure of equity of the side which is joining the two angles which are equal to their corresponding parts. Here the side QR is not given which is why we cannot conclude the congruency of both the triangles.

(i) State the three pairs of equal parts in triangles $BAC$ and $DAC$.
(ii) Is $\Delta BAC\cong \Delta DAC ?$ Give reasons.
(iii) Is $AB = AD ?$ Justify your answer.
(iv) Is $CD = CB \; ?$ Give reasons.

i)

Given in triangles $BAC$ and $DAC$

$\angle DAC=\angle BAC$

$\angle DCA=\angle BCA$

$\overline {AC}=\overline {AC}$  ( common side)

ii)

So, By ASA congruency criterion,triangles $BAC$ and $DAC$ are congruent.

$\Delta BAC\cong\Delta DAC$

iii)

Since $\Delta BAC\cong\Delta DAC$ , all corresponding parts will be equal. So

$AB = AD$.

iv)

Since $\Delta BAC\cong\Delta DAC$ , all corresponding parts will be equal. So

$CD = CB \;$

## 1.  In Fig 7.32, measures of some parts of triangles are given.By applying RHS congruence rule, state which pairs of triangles are congruent. In case of congruent triangles, write the result in symbolic form.

i) In $\Delta PQR$ and $\Delta DEF$

$\angle Q = \angle E$

$PR = DF=6cm$

$PQ\neq DE$

Hence they are not congruent.

ii)

In $\Delta ACB$ and $\Delta BDA$

$\angle D = \angle C$

$DB= AC=2cm$

$AB=BA$ ( same side )

So, by RHS congruency rule,

$\Delta ACB\cong\Delta BDA$

iii)

In $\Delta ABC$ and $\Delta ADC$

$\angle B = \angle D$

$AB= AD=3.6cm$

$AC=AC$ ( same side )

So, by RHS congruency rule,

$\Delta ABC\cong\Delta ADC$

iv)

In $\Delta PSQ$ and $\Delta PSR$

$\angle PSQ = \angle PSR$

$PQ= PR=3cm$

$PS=PS$ ( same side )

So, by RHS congruency rule,

$\Delta PSQ\cong\Delta PSR$

$\angle B= \angle P= 90^{\underline{0}}$  and $AB= RP?$

To prove congruency by RHS (Right angle, Hypotenuse, Side ) rule, we need hypotenuse and side equal to the corresponding hypotenuse and side of different angle.

So Given

$\angle B= \angle P= 90^{{0}}$  ( Right angle )

$AB= RP.$  ( Side )

So the third information we need is the equality of  Hypotenuse of both triangles. i.e.

$AC= RQ$

Hence, if this information is given then we can say,

$\bigtriangleup ABC\cong \bigtriangleup RPQ$.

(i) State the three pairs of equal parts in $\bigtriangleup CBD$ and $\bigtriangleup BCE$.
(ii) Is $\bigtriangleup CBD\cong \bigtriangleup BCE$ ? Why or why not?
(iii) Is $\angle DCB= \angle EBC$ ? Why or why not?

i) Given,  in $\bigtriangleup CBD$ and $\bigtriangleup BCE$.

$BD= CE$

$\angle CEB=\angle BDC=90^o$

$\overline{ BC} = \overline{ CB}$

ii) So, By RHS Rule of congruency, we conclude:

$\bigtriangleup CBD\cong \bigtriangleup BCE$

iii) Since both the triangle are congruent, all parts of one triangle are equal to their corresponding part from another triangle.

So.

$\bigtriangleup CBD\cong \bigtriangleup BCE$.

(i) State the three pairs of equal parts in $\bigtriangleup AD\! B$ and $\bigtriangleup ADC$.
(ii) Is $\bigtriangleup ADB\cong \bigtriangleup ADC$? Why or why not?
(iii) Is $\angle B= \angle C$? Why or why not?
(iv) Is $BD= CD$? Why or why not?

i) Given in $\bigtriangleup AD\! B$ and $\bigtriangleup ADC$.

$AB= AC$

$\angle ADB = \angle ADC=90^0$

$AD = AD$   ( Common side)

ii) So, by RHS Rule of congruency, we conclude

$\bigtriangleup ADB\cong \bigtriangleup ADC$

iii) Since both triangles are congruent all the corresponding parts will be equal.

So,

$\angle B= \angle C$

iv) Since both triangles are congruent all the corresponding parts will be equal.

So,

$BD= CD$.

## NCERT solutions for class 7 maths chapter 7 congruence of triangles exercise 7.1

(a) Two line segments are congruent if ___________.
(b) Among two congruent angles, one has a measure of $70^{o}$; the measure of the other angle is ___________.
(c) When we write $\angle A = \angle B$, we actually mean ___________.

a)Two line segments are congruent if they are identical in shape and size and which is the case when the length of two line segments are equal.

b)$70^0$  As the congruent things are a photocopy of each other.

c) When we write $\angle A = \angle B$, We mean that both the angles(A & B) are equal.

Any two things that have identical shape and size are congruent like all the same kind of pens are congruent to one another. every same kind of bench in class are congruent to one another.all the similar football is congruent to one another.

Corresponding parts of the two congruent triangles $ABC\leftrightarrow FED,$ are :

Sides:

$\overline {AB}\:\:and \:\:\overline {FE},$

$\overline {BC}\:\:and\:\:\overline {ED},$

$\overline {AC}\:\:and\:\:\overline {FD}.$

Angles:

$\angle {ABC}=\angle FED$

$\angle {BCA}=\angle EDF$

$\angle {CAB}=\angle DFE$

$(i)\; \angle E$                $(ii) \; \overline{EF}$                $(iii) \; \angle F$                 $(iv) \; \overline{DF}$

Given,

$\Delta DEF \cong \Delta BCA,$

The part of $\Delta BCA$ that correspond to

$(i)\; \angle E=\angle C$

$(ii) \; \overline{EF}=\overline{CA}$

$(iii) \; \angle F=\angle A$

$(iv) \; \overline{DF}=\overline{BA}$

## CBSE NCERT solutions for class 7 maths chapter 7 congruence of triangles exercise 7.2

$\\(a)\: Given\; AC=DF$

$AB = DE$

$BC = EF$

$So,\; \; \Delta ABC \cong \Delta DEF$

Since we are comparing all the sides of two triangles, The SSS (side, side, side) Congruent criterion is used.

$(b)\; Given:ZX=RP$

$RQ = ZY$

$\angle PRQ =\angle XZY$

$So, \Delta PQR\cong \Delta XYZ$

Since we are comparing two sides and one angle of the two triangles, the SAS (sie, angle, side) congruent criterion is used to prove them congruent.

$(c)\; Given:\angle MLN=\angle FGH$

$\angle NML=\angle GFH$

$ML = FG$

$So,\; \Delta LMN\cong \Delta GFH$

Since we are comparing two angles and one side, ASA(Angle, Side, Angle) congruency criterion is used to prove the congruency.

$(d) Given:\; EB = DB$

$AE = BC$

$\angle A = \angle C = 90^{o}$

$So, \Delta ABE\cong \Delta CDB$

Since we are comparing two sides and one angle of the two triangles, the SSA (Side, Side, Angle) congruent criterion is used to prove the congruency.

(a) If you have to use $SSS$ criterion, then you need to show

$(i)AR=$            $(ii)RT=$                $(iii)AT=$

As we know that in the criterion of proving congruent, all three corresponding sides are equal to another. So to prove the congruency, we kneed to know the following things:

$(i)AR=PE$

$(ii)RT= EN$

$(iii)AT= PN$

(b) If it is given that $\angle T=\angle N$ and you are to use SAS criterion, you need to have

$(i)RT = \; \; \; \; \; \; \; \; \; \; \; \; \; and \; \; \; \; \; \; \; \; \; \; \; \; \; (ii)PN=$

As we know in SAS  criterion the two sides and one angle are identical to their corresponding parts of another triangle. So to prove congruency we need to prove that,

$\\(i)RT = EN\; \; \; \; \; \;\; \; \; and \; \; \; \; \; \; \; \; \; \; \; \; \; \\(ii)PN= AT$

(c) If it is given that $AT=PN$ and you are to use ASA criterion, you need to have

$(i)?\; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\; \; \; \; \; (ii)?$

Given, $\Delta ART\cong \Delta PEN,$

also, $AT=PN$

Now, As we know in the ASA criterion of proving congruency, the one and side two angles are equal to their corresponding parts. So,

$(i)\:\angle{RAT}=\angle{EPN}$

$(ii)\:\angle{RTA}=\angle{ENP}$

 Steps Reasons $(i)PM= QM$ Given in the question $(ii)\angle PMA = \angle QMA$ Given in the question. $(iii)AM= AM$ the side which is common in both triangle $(iv)\bigtriangleup AMP\cong \bigtriangleup AMQ$ By SAS Congruence Rule

## 5. In the figure, the two triangles are congruent. The corresponding parts are marked. We can write $\Delta RAT\cong$?

Comparing from the figure.

$R\leftrightarrow W,A\leftrightarrow O,\:and\:\:T\leftrightarrow N$

By SAS Congruency criterion, we can say that

$\Delta RAT\cong \Delta WON$]

Comparing from the figure, we get,

$B\leftrightarrow B,A\leftrightarrow A,\:\:and\:\:C\leftrightarrow T$

So By SSS Congruency Rule,

$\Delta BCA\cong \Delta BTA$

Also,

Comparing from the figure, we get,

$P\leftrightarrow R,T\leftrightarrow Q,\:\:and\:\:Q\leftrightarrow S$

So By SSS Congruency Rule,

$\Delta QRS\cong \Delta TPQ$.

What can you say about their perimeters?

When two triangles are congruent, the corresponding parts are exactly identical so they have the same area and perimeter.

While the triangles are not congruent but have the same area, then the perimeter of both triangles are not equal.

Five pairs of congruent parts can be three pairs of sides and two pairs of angles. In that case, the SAS or ASA criterion would prove them to be congruent. Hence, such a figure is not possible.

Given $\Delta ABC \cong \Delta PQR$

One additional pair which is not given in the figure is $\overline {BC}=\overline {QR}$

We used the ASA Criterion as the two corresponding angles are given and we figured out the side by congruency.

Comparing both triangles, we have,

$\angle A = \angle F$

$\angle B = \angle E=90^0$

$\overline {BC} = \overline {ED}$

So By RHS congruency criterion,

$\Delta ABC\cong \Delta FED$.

## NCERT Solutions for Class 7 Maths - Chapter-wise

 Chapter No. Chapter Name Chapter 1 Solutions of NCERT for class 7 maths chapter 1 Integers Chapter 2 CBSE NCERT solutions for class 7 maths chapter 2 Fractions and Decimals Chapter 3 NCERT solutions for class 7 maths chapter 3 Data Handling Chapter 4 Solutions of NCERT for class 7 maths chapter 4 Simple Equations Chapter 5 CBSE NCERT solutions for class 7 maths chapter 5 Lines and Angles Chapter 6 NCERT solutions for class 7 maths chapter 6 The Triangle and its Properties Chapter 7 NCERT solutions for class 7 maths chapter 7 Congruence of Triangles Chapter 8 NCERT solutions for class 7 maths chapter 8 comparing quantities Chapter 9 CBSE NCERT solutions for class 7 maths chapter 9 Rational Numbers Chapter 10 NCERT solutions for class 7 maths chapter 10 Practical Geometry Chapter 11 Solutions of NCERT for class 7 maths chapter 11 Perimeter and Area Chapter 12 CBSE NCERT solutions for class 7 maths chapter 12 Algebraic Expressions Chapter 13 NCERT solutions for class 7 maths chapter 13 Exponents and Powers Chapter 14 Solutions of NCERT for class 7 maths chapter 14 Symmetry

## Important points of NCERT Class 7 Maths Chapter 7 Congruence of Triangles-

Questions discussed in the NCERT solutions for class 7 maths chapter 7 congruence of triangles are based on the following congruence criteria.

• SSS Congruence of two triangles- Under a given correspondence, two triangles are congruent if the three sides of the one triangle are equal in measure to the three corresponding sides of the other triangle.
• SAS Congruence of two triangles- Under a given correspondence, two triangles are congruent if two sides and the angle included between them in one of the triangles are equal in measures to the corresponding sides and the angle included between them of the other triangle.
• ASA Congruence of two triangles- Under a given correspondence, two triangles are congruent if two angles and the side included between them in one of the triangles are equal in measures to the corresponding angles and the side included between them of the other triangle.
• RHS Congruence of two right-angled triangles- Under a given correspondence, two right-angled triangles are congruent if the hypotenuse and a leg of one of the triangles are equal to the hypotenuse and the corresponding leg of the other triangle.

Same questions described in the  NCERT solutions for class 7 maths chapter 7 congruence of triangles can be expected for exams.