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NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles

Updated on Jul 28, 2020 - 12:39 p.m. IST by Ravindra Pindel

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 NCERT solutions for class 7 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 in NCERT .

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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

NCERT solutions for class 7 maths chapter 7 congruence of triangles topic 7.5

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.

Answer:

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

1. In Fig 7.14, lengths of the sides of the triangles are indicated. By applying the SSS congruence rule, state which pairs of triangles are congruent. In case of congruent triangles, write the result in symbolic form:

Answer:

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,

AD = AD

AB = AC

BD = CD

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

\Delta ADB\cong\Delta ADC .

2. In Fig 7.15, AB=AC and D is the mid-point of \overline{BC}. .

(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?

Answer:

Here in \Delta ADB and \Delta ADC.

i) Three pair of equal parts are:

AD = AD ( common side )

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 .

3. In Fig 7.16, AC=BD and AD=BC. . Which of the following statements is meaningfully written?

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

Answer:

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?

Answer:

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 .

1. Which angle is included between the sides \overline{DE} and \overline{EF} of \bigtriangleup DEF ?

Answer:

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 .

2. By applying SAS congruence rule, you want to establish that \bigtriangleup PQR\cong \bigtriangleup FED . It is given that PQ= FE and RP= DF . What additional information is needed to establish the congruence?

Answer:

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 .

3. In Fig 7.24, measures of some parts of the triangles are indicated. By applying SAS congruence rule, state the pairs of congruent triangles, if any, in each case. In case of congruent triangles, write them in symbolic form.

Answer:

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 .

1. What is the side included between the angles M and N of \bigtriangleup M\! N\! P ?

Answer:

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!)

Answer:

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} .

4. In Fig 7.25, \overline{AB} and \overline{CD} bisect each other at O .

(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

Answer:

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.

3. In Fig 7.27, measures of some parts are indicated. By applying ASA congruence rule, state which pairs triangles are congruent. In case of congruence, write the result in symoblic form.

Answer:

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}


Answer:

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.

5. In Fig 7.28, ray AZ bisects \angle DAB as well as \angle DCB .

(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.


Answer:

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 \;

NCERT solutions for class 7 maths chapter 7 congruence of triangles topic 7.7

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.

Answer:

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


2. It is to be established by RHS congruence rule that \bigtriangleup ABC\cong \bigtriangleup RPQ . What additional information is needed, if it is given that

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

Answer:

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 .

3. In Fig 7.33, BD and CE are altitudes of \bigtriangleup ABC such that BD= CE .

(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?

Answer:

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 .

4. ABC is an isosceles triangle with AB= AC and AD is one of its altitudes (Fig 7.34).

(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?

Answer:

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

1. Complete the following statements:

(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 ___________.

Answer:

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.

2. Give any two real-life examples for congruent shapes.

Answer:

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.

3. If \Delta ABC\cong \Delta FED under the correspondence ABC\leftrightarrow FED, write all the corresponding congruent parts of the triangles.

Answer:

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

4. If \Delta DEF \cong \Delta BCA, write the part(s) of \Delta BCA that correspond to

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

Answer:

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}


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

1.(a) Which congruence criterion do you use in the following?

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

AB = DE

BC = EF

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

Answer:

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

1.(b) Which congruence criterion do you use in the following?

(b)\; Given:ZX=RP

RQ = ZY

\angle PRQ =\angle XZY

So, \Delta PQR\cong \Delta XYZ

Answer:

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.

1.(c) Which congruence criterion do you use in the following?

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

\angle NML=\angle GFH

ML = FG

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

Answer:

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


1.(d) Which congruence criterion do you use in the following?

(d) Given:\; EB = DB

AE = BC

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

So, \Delta ABE\cong \Delta CDB


Answer:

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.

2.(a) You want to show that \Delta ART\cong \Delta PEN,

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

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


Answer:

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

2.(b) You want to show that \Delta ART\cong \Delta PEN,

(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=


Answer:

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

2.(c) You want to show that \Delta ART\cong \Delta PEN,

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

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

Answer:

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}



3. You have to show that \bigtriangleup AMP\cong \bigtriangleup AMQ . In the following proof, supply the missing reasons.

Steps Reasons
(i)PM= QM (i).......
(ii)\angle PMA = \angle QMA (ii).......
(iii)AM= AM (iii).......
(iv)\bigtriangleup AMP\cong \bigtriangleup AMQ (iv).......







Answer

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





4. In \bigtriangleup ABC , \angle A= 30^{o} , \angle B= 40^{o} and \angle C= 110^{o} .

In \bigtriangleup PQR , \angle P=30^{o} , \angle Q=40^{o} and \angle R=110^{o} . A student says that \bigtriangleup ABC\cong \bigtriangleup PQR by AAA congruence criterion. Is he justified? Why or why not?

Answer:

No, because it is not necessary that two triangles will be congruent if their all three corresponding angles are equal. in this case, the triangles might be zoomed copy of one another.

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


Answer:

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 ]

6. Complete the congruence statement:

Answer:

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 .

7. In a squared sheet, draw two triangles of equal areas such that
(i) the triangles are congruent.
(ii) the triangles are not congruent.

What can you say about their perimeters?

Answer:

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.

8. Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.

Answer:

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.


9. If \Delta ABC and \Delta PQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

Answer:

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.

10. Explain, why \Delta ABC\cong \Delta FED

Answer:

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

NCERT solutions for class 7 maths chapter 1 Integers

Chapter 2

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

NCERT solutions for class 7 maths chapter 4 Simple Equations

Chapter 5

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

NCERT solutions for class 7 maths chapter 9 Rational Numbers

Chapter 10

NCERT solutions for class 7 maths chapter 10 Practical Geometry

Chapter 11

NCERT solutions for class 7 maths chapter 11 Perimeter and Area

Chapter 12

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

NCERT solutions for class 7 maths chapter 14 Symmetry

Chapter 15
NCERT solutions for class 7 maths chapter 15 Visualising Solid Shapes

NCERT Solutions for Class 7 Subject wise

NCERT solutions for class 7 maths

NCERT solutions for class 7 science

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.

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles

Question: Whether the class 7 maths chapter Congruence of Triangles is useful in higher studies

Answer:

Yes, the chapter is important for higher studies in the field of maths and science and also in class 8, 9 and 10 maths also

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Consider a sphere of radius R with charge density distributed as ρ(r) = kr for r ≤ R = 0 for r >R . a) Find the electric field at all points r.
Two fixed, identical conducting plates (\alpha and \beta ), each of surface area S are charged to –Q and q, respectively, where Q > q > 0.
There is another useful system of units, besides the SI/mksA system, called the cgs (centimeter-gram-second) system. In this system Coloumb's law is given by Frr =
Two charges –q each are fixed separated by distance 2d. A third charge q of mass m placed at the mid-point is displaced slightly by x (x<<d) perpendicular to the line joining the two fixed charged
Total charge –Q is uniformly spread along length of a ring of radius R. A small test charge +q of mass m is kept at the centre of the ring and is given a gentle push along the axis of the ring.
In 1959 Lyttleton and Bondi suggested that the expansion of the universe could be explained if matter carried a net charge Suppose that the universe is made up of hydrogen atoms with a number density N which is maintained a constant
Five charges, q each are placed at the corners of a regular pentagon of side 'a' (Fig. 1.12).
Fig 1.11 shows the electric field lines around three point charges A. B and C. (a) Which charges are positive (b) Which charge has the largest magnitude Why
Two charges q and –3q are placed fixed on x-axis separated by distance 'd'. Where a third charge 2q should be placed such that it will not experience any force?
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