NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles

Download PDF

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles (Intext Questions and Exercise)

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 Chapter 7 Congruence of Triangles Class 7 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:

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 $\bigtriangleup PQR$ .

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

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:

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 $\Delta ART\cong \Delta PEN,$

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

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,

$\Delta ABC\cong \Delta FED$ .

Congruence of Triangles Chapter 7 Maths Class 7-Topics

Congruence Of Plane Figures
Congruence Of Among Line Segments
Congruence Of Angles
Congruence Of Triangles
Criteria For Congruence Of Triangles
Congruence Among Right-Angled Triangles

NCERT Solutions for Class 7 Maths Chapter Wise

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

The same questions described in the NCERT solutions for Class 7 Maths chapter 7 Congruence of Triangles can be expected for exams.

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Question (FAQs)

1. Whether the class 7 maths chapter Congruence of Triangles is useful in higher studies

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. Therefore practice congruence of triangles class 7 pdf which can be downloaded form the link given above in this article.

2. What is congruent object according to class 7th congruence of triangles?

The objects having the same shape and the same size are called congruent object. To know more in detail students can study class 7 maths chapter 7 pdf.

3. How many exercises in NCERT solution Class 7 Maths chapter 7?

There are 2 exercises in NCERT solution Class 7 Maths Chapter 7

Also Read

NCERT Solutions for Class 7 Science Chapter 18 Wastewater Story

NCERT Solutions for Class 7 Science Chapter 17 Forests Our Lifeline

NCERT Solutions for Class 7 Science Chapter 16 Water A Precious Resource

NCERT Solutions for Class 7 Science Chapter 14 Electric Current and Its Effects

NCERT Solutions for Class 7 Science Chapter 13 Motion and Time

NCERT Solutions for Class 7 Science Chapter 12 Reproduction in Plants

Articles

High Salary Courses After Class 12 Science

Jul 26, 2024

Ophthalmology Courses After 12th - Eligibility, Duration, Institutes

Jul 26, 2024

BiPC Courses After 12th - Fees, Duration & Top Institutes

Jul 25, 2024

ICSO Sample Papers 2024-25 with Answers For Classes 1 to 10

Jul 25, 2024

NMMS Chhattisgarh Result 2024-25: Check Merit List, Cut Off Marks

Jul 25, 2024

UPTAC Counselling 2024 (Started): Document Verification, Date, Registration, Seat Allotment, Rules, Guidelines

Jul 25, 2024

Essay on Diwali for Students: Short Paragraph, 10 Lines on Deepavali

Jul 24, 2024

How to Study at Home? - Check 10 Tips to Study at Home

Jul 23, 2024

IIM Courses After 12th - Eligibility, Exams, Top Colleges

Jul 19, 2024

Upcoming School Exams

National Means Cum-Merit Scholarship

Application Date:15 July,2024 - 14 August,2024

Eligibility Criteria

Application Process

Exam Pattern

National Institute of Open Schooling 12th Examination

Late Fee Application Date:22 July,2024 - 31 July,2024

Application Process

Exam Pattern

Admit Card

National Institute of Open Schooling 10th examination

Late Fee Application Date:22 July,2024 - 31 July,2024

Application Process

Exam Pattern

Mock Test

Chhattisgarh Board of Secondary Education 12th Examination

Exam Date:23 July,2024 - 12 August,2024

Exam Pattern

Admit Card

Result

Chhattisgarh Board of Secondary Education 10th Examination

Exam Date:24 July,2024 - 08 August,2024

Exam Pattern

Admit Card

Result

View All School Exams

Get answers from students and experts

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

Exams

Top Schools

Products & Resources

NCERT Solutions

Exams

Colleges

Predictors

Resources

Exams

Colleges & Courses

Predictors

Resources

Exams

Colleges

Predictors

Resources

Exams

Colleges

Predictors & E-Books

Resources

Exams

Predictors & Articles

Colleges

Resources

Exams

Colleges

Resources

Exams

Resources

Top Courses & Careers

Colleges

Exams

Colleges

Resources

Quick Links

Exams

Colleges

Resources

Exams

Colleges

Resources

Diploma Colleges

Exams

Resources

Upcoming Events

Other Exams

Exams

Colleges

Top Countries

Student Visas

Exams

Colleges

Upcoming Events

Resources

Engineering Preparation

Medical Preparation

Online Courses

Products

Top Streams

Specializations

Resources

Top Providers

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles

NCERT Solutions for Maths Chapter 7 Congruence of Triangles Class 7- Important Formulae

NCERT Solutions for Chapter 7 Maths Class 7 Congruence of Triangles - Important Points

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles (Intext Questions and Exercise)

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Topic 7.5

NCERT Solutions for Chapter 7 Congruence of Triangles Class 7 Topic 7.6

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Topic 7.7

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Exercise 7.1

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Exercise 7.2

Congruence of Triangles Chapter 7 Maths Class 7-Topics

NCERT Solutions for Class 7 Maths Chapter Wise

NCERT Solutions for Class 7 Subject Wise

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.

Frequently Asked Question (FAQs)

Also Read

Articles