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

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles are given here. Students can either practise online or download these NCERT Solutions and practise different types of questions related to this chapter and thereby achieve maximum marks in their examinations. The relation of two objects being congruent is called congruence.

Chapter 7 โ Congruence of Triangles contains 2 exercises. Subject experts prepare these NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles carefully, without any mistakes. Let us have a look at some of the concepts discussed in this chapter.

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

## Download the PDF of NCERT Solutions For Class 7 Maths Chapter 7 Congruence of Triangles

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

### Access answers to Maths NCERT Solutions For Class 7 Chapter 7 โ Congruence of Triangles

Exercise 7.1 Page: 137

**1. Complete the following statements:**

**(a) Two line segments are congruent if ___________.**

**Solution:-**

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

**(b) Among two congruent angles, one has a measure of 70 ^{o}; the measure of the other angle is ___________.**

**Solution:-**

Among two congruent angles, one has a measure of 70^{o}; the measure of the other angle is 70^{o}.

Because, if two angles have the same measure, they are congruent. Also, if two angles are congruent, their measure are same.

**(c) When we write โ A = โ B, we actually mean .**

**Solution:-**

When we write โ A = โ B, we actually mean m โ A = m โ B.

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

**Solution:-**

The two real-life example for congruent shapes are

(i) Fan feathers of same brand.

(ii) Size of chocolate in the same brand.

(iii) Size of pens in the same brand

**3. If ฮABC โ
ฮFED under the correspondence ABC โ FED, write all the corresponding congruent parts of the triangles.**

**Solution:-**

Two triangles are congruent if pairs of corresponding sides and corresponding angles are equal.

All the corresponding congruent parts of the triangles are,

โ A โ โ F, โ B โ โ E, โ C โ โ D

Correspondence between sides:

**4. If ฮDEF โ
ฮBCA, write the part(s) of ฮBCA that correspond to**

**(i) โ E (ii) (iii) โ F (iv) **

**Solution:-**

From above the figure we can say that,

The part(s) of ฮBCA that correspond to,

(i) โ E โ โ C

(ii)

(iii) โ F โ โ A

(iv)

Exercise 7.2 Page: 149

**1. Which congruence criterion do you use in the following?**

**(a) Given: AC = DF**

**AB = DE**

**BC = EF**

**So, ฮABC โ
ฮDEF**

**Solution:-**

By SSS congruence property:- Two triangles are congruent if the three sides of one triangle are respectively equal to the three sides of the other triangle.

ฮABC โ ฮDEF

**(b) Given: ZX = RP**

**RQ = ZY**

**โ PRQ = โ XZY**

**So, ฮPQR โ
ฮXYZ**

**Solution:-**

By SAS congruence property:- Two triangles are congruent if the two sides and the included angle of one are respectively equal to the two sides and the included angle of the other.

ฮACB โ ฮDEF

**(c) Given: โ MLN = โ FGH**

**โ NML = โ GFH**

**โ ML = โ FG**

**So, ฮLMN โ
ฮGFH**

**Solution:-**

By ASA congruence property:- Two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other.

ฮLMN โ ฮGFH

**(d) Given: EB = DB**

**AE = BC**

**โ A = โ C = 90 ^{o}**

**So, ฮABE โ
ฮACD**

**Solution:-**

By RHS congruence property:- Two right triangles are congruent if the hypotenuse and one side of the first triangle are respectively equal to the hypotenuse and one side of the second.

ฮABE โ ฮACD

**2. You want to show that ฮART โ
ฮPEN,**

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

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

**Solution:-**

We know that,

SSS criterion is defined as, two triangles are congruent if the three sides of one triangle are respectively equal to the three sides of the other triangle.

โด (i) AR = PE

(ii) RT = EN

(iii) AT = PN

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

**(i) RT = and (ii) PN =**

**Solution:-**

We know that,

SAS criterion is defined as, two triangles are congruent if the two sides and the included angle of one are respectively equal to the two sides and the included angle of the other.

โด (i) RT = EN

(ii) PN = AT

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

**(i) ? (ii) ?**

**Solution:-**

We know that,

ASA criterion is defined as, two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other.

Then,

(i) โ ATR = โ PNE

(ii) โ RAT = โ EPN

**3. You have to show that ฮAMP โ
ฮAMQ.**

**In the following proof, supply the missing reasons.**

Steps | Reasons |

(i) PM = QM | (i) โฆ |

(ii) โ PMA = โ QMA | (ii) โฆ |

(iii) AM = AM | (iii) โฆ |

(iv) ฮAMP โ
ฮAMQ | (iv) โฆ |

**Solution:-**

Steps | Reasons |

(i) PM = QM | (i) From the given figure |

(ii) โ PMA = โ QMA | (ii) From the given figure |

(iii) AM = AM | (iii) Common side for the both triangles |

(iv) ฮAMP โ ฮAMQ | (iv) By SAS congruence property:- Two triangles are congruent if the two sides and the included angle of one are respectively equal to the two sides and the included angle of the other. |

**4. In ฮABC, โ A = 30 ^{o}, โ B = 40^{o} and โ C = 110^{o}**

**In ฮPQR, โ P = 30 ^{o}, โ Q = 40^{o} and โ R = 110^{o}**

**A student says that ฮABC โ
ฮPQR by AAA congruence criterion. Is he justified? Why or Why not?**

**Solution:-**

No, because the two triangles with equal corresponding angles need not be congruent. In such a correspondence, one of them can be enlarged copy of the other.

**5. In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ฮRAT โ
?**

**Solution:-**

From the given figure,

We may observe that,

โ TRA = โ OWN

โ TAR = โ NOW

โ ATR = โ ONW

Hence, ฮRAT โ ฮWON

**6. Complete the congruence statement:**

**ฮBCA โ
ฮQRS โ
**

**Solution:-**

First consider the ฮBCA and ฮBTA

From the figure, it is given that,

BT = BC

Then,

BA is common side for the ฮBCA and ฮBTA

Hence, ฮBCA โ ฮBTA

Similarly,

Consider the ฮQRS and ฮTPQ

From the figure, it is given that

PT = QR

TQ = QS

PQ = RS

Hence, ฮQRS โ ฮTPQ

**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?**

**Solution:-**

**(ii)**

In the above figure, ฮABC and ฮDEF have equal areas.

And also, ฮABC โ ฮDEF

So, we can say that perimeters of ฮABC and ฮDEF are equal.

**(ii)**

In the above figure, ฮLMN and ฮOPQ

ฮLMN is not congruent to ฮOPQ

So, we can also say that their perimeters are not same.

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

**Solution:-**

Let us draw triangles LMN and FGH.

In the above figure, all angles of two triangles are equal. But, out of three sides only two sides are equal.

Hence, ฮLMN is not congruent to ฮFGH.

**9. If ฮABC and ฮPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?**

**Solution:-**

By observing the given figure, we can say that

โ ABC = โ PQR

โ BCA = โ PRQ

The other additional pair of corresponding part is BC = QR

โด ฮABC โ ฮPQR

**10. Explain, why ฮABC โ
ฮFED**

**Solution:-**

From the figure, it is given that,

โ ABC = โ DEF = 90^{o}

โ BAC = โ DFE

BC = DE

By ASA congruence property, two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other.

ฮABC โ ฮFED

## Frequently Asked Questions on NCERT Solutions for Class 7 Maths Chapter 7

### Is NCERT Solutions for Class 7 Maths Chapter 7 important from exam point of view?

Yes, all the chapters of NCERT Solutions for Class 7 Maths Chapter 7 are important from an exam perspective. As this chapter explains congruence of triangles which is important for board exams as well as for higher studies. So studying these concepts makes learning easier for students. Most of the questions in exams come from this chapter.

### Give me a short summary on NCERT Solutions for Class 7 Maths Chapter 7?

NCERT Solutions for Class 7 Maths Chapter 7 has 2 exercises. The chapter deals with different topics related to the probability that includes:

1. Congruence of Plane Figures

2.Congruence Among Line Segments.

3. Congruence of Angles

4. Congruence of Triangles

5. Criteria For Congruence of Triangles

6. Congruence Among Right-Angled Triangles.

### What are the key benefits of learning NCERT Solutions for Class 7 Maths Chapter 7?

The answers of NCERT Solutions for Class 7 Maths Chapter 7 have come in handy. This makes it easy to clear all the doubts related to congruence of triangles. Easily solve all the questions easily which have been given for assignments by the Class 7 students.