### NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2

**Ex 7.2 Class 7 Maths Question 1.**

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

**(a) Given:**

**AC = DF**

**AB = DE**

**BC = EF**

**So, ∆ABC = ∆DEF**

**(b) Given:**

**ZX = RP**

**RQ = ZY**

**∠PRQ = ∠XZY**

**So, ∆PQR ≅ ∆XYZ**

**(c) Given: ∠MLN = ∠FGH**

**∠NML = ∠GFH**

**ML = FG**

**So, ∆LMN = ∆GFH**

**(d) Given:**

**EB = DB**

**AE = BC**

**∠A = ∠C = 90°**

**∆ABE = ∆CDB**

**Solution:**

(a) ∆ABC ≅ ∆DEF (BY SSS rule)

(b) ∆PQR ≅ ∆XYZ (BY SAS rule)

(c) ∆LMN ≅ ∆GFH (BY ASA rule)

(d) ∆ABE ≅ ∆CDB (BY RHS rule)

**Ex 7.2 Class 7 Maths Question 2.**

**You want to show that ∆ART = ∆PEN,**

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

**(i) AR =**

**(ii) RT =**

**(iii) AT =**

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

**(i) RT = and**

**(ii) PN =**

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

**(i) ZA**

**(ii) ZT**

**Solution:**

(a) For SSS criterion, we need

(i) AR = PE

(ii) RT = EN

(iii) AT = PN

(b) For SAS criterion, we need

(i) RT = EN and

(ii) PN = AT

(c) For ASA criterion, we need

(i) ∠A = ∠P

(ii) ∠T = ∠N

**Ex 7.2 Class 7 Maths Question 3.**

**You have to show that ∆AMP ≅ ∆AMQ. In the fallowing proof, supply the missing reasons.**

Steps | Reasons | |

(i) PM = QM | (i) | |

(ii) ∠PMA – ∠QMA | (ii) | |

(iii) AM = AM | (iii) | |

(iv) ∆AMP = ∆AMQ | (iv) |

**Solution:**

Steps | Reasons | |

(i) PM = QM | (i) Given | |

(ii) ∠PMA = ∠QMA | (ii) Given | |

(iii) AM = AM | (iii) Common | |

(iv) ∆AMP = ∆AMQ | (iv) SAS rule |

**Ex 7.2 Class 7 Maths Question 4.**

**In ∆ABC, ∠A = 30°, ∠B = 40° and ∠C = 110°**

**In ∆PQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°.**

**A student says that ∆ABC = ∆PQR by AAA congruence criterion. Is he justified? Why or why not?**

**Solution:**

The student is not justified because there is not criterion for AAA congruence rule.

Example: In ∆ABC and ∆PQR, we have ∠A = 30°, ∠B = 40°, ∠C = 110°

∠P = 30°, ∠Q = 40°, ∠R = 110°

But ∆ABC is not congruent to ∆PQR.

**Ex 7.2 Class 7 Maths Question 5.**

**In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ∆RAT ≅ ?
**

**Solution:**

In ∆RAT and ∆WON

(Given)

(Given)

∠A = ∠O (Given)

∴ ∆RAT ≅ ∆WON (By SAS rule)

**Ex 7.2 Class 7 Maths Question 6.**

**Complete the congruence statement:
**

**Solution:**

Refer to Fig. (i)

In ∆BCA and ∆BTA

∠C = ∠T (Given)

(Given)

∠BA = ∠TBA (Given)

∴ ∆BCA = ∆BTA (by ADA rule)

Refer to Fig. (ii)

In ∆QRS and ∆TPQ

(Given)

(Given)

∠RSQ = ∠PQT (Given)

∴ ∆QRS = ∆TPQ (by SAS rule)

**Ex 7.2 Class 7 Maths Question 7.**

**In a squared sheet, draw two triangles of equal areas such that:**

**(i) the triangles are congruent.**

**(ii) the triangle are not congruent.**

**What can you say about their perimeters?**

**Solution:**

(i) On the given square sheet, we have draw two congruent triangles i.e.

∆ABC = ∆DEF

such that

On adding, we get

i.e. perimeters of ∆ABC = Perimeter of ∆DEF

(ii) On the other square sheet, we have drawn two triangles ABC and PQR which are not congruent.

Such that

Adding both sides, we get

i.e., perimeter of ∆ABC ≠ the perimeter of ∆PQR.

**Ex 7.2 Class 7 Maths Question 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:**

We have ∆PQR and ∆TSU

(Given)

(Given)

Q = ∠S (Given)

∠P = ∠T (Given)

∠R = ∠U (Given)

Since non of the criteria of congruence is relevant here.

∴ ∆PQR and ∆TSU are not congruent.

**Ex 7.2 Class 7 Maths Question 9.**

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

**Solution:**

In ∆ABC and ∆PQR

∠B = ∠Q (Given)

∠C = ∠R (Given)

For ∆ABC = ∆PQR

BC must equal to criterion that we used is ASA rule.

Hence, the additional pair of corresponding part is

**Ex 7.2 Class 7 Maths Question 10.**

**Explain, why ∆ABC = ∆FED
**

**Solution:**

In ∆ABC and ∆FED

∠B = ∠E = 90° (Given)

∠A = ∠F (Given)

∴ ∠A + ∠B = ∠E + ∠F

180° – ∠C = 180° – ∠D

[Angle sum property of triangles]

∴ ∠C =∠D

BC = ED (Given)

∴ ∆ABC = ∆FED (By ASA rule)