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?
AC = DF
AB = DE
BC = EF
So, ∆ABC = ∆DEF
ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ∆PQR ≅ ∆XYZ
(c) Given: ∠MLN = ∠FGH
∠NML = ∠GFH
ML = FG
So, ∆LMN = ∆GFH
EB = DB
AE = BC
∠A = ∠C = 90°
∆ABE = ∆CDB
(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
(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.
|(i) PM = QM||(i)|
|(ii) ∠PMA – ∠QMA||(ii)|
|(iii) AM = AM||(iii)|
|(iv) ∆AMP = ∆AMQ||(iv)|
|(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?
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 ≅ ?
In ∆RAT and ∆WON
∠A = ∠O (Given)
∴ ∆RAT ≅ ∆WON (By SAS rule)
Ex 7.2 Class 7 Maths Question 6.
Complete the congruence statement:
Refer to Fig. (i)
In ∆BCA and ∆BTA
∠C = ∠T (Given)
∠BA = ∠TBA (Given)
∴ ∆BCA = ∆BTA (by ADA rule)
Refer to Fig. (ii)
In ∆QRS and ∆TPQ
∠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?
(i) On the given square sheet, we have draw two congruent triangles i.e.
∆ABC = ∆DEF
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.
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.
We have ∆PQR and ∆TSU
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?
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
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)