NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of
Triangles Ex 7.2 are the part of NCERT Solutions for Class 7 Maths. Here you
can find the NCERT Solutions for Class 7 Maths Chapter 7 Congruence of
Triangles Ex 7.2.
- NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.1
- 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°
So, ∆ABE ≅ ∆CDB
Solution:
(a) We have, ∆ABC ≅ ∆DEF (BY SSS rule)
(b) We have, ∆PQR ≅ ∆XYZ (BY SAS rule)
(c) We have, ∆LMN ≅ ∆GFH (BY ASA rule)
(d) We have, ∆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 =
(ii) PN =
(c) If it is given that AT = PN and you are to use ASA criterion, you need to have
(i) ?
(ii) ?
Solution:
(a) For SSS criterion, we need to show
(i) AR = PE
(ii) RT = EN
(iii) AT = PN
(b) For SAS criterion, we need to have
(i) RT = EN and
(ii) PN = AT
(c) For ASA criterion, we need to have
(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.
Here, in ∆ABC and ∆PQR, we have
∠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,
AT = ON (Given)
AR = OW (Given)
∠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)
BC = BT (Given)
∠CBA = ∠TBA (Given)
∴ ∆BCA ≅ ∆BTA (By ASA rule)
Refer to fig. (ii).
In ∆QRS and ∆TPQ,
RS = PQ (Given)
QS = TQ (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 triangles are not congruent.
What can you say about their perimeters?
Solution:
(i) On the given square sheet, we have drawn two
congruent triangles, i.e.,
∆ABC ≅ ∆DEF such that
On adding, we get
AB + BC + AC = DE + EF + DF
i.e., perimeter 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
AB + BC + AC ≠ PQ + QR + PR
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
PQ = SU (Given)
PR = ST (Given)
∠Q = ∠S (Given)
∠P = ∠T (Given)
∠R = ∠U (Given)
Since, none 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 QR. Criterion that we used is ASA rule.
Hence, the additional pair of corresponding part is BC = QR.
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 + ∠C = ∠E + ∠F + ∠D
[Angle sum property of a triangle]
∴ ∠C = ∠D
BC = ED (Given)
∴ ∆ABC ≅ ∆FED (By ASA rule)
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