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

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Exercise 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
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Ex 7.2 1
(b) Given:
ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ∆PQR ≅ ∆XYZ
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Ex 7.2 2
(c) Given: ∠MLN = ∠FGH
∠NML = ∠GFH
ML = FG
So, ∆LMN = ∆GFH
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Ex 7.2 3
(d) Given:
EB = DB
AE = BC
∠A = ∠C = 90°
∆ABE = ∆CDB
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Ex 7.2 4
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 =
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Ex 7.2 5
(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.

StepsReasons
(i) PM = QM(i)
(ii) ∠PMA – ∠QMA(ii)
(iii) AM = AM(iii)
(iv) ∆AMP = ∆AMQ(iv)

NCERT Solutions for Class 7 Maths Chapter 7 Congruence Ex 7.2 6

Solution:

StepsReasons
(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.
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Ex 7.2 7
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 ≅ ?
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Ex 7.2 8
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:
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Ex 7.2 9
Solution:
Refer to Fig. (i)
In ∆BCA and ∆BTA
∠C = ∠T (Given)
BC¯¯¯¯¯¯¯=BT¯¯¯¯¯¯¯ (Given)
∠BA = ∠TBA (Given)
∴ ∆BCA = ∆BTA (by ADA 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 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
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Ex 7.2 10
such that
AB¯¯¯¯¯¯¯=DE¯¯¯¯¯¯¯,BC¯¯¯¯¯¯¯=EF¯¯¯¯¯¯¯ and AC¯¯¯¯¯¯¯=DF¯¯¯¯¯¯¯
On adding, we get
AB¯¯¯¯¯¯¯+BC+AC=DE¯¯¯¯¯¯¯+EF¯¯¯¯¯¯¯+DF¯¯¯¯¯¯¯
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
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Ex 7.2 11
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.
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Ex 7.2 12
Solution:
We have ∆PQR and ∆TSU
PQ¯¯¯¯¯¯¯=SU¯¯¯¯¯¯¯ (Given)
PR¯¯¯¯¯¯¯=ST¯¯¯¯¯¯ (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?
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Ex 7.2 13
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
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Ex 7.2 14
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)

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 Q1

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 Q1.1

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 Q2

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 Q3

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 Q4

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 Q5

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:
In some special cases (which depend on the lengths of the sides and the size of the angle involved),

SSA is enough to show congruence. However, it is not always enough. Consider the following triangles :
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 Q8

Here side AB is congruent to side DE (S) side AC is congruent to side DF(S) angle C is congruent to angle F(A)

But the triangles are not congruent, as we can see.

What happens is this : If we draw a vertical line through point A in the first triangle, we can sort of “flip” side AB around this line to get the second triangle. If we were to lay one triangle on top of the other and draw the vertical line, this how it would look.
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles 19

Clearly, side DE is just side AB flipped around the line. So, we have not changed the length of the side, and the other side AC (or DF) is unchanged, as is angle C (or F). So, these two triangles that have the same SSA information, but they are not congruent.

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 Q9

0 comments: