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NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

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Exercise 7.1 

Question 1: 

Complete the following statements: 

(a) Two line segments are congruent if _______________. 

(b) Among two congruent angles, one has a measure of 70o, the measure of other angle is _______________. 

(c) When we write ∠A = ∠B, we actually mean _______________. 

Answer 1: 

(a) they have the same length

(b) 70o

(c) mA = mB


Question 2: 

Give any two real time examples for congruent shapes. 

Answer 2: 

(i) Two footballs

(ii) Two teacher’s tables


Question 3: 

If ΔABC ≌  ΔFED under the correspondence NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7, write all the corresponding congruent parts of the triangles.

Answer 3: 

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

Given: ΔABC ≌  ΔFED.

The corresponding congruent parts of die triangles are:

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7


Question 4: 

If ΔDEF ≌  ΔBCA, write the part (s) of ΔBCA that correspond to:

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

Answer 4: 

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7
NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7


Exercise 7.2

Question 1: 

Which congruence criterion do you use in the following? 

(a) Given:    AC = DF, AB = DE, BC = EF
 So    ΔABC ≌  ΔDEF

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

(b) Given:    RP = ZX, RQ = ZY, ∠PRQ = ∠XZY
 So    ΔPQR ≌  ΔXYZ

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

(c) Given:    ∠ MLN = ∠ FGH, ∠ NML = ∠ HFG, ML = FG
 So    ΔLMN ≌ ΔGFH

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

(d) Given:    EB = BD, AE = CB, ∠A = ∠C = 90°
 So    AABE = ACDB

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

Answer 1: 

(a) By SSS congruence criterion,  since it is given that AC = DF, AB = DE, BC = EF
The three sides of one triangle are equal to the three corresponding sides of another triangle.
Therefore, ΔABC ≌  ΔDEF

(b) By SAS congruence criterion,  since it is given that RP = ZX, RQ = ZY and ∠PRQ = ∠XZY
The two sides and one angle in one of the triangle are equal to the corresponding sides and the angle of other triangle.
Therefore, ΔPQR ≌ ΔXYZ

(c) By ASA congruence criterion, since it is given that ∠MLN = ∠FGH, ∠NML = ∠HFG, ML = FG.
The two angles and one side in one of the triangle are equal to the corresponding angles and side of other triangle.
Therefore, ΔLMN ≌  ΔGFH

(d) By RHS congruence criterion, since it is given that EB = BD, AE = CB, ∠A = ∠C = 90°
Hypotenuse and one side of a right angled triangle are respectively equal to the hypotenuse and one side of another right angled triangle.
Therefore, ΔABE ≌  ΔCDB


Question 2: 

You want to show that ΔART ≌  ΔPEN:
 (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) ?    
 (ii) ?

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

Answer 2: 

(a) Using SSS criterion,    ΔART ≌  ΔPEN
(i) AR = PE    
(ii) RT= EN    
(iii) AT = PN

(b) Given: ∠T = ∠N
Using SAS criterion,    ΔART ≌  ΔPEN
(i) RT = EN    
(ii) PN = AT

(c) Given:  AT = PN
Using ASA criterion,    ΔART ≌  ΔPEN
(i) ∠RAT = ∠EPN    
(ii) ∠RTA = ∠ENP


Question 3: 

You have to show that ΔAMP = ΔAMQ. In the following proof supply the missing reasons:

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

Steps

Reasons

(i) PM = QM

(i)

(ii) ∠ PMA = ∠ QMA

(ii)

(iii) AM = AM

(iii)

(iv) ΔAMP ≌ ΔAMQ

(iv)

 

Answer 3: 

Steps

Reasons

(i) PM = QM

(i) Given

(ii) ∠PMA = ∠QMA

(ii) Given

(iii) AM = AM

(iii) Common

(iv) Δ AMP ≌ ΔAMQ

(iv) SAS congruence rule

 

Question 4: 

In ΔABC, ∠A = 30° ∠B = 40° and ∠C = 110°
 In ΔPQR, ∠P = 30° ∠Q = 40° and ∠R = 110°.
 A student says that ΔABC ≌ ΔPQR by AAA congruence criterion. Is he justified? Why or why not?

Answer 4: 

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


Question 5:

In the figure, the two triangles are congruent. The corresponding parts are marked. We can write Δ RAT ≌ ?

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

Answer 5: 

In the figure, given two triangles are congruent. So, the corresponding parts are:

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

We can write, ΔRAT ≌ ΔWON    [By SAS congruence rule]


Question 6: 

Complete the congruence statement:

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

Answer 6: 

In A BAT and ABAC, given triangles are congruent so the corresponding parts are:

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

Thus, ΔBCA ≌ ΔBTA    |By SSS congruence rule]
In ΔQRS and ΔTPQ, given triangles are congruent so the corresponding parts are:

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

Thus, ΔQRS ≌  ΔTPQ    [By SSS congruence rule]


Question 7: 

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

(i) the triangles are congruent. 

(ii) the triangles are not congruent. 

What can you say about their perimeters? 

Answer 7: 

(i) If two triangles are congruent, then all the corresponding parts of the triangles will be equal.
Let us consider two triangles, ΔABC and ΔDEF
On the given square sheet, we have drawn two congruent triangles.
such that, ∆ ABC ≅ ∆ DEF
NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7
We can say that,
AB = DE,
BC = EF
AC = DF
On adding the three relations above, we get
AB + BC + AC = DE + EF + DF
Thus, Perimeter of ∆ ABC = Perimeter of ∆ DEF
(ii) In this case, we have drawn two triangles ABC and PQR which are not congruent.
Such that
NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7
Adding the given relations, we get
AB + BC + AC ≠ PQ + QR + PR
Thus, Perimeter of ∆ ABC ≠ Perimeter of ∆ PQR.
From the above, we can say that if two triangles are not congruent then the perimeter is also not equal and if they are congruent then their perimeters are also equal.


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. 

Answer 8: 

Let us draw two triangles PQR and ABC.

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

All angles are equal, two sides are equal except one side. Hence, ΔPQR are not congruent to ΔABC.


Question 9: 

If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use? 

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

Answer 9: 

A ABC and A PQR are congruent Then one additional pair is NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

Given: ∠B = ∠Q = 90°

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

Therefore, ΔABC ≌ ΔPQR    [By ASA congruence rule]


Question 10: 

Explain, why ΔABC  ΔFED.

NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7NCERT Solutions: Congruence of Triangles - Notes | Study Mathematics (Maths) Class 7 - Class 7

Answer 10: 

Given: ∠A = ∠F, BC = ED, ∠B = ∠E
 In ΔABC and ΔFED,
 ∠B = ∠E = 90°
 ∠A = ∠F
 BC = ED
 Therefore, ΔABC ≌ ΔFED    [By RHS congruence rule]

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