NCERT Solutions for Class 7 Maths Ch 7 Congruence of Triangles
You will find Chapter 7 Congruence of Triangles NCERT Solutions that can be used to solve difficult questions in a exercise. These NCERT Solutions are useful in clearing doubts and completing your homework on time and clear your doubts. You can also download our app Studyrankers for offline NCERT Solutions and learn efficiently.
Exercise 7.1
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
(a) they have the same length
(b) 70°
(c) m∠A = m∠B
2. Give any two real time examples for congruent shapes.
Answer
(i) Two footballs
(ii) Two teacher’s tables
3. If ΔABC ≌ ΔFED under the correspondence ABC↔FED, write all the corresponding congruent parts of the triangles.
Answer
Given: ΔABC ≌ ΔFED.
The corresponding congruent parts of die triangles are:
(i) ∠A ↔ ∠F
(ii) ∠B ↔ ∠E
(iii) ∠C ↔ ∠D
4. If ΔDEF ≌ ΔBCA, write the part (s) of ΔBCA that correspond to:
Answer
Exercise 7.2
1. Which congruence criterion do you use in the following?
(a) Given: AC = DF, AB = DE, BC = EF
So ΔABC ≌ ΔDEF
So ΔPQR ≌ ΔXYZ
(c) Given: ∠ MLN = ∠ FGH, ∠ NML = ∠ HFG, ML = FG
So ΔLMN ≌ ΔGFH![]()
(d) Given: EB = BD, AE = CB, ∠A = ∠C = 90°So ΔLMN ≌ ΔGFH
So AABE = ACDB
(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
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
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
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.
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.
(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 =
(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) ?
(i) ?
(ii) ?
Answer
(i) AR = PE
(ii) RT= EN
(iii) AT = PN
(b) Given: ∠T = ∠N
Using SAS criterion, ΔART ≌ ΔPEN
(i) RT = EN
(ii) PN = AT
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
Using ASA criterion, ΔART ≌ ΔPEN
(i) ∠RAT = ∠EPN
(ii) ∠RTA = ∠ENP
3. You have to show that ΔAMP = ΔAMQ. In the following proof supply the missing reasons:
| Steps | Reasons |
| (i) PM = QM | (i) |
| (ii) ∠ PMA = ∠ QMA | (ii) |
| (iii) AM = AM | (iii) |
| (iv) ΔAMP ≌ ΔAMQ | (iv) |
Answer
| Steps | Reasons |
| (i) PM = QM | (i) Given |
| (ii) ∠ PMA = ∠ QMA | (ii) Given |
| (iii) AM = AM | (iii) Common |
| (iv) ΔAMP ≌ ΔAMQ | (iv) SAS congruence rule |
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
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.
5. In the figure, the two triangles are congruent. The corresponding parts are marked. We can write Δ RAT ≌ ?
Answer
In the figure, given two triangles are congruent. So, the corresponding parts are:
A↔O
R↔W
T↔N
We can write, ΔRAT ≌ ΔWON [By SAS congruence rule]
6. Complete the congruence statement:
Answer
In A BAT and ABAC, given triangles are congruent so the corresponding parts are:
B↔B
A↔A
T↔C
Thus, ΔBCA ≌ ΔBTA |By SSS congruence rule]
In ΔQRS and ΔTPQ, given triangles are congruent so the corresponding parts are:
P↔R
T↔Q
Q↔S
Thus, ΔQRS ≌ ΔTPQ [By SSS congruence rule]
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
In a squared sheet, draw ΔABC and ΔPQR. When two triangles have equal areas and
(i)
In the above figure, ΔABC and ΔDEF have equal areas.
And also, ΔABC≌ΔDEF
So, we can say that perimeters of ΔABC and ΔDEF are equal.
(ii)
In the above figure, ΔLMN and ΔOPQ
ΔLMN is not congruent to ΔOPQ
So, we can also say that their perimeters are not same.
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
Let us draw two triangles PQR and ABC.
All angles are equal, two sides are equal except one side. Hence, ΔPQR are not congruent to ΔABC.
9. If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?
Answer
A ABC and A PQR are congruent Then one additional pair is ![]()
Given: ∠B = ∠Q = 90°
∠C/BC = ∠R/QR
Therefore, ΔABC ≌ ΔPQR [By ASA congruence rule]
10. Explain, why ΔABC ≌ ΔFED.
Answer
Given: ∠A = ∠F, BC = ED, ∠B = ∠E
In ΔABC and ΔFED,
In ΔABC and ΔFED,
∠B = ∠E = 90°
∠A = ∠F
BC = ED
Therefore, ΔABC ≌ ΔFED [By RHS congruence rule]