NCERT Solutions for Class 7 Maths Chapter 7 - Congruence of Triangles
Recognize the importance of congruence in geometry, which defines objects with their mirror images, making them identical in shape and dimensions. Our NCERT Solutions for Class 7 Mathematics Chapter 7 delve into this vital topic, offering detailed step-by-step solutions for all the questions covered in this chapter. These comprehensive solutions are invaluable for exam preparation, guiding students through the concept of congruence with ease. In mathematics, congruence is employed to describe the similarity between objects or shapes that perfectly overlay one another. Two objects are deemed congruent when their shapes and dimensions match precisely. In the realm of geometric figures, line segments are considered congruent if they share the same length, while angles are congruent if their measures are identical.
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The NCERT Solutions for Class 7 Maths Chapter 7 - Congruence of Triangles are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2023-2024. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Access Answers to NCERT Solutions for Class 7 Maths Chapter 7 - Congruence of Triangles
Students can access the NCERT Solutions for Class 7 Maths Chapter 7 - Congruence of Triangles. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Exercise 7.1
Give any two real-life examples of congruent shapes.
The two real-life examples of congruent shapes are as follows:
(i) Fan feathers of the same brand
(ii) Size of chocolate in the same brand
(iii) Size of pens in the same brand
If ΔABC ≅ ΔFED under the correspondence ABC ↔ FED, write all the corresponding congruent parts of the triangles.
Two triangles are congruent if pairs of corresponding sides and corresponding angles are equal.
All the corresponding congruent parts of the triangles are,
∠A ↔ ∠F, ∠B ↔ ∠E, ∠C ↔ ∠D
Correspondence between sides:
If ΔDEF ≅ ΔBCA, write the part(s) of ΔBCA that correspond to
(i) ∠E
(ii) 
(iii) ∠F
(iv)
From the above figure, we can say that,
The part(s) of ΔBCA that correspond to,
(i) ∠E ↔ ∠C
(ii)
(iii) ∠F ↔ ∠A
(iv)
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 the other angle is ___________.
(c) When we write ∠A = ∠B, we actually mean ___________.
(a) Two line segments are congruent if they have the same length.
(b) Among two congruent angles, one has a measure of 70o; the measure of the other angle is 70o.
If two angles have the same measure, they are congruent. Also, if two angles are congruent, their measure is the same.
(c) When we write ∠A = ∠B, we actually mean m ∠A = m ∠B.
Exercise 7.2
Complete the congruence statement:
ΔBCA ≅ ΔQRS ≅
First, consider the ΔBCA and ΔBTA
From the figure, it is given that,
BT = BC
Then,
BA is the common side for the ΔBCA and ΔBTA
Hence, ΔBCA ≅ ΔBTA
Similarly,
Consider the ΔQRS and ΔTPQ
From the figure, it is given that
PT = QR
TQ = QS
PQ = RS
Hence, ΔQRS ≅ ΔTPQ
You want to show that ΔART ≅ ΔPEN,
(a) If you have to use the 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 the SAS criterion, you need to have
(i) RT = and (ii) PN =
(c) If it is given that AT = PN and you are to use the ASA criterion, you need to have
(i) ? (ii)?
(a) We know that,
SSS criterion is defined as two triangles being congruent if the three sides of one triangle are respectively equal to the three sides of the other triangle.
∴ (i) AR = PE
(ii) RT = EN
(iii) AT = PN
(b) We know that,
SAS criterion is defined as two triangles being congruent if the two sides and the included angle of one are respectively equal to the two sides and the included angle of the other.
∴ (i) RT = EN
(ii) PN = AT
(c) We know that,
ASA criterion is defined as two triangles being congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other.
Then,
(i) ∠ATR = ∠PNE
(ii) ∠RAT = ∠EPN
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) … |
|
Steps |
Reasons |
|
(i) PM = QM |
(i) From the given figure |
|
(ii) ∠PMA = ∠QMA |
(ii) From the given figure |
|
(iii) AM = AM |
(iii) Common side for both triangles |
|
(iv) ΔAMP ≅ ΔAMQ |
(iv) By SAS congruence property: Two triangles are congruent if the two sides and the included angle of one are respectively equal to the two sides and the included angle of the other. |
In ΔABC, ∠A = 30o, ∠B = 40o and ∠C = 110o
In ΔPQR, ∠P = 30o, ∠Q = 40o and ∠R = 110o
A student says that ΔABC ≅ ΔPQR by AAA congruence criterion. Is he justified? Why or why not?
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.
In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ΔRAT ≅ ?
From the given figure,
We may observe that,
∠TRA = ∠OWN
∠TAR = ∠NOW
∠ATR = ∠ONW
Hence, ΔRAT ≅ ΔWON
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?
(i)
In the above figure, ΔABC and ΔDEF have equal areas.
And also, ΔABC ≅ ΔDEF
So, we can say that perimeter 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 the same.
Draw a rough sketch of two triangles such that they have five pairs of congruent parts, but still, the triangles are not congruent.
Let us draw triangles LMN and FGH.
In the above figure, all angles of two triangles are equal. But, out of the three sides, only two sides are equal.
Hence, ΔLMN is not congruent to ΔFGH.
If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?
By observing the given figure, we can say that
∠ABC = ∠PQR
∠BCA = ∠PRQ
The other additional pair of corresponding parts is BC = QR
∴ ΔABC ≅ ΔPQR
Explain, why ΔABC ≅ ΔFED
From the figure, it is given that,
∠ABC = ∠DEF = 90o
∠BAC = ∠DFE
BC = DE
By ASA congruence property, two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other.
ΔABC ≅ ΔFED
Which congruence criterion do you use in the following?
(a) Given: AC = DF
AB = DE
BC = EF
So, ΔABC ≅ ΔDEF
(b) Given: ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ΔPQR ≅ ΔXYZ
(c) Given: ∠MLN = ∠FGH
∠NML = ∠GFH
∠ML = ∠FG
So, ΔLMN ≅ ΔGFH
(d) Given: EB = DB
AE = BC
∠A = ∠C = 90o
So, ΔABE ≅ ΔACD
(a) By SSS congruence property: Two triangles are congruent if the three sides of one triangle are respectively equal to the three sides of the other triangle.
ΔABC ≅ ΔDEF
(b) By SAS congruence property: Two triangles are congruent if the two sides and the included angle of one are respectively equal to the two sides and the included angle of the other.
ΔACB ≅ ΔDEF
(c) By ASA congruence property: Two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other.
ΔLMN ≅ ΔGFH
(d) By RHS congruence property: Two right triangles are congruent if the hypotenuse and one side of the first triangle are respectively equal to the hypotenuse and one side of the second.
ΔABE ≅ ΔACD
Frequently Asked Questions
The NCERT solution for Class 7 Chapter 7: Congruence of Triangles is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.
Yes, the NCERT solution for Class 7 Chapter 7: Congruence of Triangles is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. Congruence of Trianglesally, they can solve the practice questions and exercises that allow them to get exam-ready in no time.
You can get all the NCERT solutions for Class 7 Maths Chapter 7 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for Class 7 Maths Chapter 7: Congruence of Triangles as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for Class 7 Maths Chapter 7 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.