NCERT Solutions Class 7 Mathematics Chapter 7 Congruence of Triangle

Exercise 7.1

Q.1) Complete the following statements:
(a) Two line segments are congruent if __________.
(b) Among two congruent angles, one has a measure of 70° ,the measure of other angle is __________.
(c) When we write ∠ 𝐴 = ∠ 𝐵, we actually mean ___________.
Sol.1) (a) they have the same length
(b) 70°
(c) 𝑚∠ 𝐴 = 𝑚∠ 𝐵

Q.2) Give any two real time examples for congruent shapes
Sol.2) (i) Two footballs (ii) Two teacher’s tables

Q.3) If Δ 𝐴𝐵𝐶 ≅Δ 𝐹𝐸𝐷 under the correspondence 𝐴𝐵𝐶 ↔ 𝐹𝐸𝐷, write all the corresponding congruent parts of the triangles.
Sol.3) Given: Δ 𝐴𝐵𝐶 ≅ Δ 𝐹𝐸𝐷.
The corresponding congruent parts of the triangles are:
(i) ∠ 𝐴 ↔ ∠ 𝐹
(ii) ∠ 𝐵 ↔ ∠ 𝐸
(iii) ∠ 𝐶 ↔ ∠ 𝐷
(iv) 𝐴𝐵 ↔ 𝐹𝐸
(v) 𝐵𝐶 ↔ 𝐸𝐷
(vi) 𝐴𝐶 ↔ 𝐹𝐷

Q.4) If Δ DEF ≅ BCA, write the part(s) of Δ BCA that correspond to:
(i) ∠ E (ii) 𝐸̅𝐹̅ (iii) ∠ F (iv) 𝐷̅̅𝐹̅
Sol.4) Given: Δ 𝐷𝐸𝐹 ≅ Δ 𝐵𝐶𝐴.
(i) ∠ 𝐸 ↔ ∠ 𝐶
(ii) 𝐸𝐹 ↔ 𝐶𝐴
(iii) ∠ 𝐹 ↔ ∠ 𝐴
(iv) 𝐷𝐹 ↔ 𝐵𝐴

Exercise 7.2

Q.1) Which congruence criterion do you use in the following?

(a) Given: 𝐴𝐶 = 𝐷𝐹, 𝐴𝐵 = 𝐷𝐸, 𝐵𝐶 = 𝐸𝐹
So Δ 𝐴𝐵𝐶 ≅ Δ 𝐷𝐸𝐹
(b) Given: 𝑅𝑃 = 𝑍𝑋, 𝑅𝑄 = 𝑍𝑌,∠ 𝑃𝑅𝑄 = ∠ 𝑋𝑍𝑌
So Δ 𝑃𝑄𝑅 ≅ Δ 𝑋𝑌𝑍
(c) Given: ∠ 𝑀𝐿𝑁 = ∠ 𝐹𝐺𝐻, ∠ 𝑁𝑀𝐿 = ∠ 𝐻𝐹𝐺, 𝑀𝐿 = 𝐹𝐺
So Δ 𝐿𝑀𝑁 ≅ Δ 𝐺𝐹𝐻
(d) Given: 𝐸𝐵 = 𝐵𝐷, 𝐴𝐸 = 𝐶𝐵,∠ 𝐴 = ∠ 𝐶 = 90 °
So Δ 𝐴𝐵𝐸 ≅ Δ 𝐶𝐷𝐵
Sol.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 ≅ Δ 𝐷𝐸𝐹
(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, Δ 𝑃𝑄𝑅 ≅ Δ 𝑋𝑌𝑍
(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, Δ 𝐿𝑀𝑁 ≅ Δ 𝐺𝐹𝐻
(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, Δ 𝐴𝐵𝐸 ≅ Δ 𝐶𝐷𝐵

Q.2) You want to show that Δ 𝐴𝑅𝑇 ≅ Δ 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 = and (ii) PN =
(c) If it is given that AT = PN and you are to use ASA criterion, you need to have:
(i) ? (ii) ?
Sol.2) (a) Using SSS criterion, Δ 𝐴𝑅𝑇 ≅ Δ 𝑃𝐸𝑁
(i) AR = PE (ii) RT = EN (iii) AT = PN
(b) Given: ∠ 𝑇 = ∠ 𝑁 Using SAS criterion, Δ 𝐴𝑅𝑇 ≅ Δ 𝑃𝐸𝑁
(i) RT = EN (ii) PN = AT
(c) Given: AT = PN Using ASA criterion, Δ 𝐴𝑅𝑇 ≅ Δ 𝑃𝐸𝑁
(i) ∠ 𝑅𝐴𝑇 = ∠ 𝐸𝑃𝑁 (ii) ∠ 𝑅𝑇𝐴 = ∠ 𝐸𝑁𝑃

Q.3) You have to show that Δ 𝐴𝑀𝑃 ≅ Δ 𝐴𝑀𝑄. In the following proof, supply the missing reasons:
Steps Reasons
(i) PM = QM                          (i) __________
(ii) ∠ 𝑃𝑀𝐴 = ∠ 𝑄𝑀𝐴           (ii) __________
(iii) AM = AM                       (iii) __________
(iv) Δ 𝐴𝑀𝑃 ≅ Δ 𝐴𝑀𝑄         (iv) __________

Sol.3) Steps Reasons
(i) PM = QM                       (i) Given
(ii) ∠ 𝑃𝑀𝐴 = ∠ 𝑄𝑀𝐴          (ii) Given
(iii) AM = AM                    (iii) Common
(iv) Δ 𝐴𝑀𝑃 ≅ Δ 𝐴𝑀𝑄       (iv) SAS congruence rule

Q.4) In Δ𝐴𝐵𝐶, ∠ 𝐴 = 30 ,∠ 𝐵 = 40° and ∠ 𝐶 = 110° .
In Δ𝑃𝑄𝑅, ∠ 𝑃 = 30 ,∠ 𝑄 = 40° and ∠ 𝑅 = 110° .
A student says that Δ𝐴𝐵𝐶 ≅ Δ 𝑃𝑄𝑅 by 𝐴𝐴𝐴 congruence criterion. Is he justified? Why or why not?
Sol.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.

Q.5) In the figure, the two triangles are congruent. The corresponding parts are marked. We can write Δ𝑅𝐴𝑇 ≅ ?

Sol.5) In the figure, given two triangles are congruent. So, the corresponding parts are:
𝐴 ↔ 𝑂, 𝑅 ↔ 𝑊, 𝑇 ↔ 𝑁.
We can write, Δ 𝑅𝐴𝑇 ≅ Δ 𝑊𝑂𝑁 [By SAS congruence rule]

Q.6) Complete the congruence statement :

Sol.6) In Δ 𝐵𝐴𝑇 and Δ 𝐵𝐴𝐶, given triangles are congruent so the corresponding parts are:
𝐵 ↔ 𝐵, 𝐴 ↔ 𝐴, 𝑇 ↔ 𝐶
Thus, Δ 𝐵𝐶𝐴 ≅ Δ 𝐵𝑇𝐴 [By SSS congruence rule]
In Δ 𝑄𝑅𝑆 and Δ 𝑇𝑃𝑄, given triangles are congruent so the corresponding parts are:
𝑃 ↔ 𝑅, 𝑇 ↔ 𝑄, 𝑄 ↔ 𝑆
Thus, Δ 𝑄𝑅𝑆 ≅ Δ 𝑇𝑃𝑄 [By SSS congruence rule]

Q.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?
Sol.7) In a squared sheet, draw Δ𝐴𝐵𝐶 and Δ 𝑃𝑄𝑅.
When two triangles have equal areas and
(i) these triangles are congruent, i.e., Δ𝐴𝐵𝐶 ≅ Δ 𝑃𝑄𝑅 [By SSS congruence rule]
Then, their perimeters are same because length of sides of first triangle are equal to the length of sides of another triangle by SSS congruence rule.
(ii) But, if the triangles are not congruent, then their perimeters are not same because lengths of sides of first triangle are not equal to the length of corresponding sides of another triangle.

Q.8) Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.
Sol.8) Let us draw two triangles PQR and ABC.

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

Q.9) If Δ𝐴𝐵𝐶 and Δ 𝑃𝑄𝑅 are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

Sol.9) Δ𝐴𝐵𝐶 and Δ 𝑃𝑄𝑅 are congruent.
Then one additional pair is BC = QR.
Given: ∠ 𝐵 = ∠ 𝑄 = 90°
∠ 𝐶 = ∠ 𝑅 𝐵𝐶 = 𝑄𝑅
Therefore, Δ𝐴𝐵𝐶 ≅ Δ 𝑃𝑄𝑅 [By ASA congruence rule]

Q.10) Explain, why Δ𝐴𝐵𝐶 ≅ Δ 𝐹𝐸𝐷.

Sol.10) Given: ∠ A = ∠ F, BC = ED, ∠ B = ∠ E
In Δ ABC and Δ𝐹𝐸𝐷,
∠ 𝐵 = ∠ 𝐸 = 90°
∠ 𝐴 = ∠ 𝐹
𝐵𝐶 = 𝐸𝐷
Therefore, Δ𝐴𝐵𝐶 ≅ Δ𝐹𝐸𝐷 [By RHS congruence rule]