NCERT Solutions for Class 7 Maths Chapter 7: Congruence of Triangles - Exercise 7.2

Exercise 7.2

Refer the page 6 to 10 for the exercise 7.2 in the PDF

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

a) Given: \[{\mathbf{AC}} = {\mathbf{DF}},{\mathbf{AB}} = {\mathbf{DE}},{\mathbf{BC}} = {\mathbf{EF}}\]

So \[\vartriangle {\mathbf{ABC}} \cong \vartriangle {\mathbf{DEF}}\]

Ans: By \[SSS\]congruence criterion, since it is given that \[AC = DF, AB = DE,  {\text{B}}C = EF\]

The three sides of one triangle are the same as the three sides of a second triangle.

Therefore, \[\vartriangle ABC \cong \vartriangle DEF\]

b) Given: \[{\mathbf{RP}} = {\mathbf{ZX}}, {\mathbf{RQ}} = {\mathbf{ZY}}\], \[\angle {\mathbf{PRQ}} = \angle {\mathbf{XZY}}\] So \[\vartriangle {\mathbf{PQR}} \cong \vartriangle {\mathbf{XYZ}}\]

Ans: By \[SAS\] congruence criterion, since it is given that \[RP = ZX, RQ = ZY\] and \[\angle PRQ = \angle XZY\]The two sides and one angle in one of the triangles are equal to the corresponding sides and the angle of another triangle. Therefore, \[\vartriangle PQR \cong \vartriangle XYZ\]

c) Given: \[\angle {\mathbf{MLN}} = \angle {\mathbf{FGH}}, \angle {\mathbf{NML}} = \angle {\mathbf{HFG}}\], \[{\mathbf{ML}} = {\mathbf{FG}}\] So \[\vartriangle {\mathbf{LMN}} \cong \vartriangle {\mathbf{GFH}}\]

Ans: By \[ASA\]congruence criterion, since it is given that \[\angle MLN = \angle FGH,\angle NML = \angle HFG,ML = FG\]. 

The two angles and one side in one of the triangles are equal to the corresponding angles and side of another triangle. 

Therefore, \[\vartriangle LMN \cong \vartriangle GFH\]

d) Given: \[{\mathbf{EB}} = {\mathbf{BD}},{\mathbf{AE}} = {\mathbf{CB}},\angle {\mathbf{A}} = \angle {\mathbf{C}} = {\mathbf{9}}{{\mathbf{0}}^o}\] So \[\vartriangle {\mathbf{ABE}} \cong \vartriangle {\mathbf{CDB}}\]

Ans: By $RHS$ congruence criterion, since it is given that \[EB = BD,AE = CB,\angle A = \angle C = {90^o}\] Hypotenuse and one side of a right-angled triangle are respectively equal to the hypotenuse and one side of another right-angled triangle. Therefore, \[\vartriangle ABE \cong \vartriangle CDB\]

2. You want to show that\[\vartriangle \mathbf{ART}\cong \vartriangle \text{ }\mathbf{PEN}\]: 

a) If you have to use \[\mathbf{SSS}\]criterion, then you need to show: 

i) \[{\mathbf{AR}} = \]

Ans:  From \[SSS\] congruence it is clear that \[\vartriangle ART \cong \vartriangle PEN\]for a triangle. 

The three sides of one triangle are the same as the three corresponding sides of the second triangle.

Thus,\[AR = PE\]

ii) \[{\mathbf{RT}} = \]

Ans:  From \[SSS\] congruence it is clear that \[\vartriangle ART \cong \vartriangle PEN\]for a triangle. 

The three sides of one triangle are the same as the three corresponding sides of the second triangle.

Thus,\[RT = EN\]

iii) \[{\mathbf{AT}} = \]

Ans:  From \[SSS\] congruence it is clear that \[\vartriangle ART \cong \vartriangle PEN\]for a triangle. 

The three sides of one triangle are the same as the three corresponding sides of the second triangle.

Thus,\[AT{\text{ }} = {\text{ }}PN\]

b) If it is given that \[\angle {\mathbf{T}} = \angle {\mathbf{N}}\]and you are to use \[{\mathbf{SAS}}\]criterion, you need to have: 

i) \[{\mathbf{RT}}\]= 

Ans: From \[SSS\] congruence it is clear that \[\vartriangle ART \cong \vartriangle PEN\]for a triangle and given \[\angle T = \angle N\]. The three sides of one triangle are equal to the three corresponding sides of another triangle.

Thus,$RT = EN$

ii) \[\mathbf{PN}=\]

Ans: From \[SSS\] congruence it is clear that \[\vartriangle ART \cong \vartriangle PEN\]for a triangle and given \[\angle T = \angle N\]. The three sides of one triangle are equal to the three corresponding sides of another triangle.

Thus, \[PN = AT\]

c) If it is given \[\mathbf{AT}=\mathbf{PN}\] that and you are to use \[\mathbf{ASA}\] criterion, you need to have: 

i) ?

Ans:  From \[ASA\] congruence it is clear that the two angles and one side in one of the triangles are equal to the corresponding angles and side of another triangle that is \[\vartriangle ART \cong \vartriangle PEN\]

Thus, it is clear that, \[\angle RAT = \angle EPN\]

ii) ?

Ans:  From \[ASA\] congruence it is clear that two angles and one side in one of the triangles are equal to the corresponding angles and side of another triangle that is \[\vartriangle ART \cong \vartriangle PEN\] 

Thus, it is clear that, \[\angle RTA = \angle ENP\]

3. You have to show that \[\vatriangle \mathbf{AMP}\cong \vartriangle \mathbf{AMQ}\}. In the following proof, supply the missing reasons:

Ans: 

4. In \[\vartriangle {\mathbf{ABC}}\], \[\angle {\mathbf{A}} = {\mathbf{30}},\;\angle {\mathbf{B}} = {\mathbf{40}}^\circ \] and \[\angle {\mathbf{C}} = {\mathbf{110}}^\circ \].

In \[\vartriangle {\mathbf{PQR}}\], \[\angle {\mathbf{P}} = {\mathbf{30}},\;\angle {\mathbf{Q}} = {\mathbf{40}}^\circ \] and \[\angle {\mathbf{R}} = {\mathbf{110}}^\circ \].

A student says that \[\vartriangle {\mathbf{ABC}} \cong \vartriangle {\mathbf{PQR}}\] by AAA congruence criterion. Is he justified? Why or why not? 

Ans: 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 \[\vartriangle {\mathbf{RAT}} \cong ?\]

Ans: In the figure, given two triangles are congruent. 

So, the corresponding parts are: \[A=O,R=W,T=N\]

. We can write, 

\[AR = OW, \angle R= \angle W, AT= ON\]

In \[\vartriangle RAT\cong \vartriangle WON\]

(By \[SAS\]congruence rule)

6. Complete the congruence statement:\[\vartriangle {\mathbf{BCA}} \cong \]?\[\vartriangle {\mathbf{QRS}} \cong \]?

Ans:  In \[\vartriangle BAT\] and \[\vartriangle BAC\], given triangles are congruent so the corresponding parts are:

\[\angle B = \angle B, \angle A= \angle A, BT=BC\]

Thus, \[\vartriangle BCA\cong \vartriangle BTA\]

(By SAS congruence rule)

In \[\vartriangle QRS\] and \[\vartriangle TPQ\], given triangles are congruent so the corresponding parts are: 

\[ QR =TP, QS =TQ, RS=PQ\]

 Thus, \[\vartriangle QRS\cong \vartriangle TPQ\]

(By SSS congruence rule)

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

i) the triangles are congruent. 

Ans: In a squared sheet, draw \[\vartriangle ABC\]and\[\vartriangle DEF\]. 

i.e.,\[\;\vartriangle ABC \cong \vartriangle DEF\] (By SSS congruence rule).

Then, their perimeters are the same because the length of sides of the first triangle are equal to the length of sides of another triangle by SSS congruence rule.

ii) the triangles are not congruent. 

What can you say about their perimeters? 

Ans: But, if the triangles are not congruent, then their perimeters are not the same because lengths of sides of the first triangle are not equal to the length of corresponding sides of another triangle.

8. Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent. 

Ans: Let the two triangles are \[\vartriangle RPQ\] and \[\vartriangle ABC\].

For the given triangles, all the angles are equal and two sides are equal except one side. Hence, \[\vartriangle RPQ\]are not congruent to\[\vartriangle ABC\].

9. If \[\vartriangle {\mathbf{ABC}}\]and \[\vartriangle {\mathbf{PQR}}\] are to be congruent, name one additional pair of corresponding parts. What criterion did you use? 

Ans: \[\vartriangle ABC\] and \[\vartriangle PQR\] are congruent. Then one additional pair is BC = QR. 

Given: \[\angle B = \angle Q = {90^o},\angle C = \angle R,BC = QR\]

Therefore, \[\vartriangle ABC \cong \vartriangle PQR\](By ASA congruence rule)

10. Explain why \[\vartriangle {\mathbf{ABC}} \cong \vartriangle {\mathbf{FED}}\].

Ans: Given: \[\angle A = \angle F,BC = ED,\angle B{\text{ = }}\angle E\] 

In \[\vartriangle ABC\] and\[\vartriangle FED\], 

\[\angle B{\text{ = }}\angle E = {90^o}\]

\[\angle A{\text{ = }}\angle F\]

\[BC = ED\]

Therefore, \[\vartriangle ABC \cong \vartriangle FED\] (By RHS congruence rule)

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Exercise 7.2

Opting for the NCERT solutions for Ex 7.2 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 7.2 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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