NCERT Solutions for Class 7 Maths Exercise 7.2 Chapter 7 Congruence of Triangles in simple PDF are given here. Criteria for congruence of triangles and congruence among right-angled triangles are the two topics covered in this exercise of NCERT Solutions for Class 7. Students of Class 7 are suggested to solve NCERT Solutions for Class 7 Maths Chapter 7 Congruence to strengthen the fundamentals and be able to solve questions that are usually asked in the examination.
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1. Which congruence criterion do you use in the following?
(a) Given: AC = DF
AB = DE
BC = EF
So, Î”ABC â‰… Î”DEF
Solution:-
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.
Î”ABC â‰… Î”DEF
(b) Given: ZX = RP
RQ = ZY
âˆ PRQ = âˆ XZY
So, Î”PQR â‰… Î”XYZ
Solution:-
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.
Î”ACB â‰… Î”DEF
(c) Given: âˆ MLN = âˆ FGH
âˆ NML = âˆ GFH
âˆ ML = âˆ FG
So, Î”LMN â‰… Î”GFH
Solution:-
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.
Î”LMN â‰… Î”GFH
(d) Given: EB = DB
AE = BC
âˆ A = âˆ C = 90^{o}
So, Î”ABE â‰… Î”ACD
Solution:-
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.
Î”ABE â‰… Î”ACD
2. You want to show that Î”ART â‰… Î”PEN,
(a) If you have to use SSS criterion, then you need to show
(i) AR = (ii) RT = (iii) AT =
Solution:-
We know that,
SSS criterion is defined as, two triangles are congruent if the three sides of one triangle are respectively equal to the three sides of the other triangle.
âˆ´ (i) AR = PE
(ii) RT = EN
(iii) AT = PN
(b) If it is given that âˆ T = âˆ N and you are to use SAS criterion, you need to have
(i) RT = and (ii) PN =
Solution:-
We know that,
SAS criterion is defined as, 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.
âˆ´ (i) RT = EN
(ii) PN = AT
(c) If it is given that AT = PN and you are to use ASA criterion, you need to have
(i) ? (ii) ?
Solution:-
We know that,
ASA criterion is defined as, 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.
Then,
(i) âˆ ATR = âˆ PNE
(ii) âˆ RAT = âˆ EPN
3. You have to show that Î”AMP â‰… Î”AMQ.
In the following proof, supply the missing reasons.
Steps |
Reasons |
(i) PM = QM |
(i) â€¦ |
(ii) âˆ PMA = âˆ QMA |
(ii) â€¦ |
(iii) AM = AM |
(iii) â€¦ |
(iv) Î”AMP â‰… Î”AMQ |
(iv) â€¦ |
Solution:-
Steps |
Reasons |
(i) PM = QM |
(i) From the given figure |
(ii) âˆ PMA = âˆ QMA |
(ii) From the given figure |
(iii) AM = AM |
(iii) Common side for the both triangles |
(iv) Î”AMP â‰… Î”AMQ |
(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. |
4. In Î”ABC, âˆ A = 30^{o}, âˆ B = 40^{o} and âˆ C = 110^{o}
In Î”PQR, âˆ P = 30^{o}, âˆ Q = 40^{o} and âˆ R = 110^{o}
A student says that Î”ABC â‰… Î”PQR by AAA congruence criterion. Is he justified? Why or Why not?
Solution:-
No, because the two triangles with equal corresponding angles need not be congruent. In such a correspondence, one of them can be enlarged copy of the other.
5. In the figure, the two triangles are congruent. The corresponding parts are marked. We can write Î”RAT â‰… ?
Solution:-
From the given figure,
We may observe that,
âˆ TRA = âˆ OWN
âˆ TAR = âˆ NOW
âˆ ATR = âˆ ONW
Hence, Î”RAT â‰… Î”WON
6. Complete the congruence statement:
Î”BCA â‰… Î”QRS â‰…
Solution:-
First consider the Î”BCA and Î”BTA
From the figure, it is given that,
BT = BC
Then,
BA is common side for the Î”BCA and Î”BTA
Hence, Î”BCA â‰… Î”BTA
Similarly,
Consider the Î”QRS and Î”TPQ
From the figure, it is given that
PT = QR
TQ = QS
PQ = RS
Hence, Î”QRS â‰… Î”TPQ
7. 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?
Solution:-
(ii)
In the above figure, Î”ABC and Î”DEF have equal areas.
And also, Î”ABC â‰… Î”DEF
So, we can say that perimeters of Î”ABC and Î”DEF are equal.
(ii)
In the above figure, Î”LMN and Î”OPQ
Î”LMN is not congruent to Î”OPQ
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.
Solution:-
Let us draw triangles LMN and FGH.
In the above figure, all angles of two triangles are equal. But, out of three sides only two sides are equal.
Hence, Î”LMN is not congruent to Î”FGH.
9. If Î”ABC and Î”PQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?
Solution:-
By observing the given figure, we can say that
âˆ ABC = âˆ PQR
âˆ BCA = âˆ PRQ
The other additional pair of corresponding part is BC = QR
âˆ´ Î”ABC â‰… Î”PQR
10. Explain, why Î”ABC â‰… Î”FED
Solution:-
From the figure, it is given that,
âˆ ABC = âˆ DEF = 90^{o}
âˆ BAC = âˆ 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.
Î”ABC â‰… Î”FED