Congruence of Triangles
Exercise 7.2
1). Which congruence criterion do you use in the following?
(a)
Given: AC = DF
AB = DE
BC = EF
So, DABC @ DDEF
Ans: DABC @ DDEF by SSS rule
(b)
Given: ZX = RP
RQ = ZY
ÐPRQ = ÐXZY
So, DPQR @ DXYZ
Ans: DPQR @ DXYZ by SAS rule
(c)
Given: ÐMLN = ÐFGH
ÐNML = ÐGFH
ML = FG
So, DLMN @ DGFH
Ans: DLMN @ DGFH by ASA rule
(d)
Given: EB = DB
AE = BC
ÐA = ÐC = 90°
So, DABE @ DCDB
Ans: DABE @ DCDB by RHS rule
2). You want to show that DART @ DPEN,
(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 ÐT = ÐN 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) ?
Solution:
(a) If you have to use SSS criterion, then you need to show
(i) AR @ PE
(ii) RT @ EN
(iii) AT @ PN
(b) If it is given that ÐT = ÐN and you are to use SAS criterion,
you need to have
(i) RT @ EN and
(ii) PN @ AT
(c) If it is given that AT = PN and you are to use ASA criterion,
you need to have
(i) ÐA @ ÐP
(ii) ÐT @ ÐN
3). You have to show that DAMP @ DAMQ. In the following proof, supply the missing reasons.
Steps Reasons
(i) PM @ QM (i) . . . . . . . . . . . . . . .
(ii) ÐPMA @ ÐQMA (ii) . . . . . . . . . . . . . . .
(iii) AM @ AM (iii) . . . . . . . . . . . . . . .
(iv) DAMP @ DAMQ (iv) . . . . . . . . . . . . . . .
Solution:
Steps Reasons
(i) PM @ QM (i) given
(ii) ÐPMA @ ÐQMA (ii) given
(iii) AM @ AM (iii) Common Side
(iv) DAMP @ DAMQ (iv) SAS Rule
4). In DABC, ÐA = 30° , ÐB = 40° and ÐC = 110° In DPQR, ÐP = 30° , ÐQ = 40° and ÐR = 110°. A student says that DABC @ DPQR by AAA congruence criterion. Is he justified? Why or why not?
Ans: No. The student is not justified because there is no AAA congruence criterion.
5). In the figure, the two triangles are congruent. The corresponding parts are marked. We can write DRAT @ ?
In DRAT and DWON
side AR @ side OW given
ÐA @ ÐO given
side AT @ side ON given
DRAT @ DWON by SAS Rule
We can also show that the two triangles are congruent by ASA Rule
II Method
In DRAT and DWON
ÐA @ ÐO given
side AT @ side ON given
ÐT @ ÐN given
DRAT @ DWON by ASA Rule
6). Complete the congruence statement:
(i).
In DBCA and DBTA
ÐABC @ÐABT given
side BC @ side ON given
ÐC @ ÐT given
DBCA @ DBTA by ASA Rule
(ii).
In DQRS and DTPQ
ÐR @ ÐP given
side QR @ side TP given
ÐRSQ @ ÐPQT given
DQRS @ DTPQ by ASA Rule
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:
(i) the triangles are congruent.
In DABC and DPQR
side AB @ side PQ given
side AC @ side PR given
side BC @ side QR given
DABC @ DPQR by SSS Rule
AB + BC + AC = PQ + PR + QR
Perimeter of DABC = Perimeter of DPQR
(ii) the triangles are not congruent.
In DABC and DPQR
side AB ≠ side PQ given
side AC ≠ side PR given
side BC ≠ side QR given
DABC ≠ DPQR by SSS Rule
AB + BC + AC ≠ PQ + PR + QR
Perimeter of DABC ≠ Perimeter of DPQR
8). Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still, the triangles are not congruent.
9). If DABC and DPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?
In DABC and DPQR
ÐB @ ÐQ given
ÐC @ ÐR given
The additional pair of corresponding parts so that the two triangles are congruent can be
side QR @ side TP
DABC @ DPRQ by ASA Rule
10). Explain, why DABC @ DFED.
In DABC and DFED
ÐA @ ÐF given (I)
ÐB @ ÐE given (II)
In DABC
By angle sum property of triangle
ÐA + ÐB+ ÐC = 1800
ÐC = 1800 – (ÐA + ÐB) (III)
Similarly In DFED
ÐF + ÐE+ ÐD = 1800
ÐD = 1800 – (ÐF + ÐE) (IV)
From I, II, III and IV, we have
ÐC = ÐÐD (V)
In DABC and DFED
ÐB @ ÐE given
side BC @ side ED given
ÐC @ ÐD by (V)
DABC @ DFED by ASA Rule
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