.: Side opposite to right angle is hypotenuse
:. In ΔABC, hypotenuse = AC and in Δ PQR, hypotenuse=RQ

We have  [each =8 cm] [each =4cm]
\(\therefore \)The two triangles are congruent
Now, B ↔️P, A ↔️R and C ↔️ Q [By RHS congruence rule]
\(\therefore \)ΔABC≅ΔRPQ

ΔPQR and ΔLMN:
We have 


Therefore, using ASA congruence rule, the two triangles are congruent.
Now, R \(\leftrightarrow \) L,Q \(\leftrightarrow \) N and P\(\leftrightarrow \) M
\(\therefore \)ΔPQR ≅ ΔMNL

ΔPQR and ΔDEF:
In ΔPQR, ㄥP=ㄥ180° - (90° + 50°)= 40°
Also in ΔDEF, ㄥF = 180° - 90° - 50°=40°
Now in ΔPQR and ΔDEF, we have


and 
\(\therefore \)Using the ASA congruence rule, we can say that the two triangles are not congruent

In ∆ABC and ΔDCB,
AC= DB
\(\angle\) BAC = \(\angle\) CDB = 90°
BC= BC
So by R.H.S. congruency we have
ΔABC ≌ΔDBC
\(\Rightarrow\) by C.P.C T., AB = DC

In ΔABD and ΔADC, we have
AB= AC =35 cm, BD= DC = 25 cm and AD = AD [common]
Therefore, the three sides of MBD are equal to three sides of ΔADC.
So, two triangles are congruent by SSS congruence rule.
From the above equality relations, we have
A ↔️ A, B ↔️ C, and D ↔️ D
In symbolic form, ΔABD ≅ ΔACD

In ΔDEF and ΔLMN, we have
DE= MN =3.2 cm, DF= LN = 35 cm, EF = LM = 3 cm
Therefore, the three sides of ΔDEF are equal to three sides of ΔLMN.
So, two triangles are congruent by SSS congruence rule.
From the above equality relations, we have
D ↔️ N, E ↔️ M and F ↔️ L
In symbolic form, ΔDEF ≅ ΔNML

Given, ΔABC = ΔPQR
Also given ㄥB = ㄥQ and ㄥC = ㄥR
[from the given figure)
To apply the condition for congruency, included side of one triangle is equal to the included side of the other triangle.
∴ BC=QR
Hence, we use the ASA congruence criterion.

Area \(\triangle PQR=\frac { 1 }{ 2 } \times PQ\times PR=\frac { 1 }{ 2 } \times 4\times 3=6\quad cm^{ 2 }\)
Area \(\triangle PRS=\frac { 1 }{ 2 } \times PR\times ST=\frac { 1 }{ 2 } \times 4\times 3=6\quad cm^{ 2 }\quad \)
∵ Area of ΔPQR = Area of ΔPRS
But these two triangles are not congruent. Perimeter of ΔPQR
= PQ + QR + RP = 3 + 5 + 4 = 12 ern
and perimeter of ΔPRS = PR + RS + SP = 4 + 3 + 3 = 10 cm
∵ Perimeter of ΔPQR ≠ Perimeter of ΔPRS
Hence, we can say that perimeters of two congruent triangles are equal and perimeters of two non-congruent triangles need not be equal.

Area \(\triangle ABC=\frac { 1 }{ 2 } \times AB\times BC=\frac { 1 }{ 2 } \times 4\times 3=6\quad cm^{ 2 }\)
Area \(\triangle EDC=\frac { 1 }{ 2 } \times DE\times CD=\frac { 1 }{ 2 } \times 4\times 3=6\quad cm^{ 2 }\)

 Area of ΔABC = Area of ΔEDC
Also, three sides of MBC are equal to three corresponding sides of ΔEDC, so these two triangles are congruent.
Perimeter of ΔABC = AB + BC + CA = 4 + 3 + 5 = 12 cm
and perimeter of ΔEDC = CD + DE + EC = 3 + 4 + 5 = 12 cm
∴ Perimeter of ΔABC = Perimeter of ΔEDC.

According to the question, we have following figures.

Since, both triangles are isosceles.
DE = DF                              [given]
LM = LN                             [given]
where, DE = LM               [given]
EF = MN                          [given]
Also, DF = LN [∵ DE = DF, LM = LN]
Since, three sides of triangles ΔDEF and ΔLMN are equal.
So, ΔDEF ≅ ΔLMN      [by SSS congruence criterion]
If ㄥE = 40⁰  , then ㄥN = 40⁰   [by CPCT]

In the given figure,

Since, ㄥ B = ㄥP     [each 90⁰]
AB = RP            [height of right angled triangle]
If we add AC = RQ
[hypotenuse of right angled side]
Hence, ΔABC ≅ ΔRPQ
[by RHS congruence criterion]

Here, we want to establish ΔDEF ≅ ΔMNP, by using ASA congruence rule.
Given,  ㄥD = ㄥM
and ㄥF= ㄥP
So, the additional information needed to establish the congruence is
DF=MP      [side included between given angles]
Hence, ΔDEF \(≅\) ΔMNP

Given, AC = BD and AD = BC
In ΔABC and ΔABD, AC = BD [given]
BC = AD [given]
AB = AB [common]
Therefore, i.e. the three sides of ΔABC are equal to the three sides of ΔABD. So, the two triangles are congruent.
From the above equality relations, we have
A ↔️ B, B ↔️ A and C ↔️ D
(i) ΔABC ≅ ΔABD is false or meaningless.
(ii) ΔABC ≅ ΔBAD is true or meaningful.

If ΔABC ≅ ΔFED under the correspondence ABC ↔️ FED, this means A↔️ F, B ↔️ E and C ↔️D.
Therefore, all the corresponding congruent parts of ΔABC and ΔFED are ㄥB ↔️ ㄥE, ㄥC ↔️ ㄥD and \(\bar { AB } \leftrightarrow \bar { FE } \bar { ,BC } \leftrightarrow \bar { ED } ,\bar { CA } \leftrightarrow \bar { DF } \).

In Δ ABC and Δ PQR, there are side possible matchings or correspondences. Out of them, four correspondences are as follow:
(i) ABC ↔️ PRQ
(ii) ABC ↔️ RPQ
(iii) ABC ↔️ RQP
(iv) ABC ↔️ QPR
Yes, all these correspondences may lead to congruence.