NCERT Solutions for Class 7 Mathematics Chapter 7 - Congruence of Triangles

Answer:

Other four correspondences are:
ABC$\leftrightarrow$PRQ , ABC$\leftrightarrow$QPR
ABC$\leftrightarrow$RPQ , ABC$\leftrightarrow$RQP
All these correspondences will not lead to congruence.
If in any one of thse correspondences, the two triangles coincide then we say that the two triangles are congruent.

Answer:

they have the same length.

Answer:

70$°$

Answer:

m$\angle$A = m$\angle$B.

Answer:

(i) Take two stamps of same denomination.
Place one stamp over the other. We observe that one stamp covers the other completely and exactly. This means that the stamps are of the same shape and same size.

So, two stamps used are congruent to one another.
(ii) Take two blades of the same brand. When we place one blade over the other; we observe that one blade covers the other blade exactly. So; two blades are congruent to one another.

Answer:

If $∆$ABC $\cong$ $∆$FED under the correspondence ABC$\leftrightarrow$FED it follows that congruent parts (i.e., the matching parts) of congruent triangles are equal.
Therefore all the corresponding congruent parts of the triangles are:
AB = FE, BC = ED and AC = FD $\angle$A = $\angle$F, $\angle$B = $\angle$E and $\angle$C = $\angle$D

Answer:

$∆$DEF $\cong$ $∆$BCA. Therefore the correspondence is DEF$\leftrightarrow$BCA.
For better understanding of the correspondence let us use a diagram.

Answer:

In $∆$ABC and $∆$PQR, we have;
AB = PQ (= 1.5), BC = QR (= 2.5) and AC = PR = 2.2
This shows that the three sides of $∆$ABC are equal to the three sides of $∆$PQR.
So, by SSS congruence rule two triangles are congruent.
The correspondence is A$\leftrightarrow$P , B$\leftrightarrow$Q and C$\leftrightarrow$R.
So, in symbolic form $∆$ABC $\cong$ $∆$PQR

Answer:

In $∆$DEF and $∆$LMN, we have;
DE = MN (= 3.2), EF = LM (= 3) and DF = LN (= 3.5)
This shows that the three sides of $∆$DEF are equal to the three sides of $∆$LMN.
So, by SSS congruence rule two triangles are congruent.
The correspondence is D$\leftrightarrow$N, E$\leftrightarrow$M and F$\leftrightarrow$L.
So, in symbolic form $∆$DEF $\cong$ $∆$NML

Answer:

In $∆$ABC and $∆$PQR, we have
BC = PR (= 4), AC = PQ (= 5)
But AB $\ne$ QR [ $\because$ 2 $\ne$ 2.5]
So, SSS congruence rule does not applicable.
Therefore $∆$ABC is not congruent to $∆$PQR.

Answer:

In $∆$ABC and $∆$ADC, we have AB = AC (= 3.5), BD = DC (= 2.5) and AD = AD This shows that the three sides of $∆$ABD are equal to the three sides of $∆$ADC.
So, by SSS congruence rule two triangles are congruent.
The correspondence is A$\leftrightarrow$A , B$\leftrightarrow$C and D$\leftrightarrow$D.
So; in symbolic form $∆$ABD $\ne$ $∆$ACD

Answer:

Answer:

Yes; $∆$ADB $\cong$ $∆$ADC.
Reason: In $∆$ADB and $∆$ADC,
AB = AC (Given)
AD = AD (Common)
DB = DC [$\because$ D is the midpoint of BC (given)]
We observe that three sides of $∆$ADB are equal to the three sides of $∆$ADB are equal to the three sides of $∆$ADC.
So, by SSS congruence rule two triangles are congruent.
The congrunece is A$\leftrightarrow$A , D$\leftrightarrow$D and B$\leftrightarrow$C
So, $∆$ADB $\cong$ $∆$ADC

Answer:

Yes; $\angle$B = $\angle$C
Reason: In $∆$ADB and $∆$ADC,
AB = AC (Given)
AD = AD (Common)
DB = DC [$\because$ D is the midpoint of BC (given)]
We observe that three sides of $∆$ADB are equal to the three sides of $∆$ADB are equal to the three sides of $∆$ADC.
So, by SSS congruence rule two triangles are congruent.
The congrunece is A$\leftrightarrow$A , D$\leftrightarrow$D and B$\leftrightarrow$C
So, $∆$ADB $\cong$ $∆$ADC
Thus B corresponds to C (i.e. B$\leftrightarrow$C ) corresponding parts of congruent triangles are equal.
$\therefore$ $\angle$B = $\angle$C

Answer:

In $∆$ABC and $∆$ABD,
AC = BD (given)
AD = BC (given)
and AB = AB (common)

This shows that the three sides of $∆$ABC are equal to the three sides of $∆$ABD.
So, by SSS congruence rule two triangles are congruent.
The correspondence is A$\leftrightarrow$B , B$\leftrightarrow$A and C$\leftrightarrow$D .
So in symbolic form, $∆$ABC $\cong$ $∆$BAD
Hence, statement (ii) i.e., $∆$ABC $\cong$ $∆$BAD is meaningfully written.

Answer:

Three pairs of equal parts in $∆$ABC and $∆$ACB are as follows:
In $∆$ABC and $∆$ACB,
AB = AC (given)
BC = CB (same side of both triangles)
AC = AB (given)

Answer:

Yes; $∆$ABC $\cong$ $∆$ACB by SSS congruence
In $∆$ABC and $∆$ACB,
AB = AC (given)
BC = CB (same side of both triangles)
AC = AB (given)
Therefore, $∆$ABC $\cong$ $∆$ACB by SSS congruence

Answer:

Yes; $\angle$B = $\angle$C
In $∆$ABC and $∆$ACB,
AB = AC (given)
BC = CB (same side of both triangles)
AC = AB (given)
Therefore, $∆$ABC $\cong$ $∆$ACB by SSS congruence
$\therefore$ $\angle$B = $\angle$C
[Congruent parts of congruent triangles.]

Answer:

Answer:

In $∆$PQR and $∆$FED; we are given equality of two
corresponding sides as PQ = FE and RP = DF.

To establish $∆$PQR $\cong$ $∆$FED under SAS
congruence we need the equality.
Angles included between the sides PQ and RR,
FE and DF.
So, included $\angle$P should be equal to included $\angle$F.

Answer:

In $∆$ABC and $∆$DEF, we have
AB = DE (= 2.5 cm) and AC = DF (= 2.8 cm)
but included $\angle$A $\ne$ included $\angle$D
[ $\angle$A = 80$°$ and $\angle$D = 70$°$ and 80$°$ $\ne$ 70$°$]
So, SAS congruence rule does not applicable.
Therefore $∆$ABC is not congruent to $∆$DEF.

Answer:

In $∆$ABC and $∆$RPQ, we have ;
AC = RP (= 2.5 cm), BC = PQ (= 3 cm) and included $\angle$C = included $\angle$P (= 35$°$) So, by SAS congruence rule two triangles are congruent.
The correspondence is
A$\leftrightarrow$R , C$\leftrightarrow$$\angle$P and $\angle$B $\leftrightarrow$Q
So, in symbolic form $∆$ABC $\cong$ $∆$RQP.

Answer:

$∆$DEF and $∆$PQR
EF = QR (= 3 cm) and DF = PR (= 3.5 cm)
But included $\angle$F $\ne$ included $\angle$R
[$\because$ $\angle$F = 40$°$ given but measure of $\angle$R is not given as 40$°$]
So, SAS congruence rule does not applicable.
Therefore $∆$DEF is not congruent to $∆$PQR.

Answer:

In fig. PQRS is a parallelogram. Diagonal PR divides it into two triangles $∆$PQR and $∆$PRS.
Now, in $∆$PQR and $∆$PRS, we have PQ = SR (= 3.5 cm) PR = PR (common) and included $\angle$P = included $\angle$R (= 30$°$).
So, by SAS congruence rule two triangles are congruent.
The correspondence is P$\leftrightarrow$R , Q$\leftrightarrow$ S , R$\leftrightarrow$P So, in symbolic form $∆$PQR $\cong$ $∆$RSP.

Answer:

Answer:

In $∆$AOC and $∆$BOD
AO = OB,
OC = OD and
included $\angle$AOC = included $\angle$BOD.
This shows that by SAS congruence rule two triangles are congruent.
The correspondence is A$\leftrightarrow$B , O$\leftrightarrow$O , C$\leftrightarrow$D
So in symbolic form $∆$AOC $\cong$ $∆$BOD
Hence, statement (b) i.e. $∆$AOC $\cong$ $∆$BOD is true.

Answer:

The side included between the angles M and N of $∆$MNP are MN.

Answer:

We want to establish $∆$DEF $\cong$ $∆$MNP, using the ASA congruence rule for this, we have given

Also, required included sides are DF = MP

Answer:

Yes, $∆$ABC $\cong$ $∆$DEF
Reason: In $∆$ABC and $∆$FED,
$\angle$A = $\angle$F = 40$°$
AB = EF = 3.5 cm
$\angle$B = $\angle$E = 60$°$
$\therefore$ By ASA congruence rule,
$∆$ABC $\cong$ $∆$FED

Answer:

No, given triangles are not congruent.

Answer:

Yes, $∆$PRQ $\cong$ $∆$MLN
Reason: In $∆$RPQ and $∆$LMN,
$\angle$R = $\angle$L = 90$°$
RQ = LN = 6 cm
$\angle$Q = $\angle$N = 30$°$
$\therefore$ By ASA congruence rule,
$∆$RPQ $\cong$ $∆$LMN

Answer:

Yes, $∆$DAB $\cong$ $∆$CAB,
Reason: In $∆$DAB and $∆$CBA
$\angle$DAB = $\angle$CBA = 45$°$ + 30$°$ = 75$°$
AB = BA (Common side)
$\angle$DBA = $\angle$CAB = 30$°$
$\therefore$ By ASA congruence rule,
$∆$DAB $\cong$ $∆$CAB

Answer:

Yes, $∆$DEF $\cong$ $∆$PQR

Answer:

No, $∆$DEF is not congruent to $∆$PQR.

Answer:

No, $∆$DEF is not congruent to $∆$PQR.

Answer:

Three pairs of equal parts in triangles BAC and DAC are:
$\angle$BAC = $\angle$DAC [$\because$ AZ bisects $\angle$DAB]
$\angle$BCA = $\angle$DCA [$\because$ AZ bisects $\angle$DCB]
AC = CA (Common side)

Answer:

In $∆$BAC and $∆$DAC
$\angle$BAC = $\angle$DAC [$\because$ AZ bisects $\angle$DAB]
$\angle$BCA = $\angle$DCA [$\because$ AZ bisects $\angle$DCB]
AC = CA (Common side)
By ASA congruence rule,
$∆$BAC $\cong$ $∆$DAC

Answer:

Yes, AB = AD
Reason:
In $∆$BAC and $∆$DAC
$\angle$BAC = $\angle$DAC [$\because$ AZ bisects $\angle$DAB]
$\angle$BCA = $\angle$DCA [$\because$ AZ bisects $\angle$DCB]
AC = CA (Common side)
By ASA congruence rule,
$∆$BAC $\cong$ $∆$DAC
$\therefore$ AB = AD (CPCT)

Answer:

Yes, CD = CB
Reason:
In $∆$BAC and $∆$DAC
$\angle$BAC = $\angle$DAC [$\because$ AZ bisects $\angle$DAB]
$\angle$BCA = $\angle$DCA [$\because$ AZ bisects $\angle$DCB]
AC = CA (Common side)
By ASA congruence rule,
$∆$BAC $\cong$ $∆$DAC
$\therefore$ CD = CB (CPCT)

Answer:

No, $∆$PQR is not congruent to $∆$DEF.
Yes, $∆$CAB $\cong$ $∆$DAB
Reason:
$\angle$C = $\angle$D = 90$°$
AB = AB = 3.5 cm
CA = DB = 2 cm
$\therefore$ By RHS congruence rule,
$∆$CAB $\cong$ $∆$DAB

Answer:

Yes, $∆$ABC $\cong$ $∆$ADC
Reason:
$\angle$B = $\angle$D = 90$°$
CA = CA (Common side)
BA = DA = 3.6 cm
$\therefore$ By RHS congruence rule,
$∆$ABC $\cong$ $∆$ADC

Answer:

Yes, $∆$PSQ $\cong$ $∆$PSR
Reason:
$\angle$PSQ = $\angle$PSR = 90$°$
PQ = PR = 3 cm
PS = PS (Common side)
$\therefore$ By RHS congruence rule,
$∆$PSQ $\cong$ $∆$PSR

Answer:

Answer:

Three pairs of equal parts in $∆$CBD and $∆$BCE are:
$\angle$D = $\angle$E = 90$°$
CB = BC (Common side)
BD = CE (Given)

Answer:

Yes, $∆$CBD $\cong$ $∆$BCE.
Reason:
In $∆$CBD and $∆$BCE
$\angle$D = $\angle$E = 90$°$
CB = BC (Common side)
BD = CE (Given)
By RHS Congruence rule,
$\therefore$ $∆$CBD $\cong$ $∆$BCE

Answer:

Yes, $\angle$DCB $\cong$ $\angle$EBC
Reason:
In $∆$CBD and $∆$BCE
$\angle$D = $\angle$E = 90$°$
CB = BC (Common side)
BD = CE (Given)
By RHS Congruence rule,
$\therefore$ $∆$CBD $\cong$ $∆$BCE
$\therefore$ $\angle$DCB = $\angle$EBC (CPCT)

Answer:

Three pairs of equal parts in $∆$ADB and $∆$ADC are:
$\angle$ADB = $\angle$ADC = 90$°$
AB = AC (Given)
AD = AD (Common side)

Answer:

Yes, $∆$ADB $\cong$ $∆$ADC
Reason:
In $∆$ADB and $∆$ADC
$\angle$ADB = $\angle$ADC = 90$°$
AB = AC (Given)
AD = AD (Common side)
By RHS Congruence rule,
$\therefore$ $∆$ADB $\cong$ $∆$ADC

Answer:

Yes, $\angle$B = $\angle$C
Reason:
In $∆$ADB and $∆$ADC
$\angle$ADB = $\angle$ADC = 90$°$
AB = AC (Given)
AD = AD (Common side)
By RHS Congruence rule,
$\therefore$ $∆$ADB $\cong$ $∆$ADC
$\therefore$ $\angle$B = $\angle$C (CPCT)

Answer:

Yes, BD $\cong$ CD
Reason:
In $∆$ADB and $∆$ADC
$\angle$ADB = $\angle$ADC = 90$°$
AB = AC (Given)
AD = AD (Common side)
By RHS Congruence rule,
$\therefore$ $∆$ADB $\cong$ $∆$ADC
$\therefore$ BD = CD (CPCT)

Answer:

In $∆$ABC and $∆$DEF,
Given that,
AC = DF
AB = DE
and BC = EF
So, $∆$ABC $\cong$ $∆$DEF
By SSS congruence criterion.

Answer:

In $∆$RPQ and $∆$ZXY,
Given that,
RP = ZX
RQ = ZY
and $\angle$PRQ = $\angle$XZY
So, $∆$PQR $\cong$ $∆$XYZ
By SAS congruence criterion.

Answer:

In $∆$LMN and $∆$GFH,
Given that,
$\angle$MLN = $\angle$FGH
$\angle$NML = $\angle$GFH
ML = FG
So, $∆$LMN $\cong$ $∆$GFH
By ASA congruence criterion.

Answer:

In $∆$ABE and $∆$CDB,
Given that,
EB = DB
AE = BC
$\angle$A = $\angle$C = 90$°$
So, $∆$ABE $\cong$ $∆$CDB
By RHS congruence criterion.

Answer:

$∆$ART $\cong$ $∆$PEN (Given)
By SSS criterion, then
(i) AR = PE
(ii) RT = EN
(iii) AT = PN

Answer:

$∆$ART $\cong$ $∆$PEN [Given]
and $\angle$T = $\angle$N
Using SAS criterion, then
(i) RT = EN
(ii) PN = AT

Answer:

$∆$ART $\cong$ $∆$PEN (Given)
By using ASA criterion,
Also, AT = PN (Given)
(i) $\angle$RAT = $\angle$EPN
(ii) $\angle$ATR = $\angle$PNE

Answer:

Answer:

No, AAA congruence criterion is not sufficient to say that two triangles are congruent because angle shows the direction (or inclination) of any triangle and not its size.

Answer:

Given that, AT = ON
AR = OW
$\angle$A = $\angle$O
$\angle$R = $\angle$W
and $\angle$T = $\angle$N
then $∆$RAT $\cong$ $∆$WON

Answer:

(i) Given that BT = BC
AT = AC
$\angle$TBA = $\angle$CBA
$\therefore$ $∆$BCA $\cong$ $∆$BTA
(ii) Given that,
$\angle$P = $\angle$R
$\angle$T = $\angle$S
PT = QR
$\therefore$ $∆$QRS $\cong$ $∆$TPQ

Answer:

In a square sheet, draw two triangles of equal areas such that:
(i) the triangles are congruent, then their perimeters are equal.
(ii) the triangles are not congruent, then their perimeters are not equal.

Answer:

Here, $\angle$P = $\angle$A, $\angle$Q = $\angle$B, $\angle$R = $\angle$C
Also, PQ = AB, PR = BC

Answer:

For $∆$ABC and $∆$PQR to be congruent we must need BC = QR
$\because$ $\angle$B = $\angle$Q = 90$°$
and $\angle$C = $\angle$R (Given)
By ASA congruence criterion,
$\therefore$ $∆$ABC $\cong$ $∆$PQR

Answer:

Given that,
$\angle$A = $\angle$F
and $\angle$B = $\angle$E
Adding these, we get
$\angle$A + $\angle$B = $\angle$F + $\angle$E
or 180$°$ – ($\angle$A + $\angle$B) = 180$°$ – ($\angle$F + $\angle$E)
or $\angle$C = $\angle$D
Now, $\angle$B = $\angle$E = 90$°$
BC = ED (Given)
and $\angle$C = $\angle$D (Proved)
By ASA congruence criterion,
$∆$ABC ≅ $∆$FED