If $$AB=QR$$, $$BC=PQ$$ and $$CA=PR$$, then:

$$\triangle ABC\cong \triangle PQR$$

$$\triangle BAC\cong \triangle RPQ$$

$$\triangle PQR\cong \triangle BCA$$

MCQ Questions for Class 7 Maths Congruence Of Triangles Quiz 4

In the given figure,

$\triangle RTQ\cong \triangle PSQ$ by ASA congruency condition. Which of the following pairs does not satisfy the condition?

0%
$PQ=QR$
0%
$\angle P=\angle R$
0%
$\angle TQP=\angle SQR$
0%
None of these

Explanation

Given,

$\triangle RTQ \cong \triangle PSQ$
Thus,

$PQ = QR$ (by cpct)

$\angle P = \angle R$ (by cpct)

Also,

$\angle RQT = \angle PQS$ (By CPCT)

$\angle RQT - \angle TQS = \angle PQS - \angle TQS$
Thus,

$\angle TQP = \angle SQR$

In the following fig. if AB = AC and BD = DC then

$\angle ADC\, =$

0%
$60^{\circ}$
0%
$120^{\circ}$
0%
$90^{\circ}$
0%
None

Explanation

Given that, in

$\triangle ABC$ ,

$AB = AC, \ BD = DC$

To find out:

$\angle ADC$

$\triangle ADB$ and

$\triangle ADC$ ,

$AB = AC\quad \quad [given]$

$BD = DC\quad \quad [given]$

Also,

$AD = AD\quad \quad [common]$

$\therefore \$ By

$SSS$ congruence criterion,

$\Delta ADB\, \cong\, \Delta ADC$ .

Hence,

$\angle ADB=\angle ADB$
Also,

$\angle ADB\, +\, \angle ADC\, =\, 180^{\circ}\quad [linear\ pair]$

$\therefore \ \angle ADB\, =\, \angle ADC\, =\, 90^{\circ}$

Hence,

$\angle ADC=90^\circ$

$AB=QR$ ,

$BC=PQ$ and

$CA=PR$ , then:

0%
$\triangle ABC\cong \triangle PQR$
0%
$\triangle CBA\cong \triangle PQR$
0%
$\triangle BAC\cong \triangle RPQ$
0%
$\triangle PQR\cong \triangle BCA$

Explanation

$\triangle ABC$ and

$\triangle PQR$

$AB=QR$

$BC=PQ$

$CA=PR$
Thus,

$\triangle CBA \cong \triangle PQR$ (SSS rule)

In the given figure,

$OA=OB, OC=OD, \angle AOB=\angle COD.$ Which of the following statements is true

$\: ?$

0%
$AC=CD$
0%
$OA=OD$
0%
$AC=BD$
0%
$\angle OCA=\angle ODC$

Explanation

$\triangle$ s

$AOC$ and

$BOD$

$OA=OB$ (Given)

$OC=OD$ (Given)

$\angle AOB-\angle COB=\angle COD-\angle COB,$
i.e.,

$\angle AOC=\angle BOD$

$\therefore$

$\triangle AOC\cong \triangle BOD$ (SAS)

$\Rightarrow$

$AC=BD$ (cpct)

In the figure,

$\displaystyle \Delta ABC\cong \Delta ADC$ by the congruence postulate

0%
SSS
0%
RHS
0%
SAS
0%
ASA

In the figure,

$\displaystyle AD\parallel BC$ and

$\displaystyle AD=BC$ . Then

$\displaystyle \Delta ABD\cong \Delta CDB$ by congruence postulate.

0%
SSS
0%
ASA
0%
AAS
0%
SAS

Explanation

$\displaystyle \Delta ABD$ and

$\displaystyle \Delta CDB$

$\displaystyle AD=BC$

$\displaystyle \angle BDA=\angle DBC$ (Alternate angles)

$\displaystyle BD=BD$

$\displaystyle \therefore \quad \Delta ABD\cong \Delta CDB$ (By SAS postulate)

$\overline{AC}$ and

$\overline{BD}$ intersect at

$O$ such that

$AO = CO$ and

$BO = DO$ , then:

0%
$AB = CD$
0%
$AB \parallel CD$ and $AB = CD$
0%
$AB \parallel CD$
0%
None of these

Explanation

$\triangle AOB$ and

$\triangle DOC$ ,

side

$OA =$ side

$OC$ ... given

side

$OB =$ side

$OD$ ... given

$\angle AOB \cong \angle COD$ ...Vertically opposite angles

$\therefore \triangle AOB \cong \triangle COD$ ...SAS test of congruence.

$\therefore$ side

$AB$

$\cong$ side

$CD$ ...c.s.c.t

$\therefore \angle BAO \cong \angle DCO$ ....c.a.c.t

i.e.

$\angle BAC \cong \angle DCA$ ....

$A-O-C$ and

$B-O-D$

$\therefore AB \parallel CD$ ....by converse of alternate angles test

So, option B is correct.

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DBC,AC=BD$ and

$\displaystyle \angle ABC=\angle BCD={ 90 }^{ o }$ . Then

$\displaystyle \Delta ABC\cong \Delta DCB$ by the postulate

0%
SSS
0%
SAS
0%
RHS
0%
ASA

Explanation

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DBC$ ,

$\displaystyle \angle ABC=\angle BCD$

$\displaystyle AC=BD$ (Givenn)

$\displaystyle BC=BC$ (Common Side)

$\displaystyle \Delta ABC\cong \Delta DCB$ WCB (By RHS postulate)

In the given figure,

$AD=BC, AC=BD.$ Then

$\triangle PAB$ is

0%
equilateral
0%
right angled
0%
scalene
0%
isosceles

Explanation

$\triangle ADB$ and

$\triangle ACB$

$AD=BC$ (Given)

$AC=BD$

$AB=BA$ (Common)

$\therefore$

$\triangle ADB\cong \triangle ACB$ (SSS)

$\Rightarrow$

$\angle ABD=\angle CAB$ (cpct)

$\Rightarrow$

$\angle ABP=\angle PAB$

$\Rightarrow$

$PA=PB$ (Sides opp. equal angles are equal)

$\Rightarrow$

$\triangle PAB$ is isosceles.

It is given that

$AB=BC$ and

$AD=EC.$ The

$\triangle ABE\cong \triangle CBD$ by __________ congruency.

0%
$SSS$
0%
$ASA$
0%
$SAS$
0%
$AAS$

Explanation

Given,

$AD=EC$

$\Rightarrow$

$AD+DE=DE+EC$

$\Rightarrow$

$AE=DC$
Also,

$AB=BC$

$\Rightarrow$

$\angle BCA=\angle BAC$ (isos.

$\triangle$ property)

$\Rightarrow$

$\angle BCD=\angle BAE$

$\therefore$ In

$\triangle$ s

$ABE$ and

$CBD,$

$AB=CB$ (Given)

$AE=DC$ (proved above)

$\angle BAE=\angle BCD$ (Proved above)

$\therefore$

$\triangle ABE\cong \triangle CBD$ (SAS)

In the figure,

$\displaystyle PS=QR$ and

$PQ=SR$ , then choose the correct option.

0%
$\displaystyle \Delta QRP\cong \Delta PSR$
0%
$\displaystyle \Delta PQR\cong \Delta PRS$
0%
$\displaystyle \Delta PQR\cong \Delta RSP$
0%
$\displaystyle \Delta RQP\cong \Delta RSP$

Explanation

$\displaystyle \Delta PQR$ and

$\displaystyle \Delta PSR$

$\displaystyle PS=QR,PQ=SR,PR=PR$

$\displaystyle \therefore \quad \Delta RQP\cong \Delta RSP$ (By SSS postulate)

The two quadrilaterals are congruent.
Which angle in quadrilateral WXYZ corresponds to

$\displaystyle \angle BCD$ in quadrilateral ABCD?

0%
$\displaystyle \angle XYZ$
0%
$\displaystyle \angle XWZ$
0%
$\displaystyle \angle WZY$
0%
$\displaystyle \angle WXY$

Explanation

Clearly,

$\angle WXY$ corresponds to

$\angle BCD$ because

$BC$ faces the angles marked with one arc and three arc and

$XY$ also faces the angles marked with one arc and three arcs.

So,

$\text{D}$ is the correct option.

$\displaystyle \bigtriangleup ABC,AD\perp BC,\angle B=\angle C$ and AB = AC State by which property

$\displaystyle \Delta ADB\cong \Delta ADC$ ?

0%
RHS property
0%
SSS property
0%
SAS property
0%
ASA property

For a

$\displaystyle \Delta XYZ, ZY = 12\ m, YX = 8\ m$ and

$XZ = 10\ m$ . If

$\displaystyle \Delta ZYX\cong \Delta ABC$ , then

$AC =$ _____

$m$ .

0%
$10$
0%
$12$
0%
$8$
0%
$1$

Explanation

Given :

$\displaystyle \Delta XYZ, ZY = 12\ m, YX = 8\ m$ and

$XZ = 10\ m$ .

Also,

$\displaystyle \Delta ZYX\cong \Delta ABC$ .

We know, congruent triangles have their sides congruent.

$\implies AB=ZY, BC=YX$ and

$AC=ZX$ .

$\implies \displaystyle ZX=AC$ .

$\displaystyle \therefore AC=10\ m$ .

Therefore, option

$A$ is correct.

In the figure given below,

$\displaystyle \Delta ABC\cong \Delta MBC.$ The corresponding side to

$BA$ is ......

0%
$BM$
0%
$AB$
0%
$AC$
0%
$MC$

Explanation

Given $\triangle ABC \cong \triangle MBC$ .

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $AB=MB$ , $BC=BC$ , $AC=MC$ , $\angle A=\angle M$ , $\angle B=\angle B$ and $\angle C=\angle C$ .

Thus, $AB=MB$ $\implies$ $BA=BM$ .

Hence, option $A$ is correct.

In the figure given below,

$\displaystyle \Delta ABC\cong \Delta XYZ$ . The corresponding side to

$XZ$ is .......

0%
$AB$
0%
$BC$
0%
$AC$
0%
$YZ$

Explanation

Given $\triangle ABC \cong \triangle XYZ$ .

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $AB=XY$ , $BC=YZ$ , $AC=XZ$ , $\angle A=\angle X$ , $\angle B=\angle Y$ and $\angle C=\angle Z$ .

Thus, $AC=XZ$ .

Hence, option $C$ is correct.

In the triangles

$ABC$ and

$PQR,$

$AC=QP, BC=RP$ and

$AB=QR.$ The measure of

$\angle A$ is ___.

0%
$\displaystyle { 30 }^{ \circ }$
0%
$\displaystyle { 120 }^{ \circ }$
0%
$\displaystyle { 20 }^{ \circ }$
0%
$\displaystyle { 40 }^{ \circ }$

Explanation

Given: In

$\triangle ABC$ and

$\triangle PQR,$

$\angle C=120^\circ$ and

$\angle R=40^\circ$

Now,

$AC=QP$ ...(1)

and

$BC=RP$ ...(2)

and

$AB=QR$ ...(3)

Hence

$ABC \cong PQR$

$($ by

$SSS$ congruency rule

$)$

By CPCT,

$\angle B=\angle R=40^\circ$

and

$\angle C=\angle P={ 120 }^{ \circ }$

and

$\angle A=\angle Q$

Now,

$\displaystyle \angle A+\angle B+\angle C={ 180 }^{ o }$

$\displaystyle \Rightarrow \angle A+{ 40 }^{ \circ }+{ 120 }^{ \circ }={ 180 }^{ \circ }$

$\displaystyle \Rightarrow \angle A={ 20 }^{ \circ}$

In the figure given below,

$\displaystyle \Delta BCA\cong \Delta BCD$ . Corresponding angle to

$\displaystyle \angle D$ is ...........

0%
$\displaystyle \angle B$
0%
$\displaystyle \angle C$
0%
$\displaystyle \angle D$
0%
$\displaystyle \angle A$

Explanation

Given $\triangle BCA \cong \triangle BCD$ .

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $BC=BC$ , $CA=CD$ , $BA=BD$ , $\angle B=\angle B$ , $\angle C=\angle C$ and $\angle A=\angle D$ .

Thus, $\angle A=\angle D$ .

Hence, option

$D$ is correct.

In the figure given below, both the triangles are congruent. The corresponding side to

$BC$ is .......

0%
$EF$
0%
$AC$
0%
$DF$
0%
$DE$

Explanation

Given $\triangle ABC \cong \triangle DEF$ .

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $AB=DE$ , $BC=EF$ , $AC=DF$ , $\angle A=\angle D$ , $\angle B=\angle E$ and $\angle C=\angle F$ .

Thus, $BC=EF$ .

Hence, option

$A$ is correct.

Which condition for congruence is used to prove that the following pairs of triangles are congruent

0%
$ASA$
0%
$AAS$
0%
$SAS$
0%
$SSS$

Explanation

In the given

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$ ,

we can clearly see that sides

$AB = DE, BC = EF$ and

$AC = DF.$

Thus, by

$SSS$ condition of congruence, the given

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$ are congruent.

In the figure given below, both the triangles are congruent. The corresponding side to

$AB$ is ___.

0%
$AB$
0%
$DE$
0%
$BC$
0%
$DF$

Explanation

Given $\triangle ABC \cong \triangle DEF$ .

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $AB=DE$ ,

$BC=EF$ ,

$AC=DF$ ,

$\angle A=\angle D$ ,

$\angle B=\angle E$

$\angle C=\angle F$ .

Thus, $AB=DE$ .

Hence, option $B$ is correct.

It is given that

$\Delta ABC\cong \Delta DEF$ . In the given figure the corresponding angle to

$\displaystyle \angle C$ is ........

0%
$\displaystyle \angle B$
0%
$\displaystyle \angle C$
0%
$\displaystyle \angle F$
0%
$\displaystyle \angle A$

Explanation

Given $\triangle ABC \cong \triangle DEF$ .

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $AB=DE$ , $BC=EF$ , $AC=DF$ , $\angle A=\angle D$ , $\angle B=\angle E$ and $\angle C=\angle F$ .

Thus, $\angle C=\angle F$ .

Hence, option

$C$ is correct

It is given that

$\Delta ABC\cong \Delta DEF$ . In the given figure the corresponding angle to

$\displaystyle \angle B$ is___.

0%
$\displaystyle \angle A$
0%
$\displaystyle \angle C$
0%
$\displaystyle \angle E$
0%
$\displaystyle \angle F$

Explanation

Given $\triangle ABC \cong \triangle DEF$ .

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $AB=DE$ , $BC=EF$ , $AC=DF$ , $\angle A=\angle D$ , $\angle B=\angle E$ and $\angle C=\angle F$ .

Thus,

$\angle B=\angle E$ .

Hence, option

$C$ is correct.

Referring the figure below, for

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$ ,

$AB=DE, BC=EF$ and

$\displaystyle \angle B=\angle E.$ The value of

$\displaystyle \angle E$ is .........

0%
$\displaystyle { 30 }^{ o }$
0%
$\displaystyle { 60 }^{ o }$
0%
$\displaystyle { 90 }^{ o }$
0%
$\displaystyle { 45 }^{ o }$

Explanation

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$

$\displaystyle AB=DE,BC=EF$ and

$\displaystyle \angle B=\angle E$

$\displaystyle \therefore \quad \Delta ABC\cong \Delta DEF$ (S.A.S condition for congruence)
As,

$\displaystyle AB=DE\Rightarrow \angle C=\angle F={ 60 }^{ o }$

$\displaystyle BC=EF\Rightarrow \angle A=\angle D={ 60 }^{ o }$

$\displaystyle AC=DF\Rightarrow \angle B=\angle E$
Now,

$\displaystyle \angle D+\angle E+\angle F={ 180 }^{ o }$

$\displaystyle \therefore \quad { 120 }^{ o }+\angle E={ 180 }^{ o }$

$\displaystyle \therefore \quad \angle E={ 60 }^{ o }$

In a rectangle

$PQRS$ ,

$QS$ is a diagonal. Then

$\displaystyle \Delta PQS......\Delta RSQ$

0%
is similar to
0%
is not equal to
0%
is congruent to
0%
is not congruent to

Explanation

In a rectangle

$PQRS$ ,

$QS$ is a diagonal.
We know that , in a rectangle opposite sides are equal.

$\displaystyle \therefore PQ=SR$
Also,

$\displaystyle \angle R=\angle P={ 90 }^{ o }$
Ans

$QS$ is the hypotenuse for both right angled triangles

$PQS$ and

$RSQ$ .

$\displaystyle \therefore$ according to R.H.S condition for congruence,

$\displaystyle \Delta PQS\cong \Delta RSQ$ .

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF,\angle A={ 47 }^{ o },\angle E={ 53 }^{ o }$ and

$\displaystyle AB=DE=16 \ cm\\$ .

$\displaystyle \angle A=\angle D$ and

$\displaystyle \angle B=\angle E$ , then

$\displaystyle \angle C=$ ..............

0%
$\displaystyle { 47 }^{ o }$
0%
$\displaystyle { 53 }^{ o }$
0%
$\displaystyle { 80 }^{ o }$
0%
$\displaystyle { 100 }^{ o }$

Explanation

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$

$\displaystyle \angle A=\angle D,\angle B=\angle E$ and

$\displaystyle AB=DE$

$\displaystyle \therefore$ According to A.S.A condition for congruence,

$\displaystyle \Delta ABC\cong \Delta DEF$

$\displaystyle \angle A=\angle D={ 47 }^{ o }$

$\displaystyle \angle B=\angle E={ 53 }^{ o }$

$\displaystyle \angle C=\angle F$

$\displaystyle \therefore \quad \angle D+\angle E+\angle F=\angle A+\angle B+\angle C={ 180 }^{ o }\\$

$\displaystyle \therefore \quad \angle C={ 180 }^{ o }-{ 100 }^{ o }$

$\displaystyle \therefore \quad \angle C={ 80 }^{ o }$

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta PQR$ ,

$AB=PQ, AC=PR,$ and

$\displaystyle \angle A=\angle P,$ then:

0%
$\displaystyle \Delta BAC\cong \Delta PRQ$
0%
$\displaystyle \Delta BAC\cong \Delta PQR$
0%
$\displaystyle \Delta ABC\cong \Delta PRQ$
0%
$\displaystyle \Delta ABC\cong \Delta PQR$

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF,$

$AB=FE, BC=ED$ and

$\displaystyle \angle B=\angle E$ . Therefore ............

0%
$\displaystyle \Delta ABC\cong \Delta DEF$
0%
$\displaystyle \Delta ABC\cong \Delta EDF$
0%
$\displaystyle \Delta BCA\cong \Delta DEF$
0%
$\displaystyle \Delta BCA\cong \Delta EDF$

Explanation

For,

$\displaystyle \Delta BCA$ and

$\displaystyle \Delta EDF,$

$\displaystyle AB=FE\quad \therefore \angle C=\angle D$

$\displaystyle BC=ED\quad \therefore \angle A=\angle F$

$\displaystyle \angle B=\angle E$

$\displaystyle \Delta BCA\cong \Delta DEF$
(By

$SAS$ condition for congruence).

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$ ,

$\displaystyle \angle B=\angle D,\angle C=\angle F$ and

$BC=DF$ . Therefore which of the following is correct?

0%
$\displaystyle \Delta ABC\cong \Delta DEF$
0%
$\displaystyle \Delta BCA\cong \Delta DFE$
0%
$\displaystyle \Delta BCA\cong \Delta DEF$
0%
$\displaystyle \Delta ABC\cong \Delta DFE$

Explanation

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$ , we have

$\displaystyle \angle B=\angle D,\angle C=\angle F$ .....Given

$BC=DF$ .....Given
Therefore, according to A.S.A condition for congruence,

$\displaystyle \Delta BCA\cong \Delta DFE$

Hence, option B is correct.

$\Delta ABC$ and

$\Delta DEF$ are congruent, then find the value of

$x$ and

$y$ .

0%
$x = 40, y = 60$
0%
$x = 30, y = 60$
0%
$x = 60, y = 60$
0%
$x = 60, y = 40$

Explanation

We know

$\Delta ABC \cong \Delta DEF$ by ASA congruence.
Therefore,

$\angle A = \angle D=40$

$\therefore x = 40$

$\angle B = \angle F=60$

$\therefore y=60$
So,

$x = 40, y = 60$

CBSE Questions for Class 7 Maths Congruence Of Triangles Quiz 4 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Congruence Of Triangles Quiz 4 - MCQExams.com

0:0:1

Practice Class 7 Maths Quiz Questions and Answers

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