MCQ Questions for Class 7 Maths Congruence Of Triangles Quiz 1

Given that

$\Delta ABC = \Delta FDE$ and

$AB = 5 cm, \angle B = 40^{0}$ and

$\angle A = 80^{0}$ then which of the following relation is true?

0%
$DF=5cm,\angle F=60^{0}$
0%
$DF=5cm,\angle E=60^{0}$
0%
$DE=5cm,\angle E=60^{0}$
0%
$DE=5cm,\angle D=40^{0}$

ABC is an isosceles triangle with AB

$=$ AC and D is a point on BC such that

$AD \perp BC$ (Fig. 7.13). To prove that

$\angle BAD = \angle CAD,$ a student proceeded as follows:

$\Delta ABD$ and

$\Delta ACD,$
AB

$=$ AC (Given)

$\angle B = \angle C$ (because AB

$=$ AC)
and

$\angle ADB = \angle ADC$
Therefore,

$\Delta ABD \cong \Delta ACD (AAS)$
So,

$\angle BAD = \angle CAD (CPCT)$
What is the defect in the above arguments?

0%
It is defective to use $\angle ABD = \angle ACD$ for proving this result.
0%
It is defective to use $\angle ADB = \angle ADC$ for proving this result.
0%
It is defective to use $\angle BAD = \angle DCA$ for proving this result.
0%
Cannot be determined

Explanation

$\bigtriangleup ABD$ and

$\bigtriangleup ACD$

AB=AC (given )

Then

$\angle ABD=\angle ACD$ ( because AB=AC )

and

$\angle ADB=\angle ADC=90$ ( because AD⊥BC )

$\therefore \bigtriangleup ABD=\bigtriangleup ACD$

$\angle BAD=\angle CAD$

It is defective to use

$\angle ABD=\angle ACD$ for proving this result

Given

$\Delta OAP \cong \Delta OBP$ in figure, the criteria by which the triangles are congruent is:

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

Explanation

$\triangle OAP$ and

$\triangle OBP$

$OA=OB$ (given)

$\angle AOP=\angle BOP$ (given)

$OP=OP$ (common side)

$\therefore \triangle OAP\cong \triangle OBP$ by

$SAS$ congruent rule as two corresponding sides and the included angles of the triangles are equal.

$\Delta ABC \cong \Delta DEF$ by SSS congruence rule then

0%
$AB=EF,BC=FD,CA=DE$
0%
$AB=FD,BC=DE,CA=EF$
0%
$AB=DE,BC=EF,CA=FD$
0%
$AB=DE,BC=EF,\angle C = \angle F$

Explanation

If triangles are congruent by

$SSS$ then their corresponding sides are equal , therefore

$AB=DE$

$BC=EF$

$AC=DF$ or

$CA=FD$

So option

$C$ is correct.

$\Delta ABC$ and

$\Delta DEF$ , AB = DF and

$\angle A = \angle D$ . The two triangles will be congruent by SAS axiom if :

0%
BC = EF
0%
AC = DE
0%
BC = DE
0%
AC = EF

Explanation

For triangle to be congruent

$SAS$ (Side angle Side) axiom we need to have two equal sides and the included angle same.

$\Delta ABC$ and

$\Delta DFE$

We have

$AB=DF$

and

$\angle A=\angle D$

But

$\angle A$ is formed by the two sides AB and AC of

$\Delta ABC$

But

$\angle D$ is formed by the two sides DE and DF of

$\Delta DFE$

So we need

$AC=DE$ for the two triangles to be congruent.

In triangles ABC and DEF, AB

$=$ FD and

$\angle A = \angle D$ . The two triangles will be congruent by
SAS axiom if :

0%
BC $=$ EF
0%
AC $=$ DE
0%
AC $=$ EF
0%
BC $=$ DE

Explanation

For triangle to be congruent by

$SAS$ axiom two of the sides and the included angle of the triangles must be equal.

Given

$AB=DF$ and

$\angle A=\angle D$

No for

$\triangle ABC\cong \triangle DFE$ by

$SAS$ axiom we need

$AC=DE$

So option

$B$ is correct.

$\Delta ABC$ and

$\Delta DEF$ ,

$AB=FD$ ,

$\angle A= \angle D$ . The two triangles will be congruent by SAS axiom if :

0%
$BC=DE$
0%
$AC=EF$
0%
$BC=EF$
0%
$AC=DE$

Explanation

For the triangles to be congruent by

$SAS$ axiom we need two sides and included angle of given triangles to be equal.

Given:

$AB=FD$ and

$\angle A=\angle D$

Now from the figures, we can say that we need the second pair of sides which is

$AC$ and

$DE$ according to the given definition.

So for given triangles to be congruent by

$SAS$ criteria

we need

$AC=DE$

Option

$D$ is correct.

In the given figure, If OA=OB,OD=OC, then

$\Delta AOD \cong \Delta BOC$ by congruence rule:

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

Explanation

$\triangle AOD$ and

$\triangle BOC$

$OA=OB$ (given)

$OD=OC$ (given)

$\angle AOD=\angle BOC$ (vertically opposite angle)

$\triangle AOD\cong \triangle BOC$ by

$SAS$ congruence rule.

$\Delta ABC \cong \Delta DEF$ by SSS congruence rule then :

0%
$AB=EF, BC=FD, CA=DE$
0%
$AB=FD, BC=DE, CA=EF$
0%
$AB=DE, BC=EF, CA=FD$
0%
$AB=DE, BC=EF, \angle C=\angle F$

Explanation

If triangles are congruent by

$SSS$ congruence rule then their corresponding sides are equal.

Here

$\triangle ABC\cong \triangle DEF$

$\therefore AB=DE,BC=EF,AC=DF$ or

$(CA=FD)$

So option

$C$ is correct.

In the given figure, if

$AB=DC$ ,

$\angle ABD= \angle CDB$ , which congruence rule would you apply to prove

$\Delta ABD$

$\cong \Delta CDB$ ?

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

Explanation

$\triangle ABD$ and

$\triangle CDB$

$AB=DC$ (given)

$\angle ABD=\angle CDB$ (given)

$BD=DB$ (common side)

$\therefore \triangle ABD\cong \triangle CBD$ by

$SAS$ congruence rule.

Hence option (A) is correct

State true or false:

If two sides and an angle of a triangle are respectively equal to two sides and an angle of another triangle,then the two triangles are always congruent.

0%
True
0%
False

Explanation

If two sides and one angle of one triangle is equal to the two sides and one angle of another triangle, then the two triangles may or may not be congruent.
For the two triangles to be congruent, the angle must be contained by the two sides. Thus, by

$SAS$ rule the two triangles will be congruent.

So, the triangles are not always congruent.

Hence,

$False$

By which congruency are the following pairs of triangles congruent:

In $\Delta\,ABC$ and $\Delta PQR$ , $AB=PQ$ , $BC=QR$ and $AC=PR$

0%
$ASA$
0%
$SSS$
0%
$AAA$
0%
$ASS$

Explanation

In triangles

$ABC$ and

$PQR$ , we have

$AB = PQ$ ....(Given)

$AC = PR$ ....(Given)

$BC = QR$ ....(Given)
Thus,

$\triangle ABC \cong \triangle PQR$ ..... (SSS Postulate)

Angles opposite to the equal sides of a triangle are equal.

0%
True
0%
False
0%
Incomplete information
0%
None

Explanation

Consider

$\triangle ABC$ ,

$\angle B = \angle C$
Construction: Draw a median AD on BC.

Hence,

$BD = DC$
Now, In

$\triangle ABD$ and

$\triangle ACD$

$\angle ADB = \angle ADC$ .... Common angle

$AD = AD$ ..... (Common)

$BD = DC$
Thus,

$\triangle ABD \cong \triangle ACD$ .... (SAS rule)
Thus,

$\angle B = \angle C$ .... (By CPCT)

State true/false:

If in $\Delta\, ABC$ and $\Delta \,DEF$ , AB = DE, BC = DE, BC = EF and $\angle\,B\,=\, \angle\,E$ , then both the triangles are congruent.

0%
True
0%
False

Explanation

$\triangle ABC$ and

$DEF$ ,

$AB = DE$ .......... (Given)

$\angle B = \angle E$ ......... (Given)

$BC = EF$ ............. (given)
Thus,

$\triangle ABC \cong \triangle DEF$ (SAS rule)

In a triangle

$ABC$ and

$DEF$ ,

$AB=FD$ and

$\angle A=\angle D$ . The two triangles will be congruent by SAS axiom if:

0%
$BC=EF$
0%
$AC=DE$
0%
$AC=EF$
0%
$BC=DE$

Explanation

$\triangle ABC$ and

$\triangle DEF$ ,

$\angle A = \angle D$ ....(Given)

$AB = FD$ ....(Given)
and

$\triangle ABC \cong \triangle DFE$ ....(SAS rule)
It is possible, if corresponding sides are equal,

$AC = ED$

If in two triangles

$ABC$ and

$PQR$ ,

$AB=\, QR$ ,

$\angle A=\angle Q$ and

$\angle B=\angle R$ , then

$\triangle ABC\cong$

$\triangle$

0%
$QRP$
0%
$PQR$
0%
$PRQ$
0%
None of the above.

Explanation

Given:

$AB=\, QR$ ,

$\angle A=\angle Q$ and

$\angle B=\angle R$

Now, In

$\triangle ABC$ and

$\triangle PQR$

$AB = QR$

$\angle A = \angle Q$

$\angle B = \angle R$
Thus, by

$ASA$ congruence rule,

$\triangle ABC \cong \triangle QRP$

Hence,

$Op-A$ is correct.

If in two triangles

$ABC$ and

$DEF$ ,

$AB=\,DF$ ,

$BC=\,DE$ and

$\angle B=\angle D$ , then

$\triangle ABC\cong$

$\triangle$ ____.

0%
$FDE$
0%
$DEF$
0%
$EDF$
0%
$EFD$

Explanation

Given:

In

$\triangle ABC$ and

$\triangle DEF,$

$AB = DF$

$\angle B = \angle D$

$BC = DE$

Thus,

$\triangle ABC \cong \triangle FDE$ (SAS rule)

If the two sides and the ____ angle of one triangle are respectively equal to two sides and the included angle of the other triangle, then the triangles are congruent.

0%
Included
0%
Excluded
0%
Adjacent
0%
Any

Explanation

Two sides and an included angle will fulfill the condition of the

$SAS$ rule of congruence so, the two triangles will be equal. Hence,

Two triangles are congruent if the two sides and the

$\underline{\text{included}}$ angle are equal to the two sides and the included angle of another triangle.

Which of the following pairs of triangles are congruent?

0%
$\triangle ABC$ and $\triangle DEF$ in which : $BC=\,EF$ , $AC=\,DF$ and $\angle C=\angle F$
0%
$\triangle ABC$ and $\triangle DEF$ in which : $AB=\,PQ$ , $BC=\,QR$ and $\angle C=\angle R$
0%
$\triangle ABC$ and $\triangle LMN$ in which : $\angle A=\angle L={ 90 }^{ 0 }$ , $AB=\,LM$ , $\angle C={ 40 }^{ 0 }$ and $\angle M={ 50 }^{ 0 }$
0%
$\triangle ABC$ and $\triangle DEF$ in which : and $\angle B=\angle E={ 50 }^{ 0 }$ and $AC=\,DF$

Explanation

From Option A,
Given:

$\triangle ABC$ and

$\triangle DEF$

$BC = EF$

$AC = DF$

$\angle C = \angle F$
Thus,

$\triangle ABC \cong \triangle DEF$ (SAS rule)

From Option C,
In

$\triangle ABC$ ,
Sum of angles = 180

$\angle A + \angle B + \angle C =180$

$90 + \angle B + 40= 180$

$\angle B = 50^{\circ}$

Given,

$\triangle ABC$ and

$\triangle LMN$ ,

$\angle A = \angle L = 90^{\circ}$

$AB = LM$

$\angle B = \angle M = 50^{\circ}$
Thus,

$\triangle ABC \cong \triangle LMN$ (ASA rule)

If ____sides of a triangle are respectively equal to the ____ sides of the other triangle, then the triangles are congruent.

0%
Three
0%
Two
0%
One
0%
None

Which of the following statements is true when

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

0%
$\displaystyle \angle A=\angle D$
0%
$\displaystyle \angle A=\angle E$
0%
$\displaystyle \angle A=\angle F$
0%
none of these

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,

$\angle A=\angle D$ .

Hence, option

$A$ is correct.

When two triangles have corresponding sides equal in length, then the two triangles are congruent.

0%
SAS congruency Theorem
0%
AA congruency Theorem
0%
AAA congruency Theorem
0%
SSS congruency Theorem

Explanation

All

$3$ sides of a triangle are equal with all

$3$ sides of another triangle then these two triangles are said to be congruent SSS congruency Theorem.
Therefore, D is the correct answer.

$\Delta ABC$ , AB = AC and AD is perpendicular to BC. State the property by which

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

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

Explanation

$\triangle ADB$ and

$\triangle ADC$

$AB=AC$ (given)

$AD=AD$ (common side)

$\angle ADB=\angle ADC=90^{\circ}$ (As

$AD\perp BC)$

$\therefore \triangle ADB\cong \triangle ADC$ by

$RHS$ property.

In the following fig. if

$AB=AC$ and

$BD= DC$ then

${\angle ADC}$ =

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

Explanation

$\triangle ABD$ and

$\triangle ACD$

$AB=AC$ (Given)

$BD=DC$ (Given)

$AD=AD$ (Common side)

$\therefore \triangle ABD\cong ACD$ by

$SSS$ criteria.

Now corresponding angles of congruent triangles are equal

$\Rightarrow \angle ADB=\angle ADC$

Now

$\angle ADB+\angle ADC=180^{\circ}$ (Adjacent angles on staraight line)

$\Rightarrow \angle ADC+\angle ADC={ 180 }^{ \circ }\\ \Rightarrow 2\angle ADC={ 180 }^{ \circ }\\ \Rightarrow \angle ADC={ 90 }^{ \circ }\\$

In the figure, AB and CD are two walls of equal height. A ladder BO is shifted to reach the other wall at D. Then, the triangles are congruent by the postulate

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

State the property by which

${\triangle }$ ADB

${\cong }$

${\triangle }$ ADC in the following figure.

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

Explanation

$AB=AC$ (Given)

$\angle BAD=\angle CAD$ (given)

$AD=AD$ (common)

$\therefore \triangle ADB\cong \triangle ADC$

$SAS$ property.

In the figure,

$\displaystyle AB=CD$ and

$\displaystyle \angle A={ 90 }^{ o }=\angle D$ . Then

0%
$\displaystyle \Delta ABC\cong \Delta DBC$ by SAS postulate
0%
$\displaystyle \Delta ABC\cong \Delta DCB$ by RHS postulate
0%
$\displaystyle \Delta ABC\cong \Delta DBC$ by AAS postulate
0%
$\displaystyle \Delta ABC\cong \Delta DCB$ by SSS postulate

Explanation

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DBC$

$\displaystyle \angle A={ 90 }^{ o }=\angle D$ given

$\displaystyle BC=BC$ common side

$\displaystyle AB=CD$ given

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

In the figure,

$AD$ and

$BC$ are perpendicular to

$AB$ , and

$AD=BC$ . Then , by

$SAS$ congruence postulate,

$\displaystyle \Delta ABC\cong$

0%
$\displaystyle \Delta ABD$
0%
$\displaystyle \Delta BDA$
0%
$\displaystyle \Delta BAD$
0%
$\displaystyle \Delta ADB$

Explanation

$AD\perp AB$ and

$BC\perp AB$ [ Given ]

$\therefore$

$\angle BAD=\angle ABC=90^o$ ---- ( 1 )

$\triangle ABC$ and

$\triangle BAD$

$BC=AD$ [ Given ]

$\angle ABC=\angle BAD=90^0$ [ From ( 1 ) ]

$AB=BA$ [ Common side ]

$\therefore$

$\triangle ABC\cong\triangle BAD$

1358571_378258_ans_cb212ca0e8b84df88d0ca76d5cbe5239.png

If three sides of a triangle are equal to three sides of another triangle, then the two triangles are congruent. This is _____ condition for congruence.

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

Explanation

Side-Side-Side

$(S.S.S)$ condition states that if three sides of one triangle are equal to three sides of another triangle then the two triangles are congruent.

If the hypotenuse and one of the other two sides of a right angles triangle is equal to the hypotenuse and one of the other two sides of the other right-angled triangle respectively, then the two right-angled triangles are ___.

0%
congruent
0%
unequal
0%
equilateral
0%
None of the these

CBSE Questions for Class 7 Maths Congruence Of Triangles Quiz 1 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Congruence Of Triangles Quiz 1 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

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