MCQ Questions for Class 7 Maths Congruence Of Triangles Quiz 3

In right triangle

$ABC$ , right-angled at

$C, M$ is the mid-point of hypotenuse

$AB.\ C$ is joined to

$M$ and produced to a point

$D$ such that

$DM = CM$ . Point

$D$ is joined to point

$B$ .Which of the following is correct.

0%
$\Delta \mathrm { AMC } \cong \Delta \mathrm { BMD }$
0%
$\angle D B C$ is a right angle.
0%
$\Delta D B C \equiv \Delta A BC$
0%
$\mathrm { CM } = \dfrac { 1 } { 2 } \mathrm { AB }$

Explanation

REF.Image

Given

$\triangle ABC$ is right-angled at

$C$

$M$ is the midpoint of side

$AB$

$\Rightarrow CM \perp AB$ (As M is the midpoint and

$\triangle ABC$ is right-angled)

$\Rightarrow \angle CMA = \angle DMB \rightarrow (1)$

$\triangle ABC$ and

$\triangle DBC$

$BC= BC$

$\angle DBC= \angle ACB$

$\angle DCB= \angle ABC$

$\Rightarrow \triangle ABC$ is concorent to

$\triangle DBC$

$\therefore$ In

$\triangle DBM$ and

$\triangle MAC$

$DB= AC$

$\angle CMA= \angle DMB$ and

$MB= AM$

$\therefore \triangle DBM$ is congruent to

$\triangle CMA$

If two legs of a right triangle are equal to two legs of another right triangle. then the right triangles are congruent.

0%
True
0%
False

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

0%
True
0%
False

In Figure 6.28, two triangles are congruent by RHS.

0%
True
0%
False

If two triangles are congruent, then the corresponding angles are equal.

0%
True
0%
False

If two angles and a side of a triangle are equal to two angles and a side of another triangle, then the triangles are congruent.

0%
True
0%
False

If the hypotenuse of one right triangle is equal to the hypotenuse of another right triangle, then the triangles are congruent.

0%
True
0%
False

In the above 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)

$\angle AOD=\angle BOC$

$\therefore \triangle AOD\cong \triangle BOC$ by

$SAS$ criteria.

State true/false:

If in

$\bigtriangleup ABC and \bigtriangleup QRP, AB =QR,\angle B=\angle R\:and \: \angle C=\angle P.$

Then, by A.S.A criteria,

$\triangle ABC$ and

$\triangle QRP$ are congruent.

0%
True
0%
False

Explanation

$\triangle$ s, ABC and QRP,
Given,

$BC = RP$

$\angle B = \angle R$

$\angle C = \angle P$
Thus,

$\triangle ABC \cong \triangle QRP$ (By

$ASA$ criteria)

If in two triangles

$\Delta ABC$ and

$\Delta PQR$ ,

$AB = QR, BC = PR$ and

$CA = PQ,$ then :

0%
$\Delta ABC\cong \Delta PQR$
0%
$\Delta CBA\cong \Delta PRQ$
0%
$\Delta BAC\cong \Delta RPQ$
0%
$\Delta PQR\cong \Delta BCA$

Explanation

$\triangle ABC$ and

$\triangle PQR$ , we are given

$AB=QR$ ,

$BC=PR$

and

$CA=PQ$

$\therefore \Delta CBA\cong \Delta PRQ$ .... SSS test

State whether the given statement is true or false:

$\displaystyle \bigtriangleup ABC$ and

$\bigtriangleup DEF, AB=DE,BC=EF$ and

$\angle B=\angle E$ . The triangles are congruent by

$SSS$ test.

0%
True
0%
False

Explanation

$\triangle$ ABC and

$\triangle$ DEF,

$AB = DE$ (Given)

$BC = EF$ (Given)

$\angle B = \angle E$ (Given)
Hence,

$\triangle ABC \cong \triangle DEF$ (By SAS postulate)

So, the given statement is False.

In figure, if

$AB=AC$ and

$AP=AQ$ , then by which congruence criterion

$\Delta PBC \cong \Delta QCB$ ?

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

Explanation

In the given triangle:

$AB=AC$

$\Rightarrow \angle B=\angle C$ (As angles opposite to equal to opposite sides are equal.

$AB=AC$

$AP+PB=AQ+QC$

Now

$AP=AQ$

$\Rightarrow PB=QC$

Now in

$\triangle PBC$ and

$\triangle QCB$

$PB=QC$ (Proved above)

$\angle B=\angle C$ (Proved above)

$BC=CB$ (Common side)

$\therefore \triangle PBC\cong \triangle QCB$ by

$SAS$ criteria.

$D$ and

$E$ are the mid-points of the sides

$AB$ and

$AC$ respectively of

$\triangle ABC$ .

$DE$ is produced to

$F$ . To prove that

$CF$ is equal and parallel to

$DA$ , we need an additional information which is:

0%
$\triangle DAE$ = $\triangle EFC$
0%
$AE = EF$
0%
$DE = EF$
0%
$\triangle ADE = \triangle ECF$

Explanation

Given

$CF$ is equal and parallel to

$DA$

$D$ is the midpoint of

$AB$

So,

$AD$ =

$DB$ =

$CF$ =

$\cfrac { c }{ 2 }$

Similarly

$AE$ =

$EC$ =

$\cfrac { b }{ 2 }$

Let

$\angle AED=\alpha$

$ADE=\beta \quad$ and

$\quad DAE=\gamma$

Since

$AD$ is parallel to

$FC$

$\angle FEC = \alpha \angle EFC=\beta$ and

$\angle ECF=\gamma$

Applying

$S.A.S$ Congruency for triangle

$ADE$ and

$CFE$ ,

$AD= CF$ ,

$AE= AC$

$\angle DAE=\angle ECF$

${ \Delta }^{ k } DAE$ and

${ \Delta }^{ k } FCE$ are congruent

$DE = EF$

It is given that

$DE= EF$ ,then by

$S.S.S.$ congruency all angles will be equal and

$CF$ and parallel to

$DA$

$\therefore$ Additionsl information needed is

$DE= EF$ .

$\Delta ABC$ is congruent of

$\Delta DEF$ , if

0%
$AB \neq DE$ , $AC = 6 = DF$ , $\angle A = \angle D$
0%
$AB = DE$ , $AC = DF$ , $\angle B =\angle E$
0%
$AB = DE$ , $AC \neq DF$ , $\angle C =\angle F$
0%
$AB = DE$ , $AC = DF$ , $\angle C \neq \angle F$

Explanation

We say that triangle

$\triangle ABC$ is congruent to

$\triangle DEF$ if -

AB = DF
BC = EF
CA = FD
$\angle A = \angle D$
$\angle B = \angle E$
$\angle C = \angle F$

For this equation all options satisfies the above conditions.

Therefore for this question,

$\triangle ABC$ is congruent of

$\triangle DEF$ if all the options are correct.

In the given figure,

$AB= AC$ and

$\angle DBC= \angle ECB= 90^{\circ}$ , then

$AD= AE$ .

0%
True
0%
False

Explanation

In triangle ABC,

$AB = AC$ thus,

$\angle ABC = \angle ACB$ (I) (Opposite angles of equal sides of triangle are equal)
Now, in

$\triangle$ s DBC and ECB,

$BC = BC$ (Common)

$\angle DBC = \angle ECB = 90$ (Given)

$\angle EBC = \angle DCB$ (From I)
Hence,

$\triangle DBC \cong \triangle ECB$ (ASA Congruency)
or

$DC = BE$ (equal sides of triangle of opposite angle are equal)
Also, Given

$AB = AC$

$DC - AC = BE - AB$

$AD = AE$

$\triangle ABC$ and

$\triangle DEF$ ,

$AB = FD$ 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

$SAS=$ If two pairs of sides of two triangles are equal in length, and the included angles are equal in measurement, then the triangles are congruent.

Here,

$AB = FD$ ,

$\angle A = \angle D$

So,

$AC = DE$

Hence, option B is correct.

In the given figure,

$AB= AC$ and

$\angle DBC= \angle ECB= 90^{\circ}$ ,then

$BD= CE$ .

0%
True
0%
False

Explanation

In triangle ABC,

$AB = AC$ thus,

$\angle ABC = \angle ACB$ (I) (Opposite angles of equal sides of triangle are equal)
Now, in

$\triangle$ s DBC and ECB,

$BC = BC$ (Common)

$\angle DBC = \angle ECB = 90$ (Given)

$\angle EBC = \angle DCB$ (From I)
Hence,

$\triangle DBC \cong \triangle ECB$ (ASA Congruency)
or

$BD = CE$ (equal sides of triangle of opposite angle are equal)

In triangle

$ABC$ , bisector of angle

$BAC$ meets opposite side

$BC$ at point

$D$ . If

$AB=AC$ , then

$BD=CD.$

0%
True
0%
False

Explanation

$\triangle ABD$ and

$\triangle ACD$ ,

$AB = AC$ [Given]

$AD = AD$ [Common]

$\angle BAD = \angle CAD$ [

$BD$ is bisector of

$\angle A$ ]

So, by

$SAS$ rule of congruence,

$\triangle ABD \cong \triangle ACD$

$\Rightarrow BD = CD$

By which congruency are the following pair of triangles congruent:

In $\Delta\,ABC$ and $\Delta \,DEF$ , $\angle\,B = \angle\,E = 90\,^{\circ}, AC = DF$ and $BC = EF.$

0%
RHS
0%
SAS
0%
AAA
0%
SSS

Explanation

$\triangle ABC$ and

$DEF$ ,

$\angle B = \angle E = 90^{\circ}$ .......... (Given)

$AC = DF$ ........ (Given)

$BC = EF$ ......... (Given)

Thus,

$\triangle ABC \cong \triangle DEF$ (RHS postulate)

In the figure

$\angle ABC=135^{\circ},\angle ABX=90^{\circ} ,\angle XCD=55^{\circ}, \angle BCD=100^{\circ}$ ,then determine whether

$\angle XBC$ and

$\angle XCB$ are congruent to each other.

0%
$\angle XBC = 45^{\circ}, \angle XCB = 46{\circ}$
0%
$\angle XBC = 45^{\circ}, \angle XCB = 45^{\circ}$
0%
$\angle XBC = 45^{\circ}, \angle XCB = 60^{\circ}$
0%
$\angle XBC = 45^{\circ}, \angle XCB = 25^{\circ}$

Explanation

Given,

$\angle ABC={ 135 }^{ o },\quad ABX={ 90 }^{ o },\quad XCD={ 55 }^{ o },\quad BCD={ 100 }^{ o }$ .

Here,

$\angle ABC=\angle ABX+\angle XBC\Rightarrow XBC=\angle ABC-ABX={ 135 }^{ o }-{ 90 }^{ o }\Rightarrow XBC=45^o.$

Again

$\angle BCD=\angle XCB+\angle XCD\Rightarrow \angle XCB=\angle BCD-\angle XCD\Rightarrow \angle XCB={ 100 }^{ o }-{ 55 }^{ o }$

$=45^o$ .

Since,

$\angle XBC=\angle XCB=45^o$ , they are congruent.

Hence, option

$B$ is correct.

From the following figure, we can say:

$AD= CE$

State TRUE or FALSE

0%
True
0%
False

Explanation

$\triangle$ ABC,

$AB = AC$
thus,

$\angle ABC = \angle ACB = x$ (Opposite angles of equal sides are equal)
Now,

$\angle ABC + \angle CBE = 180$ (Angles on a straight line)

$x + \angle CBE = 180$

$\angle CBE = 180 - x$ (I)
Also,

$\angle ACB + \angle ACD = 180$ (Angles on a straight line)

$x + \angle ACD = 180$

$\angle ACD = 180 - x$ (II)
Hence, from I and II

$\angle ACD = \angle CBE$
In triangles ACD and CBE

$\angle ACD = \angle CBE$

$CD = CB$ (Given)

$AC = BE$ (Given)
Thus,

$\triangle ACD \cong \triangle ECB$
Hence,

$AD = CE$

State true or false:

In the given figure, the diagonals

$AC$ and

$BD$ intersect at point

$O$ . If

$OB =OD$ and

$AB//DC$ , then

$Area \left ( \bigtriangleup \: DOC \right )=Area \left ( \bigtriangleup \: AOB \right )$ .

0%
True
0%
False

Explanation

Given, the diagonals AC and BD intersect at point O such that OB

$=$ OD and AB

$\parallel$ DC.
Now, In

$\triangle DOC$ and

$\triangle AOB$

$\angle BDC=\angle DBA$ (Alternate angles, As AB

$\parallel$ CD and BD intersects it.)

$OD= OB$ (Given)

$\angle DOC=\angle AOB$ ( Vertically opposite angles)
So,

$\triangle DOC\quad \cong \quad \triangle AOB$ (By ASA concurrence Property)
Since, Concurrent triangle have equal areas.

$\therefore$ Area of

$\triangle DOC =$ Area of

$\triangle AOB$

$ABC$ and

$DBC$ are two isosceles triangle on the same side of

$BC$ . Then,

$\angle BDA= \angle CDA$ .

0%
True
0%
False

Explanation

Join AD.
In

$\triangle$ ABD and

$\triangle$ ACD,

$AB = AC$ (Given)

$BD = CD$ (Given)

$AD = AD$ (Common side)

$\therefore$

$\triangle$

$ABD \cong \triangle ACD$ (By SSS congruence rule)
Hence,

$\angle BDA = \angle CDA$ (Corresponding angles of congruent triangles)

State the property by which

$\Delta ADB\, \cong\, \Delta ADC$ in the following figure.,

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

Explanation

From figure, we have

$AB = AC$ ....Given

$\angle BAD\, =\, \angle CAD$

$AD = AD$ ....Common sides

$\therefore$ By SAS property,

$\Delta ADB \cong \Delta ADC$ .

$l$ and

$m$ are two parallel lines intersected by another pair of parallel lines

$p$ and

$q$ . Then:

0%
$\triangle ABC\cong \triangle CDA$ .
0%
$\triangle ABC\cong \triangle ADC$ .
0%
$\triangle ABC\cong \triangle DCA$ .
0%
$\triangle ABC\cong \triangle CAD$ .

Explanation

$\triangle ABC$ and

$\triangle CDA$

$\angle CAB=\angle ACD$ (Pair of alternate angle)

$\angle BCA=\angle DAC$ (Pair of alternate angle)

$AC=AC$ (Common side)

$\triangle ABC\cong \triangle CDA$ (ASA criteria)

If in two triangles

$PQR$ and

$DEF$ ,

$PR=\,EF$ ,

$QR=\,DE$ and

$PQ=\,FD$ , then

$\triangle PQR\cong$

$\triangle$ ___.

0%
$FDE$
0%
$DEF$
0%
$FED$
0%
$DFE$

Explanation

Given:

$PR=\,EF$ ,

$QR=\,DE$ and

$PQ=\,FD$
Now, In

$\triangle PRQ$ and

$\triangle DEF$

$PR = EF$

$QR = DE$

$PQ = FD$
Thus,

$\triangle PQR \cong \triangle FDE$ (SSS rule).

Ankita wants to prove

$\Delta ABC\cong \Delta DEF$ using

$SAS$ . She knows

$AB=DE$ and

$AC=DF$ . What additional piece of information does she need?

0%
$\;\angle A=\angle D$
0%
$\;\angle C=\angle F$
0%
$\;\angle B=\angle E$
0%
$\;\angle A=\angle B$

Explanation

By SAS, two corresponding sides and their included angles should be congruent.
Since

$AB=DE$ and

$AC=DF$ ,

The included angles of

$AB$ and

$AC = \angle A$

and included angle of

$DE$ and

$DF = \angle D.$

$\Rightarrow \angle A = \angle D$ is needed.

By which congruency property, the two triangles connected by the following figure are congruent.

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

Explanation

Separate the given figure into two triangles.

$\Delta ABC$ and

$\Delta ABD$

We can observe that,

$AC = AD$ {Both are

$x \,\,cm$ }

$BC = BD$ {Both are

$y\,\,cm$ }

$AB = AB$ {Common Side}

$\Delta ABC\, \cong\, \Delta ABD$ by SSS property.

In the given fig. if AD = BC and AD || BC, then:

0%
AB = AD
0%
AB = DC
0%
BC = CD
0%
None

Explanation

In the give fig. AD = BC, AC = AC,

$\angle ADC\, =\, \angle ABC$ (

$\because$ AD || BC)
By SAS theorem

$\Delta ABC\, \cong\, \Delta CDA$
So AB = DC.

ABC is an isosceles triangle in which the altitudes

$BE$ and

$CF$ are drawn to the equal sides

$AC$ and

$AB$ respectively. Then

0%
$BE = CF$
0%
$BE = AB$
0%
$AB = BC$
0%
$AC = BC$

Explanation

$\triangle ABE$ and

$\triangle ACF$ ,

we have

$\angle BEA=\angle CFA$ (Each

${ 90 }^{ 0 }$ )

$\angle A= \angle A$ (Common angle)

$AB=\,AC$ (Given)

$\therefore \triangle ABE \cong \triangle ACF$ (By SAS congruence criteria)

$\therefore BF=\,CF$ [C.S.C.T]

CBSE Questions for Class 7 Maths Congruence Of Triangles Quiz 3 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Congruence Of Triangles Quiz 3 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

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