MCQ Questions for Class 7 Maths Congruence Of Triangles Quiz 5

In the given figure, which of the following is correct?

0%
$\Delta PQR \cong \Delta RSP$
0%
$\Delta PQR \cong \Delta SPR$
0%
$\Delta PQR \cong \Delta RPS$
0%
$\Delta PQR \cong \Delta PSR$

Explanation

$\triangle PQR$ and

$\triangle RSP$

$\angle QPR = \angle SRP = 45^o$

$PQ = RS = 5.5 cm$

$PR = RP$ (common)

$\triangle PQR \cong \triangle RSP$ [By SAS congruency]

Hence option (A) is correct

Referring the figure given below, for

$\displaystyle \Delta PQR$ and

$\displaystyle \Delta XYZ$ ,

$PQ=YZ, QR=XZ$ and

$\displaystyle \angle Q=\angle Z$ . The value of

$\displaystyle \angle P$ is ___.

0%
$\displaystyle { 42 }^{ \circ }$
0%
$\displaystyle { 75 }^{ \circ }$
0%
$\displaystyle { 63 }^{ \circ }$
0%
Cannot be determined

Explanation

$\displaystyle \Delta PQR$ and

$\displaystyle \Delta XYZ,$

$\displaystyle PQ=YZ$

$\displaystyle QR=XZ$

$\displaystyle \angle Q=\angle Z$

$\displaystyle \therefore \Delta PQR\cong \Delta XYZ$ (

$S.A.S.$ condition for congruence)

$\displaystyle \therefore \angle R=\angle X={ 63 }^{ \circ }$

$\displaystyle \Rightarrow \angle P=\angle Y$

$\displaystyle \Rightarrow \angle Q=\angle Z={ 75 }^{ \circ }$

Now,

$\displaystyle \angle P+\angle Q+\angle R={ 180 }^{ \circ }$

$\displaystyle \therefore \angle P+{ 75 }^{ \circ }+{ 63 }^{ \circ }={ 180 }^{ \circ }$

$\displaystyle \Rightarrow \angle P={ 42 }^{ \circ }$

In a

$\triangle ABC$ ,

$D$ is the midpoint of the side

$BC. DE$ is perpendicular to

$AB$ and

$DF$ is perpendicular to

$AC$ . Also,

$DE=DF$ , Then

$\displaystyle \angle B$ =

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

Explanation

Given:

$D$ is the midpoint of

$BC$

$\displaystyle \therefore \quad BD=DC$

Also,

$DE = DF$

$DE$ is perpendicular to

$AB$

$\displaystyle \therefore \angle E={ 90 }^{ o }$

$DF$ is perpendicular to

$AC$

$\displaystyle \therefore \angle F={ 90 }^{ o }$

If we consider two triangles so formed,

$\displaystyle \Delta DEB$ and

$\displaystyle \Delta CFD$ , then their hypotenuse

$BD$ and

$CD$ are equal.

$\displaystyle \angle E=\angle F={ 90 }^{ o }$ . Also

$DE=CF$ .

Thus according to R.H.S condition of congruence

$\displaystyle \Delta DEB\cong \Delta CFD$

Thus, the corresponding angles

$\displaystyle \angle B$ and

$\angle C$ are equal, i.e

$\displaystyle \angle B=\angle C$

In the given figures, which of the following satisfies A.S.A. condition for congruence?

0%
Figure II
0%
Figure I and II
0%
Figure I
0%
None of these

Explanation

Figure 1 satisfy the condition of A.S.A

Here

$\angle ACB=\angle DCE$ {vertical angle|

$DC=BC$ and

$\angle EDC=\angle ABC$ [Alternate angles]

$\therefore$ It satisfy A.S.A Congruency.

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$ ,

$\displaystyle \angle B=\angle D={ 90 }^{ o }$ ,

$\displaystyle AC=EF$ and

$\displaystyle AB=ED$ . Then

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

0%
$\displaystyle \Delta EDF$
0%
$\displaystyle \Delta DEF$
0%
$\displaystyle \Delta EFD$
0%
$\displaystyle \Delta FED$

Explanation

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$ ,

$\displaystyle \angle B=\angle D={ 90 }^{ o }$ ,

$\displaystyle AC=EF$ and

$\displaystyle AB=ED$

$\displaystyle \therefore$ According to R.H.S condition for congruence, both triangles are

$\displaystyle \therefore \Delta ABC\cong \Delta EDF$

For

$\displaystyle \Delta ABC \cong \displaystyle \Delta DEF$ . Then according to the given figure,

$\angle D = 40^o$ ,

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

0%
$\displaystyle { 40 }^{ o }$
0%
$\displaystyle { 90 }^{ o }$
0%
$\displaystyle { 60 }^{ o }$
0%
$\displaystyle { 50 }^{ o }$

Explanation

According to the given figure,

$\displaystyle \angle B=\angle E={ 90 }^{ o }$

$\Rightarrow AC=DE=5$ cm (hypotenuse) and

$BC=EF=2$ cm

$\displaystyle \therefore \Delta ABC\cong \Delta DEF$ ..... (R.H.S condition for congruence)

$\displaystyle \Rightarrow \angle A=\angle D={ 40 }^{ o }$

$\displaystyle \Rightarrow \angle B={ 90 }^{ o }$

$\displaystyle \angle C=?$
We know

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

$\displaystyle \therefore \angle C={ 180 }^{ o }-{ 90 }^{ o }-{ 40 }^{ o }$

$\displaystyle \therefore \angle C={ 50 }^{ o }$

Given are two congruent triangles

$ABC$ and

$XYZ.$ Which of the following statements is incorrect?

0%
$\displaystyle \angle A=\angle X$
0%
$\displaystyle AB=XY$
0%
$\displaystyle \Delta ABC\cong \Delta XYZ$
0%
$\displaystyle AC=YZ$

Explanation

Given,

$\triangle ABC$ and

$\triangle XYZ$ are congruent

From figure, we have

$AC=XZ, BC=YZ$ and

$\angle B=\angle Y=90^o$

$\displaystyle \therefore AC=YZ$ is an incorrect statement.

Which condition of congruency proves that the following pair of triangles are congruent

0%
$A.A.S$
0%
$S.S.S$
0%
$S.A.S$
0%
$A.S.A$

Explanation

From the given figure, we have

$DE=EG, DF=GP$ and

$FE=PE.$

Thus, from

$S.S.S$ condition of congruency,

$\displaystyle \Delta DEF\cong \Delta GEP$

Complete the correspondence statement:

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

0%
$\displaystyle \Delta DEF$
0%
$\displaystyle \Delta DFE$
0%
$\displaystyle \Delta FDE$
0%
$\displaystyle \Delta EFD$

Explanation

Here,

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

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

Also,

$AB=DF$ ,

$BC=FE$ and

$AC=DE$ .

Since, all the corresponding parts are equal,

$\displaystyle \Delta ABC\cong \Delta DFE$ .

Therefore, option

$B$ is correct.

In the figure ,

$\angle A = \angle C$ and

$AB = BC$ . Then

$\Delta ABD \cong \Delta CBE$

0%
True
0%
False

Explanation

Since,

$\angle B$ is common to both Triangles

Hence,

$\angle B=\angle B\\\angle A=\angle C\\AB=AC$

From ASA Congruency Criterion

$\triangle ABD\cong\triangle CBE$

$\displaystyle \Delta BAC\cong \Delta DEC,AB\parallel DE$ , if

$\displaystyle \angle B$ is

$\displaystyle { 90 }^{ o }$ and

$\displaystyle \angle C$ is

$\displaystyle { 50 }^{ o }$ ,
Find

$\displaystyle \angle A$ and

$\displaystyle \angle D$ =.......

0%
$\displaystyle { 35 }^{ o }$
0%
$\displaystyle { 50 }^{ o }$
0%
$\displaystyle { 40 }^{ o }$
0%
$\displaystyle { 55 }^{ o }$

Explanation

Given $\triangle ABC \cong \triangle DEC$ .

Then the corresponding parts will also be equal.

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

Then,

$AB=DE$ ,

$BC=EC$ ,

$AC=DC$ ,

$\angle A=\angle D$ ,

$\angle B=\angle E$ and

$\angle C=\angle C$ .

Also given,

$\angle B=90^o$ and

$\angle C=50^o$ .

$\triangle ABC$ ,

by angle sum property,

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

$\implies$

$\angle A+90^o + 50^o=180^o$

$\implies$

$\angle A+140^o=180^o$

$\implies$

$\angle A=180^o-140^o$

$\implies$

$\angle A=40^o$ .

Since,

$\angle A=\angle D$

Therefore,

$\angle A=\angle D$

$=40^o$ .

Hence, option

$C$ is correct.

In figure,

$BD$ is a bisector of

$\angle B, CD = DA$
In order to show the

$\triangle BDA$ and

$\triangle BDC$ are congruent, which property of triangle applies here?

0%
$SAS$
0%
$SSS$
0%
$AAS$
0%
None

Explanation

$\triangle ABD$ and

$\triangle CBD$

$BD$ is a bisector of

$\angle B$

$BD$ is common side.

$\angle ABD=\angle CBD$

$CD=DA$

Two triangles are congruent from

$SAS$ postulate

Hence option A is correct.

In the figure,

$\displaystyle \Delta ACB\cong \Delta DCB$ , then the value of

$x$ and

$y$ are:

0%
$14$ cm, $\displaystyle { 20 }^{ o }$
0%
$28$ cm, $\displaystyle { 20 }^{ o }$
0%
$\displaystyle { 14 }^{ o }$ , $20$ cm
0%
$\displaystyle { 28 }^{ o }$ , $20$ cm

Explanation

From the figure, we have,

$\angle D= 20^o, BD=28$ cm.

We need to find value of

$x$ and

$y$ .

$\displaystyle \Delta ACB\cong \Delta DCB$ , we have,

Then the corresponding parts will also be equal.

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

Then,

$AC=DC$ ,

$CB=CB$ ,

$AB=DB$ ,

$\angle A=\angle D$ ,

$\angle C=\angle C$ and

$\angle B=\angle B$ .

$\displaystyle \therefore AB=BD$ cm

$\displaystyle \therefore 2x=28$ cm

$\displaystyle \therefore x=14$ cm.

Also,

$\displaystyle \angle A=\angle D$

$\displaystyle \therefore y={ 20 }^{ o }$ .

Therefore, option

$A$ is correct.

Which condition of congruence proves that the following pair of triangles are congruent

0%
$A.A.S$
0%
$S.S.S$
0%
$S.A.S$
0%
$A.S.A$

Explanation

From the given figure, we have

$AC=DF, BC=ED$ and

$\displaystyle \angle C=\angle D$

$\displaystyle \therefore$ According to

$S.A.S$ condition of congruence,

$\displaystyle \Delta ABC\cong \Delta FED$

Which condition of congruency proves that the following pair of triangles are congruent

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

Explanation

According to the given figure, we have

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

$BC=FG$

$\displaystyle \therefore \Delta ABC\cong \Delta EFG$ ....

$(ASA$ condition for congruence

$)$

Complete the correspondence statement:

$\displaystyle \Delta GHK\cong$ ____.

0%
$\displaystyle \Delta GTK$
0%
$\displaystyle \Delta KTG$
0%
$\displaystyle \Delta GKT$
0%
$\displaystyle \Delta TGK$

Explanation

Here,

$\displaystyle \angle K={ 90 }^{ o }$ (common angle)

$GH=GT$ and

$KH=KT$

$\displaystyle \therefore$ According to

$R.H.S$ condition of congruency of triangles,

$\displaystyle \Delta GHK\cong \Delta GTK$ .

Therefore, option

$A$ is correct.

$AD$ is the bisector of BC.
Which property of triangle applies here?

0%
$SAS$
0%
$SSS$
0%
$AAS$
0%
none

Explanation

$\triangle ABD$ &

$\triangle ADC$

$AD = AD$

$BD = DC$

$\angle ADB = \angle ADC$

$\triangle ABD \cong \triangle ADC$

$\therefore$ Both triangles are congruent by

$SAS$ congruence condition.

$\Delta ABC \cong \Delta DEF$ by ASA congruence condition. Find the value of

$x$ .

0%
$x = 20$
0%
$x = 30$
0%
$x = 40$
0%
$x = 50$

Explanation

We know

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

$\angle B = \angle F$

$\Rightarrow 60 = x + 30$

$\Rightarrow x = 60 - 30$

$\Rightarrow x = 30$

Which condition will prove these triangles as congruent?

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

Explanation

$\triangle$

$HJG$ and

$\triangle$

$MNK$ ,

$\angle$

$HJG$

$=$

$\angle$

$MNK$ (given)

$JG = NK$ (given)

$\angle$

$JGH =$

$\angle$

$NKM$ (given).

Therefore,

$\triangle$

$HJG$

$\cong$

$\triangle$

$MNK$ (By

$ASA$ congruency criterion).

Hence, option

$D$ is correct.

Which of the following pairs of triangle are congruent ?

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

Given two right angled triangle ABC and PQR, such that

$<A=20^o, <Q=20^o$ and

$AC=QP$ . Write the correspondence if triangles are congruent.

0%
$\Delta ABC\cong \Delta PQR$
0%
$\Delta ABC\cong \Delta PRO$
0%
$\Delta ABC\cong \Delta ROP$
0%
$\Delta ABC\cong \Delta QRP$

$\triangle PQR$ is right angled at

$Q$ ,

$PR=5cm$ and

$QR=4cm$ . If the lengths of sides of another triangle

$ABC$ are

$3cm$ ,

$4cm$ ,

$5cm$ , then which one of the following is correct?

0%
Area of $\triangle PQR$ is double that of $\triangle ABC$ .
0%
Area of $\triangle ABC$ is double that of $\triangle PQR$ .
0%
$\sqrt { B } =\cfrac { \sqrt { Q } }{ 2 }$
0%
Both triangles are congruent.

Explanation

Given:

$\Delta PQR$ is right angled at

$Q$ ,

$PR = 5 cm, QR = 4 cm$ .

Length of sides of another triangle are

$3 cm, 5 cm, 4 cm$ .

According to question,

Ref. image

Applying Pythagoras theorem in

$\Delta PQR$ ,

$H^2 = P^2 + B^2$

$5^2 = P^2 + 4^2$

$25 - 16 = P^2$

$9 = P^2$

$P = 3 cm, PQ = 3cm$

Now we can see that both the triangles have equal sides.

i.e.

$PQ = AB = 3cm$

$PR = AC = 5 cm$

$QR = BC = 4 cm$

It means by SSS congruence property both triangles are congruent.

{SSS congruence property: Two triangles are congruent if the three sides of one triangle are respectively equal to the three sides of other triangle.}

i.e.

$\Delta PQR \cong \Delta ABC$

The correct option is (D). i.e. both triangles are congruent.

2106402_551937_ans_f3067abf757246458fd5df7a2d336307.png

In quadrilateral

$ACBD, AC = AD$ and

$AB$ bisects

$\angle CAD.$ Show that

$\triangle ABC \cong \triangle ABD$ . Can you say that

$BC$ =

$BD$ ?

0%
True
0%
False

Explanation

$\triangle ABC$ and

$\triangle ABD,$

$\angle CAB=\angle BAD$ [Given]

$AC=AD$ [Given]

$AB=AB$ [Common]

So, by

$SAS$ rule of congruence,

$\triangle ABC \cong \triangle ABD$

As corresponding sides in congruent triangles are equal so,

$BC=BD.$

$\triangle ABC$ , the bisector of

$\angle {A}$ intersects

$\overline { BC }$ at a point

$D$ . Then:

0%
$BD\times AC=BC\times AB$
0%
$BD\times AB=DC\times AC$
0%
$AC\times AB=DC\times BC$
0%
$BD\times AC=DC\times AB$

Explanation

$\triangle ABC$ , the bisector of

$\angle A$ intersects

$\overline BC$ at

$D$ .

$\therefore$

$AD$ bisects

$BC$

$\therefore BD=DC$ ........

$(i)$

In triangles,

$ABD$ and

$ADC$

$BD=CD$ ......... From

$(i)$

$\angle BAD=\angle DAC$ ....... (Given)

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

$\therefore$

$\triangle ABD\cong \triangle ADC$ ........ [By S.A.S criterion]

$\implies AB=AC$

Thus,

$BD\times AC=DC\times AB$

The two triangles in the figure are congruent by the congruence theorem. Here, it is given

$OQ=OR$ . Which of the following condition, along with the given condition, is sufficient to prove that the two triangles are congruent to each other?

0%
$\angle P=\angle S$
0%
$\angle Q=\angle R$
0%
$OP=OS$
0%
$PQ=SR$

Explanation

Given

$OQ=OR$ and

$\triangle POQ\cong \triangle ROS$
We know that, congruent parts of congruent triangles are congruent

$\angle POQ\cong \angle ROS$ (vertically opposite angles)
If

$OP=OS$ then by using the (SAS) congruent, we can conclude the congruency of two triangles.

Hence, option C is sufficient to prove the congruency.

Which congruence criteria can be used to state that

$\triangle$ XOY

$\cong$

$\triangle$ POQ?

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

Explanation

$\triangle XOY$ and

$\triangle POQ$

$\Rightarrow$

$\angle YXO=\angle QPO=65^o$ [Given]

$\Rightarrow$

$OX=OP$ [Given]

$\Rightarrow$

$\angle XOY=\angle POQ$ [Vertically opposite angles]

$\therefore$

$\triangle XOY\cong\triangle POQ$ [By ASA criteria]

$\triangle FGC$ is an isosceles triangle, which of the following method will prove

$\triangle ABC$ congruent to

$\triangle DEF$ ?

0%
SSS
0%
RHS
0%
AAS
0%
SAS

Explanation

$\triangle ABC$ and

$\triangle DEF$

Given,

$AB=DE$

$\angle ABC=\angle DEF=90^\circ$

$\triangle FCG$ is isoceles,

$GF=GC$

$\implies FD=CA$ ...... (as

$AB||DE$ ,

$GD$ should be equal to

$GA$ )

$AC=FD$

Hence,

$\triangle ABC \cong \triangle DEF$ by

$RHS \ postulate$

To prove that

$\Delta DEF$ and

$\Delta ABC$ are congruent by SAS, what additional information is needed?

0%
$EF \cong BC$
0%
$\angle DFE \cong \angle ABC$
0%
$DE \cong AB$
0%
$\angle DFE \cong \angle ACB$

Explanation

Given

$FD=CA$ and

$\angle D=\angle A$

Now for triangles to be congruent by

$SAS$ we need a third side that include the given angles with the given equal sides.

$DE=AB$

$\implies DE\cong AB$

Hence, option

$C$ is correct.

In given figure

$\triangle PQR \cong \triangle XYZ$ by _______ congruency rule.

0%
$SSS$
0%
$AAA$
0%
$SAS$
0%
$ASA$

Explanation

$\triangle PQR$ and

$\triangle XYZ$ ,

$\angle$

$QPR$

$=$

$\angle$

$YXZ$ (given)

$PQ = XY$ (given)

$\angle$

$PQR$

$=$

$\angle$

$XYZ$ (given)

Therefore,

$\triangle PQR \cong \triangle XYZ$ by

$ASA$ rule of congruency.

In the given figure, the line segment

$AB$ is parallel to another line segment

$RS$ and

$O$ is the midpoint of

$AS$ , Then

$\Delta AOB \cong \Delta SOR \ ?$

0%
True
0%
False

Explanation

$AO=OS$

$\\\angle ROS=\angle AOB(\ Vertically\ Opposite\ Angle)$

$\\\angle RSA=\angle BAS\ (\ Alternating\ Angle )$

Hence, According to ASA Congruence Criterion

$\triangle AOB\cong\triangle SOR$

CBSE Questions for Class 7 Maths Congruence Of Triangles Quiz 5 - MCQExams.com

0:0:1

Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Congruence Of Triangles Quiz 5 - MCQExams.com

0:0:1

Practice Class 7 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.