If $$AB=QR,BC=PR$$ and $$CA=PQ$$ 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 6

In given figure,
CF and AE are equal perpendiculars on BD, BF = FE = ED.

$\angle$ BAE = ______.

0%
$\angle$ BCD
0%
$\angle$ CBA
0%
$\angle$ ADC
0%
$\angle$ DCF

Explanation

$\triangle ABE$ and

$\triangle CDF$

$\Rightarrow$

$AE = CF$ [given]

$\Rightarrow$

$\angle AEB=\angle CFD$ [given]

$\Rightarrow$

$BF + FE = DE + EF$

$\Rightarrow$

$BE = DF$

$\therefore$

$\triangle ABE \cong \triangle CDF$ [By SAS criteria]

$\therefore$

$\angle BAE=\angle DCF$ [By CPCT]

In the figure, ABC is a triangle in which AB = AC. X and Y are points on AB and AC such that AX =AY. Then

$\Delta ABY \cong \Delta ACX$

0%
True
0%
False

Explanation

$The\quad statement\quad is\quad true$

$Consider\triangle ABY\quad and\quad \triangle ACX$

$AB=AC\quad (given)$

$AX=AY\quad (given)$

$\angle A=\angle A\quad (common)$

$\therefore \triangle ABY\cong \triangle ACX\quad \{ By\quad SAS\quad congruency\}$

$\Delta ABC$ , E and D are the two points on BC such that CE = DB and

$\angle AEC= \angle ADB$ . Then the statement

$\Delta CAD \cong \Delta BAE$ is

0%
True
0%
False

Two line segments are congruent if they have the same length.

0%
True
0%
False

Explanation

We know, two shapes are said to be congruent when they exactly overlap each other when one is put over the other.

Similarly, when length of two line segments are equal, they overlap each other.

Therefore, we can say that, two line segments are congruent if they have the same length.

Hence they are congruent.

Therefore, the statement is true and and option

$A$ is correct.

In right triangle ABC, the right angle is 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 (see figure).

State whether the statement is true/false

$\Delta AMC \cong \Delta BMD$

0%
True
0%
False

Explanation

$In\triangle AMC\& \triangle DMB$

$\quad AM=MB=x$ (given)

$\quad DM=CN=y$ (given)

$\quad \angle DMB=\angle AMC$ (vertically opposite)

$\therefore \triangle AMC\cong \triangle DMB$ (by S.A.S)

1003231_1040288_ans_5f960e552d644887ad245eb047924f6c.png

$ABCD$ is a square and

$\Delta PAD$ is equilateral then

$PB = 2PC$ .

0%
True
0%
False

Explanation

$\Delta PAB$ &

$\Delta PDC$

$PA = PD$ [ Side of equilateral

$\Delta$ ]

$AB = CD$ [Side of square]

$\angle BAP = \angle CDP = 150^{\circ}$

$\Delta PAB \cong \Delta PDC$

$PB = PC$

1323457_1063214_ans_441f8a70c2784b7a9c1cc1af3bddf5bb.png

In the given figure,

$ABC$ is an isosceles triangle in which

$AB = AD$ . Also,

$D$ is a point such that

$BD = CD$ . then "

$AD$ bisects

$\angle A$ and

$\angle D$ ." is ?

0%
True
0%
False

Explanation

Given:

$\triangle ABC$ is an isosceles triangle.

Consider

$\triangle ABD$ and

$\triangle ACD$

$\Rightarrow$

$AB=AC$ [ Given ]

$\Rightarrow$

$BD=CD$ [ Given ]

$\Rightarrow$

$AD=AD$ [ Common side ]

$\therefore$

$\triangle ABD\cong\triangle ACD$ [ By

$SSS$ congruence property ]

$\Rightarrow$

$\angle BAD=\angle CAD$ [ C.P.C.T ]

$\Rightarrow$

$\angle BDA=\angle CDA$ [ C.P.CT. ]

Hence, we have proved that

$AD$ bisects

$\angle A$ and

$\angle D.$

PQR is a right angled triangle in which

$\angle R=90^{\circ}$ . If RS

$\perp$ PQ, PR=3 cm, and RQ=4 cm, then what is the value of RS(in cm)?

0%
$\dfrac {12} 5$
0%
$\dfrac {36} 5$
0%
$5$
0%
$2.5$

State true or false.

In the given figure,

$AB=AC$ , then

$\begin{array} { l } { \text {} \mathrm { DP } = \mathrm { DQ } } \end{array}.$

0%
True
0%
False

Explanation

$AB$ &

$AC$ are equal in length
So, in

$\triangle ABC$ ,

$\angle B=\angle C$

From

$\triangle BPD$ &

$\triangle DQC$

$i) \angle BPD=\angle CQD=90^0$

$ii) BD=DC$

$iii) \angle B=\angle C$

$\triangle BPD \cong \triangle CQD (by \ ASA \ criteria)$

by CPCT

$DP=DQ$

$\Delta PQR$ and

$\Delta SQR$ ,

$\overline{PQ}$ =

$\overline{SR}$ and

$\angle PQR = \angle QRS$ then:

0%
$\Delta PQR \sim \Delta SRQ$
0%
$\Delta PQR \equiv \Delta SRQ$
0%
$\Delta PQR = \Delta SRQ$
0%
None of these

Explanation

$\Delta PQR$ and

$\Delta SQR$ ,

$\overline{PQ} =\overline{SR}$ (Given)

And

$\angle PQR = \angle QRS$

$QR=QR$ [common side]

$\Delta PQR \sim \Delta SRQ$ (SAS property)

$\triangle APQ$ is such that perpendicular bisector of side

$PQ$ bisect it at

$M$ and passes from vertex

$A$ . Then by using which of the following congruence rule it can be proved that

$\triangle \mathrm{AMP} \cong \triangle \mathrm{AMQ}$ .

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

$\Delta ABC$ and

$\Delta XYZ$ are equilateral triangles and

$AB = XY,$ the condition under which

$\Delta ABC \cong\Delta XYZ$ is

0%
$ASA$
0%
$RHS$
0%
$SSS$
0%
$AAS$

Explanation

Given that

$\Delta ABC$ and

$\Delta XYZ$ are equilateral triangles.

i.e.,

$AB=BC=CA$

and

$XY=YZ=ZX$ …………….

$(1)$

and

here

$AB=XY$ ………………

$(2)$

By equation

$(1)$ &

$(2)$

$BC=YZ$ and

$CA=ZX$ …………….

$(3)$

We have to show that

$\Delta ABC \cong\Delta XYZ$

$\Delta ABC$ and

$\Delta XYZ$

$AB=XY$ (Given)

$BC=YZ$ (by equation

$(3)$ )

$CA=ZX$ (by equation

$(3)$ )

$\Delta ABC \cong\Delta XYZ$ (By

$SSS$ rules).

1221372_1450931_ans_0574deedf4ee4e5586fe20f3a179e7f1.png

$\Delta PQR$ in congruent to

$\Delta STU$ , then what is the length of

$TU$ ?

0%
$5 \,cm$
0%
$6 \,cm$
0%
$7 \,cm$
0%
cannot be determined

Explanation

Given,

$\Delta PQR \cong \Delta STU$ .

Then the corresponding parts will also be equal.

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

Then, $PQ=ST$ , $QR=TU$ , $PR=SU$ , $\angle P=\angle S$ , $\angle Q=\angle T$ and $\angle R=\angle U$ .

Thus, $QR=TU$ .

$\therefore TU = 6 \ cm$ .

Therefore, option

$B$ is correct.

By which congruency criterion, the two triangles in Fig. are congruent?

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

Explanation

In triangle

$PQR$ and

$PQS$ ,

$PQ = PQ$ ...Common

$PR = PS = a \,cm$ ...Given

$QR = QS = b \,cm$ ...Given

$\therefore \Delta PSQ = \Delta PRQ$ ...

$SSS$ criterion

AS, all three sides are same so, it is congruent by

$SSS$ test.

So, option

$C$ is correct.

$\Delta ABC$ and

$\Delta DBC$ are on the same base

$BC, AB = DC$ and

$AC = DB$ , then which of the following gives a congruence relationship?

0%
$\Delta ABC\cong \Delta DBC$
0%
$\Delta ABC \cong \Delta CBD$
0%
$\Delta ABC\cong \Delta DCB$
0%
$\Delta ABC\cong \Delta BCD$

Explanation

$\triangle ABC$ and

$\triangle DCB,$

$AC = DB$ [Given]

$AB = DC$ [Given]

$BC = CB$ [Common]

So, by

$SSS$ rule of congruence,

$\Delta ABC = \Delta DCB$

Hence, option

$C$ is correct.

All congruent triangles are equal in area.

0%
True
0%
False

All triangles are congruent.

0%
True
0%
False

Write true or False and justify your answer in each of the following:
If the angle between two tangents drawn from a point P to a circle of radius 'a' and centre O is

$60^\circ,$ then

$OP = a\sqrt{3}.$

0%
True
0%
False

Explanation

PT and OT are tangent and radius respectively at contact point T.

$\therefore \angle OTP = 90^\circ$

$\Rightarrow \Delta OTP$ is right angle

$\Delta$ at T

$\Delta OPT \cong \Delta OPR$ [By SSS criterion of congruence]

$\Rightarrow \angle OPT = \angle OPR = \dfrac{1}{2} \times 60^\circ = 30^\circ$

$\therefore$ In right angle

$\Delta OPT,$

$sin 30^\circ = \dfrac{OT}{OP} \Rightarrow$

$\Rightarrow OP = 2a$

$AB=QR,BC=PR$ and

$CA=PQ$ then

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

Explanation

$\triangle ABC$ and

$\triangle QRP$

AB=QR

BC=PRAB=QR,BC=PR

and CA=PQCA=PQ

By SSS congruency,

$\triangle ABC\cong \triangle QRP$

$\triangle CBA\cong \triangle PRQ$

In triangles

$ABC$ and

$DEF$ ,

$AB=FD$ and

$\angle A=\angle D$ . Two triangles are congruent by SAS criterion if

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

Explanation

$\triangle ABC$ and

$\triangle DEF$

$AB=DF$

$\angle A=\angle D$

for congruence by SAS rule

$AC=DE$

1863856_1876416_ans_6f62c0688d2d4528851274d53898994f.PNG

In two triangles

$ABC$ and

$DEF$ , if

$AC = DF, BC = EF$ and

$\angle ABC = \angle DEF = 90^{0}$ then two triangles are said to be congruent by:

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

$AB = QR; BC = PR$ and

$CA = PQ$ , then

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

In fig 3, if AB

$=$ AC and AP

$=$ AQ, then by which congruence criterion

$\Delta PBC \cong \Delta QCB$

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

$\Delta ABC\cong \Delta DEF$ write the part of

$\Delta ABC$ that correspond to

0%
DE
0%
$\angle E$
0%
DF
0%
EF
0%
$\angle F$

if ABC and DEF are congruent triangles such that

$\angle A={ 47 }^{ \circ }\quad and\quad \angle E={ 83 }^{ \circ },\quad then\quad \angle C=$

0%
${ 100 }^{ \circ }$
0%
${ 50 }^{ \circ }$
0%
${ 90 }^{ \circ }$
0%
NONE OF THESE

CBSE Questions for Class 7 Maths Congruence Of Triangles Quiz 6 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Congruence Of Triangles Quiz 6 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

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