Congruence of Triangles Test

Result

Congruence of Triangles Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

If $$\Delta ABC$$ and $$\Delta XYZ$$ are equilateral triangles and $$AB = XY,$$ the condition under which $$\Delta ABC \cong\Delta XYZ$$ is

A
$$ASA$$

B
$$RHS$$

C
$$SSS$$

D
$$AAS$$

Solution

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

$$\Delta ABC \cong\Delta XYZ$$

In $$\Delta ABC$$ and $$\Delta XYZ$$
$$AB=XY$$ (Given)
$$BC=YZ$$ (by equation $$(3)$$)
$$CA=ZX$$ (by equation $$(3)$$)
So $$\Delta ABC \cong\Delta XYZ$$ (By $$SSS$$ rules).
Question 2
1 / -0

If $$\Delta PQR$$ in congruent to $$\Delta STU$$, then what is the length of $$TU$$?

A
$$5 \,cm$$

B
$$6 \,cm$$

C
$$7 \,cm$$

D
cannot be determined

Solution

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.
Question 3
1 / -0

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

A
$$RHS$$

B
$$ASA$$

C
$$SSS$$

D
$$SAS$$

Solution

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.
Question 4
1 / -0

If $$\Delta ABC$$ and $$\Delta DBC$$ are on the same base $$BC, AB = DC$$ and $$AC = DB$$, then which of the following gives a congruence relationship?

A
$$\Delta ABC\cong \Delta DBC$$

B
$$\Delta ABC \cong \Delta CBD$$

C
$$\Delta ABC\cong \Delta DCB$$

D
$$\Delta ABC\cong \Delta BCD$$

Solution

In $$\triangle ABC$$ 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.
Question 5
1 / -0

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

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

B
$$\triangle CBA\cong \triangle PRQ$$

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

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

Solution

In$$\triangle ABC$$ and $$\triangle QRP$$
AB=QR
BC=PR
and CA=PQ
By SSS congruency,
$$\triangle ABC\cong \triangle QRP$$
or $$\triangle CBA\cong \triangle PRQ$$
Question 6
1 / -0

In triangles $$ABC$$ and $$DEF$$, $$AB=FD$$ and $$\angle A=\angle D$$. Two triangles are congruent by SAS criterion if

A
$$BC=EF$$

B
$$AC=DE$$

C
$$AC=EF$$

D
$$BC=DE$$

Solution

In $$\triangle ABC$$ and $$\triangle DEF$$
$$AB=DF$$
$$\angle A=\angle D$$
for congruence by SAS rule $$AC=DE$$
Question 7
1 / -0

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:

A
RHS property

B
SAS property

C
ASA property

D
SSS property

Solution
Question 8
1 / -0

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

A
$$\Delta ABC=\Delta PQR$$

B
$$\Delta CBA=\Delta PRQ$$

C
$$\Delta BAC=\Delta RPQ$$

D
$$\Delta PQR=\Delta BCA$$

Solution
Question 9
1 / -0

In fig 3, if AB $$=$$ AC and AP $$=$$ AQ, then by which congruence criterion $$\Delta PBC \cong \Delta QCB$$

A
SSS

B
ASA

C
SAS

D
RHS
Question 10
1 / -0

If $$\Delta ABC\cong \Delta DEF$$ write the part of $$\Delta ABC$$ that correspond to

A
DE

B
$$\angle E$$

C
DF

D
EF

E
$$\angle F$$