Congruence of Triangles Test

Result

Congruence of Triangles Test - 14

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Question 1
1 / -0

Which pair of triangles shows congruency by the SSS postulate?

A
figure C

B
figure A

C
figure B

D
figure D

Solution

From figure C, triangles are congruent by the SSS postulate because the three sides of one triangle are congruent to three sides of another triangle.
Therefore, the given triangles are congruent by the SSS congruency postulate.
Question 2
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In $$\Delta JKL$$ and $$\Delta MNO$$, $$\displaystyle { JL } \cong { ON } $$ and $$\displaystyle { KL } \cong { OM } $$, then the condition which will make both triangles congruent is:

A
$$\displaystyle \angle L\cong \angle O$$

B
$$\displaystyle \angle K\cong \angle M$$

C
$$\displaystyle \angle J\cong M$$

D
$$\displaystyle \angle J\cong \angle N$$

Solution

Two triangles can be congruent by the SAS test, since here we are equating two sides and one angle.
For the SAS test of congruency, the condition is that the angle must be included between the two sides to ensure correctness of the test.
Similarly, here the two included angles must be congruent for the triangles to be congruent.
They are $$\angle L \cong \angle O$$ since they are the included angles of sides $$JL-KL$$ and $$ON-OM$$ respectively.
Question 3
1 / -0

In $$\triangle ABC$$ and $$\triangle DEF$$, $$\angle B=\angle E,AB=DE,BC=EF$$. The two triangles are congruent under ............. axiom.

A
SSS

B
AAA

C
SAS

D
ASA

Solution

Two sides and included angle of $$\triangle ABC$$ and $$\triangle DEF$$ are equal.
Therefore the triangles are congruent by $$SAS$$ axiom.
Option $$C$$ is correct.
Question 4
1 / -0

Two plane figures are said to be congruent if they have_____.

A
same size

B
same shape

C
same size and same shape

D
none

Solution

Two plane figures are congruent to each other, if trace copy of one of the figures covers the other figure completely.

Therefore, option $$C$$ is correct.
Question 5
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Which of the following is congruent to the above figure?

A

B

C

D
None of these

Solution

Two figures are said to be congruent if they have exactly same shape and size or one can exactly overlap the other.
Here, if we rotate the figure in option $$C$$, and place it over each other, we get the same figure given, i.e. they will overlap.

Therefore, option $$C$$ is correct.
Question 6
1 / -0

Which of the following is congruent to the above figure?

A

B

C

D
None of these

Solution

Two figures are said to be congruent if they have exactly same shape and size or one can exactly overlap the other.
Here, if we rotate the figure in option $$A$$, and place it over each other, we get the same figure given, i.e. they will overlap.

Therefore, option $$A$$ is correct.
Question 7
1 / -0

In the given figure, $$PA$$ $$\perp$$ $$AB$$, $$QB$$ $$\perp$$ $$AB$$ and $$\Delta$$ $$OAP$$ $$\cong \Delta$$ $$OBQ$$, then:

A
$$PA$$ $$=$$ $$OB$$

B
$$AP$$ $$=$$ $$QB$$

C
$$OP$$ $$=$$ $$BQ$$

D
$$OA$$ $$=$$ $$OQ$$

Solution

It is given that $$\Delta OAP $$ $$\cong $$ $$\Delta OBQ$$ both triangles are congruent.
That is by CPCT rule, corresponding parts of congruent triangles are equal.
Then, $$OA = OQ , AP = QB , OP = OQ$$, $$\angle O=\angle O$$, $$\angle A=\angle B$$ and $$\angle P=\angle Q$$.

So, option $$B$$ is correct.
Question 8
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If for $$\triangle$$ABC and $$\triangle$$DEF, the correspondence CAB $$\leftrightarrow$$ EDF gives a congruence, then which of the following is NOT true?

A
AC = DE

B
AB = EF

C
$$\angle$$A = $$\angle$$D

D
$$\angle$$C = $$\angle$$E

Solution

In $$\triangle ABC$$ and $$\triangle DEF$$
$$\Rightarrow$$ The correspondence $$CAB\leftrightarrow EDF$$ gives a congruence.
$$\Rightarrow$$ So, $$CA=ED,\,AB=DF,\,CB=EF$$
$$\Rightarrow$$ $$\angle CAB=\angle EDF,\,\angle ACB=\angle DEF,\,\angle ABC=\angle DFE$$
$$\therefore$$ $$AB=EF$$ is not true.
Question 9
1 / -0

Which of the following pair of triangles are congruent by RHS criterion?

A
(i) and (ii)

B
(iii) and (iv)

C
(i) and (iii)

D
(ii) and (iv)

Solution

In $$\triangle ABC$$ and $$\triangle QRP$$
$$\Rightarrow$$ $$AB=QR=4\,cm$$
$$\Rightarrow$$ $$BC=RP=5\,cm$$ [Hypotenuse]
$$\Rightarrow$$ $$\angle A=\angle Q=90^o$$
$$\therefore$$ $$\triangle ABC\cong\triangle QRP$$ [By RHS criteria]
Question 10
1 / -0

In the given figure, triangles $$ABC$$ and $$DCB$$ are right angled at $$A$$ and $$D$$ respectively and $$AC$$ $$=$$ $$DB$$, then $$\Delta$$ $$ABC$$ $$\cong\Delta$$ $$DCB$$ of from.

A
$$AAA$$

B
$$SAS$$

C
$$RHS$$

D
None of these

Solution

In $$\Delta ABC$$ and $$\Delta DCB,$$
$$AC = DB$$ [Given]
$$ \angle BAC = \angle CDB = 90^{\circ}$$
$$BC=BC$$ [Common Hypotenuse ]

So, by $$RHS$$ rule of congruence,
$$\Delta ABC $$ $$\cong $$ $$\Delta DCB$$

Hence, option $$C$$ is correct.