Congruence of Triangles Test

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Congruence of Triangles Test - 16

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Question 1
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$$\Delta ABC$$ is congruent of $$\Delta DEF$$, if

A
$$AB \neq DE$$, $$AC = 6 = DF$$, $$\angle A = \angle D$$

B
$$AB = DE$$, $$AC = DF$$, $$\angle B =\angle E$$

C
$$AB = DE$$, $$AC \neq DF$$, $$\angle C =\angle F$$

D
$$AB = DE$$, $$AC = DF$$, $$\angle C \neq \angle F$$
Solution
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.
Question 2
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If $$D$$ 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:

A
$$\triangle DAE$$ = $$\triangle EFC$$

B
$$AE = EF$$

C
$$DE = EF$$

D
$$\triangle ADE = \triangle ECF$$

Solution

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 $$ CFE$$, $$AD= CF$$, $$AE= AC$$
$$\angle DAE=\angle ECF$$
So $${ \Delta }^{ k } DAE$$ and$${ \Delta }^{ k } FCE$$ are congruent
So $$ DE = EF$$
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 $$ DE= EF$$.
Question 3
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In $$\triangle ABC$$ and $$\triangle DEF$$, $$AB = FD$$ and $$\angle A = \angle D.$$ The two triangles will be congruent by $$SAS$$ axiom, if:

A
$$BC = EF$$

B
$$AC = DE$$

C
$$AC = EF $$

D
$$ BC = DE$$

Solution

$$SAS=$$ 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.
Question 4
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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 $$BC = EF.$$

A
RHS

B
SAS

C
AAA

D
SSS

Solution

In $$\triangle ABC$$ and $$DEF$$,

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

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

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

Thus, $$\triangle ABC \cong \triangle DEF$$ (RHS postulate)
Question 5
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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.

A
$$\angle XBC = 45^{\circ}, \angle XCB = 46{\circ}$$

B
$$\angle XBC = 45^{\circ}, \angle XCB = 45^{\circ}$$

C
$$\angle XBC = 45^{\circ}, \angle XCB = 60^{\circ}$$

D
$$\angle XBC = 45^{\circ}, \angle XCB = 25^{\circ}$$

Solution

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.
Question 6
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Which pair of triangles is not congruent by $$SAS$$ congruence criterion

A

B

C

D
None of these

Solution

In option (3), we have

$$\angle BAC = \angle DFE = 120^{\circ}$$

$$BC = DE = 5$$

$$AC = DF = 3$$

For SAS to hold true corresponding angle included by two congruent sides must be congruent

here, angle A and F have not included angles by the congruent sides.
Hence, $$\triangle ABC \not\cong \triangle DEF$$
Question 7
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If $$AB=QR$$, $$BC=PQ$$ and $$CA=PR$$, then:

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

B
$$\triangle CBA\cong \triangle PQR$$

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

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

Solution

In $$\triangle ABC$$ and $$\triangle PQR$$
$$AB=QR$$
$$BC=PQ$$
$$CA=PR$$
Thus, $$\triangle CBA \cong \triangle PQR$$ (SSS rule)
Question 8
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$$l$$ and $$m$$ are two parallel lines intersected by another pair of parallel lines $$p$$ and $$q$$. Then:

A
$$\triangle ABC\cong \triangle CDA$$.

B
$$\triangle ABC\cong \triangle ADC$$.

C
$$\triangle ABC\cong \triangle DCA$$.

D
$$\triangle ABC\cong \triangle CAD$$.

Solution

In $$\triangle ABC$$ 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)
Question 9
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State the property by which $$\Delta ADB\, \cong\, \Delta ADC$$ in the following figure.,

A
SAS property

B
SSS property

C
RHS property

D
ASA property

Solution

From figure, we have
$$AB = AC $$ ....Given
$$\angle BAD\, =\, \angle CAD$$
$$AD = AD$$ ....Common sides
$$\therefore$$ By SAS property, $$\Delta ADB \cong \Delta ADC$$.
Question 10
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If in two triangles $$PQR$$ and $$DEF$$, $$PR=\,EF$$, $$QR=\,DE$$ and $$PQ=\,FD$$, then $$\triangle PQR\cong$$ $$\triangle$$ ___.

A
$$FDE$$

B
$$DEF$$

C
$$FED$$

D
$$DFE$$

Solution

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).