Triangles Test

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Triangles Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

In the figure given below, $$\displaystyle \Delta ABC\cong \Delta MBC.$$ The corresponding side to $$BA$$ is ......

A
$$BM$$

B
$$AB$$

C
$$AC$$

D
$$MC$$

Solution

Given $$\triangle ABC \cong \triangle MBC$$.

Then the corresponding parts will also be equal.

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

Then, $$AB=MB$$, $$BC=BC$$, $$AC=MC$$, $$\angle A=\angle M$$, $$\angle B=\angle B$$ and $$\angle C=\angle C$$.

Thus, $$AB=MB$$ $$\implies$$ $$BA=BM$$.

Hence, option $$A$$ is correct.
Question 2
1 / -0

In the figure given below, $$\displaystyle \Delta ABC\cong \Delta XYZ$$. The corresponding side to $$XZ$$ is .......

A
$$AB$$

B
$$BC$$

C
$$AC$$

D
$$YZ$$

Solution

Given $$\triangle ABC \cong \triangle XYZ$$.

Then the corresponding parts will also be equal.

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

Then, $$AB=XY$$, $$BC=YZ$$, $$AC=XZ$$, $$\angle A=\angle X$$, $$\angle B=\angle Y$$ and $$\angle C=\angle Z$$.

Thus, $$AC=XZ$$.

Hence, option $$C$$ is correct.
Question 3
1 / -0

In the triangles $$ABC$$ and $$PQR,$$ $$AC=QP, BC=RP$$ and $$AB=QR.$$ The measure of $$\angle A$$ is ___.

A
$$\displaystyle { 30 }^{ \circ }$$

B
$$\displaystyle { 120 }^{ \circ }$$

C
$$\displaystyle { 20 }^{ \circ }$$

D
$$\displaystyle { 40 }^{ \circ }$$

Solution

Given: In $$\triangle ABC$$ and $$\triangle PQR,$$

$$\angle C=120^\circ$$ and $$\angle R=40^\circ$$

Now, $$AC=QP$$ ...(1)

and $$ BC=RP$$ ...(2)

and $$AB=QR$$ ...(3)

Hence $$ABC \cong PQR$$ $$($$by $$SSS$$ congruency rule$$)$$

By CPCT, $$ \angle B=\angle R=40^\circ$$

and $$ \angle C=\angle P={ 120 }^{ \circ }$$

and $$ \angle A=\angle Q$$

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

$$\displaystyle \Rightarrow \angle A+{ 40 }^{ \circ }+{ 120 }^{ \circ }={ 180 }^{ \circ }$$

$$\displaystyle \Rightarrow \angle A={ 20 }^{ \circ}$$
Question 4
1 / -0

In the figure given below, $$\displaystyle \Delta BCA\cong \Delta BCD$$. Corresponding angle to $$\displaystyle \angle D$$ is ...........

A
$$\displaystyle \angle B$$

B
$$\displaystyle \angle C$$

C
$$\displaystyle \angle D$$

D
$$\displaystyle \angle A$$

Solution

Given $$\triangle BCA \cong \triangle BCD$$.

Then the corresponding parts will also be equal.

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

Then, $$BC=BC$$, $$CA=CD$$, $$BA=BD$$, $$\angle B=\angle B$$, $$\angle C=\angle C$$ and $$\angle A=\angle D$$.

Thus, $$\angle A=\angle D$$.
Hence, option $$D$$ is correct.
Question 5
1 / -0

In the figure given below, both the triangles are congruent. The corresponding side to $$BC$$ is .......

A
$$EF$$

B
$$AC$$

C
$$DF$$

D
$$DE$$

Solution

Given $$\triangle ABC \cong \triangle DEF$$.

Then the corresponding parts will also be equal.

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

Then, $$AB=DE$$, $$BC=EF$$, $$AC=DF$$, $$\angle A=\angle D$$, $$\angle B=\angle E$$ and $$\angle C=\angle F$$.

Thus, $$BC=EF$$.
Hence, option $$A$$ is correct.
Question 6
1 / -0

Which condition for congruence is used to prove that the following pairs of triangles are congruent

A
$$ASA$$

B
$$AAS$$

C
$$SAS$$

D
$$SSS$$

Solution

In the given $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta DEF$$,
we can clearly see that sides $$AB = DE, BC = EF$$ and $$AC = DF.$$
Thus, by $$SSS$$ condition of congruence, the given $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta DEF$$ are congruent.
Question 7
1 / -0

In the figure given below, both the triangles are congruent. The corresponding side to $$AB$$ is ___.

A
$$AB$$

B
$$DE$$

C
$$BC$$

D
$$DF$$

Solution

Given $$\triangle ABC \cong \triangle DEF$$.

Then the corresponding parts will also be equal.

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

Then, $$AB=DE$$,
$$BC=EF$$,
$$AC=DF$$,
$$\angle A=\angle D$$,
$$\angle B=\angle E$$
$$\angle C=\angle F$$.

Thus, $$AB=DE$$.

Hence, option $$B$$ is correct.
Question 8
1 / -0

It is given that $$\Delta ABC\cong \Delta DEF$$. In the given figure the corresponding angle to $$\displaystyle \angle B$$ is___.

A
$$\displaystyle \angle A$$

B
$$\displaystyle \angle C$$

C
$$\displaystyle \angle E$$

D
$$\displaystyle \angle F$$

Solution

Given $$\triangle ABC \cong \triangle DEF$$.
Then the corresponding parts will also be equal.
That is by CPCT rule, corresponding parts of congruent triangles are equal.
Then, $$AB=DE$$, $$BC=EF$$, $$AC=DF$$, $$\angle A=\angle D$$, $$\angle B=\angle E$$ and $$\angle C=\angle F$$.

Thus, $$\angle B=\angle E$$.
Hence, option $$C$$ is correct.
Question 9
1 / -0

It is given that $$\Delta ABC\cong \Delta DEF$$. In the given figure the corresponding angle to $$\displaystyle \angle C$$ is ........

A
$$\displaystyle \angle B$$

B
$$\displaystyle \angle C$$

C
$$\displaystyle \angle F$$

D
$$\displaystyle \angle A$$

Solution

Given $$\triangle ABC \cong \triangle DEF$$.

Then the corresponding parts will also be equal.

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

Then, $$AB=DE$$, $$BC=EF$$, $$AC=DF$$, $$\angle A=\angle D$$, $$\angle B=\angle E$$ and $$\angle C=\angle F$$.

Thus, $$\angle C=\angle F$$.
Hence, option $$C$$ is correct
Question 10
1 / -0

Referring the figure below, for $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta DEF$$, $$ AB=DE, BC=EF$$ and
$$\displaystyle \angle B=\angle E.$$ The value of $$\displaystyle \angle E$$ is .........

A
$$\displaystyle { 30 }^{ o }$$

B
$$\displaystyle { 60 }^{ o }$$

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

D
$$\displaystyle { 45 }^{ o }$$

Solution

For $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta DEF$$
$$\displaystyle AB=DE,BC=EF$$ and $$\displaystyle \angle B=\angle E$$
$$\displaystyle \therefore \quad \Delta ABC\cong \Delta DEF$$ (S.A.S condition for congruence)
As, $$\displaystyle AB=DE\Rightarrow \angle C=\angle F={ 60 }^{ o }$$
$$\displaystyle BC=EF\Rightarrow \angle A=\angle D={ 60 }^{ o }$$
$$\displaystyle AC=DF\Rightarrow \angle B=\angle E$$
Now, $$\displaystyle \angle D+\angle E+\angle F={ 180 }^{ o }$$
$$\displaystyle \therefore \quad { 120 }^{ o }+\angle E={ 180 }^{ o }$$
$$\displaystyle \therefore \quad \angle E={ 60 }^{ o }$$