Congruence of Triangles Test

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Congruence of Triangles Test - 20

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Question 1
1 / -0

For $$\displaystyle \Delta ABC \cong \displaystyle \Delta DEF$$. Then according to the given figure,$$\angle D = 40^o$$, $$\displaystyle \angle C=$$.....

A
$$\displaystyle { 40 }^{ o }$$

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

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

D
$$\displaystyle { 50 }^{ o }$$

Solution

According to the given figure,
$$\displaystyle \angle B=\angle E={ 90 }^{ o }$$
$$\Rightarrow AC=DE=5 $$ cm (hypotenuse) and $$BC=EF=2 $$ cm
$$\displaystyle \therefore \Delta ABC\cong \Delta DEF$$ ..... (R.H.S condition for congruence)
$$\displaystyle \Rightarrow \angle A=\angle D={ 40 }^{ o }$$
$$\displaystyle \Rightarrow \angle B={ 90 }^{ o }$$
$$\displaystyle \angle C=?$$
We know $$\displaystyle \angle A+\angle B+\angle C={ 180 }^{ o }$$
$$\displaystyle \therefore \angle C={ 180 }^{ o }-{ 90 }^{ o }-{ 40 }^{ o }$$
$$\displaystyle \therefore \angle C={ 50 }^{ o }$$
Question 2
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For $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta PQR$$, $$AB=PQ, AC=PR,$$ and $$\displaystyle \angle A=\angle P,$$ then:

A
$$\displaystyle \Delta BAC\cong \Delta PRQ$$

B
$$\displaystyle \Delta BAC\cong \Delta PQR$$

C
$$\displaystyle \Delta ABC\cong \Delta PRQ$$

D
$$\displaystyle \Delta ABC\cong \Delta PQR$$
Question 3
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For $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta DEF$$, $$\displaystyle \angle B=\angle D,\angle C=\angle F$$ and $$BC=DF$$. Therefore which of the following is correct?

A
$$\displaystyle \Delta ABC\cong \Delta DEF$$

B
$$\displaystyle \Delta BCA\cong \Delta DFE$$

C
$$\displaystyle \Delta BCA\cong \Delta DEF$$

D
$$\displaystyle \Delta ABC\cong \Delta DFE$$

Solution

For $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta DEF$$, we have
$$\displaystyle \angle B=\angle D,\angle C=\angle F$$ .....Given
$$BC=DF$$ .....Given
Therefore, according to A.S.A condition for congruence,
$$\displaystyle \Delta BCA\cong \Delta DFE$$
Hence, option B is correct.
Question 4
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Referring the figure given below, for $$\displaystyle \Delta PQR$$ and $$\displaystyle \Delta XYZ$$, $$PQ=YZ, QR=XZ$$ and $$\displaystyle \angle Q=\angle Z$$. The value of $$\displaystyle \angle P$$ is ___.

A
$$\displaystyle { 42 }^{ \circ }$$

B
$$\displaystyle { 75 }^{ \circ }$$

C
$$\displaystyle { 63 }^{ \circ }$$

D
Cannot be determined

Solution

In $$\displaystyle \Delta PQR$$ and $$\displaystyle \Delta XYZ,$$

$$\displaystyle PQ=YZ$$

$$\displaystyle QR=XZ$$

$$\displaystyle \angle Q=\angle Z$$

$$\displaystyle \therefore \Delta PQR\cong \Delta XYZ$$ ($$S.A.S.$$ condition for congruence)

$$\displaystyle \therefore \angle R=\angle X={ 63 }^{ \circ }$$

$$\displaystyle \Rightarrow \angle P=\angle Y$$

$$\displaystyle \Rightarrow \angle Q=\angle Z={ 75 }^{ \circ }$$

Now, $$\displaystyle \angle P+\angle Q+\angle R={ 180 }^{ \circ }$$

$$\displaystyle \therefore \angle P+{ 75 }^{ \circ }+{ 63 }^{ \circ }={ 180 }^{ \circ }$$

$$\displaystyle \Rightarrow \angle P={ 42 }^{ \circ }$$
Question 5
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In a $$\triangle ABC$$, $$D$$ is the midpoint of the side $$BC. DE$$ is perpendicular to $$AB$$ and $$DF$$ is perpendicular to $$AC$$. Also, $$DE=DF$$, Then $$\displaystyle \angle B$$ =

A
$$\displaystyle \angle E$$

B
$$\displaystyle \angle F$$

C
$$\displaystyle \angle C$$

D
$$\displaystyle \angle A$$

Solution

Given: $$D$$ is the midpoint of $$BC$$

$$\displaystyle \therefore \quad BD=DC$$

Also, $$DE = DF$$

$$DE$$ is perpendicular to $$AB$$

$$\displaystyle \therefore \angle E={ 90 }^{ o }$$

$$DF$$ is perpendicular to $$AC$$

$$\displaystyle \therefore \angle F={ 90 }^{ o }$$

If we consider two triangles so formed, $$\displaystyle \Delta DEB$$ and $$\displaystyle \Delta CFD$$, then their hypotenuse $$BD$$ and $$CD$$ are equal.

$$\displaystyle \angle E=\angle F={ 90 }^{ o }$$. Also $$DE=CF$$.

Thus according to R.H.S condition of congruence $$\displaystyle \Delta DEB\cong \Delta CFD$$

Thus, the corresponding angles $$\displaystyle \angle B$$ and $$ \angle C$$ are equal, i.e $$\displaystyle \angle B=\angle C$$
Question 6
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In the given figures, which of the following satisfies A.S.A. condition for congruence?

A
Figure II

B
Figure I and II

C
Figure I

D
None of these

Solution

Figure 1 satisfy the condition of A.S.A
Here $$\angle ACB=\angle DCE$$ {vertical angle|
$$DC=BC$$ and $$\angle EDC=\angle ABC$$ [Alternate angles]
$$\therefore$$ It satisfy A.S.A Congruency.
Question 7
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For $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta DEF$$, $$\displaystyle \angle B=\angle D={ 90 }^{ o }$$, $$\displaystyle AC=EF$$ and $$\displaystyle AB=ED$$. Then $$\displaystyle \Delta ABC\cong $$..........

A
$$\displaystyle \Delta EDF$$

B
$$\displaystyle \Delta DEF$$

C
$$\displaystyle \Delta EFD$$

D
$$\displaystyle \Delta FED$$

Solution

For $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta DEF$$,
$$\displaystyle \angle B=\angle D={ 90 }^{ o }$$, $$\displaystyle AC=EF$$ and $$\displaystyle AB=ED$$
$$\displaystyle \therefore $$ According to R.H.S condition for congruence, both triangles are
$$\displaystyle \therefore \Delta ABC\cong \Delta EDF$$
Question 8
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$$AD$$ is the bisector of BC.
Which property of triangle applies here?

A
$$SAS$$

B
$$SSS$$

C
$$AAS$$

D
none

Solution

In $$\triangle ABD$$ & $$\triangle ADC$$
$$AD = AD$$
$$BD = DC$$
$$\angle ADB = \angle ADC$$
$$\triangle ABD \cong \triangle ADC$$
$$\therefore$$ Both triangles are congruent by $$SAS$$ congruence condition.
Question 9
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Given are two congruent triangles $$ABC$$ and $$XYZ.$$ Which of the following statements is incorrect?

A
$$\displaystyle \angle A=\angle X$$

B
$$\displaystyle AB=XY$$

C
$$\displaystyle \Delta ABC\cong \Delta XYZ$$

D
$$\displaystyle AC=YZ$$

Solution

Given, $$\triangle ABC$$ and $$\triangle XYZ$$ are congruent

From figure, we have

$$AC=XZ, BC=YZ$$ and $$\angle B=\angle Y=90^o$$

$$\displaystyle \therefore AC=YZ$$ is an incorrect statement.
Question 10
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In the figure, $$\displaystyle \Delta ACB\cong \Delta DCB$$, then the value of $$x$$ and $$y$$ are:

A
$$14 $$ cm, $$\displaystyle { 20 }^{ o }$$

B
$$28 $$ cm, $$\displaystyle { 20 }^{ o }$$

C
$$\displaystyle { 14 }^{ o }$$, $$20 $$ cm

D
$$\displaystyle { 28 }^{ o }$$, $$20 $$ cm

Solution

From the figure, we have,
$$\angle D= 20^o, BD=28 $$ cm.
We need to find value of $$x$$ and $$y$$.

In $$\displaystyle \Delta ACB\cong \Delta DCB$$, we have,
Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.
Then, $$AC=DC$$, $$CB=CB$$, $$AB=DB$$, $$\angle A=\angle D$$, $$\angle C=\angle C$$ and $$\angle B=\angle B$$.
$$\displaystyle \therefore AB=BD $$ cm
$$\displaystyle \therefore 2x=28 $$ cm
$$\displaystyle \therefore x=14 $$ cm.

Also, $$\displaystyle \angle A=\angle D$$
$$\displaystyle \therefore y={ 20 }^{ o }$$.

Therefore, option $$A$$ is correct.