Triangles Test

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Triangles Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

In a rectangle $$PQRS$$, $$QS$$ is a diagonal. Then $$\displaystyle \Delta PQS......\Delta RSQ$$

A
is similar to

B
is not equal to

C
is congruent to

D
is not congruent to

Solution

In a rectangle $$PQRS$$, $$QS$$ is a diagonal.
We know that , in a rectangle opposite sides are equal.
$$\displaystyle \therefore PQ=SR$$
Also, $$\displaystyle \angle R=\angle P={ 90 }^{ o }$$
Ans $$QS$$ is the hypotenuse for both right angled triangles $$PQS$$ and $$RSQ$$.
$$\displaystyle \therefore $$ according to R.H.S condition for congruence, $$\displaystyle \Delta PQS\cong \Delta RSQ$$.
Question 2
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For $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta DEF,\angle A={ 47 }^{ o },\angle E={ 53 }^{ o }$$ and $$\displaystyle AB=DE=16 \ cm\\ $$. $$\displaystyle \angle A=\angle D$$ and $$\displaystyle \angle B=\angle E$$, then $$\displaystyle \angle C= $$..............

A
$$\displaystyle { 47 }^{ o }$$

B
$$\displaystyle { 53 }^{ o }$$

C
$$\displaystyle { 80 }^{ o }$$

D
$$\displaystyle { 100 }^{ o }$$

Solution

For $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta DEF$$
$$\displaystyle \angle A=\angle D,\angle B=\angle E$$ and $$\displaystyle AB=DE$$
$$\displaystyle \therefore $$ According to A.S.A condition for congruence,
$$\displaystyle \Delta ABC\cong \Delta DEF$$
$$\displaystyle \angle A=\angle D={ 47 }^{ o }$$
$$\displaystyle \angle B=\angle E={ 53 }^{ o }$$
$$\displaystyle \angle C=\angle F$$
$$\displaystyle \therefore \quad \angle D+\angle E+\angle F=\angle A+\angle B+\angle C={ 180 }^{ o }\\ $$
$$\displaystyle \therefore \quad \angle C={ 180 }^{ o }-{ 100 }^{ o }$$
$$\displaystyle \therefore \quad \angle C={ 80 }^{ o }$$
Question 3
1 / -0

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 4
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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 5
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For $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta DEF,$$ $$AB=FE, BC=ED$$ and $$\displaystyle \angle B=\angle E$$. Therefore ............

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

B
$$\displaystyle \Delta ABC\cong \Delta EDF$$

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

D
$$\displaystyle \Delta BCA\cong \Delta EDF$$

Solution

For, $$\displaystyle \Delta BCA$$ and $$\displaystyle \Delta EDF,$$
$$\displaystyle AB=FE\quad \therefore \angle C=\angle D$$
$$\displaystyle BC=ED\quad \therefore \angle A=\angle F$$
$$\displaystyle \angle B=\angle E$$
$$\displaystyle \Delta BCA\cong \Delta DEF$$
(By $$SAS$$ condition for congruence).
Question 6
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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 7
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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 8
1 / -0

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 9
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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 10
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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 $$\displaystyle \angle A+\angle B+\angle C={ 180 }^{ o }$$
$$\displaystyle \therefore \angle C={ 180 }^{ o }-{ 90 }^{ o }-{ 40 }^{ o }$$
$$\displaystyle \therefore \angle C={ 50 }^{ o }$$