Triangles Test

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Triangles Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

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 2
1 / -0

Which condition of congruency proves that the following pair of triangles are congruent

A
$$A.A.S$$

B
$$S.S.S$$

C
$$S.A.S$$

D
$$A.S.A$$

Solution

From the given figure, we have
$$DE=EG, DF=GP$$ and $$FE=PE.$$
Thus, from $$S.S.S$$ condition of congruency,
$$\displaystyle \Delta DEF\cong \Delta GEP$$
Question 3
1 / -0

Complete the correspondence statement:
$$\displaystyle \Delta ABC\cong $$.............

A
$$\displaystyle \Delta DEF$$

B
$$\displaystyle \Delta DFE$$

C
$$\displaystyle \Delta FDE$$

D
$$\displaystyle \Delta EFD$$

Solution

Here, $$\displaystyle \angle A=\angle D,\angle B=\angle F$$ and $$\displaystyle \angle C=\angle E$$.
Also, $$AB=DF$$, $$BC=FE$$ and $$AC=DE$$.

Since, all the corresponding parts are equal,
$$\displaystyle \Delta ABC\cong \Delta DFE$$.

Therefore, option $$B$$ is correct.
Question 4
1 / -0

$$\displaystyle \Delta BAC\cong \Delta DEC,AB\parallel DE$$, if $$\displaystyle \angle B$$ is $$\displaystyle { 90 }^{ o }$$ and $$\displaystyle \angle C$$ is $$\displaystyle { 50 }^{ o }$$,
Find $$\displaystyle \angle A$$ and $$\displaystyle \angle D$$=.......

A
$$\displaystyle { 35 }^{ o }$$

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

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

D
$$\displaystyle { 55 }^{ o }$$

Solution

Given $$\triangle ABC \cong \triangle DEC$$.

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.
Then, $$AB=DE$$, $$BC=EC$$, $$AC=DC$$, $$\angle A=\angle D$$, $$\angle B=\angle E$$ and $$\angle C=\angle C$$.

Also given, $$\angle B=90^o$$ and $$\angle C=50^o$$.

In $$\triangle ABC$$,
by angle sum property,
$$\angle A+\angle B +\angle C=180^o$$
$$\implies$$ $$\angle A+90^o + 50^o=180^o$$
$$\implies$$ $$\angle A+140^o=180^o$$
$$\implies$$ $$\angle A=180^o-140^o$$
$$\implies$$ $$\angle A=40^o$$.

Since, $$\angle A=\angle D$$
Therefore, $$\angle A=\angle D$$ $$=40^o$$.

Hence, option $$C$$ is correct.
Question 5
1 / -0

In figure,
$$BD $$ is a bisector of $$\angle B, CD = DA$$
In order to show the $$\triangle BDA$$ and $$\triangle BDC$$ are congruent, which property of triangle applies here?

A
$$SAS$$

B
$$SSS$$

C
$$AAS$$

D
None

Solution

In $$\triangle ABD$$ and $$\triangle CBD$$

$$BD$$ is a bisector of $$\angle B$$
$$BD$$ is common side.
$$\angle ABD=\angle CBD$$
$$CD=DA$$

Two triangles are congruent from $$SAS$$ postulate

Hence option A is correct.
Question 6
1 / -0

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.
Question 7
1 / -0

Which condition of congruence proves that the following pair of triangles are congruent

A
$$A.A.S$$

B
$$S.S.S$$

C
$$S.A.S$$

D
$$A.S.A$$

Solution

From the given figure, we have
$$AC=DF, BC=ED$$ and $$\displaystyle \angle C=\angle D$$
$$\displaystyle \therefore $$ According to $$S.A.S$$ condition of congruence,
$$\displaystyle \Delta ABC\cong \Delta FED$$
Question 8
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Which condition of congruency proves that the following pair of triangles are congruent

A
$$RHS$$

B
$$SSS$$

C
$$ASA$$

D
$$SAS$$

Solution

According to the given figure, we have
$$\displaystyle \angle B=\angle F,\angle C=\angle G$$ and $$BC=FG$$
$$\displaystyle \therefore \Delta ABC\cong \Delta EFG$$ ....$$(ASA$$ condition for congruence$$)$$
Question 9
1 / -0

Complete the correspondence statement:
$$\displaystyle \Delta GHK\cong $$____.

A
$$\displaystyle \Delta GTK$$

B
$$\displaystyle \Delta KTG$$

C
$$\displaystyle \Delta GKT$$

D
$$\displaystyle \Delta TGK$$

Solution

Here,
$$\displaystyle \angle K={ 90 }^{ o }$$ (common angle)
$$GH=GT$$ and $$KH=KT$$
$$\displaystyle \therefore $$ According to $$R.H.S$$ condition of congruency of triangles,
$$\displaystyle \Delta GHK\cong \Delta GTK$$.

Therefore, option $$A$$ is correct.
Question 10
1 / -0

Which condition will prove these triangles as congruent?

A
SSS

B
AAS

C
SAS

D
ASA

Solution

In $$\triangle$$ $$HJG$$ and $$\triangle$$ $$MNK$$,

$$\angle$$ $$HJG$$ $$=$$ $$\angle$$ $$MNK$$ (given)

$$JG = NK$$ (given)

$$\angle$$ $$JGH =$$ $$\angle$$ $$NKM$$ (given).

Therefore, $$\triangle$$ $$HJG$$ $$\cong$$ $$\triangle$$ $$MNK$$ (By $$ASA$$ congruency criterion).

Hence, option $$D$$ is correct.