Congruence of Triangles Test

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Congruence of Triangles Test - 12

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Question 1
1 / -0

By which congruency are the following pairs of triangles congruent:
In $$\Delta\,ABC$$ and $$\Delta PQR$$, $$AB=PQ$$, $$BC=QR$$ and $$AC=PR$$

A
$$ASA$$

B
$$SSS$$

C
$$AAA$$

D
$$ASS$$

Solution

In triangles $$ABC$$ and $$PQR$$, we have
$$AB = PQ$$ ....(Given)
$$AC = PR$$ ....(Given)
$$BC = QR$$ ....(Given)
Thus, $$\triangle ABC \cong \triangle PQR$$ ..... (SSS Postulate)
Question 2
1 / -0

Angles opposite to the equal sides of a triangle are equal.

A
True

B
False

C
Incomplete information

D
None

Solution

Consider $$\triangle ABC$$, $$\angle B = \angle C$$
Construction: Draw a median AD on BC.
Hence, $$BD = DC$$
Now, In $$\triangle ABD$$ and $$\triangle ACD$$
$$\angle ADB = \angle ADC$$ .... Common angle
$$AD = AD$$ ..... (Common)
$$BD = DC$$
Thus, $$\triangle ABD \cong \triangle ACD$$ .... (SAS rule)
Thus, $$\angle B = \angle C$$ .... (By CPCT)
Question 3
1 / -0

In a triangle $$ABC$$ and $$DEF$$, $$AB=FD$$ 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

In $$\triangle ABC$$ and $$\triangle DEF$$,
$$\angle A = \angle D$$ ....(Given)
$$AB = FD$$ ....(Given)
and $$\triangle ABC \cong \triangle DFE$$ ....(SAS rule)
It is possible, if corresponding sides are equal, $$AC = ED$$
Question 4
1 / -0

If in two triangles $$ABC$$ and $$PQR$$, $$AB=\, QR$$, $$\angle A=\angle Q$$ and $$\angle B=\angle R$$, then $$\triangle ABC\cong$$ $$\triangle $$

A
$$QRP$$

B
$$PQR$$

C
$$PRQ$$

D
None of the above.

Solution

Given: $$AB=\, QR$$, $$\angle A=\angle Q$$ and $$\angle B=\angle R$$

Now, In $$\triangle ABC$$ and $$\triangle PQR$$
$$AB = QR$$
$$\angle A = \angle Q$$
$$\angle B = \angle R$$
Thus, by $$ASA$$ congruence rule,
$$\triangle ABC \cong \triangle QRP$$

Hence, $$Op-A$$ is correct.
Question 5
1 / -0

If in two triangles $$ABC$$ and $$DEF$$, $$AB=\,DF$$, $$BC=\,DE$$ and $$\angle B=\angle D$$, then $$\triangle ABC\cong $$ $$\triangle $$____.

A
$$FDE$$

B
$$DEF$$

C
$$EDF$$

D
$$EFD$$

Solution

Given:

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

$$AB = DF$$

$$\angle B = \angle D$$

$$BC = DE$$

Thus, $$\triangle ABC \cong \triangle FDE$$ (SAS rule)
Question 6
1 / -0

If the two sides and the ____ angle of one triangle are respectively equal to two sides and the included angle of the other triangle, then the triangles are congruent.

A
Included

B
Excluded

C
Adjacent

D
Any

Solution

Two sides and an included angle will fulfill the condition of the $$SAS$$ rule of congruence so, the two triangles will be equal. Hence,
Two triangles are congruent if the two sides and the $$\underline{\text{included}}$$ angle are equal to the two sides and the included angle of another triangle.
Question 7
1 / -0

If ____sides of a triangle are respectively equal to the ____ sides of the other triangle, then the triangles are congruent.

A
Three

B
Two

C
One

D
None

Solution

If all the three sides of the triangle are equal to corresponding three sides of other triangle. Both the triangles will be congruent by SSS rule.
Question 8
1 / -0

Which of the following statements is true when $$\displaystyle \Delta ABC\cong \Delta DEF.$$

A
$$\displaystyle \angle A=\angle D$$

B
$$\displaystyle \angle A=\angle E$$

C
$$\displaystyle \angle A=\angle F$$

D
none of these

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, $$\angle A=\angle D$$.
Hence, option $$A$$ is correct.
Question 9
1 / -0

In $$\Delta ABC$$, AB = AC and AD is perpendicular to BC. State the property by which $$\Delta ADB\, \cong\, \Delta ADC$$.

A
SAS property

B
SSS property

C
RHS property

D
ASA property

Solution

In $$\triangle ADB$$ and $$\triangle ADC$$
$$AB=AC$$ (given)
$$AD=AD$$ (common side)
$$\angle ADB=\angle ADC=90^{\circ}$$ (As$$AD\perp BC)$$
$$\therefore \triangle ADB\cong \triangle ADC$$ by $$RHS$$ property.
Question 10
1 / -0

In the following fig. if $$AB=AC$$ and $$BD= DC$$ then $${\angle ADC}$$ =

A
$${60^ 0}$$

B
$${120^ 0}$$

C
$${90^ 0}$$

D
None

Solution

In $$\triangle ABD$$ and $$\triangle ACD$$
$$AB=AC$$ (Given)
$$BD=DC$$ (Given)
$$AD=AD$$ (Common side)
$$\therefore \triangle ABD\cong ACD$$ by $$SSS$$ criteria.
Now corresponding angles of congruent triangles are equal
$$\Rightarrow \angle ADB=\angle ADC$$
Now $$\angle ADB+\angle ADC=180^{\circ}$$ (Adjacent angles on staraight line)
$$\Rightarrow \angle ADC+\angle ADC={ 180 }^{ \circ }\\ \Rightarrow 2\angle ADC={ 180 }^{ \circ }\\ \Rightarrow \angle ADC={ 90 }^{ \circ }\\ $$