Congruence of Triangles Test

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Congruence of Triangles Test - 11

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Question 1
1 / -0

In figure, if $$AB=AC$$ and $$AP=AQ$$, then by which congruence criterion $$\Delta PBC \cong \Delta QCB$$?

A
$$SSS$$

B
$$ASA$$

C
$$SAS$$

D
$$RHS$$

Solution

In the given triangle:
$$AB=AC$$
$$\Rightarrow \angle B=\angle C$$ (As angles opposite to equal to opposite sides are equal.
$$AB=AC$$
$$AP+PB=AQ+QC$$
Now $$AP=AQ$$
$$\Rightarrow PB=QC$$
Now in $$\triangle PBC$$ and $$\triangle QCB$$
$$PB=QC$$ (Proved above)
$$\angle B=\angle C$$ (Proved above)
$$BC=CB$$ (Common side)
$$\therefore \triangle PBC\cong \triangle QCB$$ by $$SAS$$ criteria.
Question 2
1 / -0

In $$\Delta ABC$$ and $$\Delta DEF$$, AB = DF and $$\angle A = \angle D$$. The two triangles will be congruent by SAS axiom if :

A
BC = EF

B
AC = DE

C
BC = DE

D
AC = EF

Solution

For triangle to be congruent $$SAS$$(Side angle Side) axiom we need to have two equal sides and the included angle same.
In $$\Delta ABC$$ and $$\Delta DFE$$
We have $$AB=DF$$
and $$\angle A=\angle D$$
But $$\angle A$$ is formed by the two sides AB and AC of $$\Delta ABC$$
But $$\angle D$$ is formed by the two sides DE and DF of $$\Delta DFE$$
So we need $$AC=DE$$ for the two triangles to be congruent.
Question 3
1 / -0

If $$\Delta ABC \cong \Delta DEF$$ by SSS congruence rule then :

A
$$AB=EF, BC=FD, CA=DE$$

B
$$AB=FD, BC=DE, CA=EF$$

C
$$AB=DE, BC=EF, CA=FD$$

D
$$AB=DE, BC=EF, \angle C=\angle F$$

Solution

If triangles are congruent by $$SSS$$ congruence rule then their corresponding sides are equal.
Here $$\triangle ABC\cong \triangle DEF$$
$$\therefore AB=DE,BC=EF,AC=DF$$ or $$(CA=FD)$$
So option $$C$$ is correct.
Question 4
1 / -0

In the given figure, If OA=OB,OD=OC, then $$\Delta AOD \cong \Delta BOC$$ by congruence rule:

A
SSS

B
ASA

C
SAS

D
RHS

Solution

In $$\triangle AOD$$ and $$\triangle BOC$$
$$OA=OB$$ (given)
$$OD=OC$$ (given)
$$\angle AOD=\angle BOC$$ (vertically opposite angle)
$$\triangle AOD\cong \triangle BOC$$ by $$SAS$$ congruence rule.
Question 5
1 / -0

In triangles 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

For triangle to be congruent by $$SAS$$ axiom two of the sides and the included angle of the triangles must be equal.
Given $$AB=DF$$ and $$\angle A=\angle D$$

No for $$\triangle ABC\cong \triangle DFE$$ by $$SAS$$ axiom we need $$AC=DE$$
So option $$B$$ is correct.
Question 6
1 / -0

In $$\Delta ABC$$ and $$\Delta DEF$$, $$AB=FD$$, $$\angle A= \angle D$$. The two triangles will be congruent by SAS axiom if :

A
$$BC=DE$$

B
$$AC=EF$$

C
$$BC=EF$$

D
$$AC=DE$$

Solution

For the triangles to be congruent by $$SAS$$ axiom we need two sides and included angle of given triangles to be equal.
Given: $$AB=FD$$ and $$\angle A=\angle D$$
Now from the figures, we can say that we need the second pair of sides which is $$AC$$ and $$DE$$ according to the given definition.
So for given triangles to be congruent by $$SAS$$ criteria
we need $$AC=DE$$
Option $$D$$ is correct.
Question 7
1 / -0

In the given figure, if $$AB=DC$$, $$\angle ABD= \angle CDB$$, which congruence rule would you apply to prove $$\Delta ABD$$ $$\cong \Delta CDB$$ ?

A
SAS

B
ASA

C
RHS

D
SSS

Solution

In $$\triangle ABD$$ and $$\triangle CDB$$
$$AB=DC$$ (given)
$$\angle ABD=\angle CDB$$ (given)
$$BD=DB$$ (common side)
$$\therefore \triangle ABD\cong \triangle CBD$$ by $$SAS$$ congruence rule.

Hence option (A) is correct
Question 8
1 / -0

ABC is an isosceles triangle with AB $$= $$AC and D is a point on BC such that $$AD \perp BC$$ (Fig. 7.13). To prove that $$\angle BAD = \angle CAD,$$ a student proceeded as follows:

$$\Delta ABD$$ and $$ \Delta ACD,$$
AB $$=$$ AC (Given)
$$\angle B = \angle C$$ (because AB $$=$$ AC)
and $$\angle ADB = \angle ADC$$
Therefore, $$\Delta ABD \cong \Delta ACD (AAS)$$
So, $$\angle BAD = \angle CAD (CPCT)$$
What is the defect in the above arguments?

A
It is defective to use $$\angle ABD = \angle ACD$$ for proving this result.

B
It is defective to use $$\angle ADB = \angle ADC$$ for proving this result.

C
It is defective to use $$\angle BAD = \angle DCA$$ for proving this result.

D
Cannot be determined

Solution

$$\bigtriangleup ABD$$ and $$\bigtriangleup ACD$$
AB=AC (given )
Then $$\angle ABD=\angle ACD$$ ( because AB=AC )
and $$\angle ADB=\angle ADC=90$$( because AD⊥BC )
$$\therefore \bigtriangleup ABD=\bigtriangleup ACD$$
$$\angle BAD=\angle CAD$$
It is defective to use $$\angle ABD=\angle ACD$$ for proving this result
Question 9
1 / -0

Given $$\Delta OAP \cong \Delta OBP$$ in figure, the criteria by which the triangles are congruent is:

A
$$SAS$$

B
$$SSS$$

C
$$RHS$$

D
$$ASA$$

Solution

In $$\triangle OAP$$ and $$\triangle OBP$$
$$OA=OB$$ (given)
$$\angle AOP=\angle BOP$$ (given)
$$OP=OP$$ (common side)
$$\therefore \triangle OAP\cong \triangle OBP$$ by $$SAS$$ congruent rule as two corresponding sides and the included angles of the triangles are equal.
Question 10
1 / -0

If $$\Delta ABC \cong \Delta DEF$$ by SSS congruence rule then

A
$$AB=EF,BC=FD,CA=DE$$

B
$$AB=FD,BC=DE,CA=EF$$

C
$$AB=DE,BC=EF,CA=FD$$

D
$$AB=DE,BC=EF,\angle C = \angle F$$

Solution

If triangles are congruent by $$SSS$$ then their corresponding sides are equal , therefore
$$AB=DE$$
$$BC=EF$$
$$AC=DF$$ or $$CA=FD$$
So option $$C$$ is correct.