Congruence of Triangles Test

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Congruence of Triangles Test - 22

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Question 1
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$$\triangle PQR$$ is right angled at $$Q$$, $$PR=5cm$$ and $$QR=4cm$$. If the lengths of sides of another triangle $$ABC$$ are $$3cm$$, $$4cm$$, $$5cm$$, then which one of the following is correct?

A
Area of $$\triangle PQR$$ is double that of $$\triangle ABC$$.

B
Area of $$\triangle ABC$$ is double that of $$\triangle PQR$$.

C
$$\sqrt { B } =\cfrac { \sqrt { Q } }{ 2 } $$

D
Both triangles are congruent.

Solution
Question 2
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In $$\triangle ABC$$, the bisector of $$\angle {A}$$ intersects $$\overline { BC } $$ at a point $$D$$. Then:

A
$$BD\times AC=BC\times AB$$

B
$$BD\times AB=DC\times AC$$

C
$$AC\times AB=DC\times BC$$

D
$$BD\times AC=DC\times AB$$

Solution

In $$\triangle ABC$$, the bisector of $$\angle A$$ intersects $$\overline BC$$ at $$D$$.
$$\therefore$$ $$AD$$ bisects $$BC$$
$$\therefore BD=DC$$ ........ $$(i)$$
In triangles, $$ABD$$ and $$ADC$$
$$BD=CD$$ ......... From $$(i)$$
$$\angle BAD=\angle DAC$$ ....... (Given)
$$AD=AD$$ ........... (Common side)
$$\therefore$$ $$\triangle ABD\cong \triangle ADC$$ ........ [By S.A.S criterion]
$$\implies AB=AC$$
Thus, $$BD\times AC=DC\times AB$$
Question 3
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$$\triangle FGC$$ is an isosceles triangle, which of the following method will prove $$\triangle ABC$$ congruent to $$\triangle DEF$$ ?

A
SSS

B
RHS

C
AAS

D
SAS

Solution

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

Given,
$$AB=DE$$
$$\angle ABC=\angle DEF=90^\circ$$

$$\triangle FCG$$ is isoceles, $$GF=GC$$ $$\implies FD=CA$$ ...... (as $$AB||DE$$, $$GD$$ should be equal to $$GA$$)

$$AC=FD$$

Hence, $$\triangle ABC \cong \triangle DEF$$ by $$RHS \ postulate$$
Question 4
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The two triangles in the figure are congruent by the congruence theorem. Here, it is given $$OQ=OR$$. Which of the following condition, along with the given condition, is sufficient to prove that the two triangles are congruent to each other?

A
$$\angle P=\angle S$$

B
$$\angle Q=\angle R$$

C
$$OP=OS$$

D
$$PQ=SR$$

Solution

Given $$OQ=OR$$ and $$\triangle POQ\cong \triangle ROS$$
We know that, congruent parts of congruent triangles are congruent
$$\angle POQ\cong \angle ROS$$ (vertically opposite angles)
If $$OP=OS$$ then by using the (SAS) congruent, we can conclude the congruency of two triangles.
Hence, option C is sufficient to prove the congruency.
Question 5
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To prove that $$\Delta DEF$$ and $$\Delta ABC$$ are congruent by SAS, what additional information is needed?

A
$$EF \cong BC$$

B
$$ \angle DFE \cong \angle ABC$$

C
$$DE \cong AB$$

D
$$\angle DFE \cong \angle ACB$$

Solution

Given $$FD=CA$$ and $$\angle D=\angle A$$
Now for triangles to be congruent by $$SAS$$ we need a third side that include the given angles with the given equal sides.
So $$DE=AB$$ $$\implies DE\cong AB$$

Hence, option $$C$$ is correct.
Question 6
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In given figure $$\triangle PQR \cong \triangle XYZ$$ by _______ congruency rule.

A
$$SSS$$

B
$$AAA$$

C
$$SAS$$

D
$$ASA$$

Solution

In $$\triangle PQR$$ and $$\triangle XYZ$$,

$$\angle$$ $$QPR$$ $$=$$ $$\angle$$ $$YXZ$$ (given)
$$PQ = XY $$ (given)
$$\angle$$ $$PQR$$ $$=$$ $$\angle$$ $$XYZ$$ (given)

Therefore, $$\triangle PQR \cong \triangle XYZ$$ by $$ASA$$ rule of congruency.
Question 7
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Which congruence criteria can be used to state that $$\triangle$$XOY $$\cong$$ $$\triangle$$POQ?

A
ASA

B
SAS

C
SSS

D
RHS

Solution

In $$\triangle XOY$$ and $$\triangle POQ$$

$$\Rightarrow$$ $$\angle YXO=\angle QPO=65^o$$ [Given]

$$\Rightarrow$$ $$OX=OP$$ [Given]

$$\Rightarrow$$ $$\angle XOY=\angle POQ$$ [Vertically opposite angles]

$$\therefore$$ $$\triangle XOY\cong\triangle POQ$$ [By ASA criteria]
Question 8
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In the given figure, which of the following is correct?

A
$$\Delta PQR \cong \Delta RSP$$

B
$$\Delta PQR \cong \Delta SPR$$

C
$$\Delta PQR \cong \Delta RPS$$

D
$$\Delta PQR \cong \Delta PSR$$

Solution

In $$\triangle PQR$$ and $$\triangle RSP$$
$$\angle QPR = \angle SRP = 45^o$$
$$PQ = RS = 5.5 cm$$
$$PR = RP$$ (common)
$$\triangle PQR \cong \triangle RSP$$ [By SAS congruency]
Hence option (A) is correct
Question 9
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In given figure,
CF and AE are equal perpendiculars on BD, BF = FE = ED.
$$\angle$$BAE = ______.

A
$$\angle$$BCD

B
$$\angle$$CBA

C
$$\angle$$ADC

D
$$\angle$$DCF

Solution

In $$\triangle ABE$$ and $$\triangle CDF$$
$$\Rightarrow$$ $$AE = CF$$ [given]
$$\Rightarrow$$ $$\angle AEB=\angle CFD$$ [given]
$$\Rightarrow$$ $$BF + FE = DE + EF$$
$$\Rightarrow$$ $$BE = DF$$
$$\therefore$$ $$ \triangle ABE \cong \triangle CDF$$ [By SAS criteria]
$$\therefore$$ $$\angle BAE=\angle DCF$$ [By CPCT]
Question 10
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In $$\Delta PQR$$ and $$\Delta SQR$$, $$\overline{PQ}$$ =$$\overline{SR}$$ and $$\angle PQR = \angle QRS$$ then:

A
$$\Delta PQR \sim \Delta SRQ$$

B
$$\Delta PQR \equiv \Delta SRQ$$

C
$$\Delta PQR = \Delta SRQ$$

D
None of these

Solution

In $$\Delta PQR$$ and $$\Delta SQR$$,
$$\overline{PQ} =\overline{SR}$$ (Given)
And $$\angle PQR = \angle QRS$$
$$QR=QR$$ [common side]
$$\Delta PQR \sim \Delta SRQ$$ (SAS property)