Congruence of Triangles Test

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Congruence of Triangles Test - 15

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Question 1
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Direction (14 - 15) : Study the figure and information given below carefully and answer the following questions.
CF and AE are equal perpendiculars on BD, BF = FE = ED.
$$\triangle$$ABE is congruent to

A
$$\triangle$$AED

B
$$\triangle$$BFC

C
$$\triangle$$CDF

D
$$\triangle$$BCD

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]
Question 2
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By which congruency criterion, $$\triangle$$PQR $$\cong$$ $$\triangle$$PQS?

A
RHS

B
ASA

C
SSS

D
SAS

Solution

In $$\triangle PQR$$ and $$\triangle PQS$$
$$\Rightarrow$$ $$PR = PS = a \,cm$$ [given]
$$\Rightarrow$$ $$RQ = SQ = b\,cm $$ [given]
$$\Rightarrow$$ $$PQ = PQ$$ [common side]
$$\Rightarrow$$ $$\triangle PQR \cong \triangle PQS$$ (By SSS)
Question 3
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Which of the following is congruent to the above figure?

A

B

C

D
None of these

Solution

Two figures are said to be congruent if they have exactly same shape and size or one can exactly overlap the other.
Here, if we rotate the figure in option $$B$$, and place it over each other, we get the same figure given, i.e. they will overlap.

Therefore, option $$B$$ is correct.
Question 4
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The ________ criterion is used to construct a triangle congruent to another triangle whose length of three sides are given.

A
$$SAS$$

B
$$SSS$$

C
$$RHS$$

D
$$ASA$$

Solution

When the length of all three sides of a triangle are given, then by Side-Side-Side i.e. $$SSS$$ criterion we can say that the sides of the other triangle will be equal to that by $$CPCT$$.

Hence, the answer is $$SSS$$.
Question 5
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In $$\triangle ABC$$, if $$AB = 7$$ cm, $$\angle A= 40^o$$ and $$\angle B = 70^o$$, which criterion can be used to construct this triangle?

A
ASA

B
SSS

C
SAS

D
RHS

Solution

Given : In a triangle $$ABC, AB = 7 cm, \angle A=40^{o}$$ and $$\angle B=70^{o}$$
So, here we know two of the angles and a side including these angles.
Hence, Angle Side Angle i.e ASA criterion can be used to construct this triangle.
Question 6
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Two triangles are congruent, if two angles and the side included between them in one triangle is equal to the two angles and the side included between them of the other triangle.This is known as

A
RHS congruence criterion

B
ASA congruence criterion

C
SAS congruence criterion

D
SSS congruence criterion

Solution

When two angles and side included in between these angles of one triangle are same as another one, both these triangle are Congruent.
This criteria of congruency is knows as ASA criterion as a side included between two angles are all same.
Question 7
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If $$ \triangle ABC$$ and $$\triangle DBC$$ are on the same base BC,AB=DC and AC=DB,then which of the following gives a CORRECT congruence relationship?

A
$$\triangle ABC=\triangle DBC$$

B
$$\triangle ABC=\triangle CBD$$

C
$$\triangle ABC=\triangle DCB$$

D
$$\triangle ABC=\triangle BCD$$

Solution

In triangle $$\Delta ABC$$ and $$\Delta DCB$$ we have
$$AB=DC$$ , $$AC=DB$$ and $$BC=BC$$
then from $$SSS$$ congruence ,$$\Delta ABC=\Delta DCB$$
Question 8
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Which of the following triangles is congruent to the given triangle?

A

B

C

D

Solution

Triangle in option (C) is congruent to given triangle by SAS congruene criteria.
Question 9
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If in two triangles $$\Delta ABC$$ and $$\Delta PQR$$, $$AB = QR, BC = PR$$ and $$CA = PQ,$$ then :

A
$$\Delta ABC\cong \Delta PQR$$

B
$$\Delta CBA\cong \Delta PRQ$$

C
$$\Delta BAC\cong \Delta RPQ$$

D
$$\Delta PQR\cong \Delta BCA$$

Solution

In $$\triangle ABC$$ and $$\triangle PQR$$, we are given
$$AB=QR$$,
$$BC=PR$$
and $$CA=PQ$$
$$\therefore \Delta CBA\cong \Delta PRQ$$ .... SSS test
Question 10
1 / -0

In the above 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)

$$\angle AOD=\angle BOC$$

$$\therefore \triangle AOD\cong \triangle BOC$$ by $$SAS$$ criteria.