Congruence of Triangles Test

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Congruence of Triangles Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

In the figure, $$\displaystyle AD\parallel BC$$ and $$\displaystyle AD=BC$$. Then $$\displaystyle \Delta ABD\cong \Delta CDB$$ by congruence postulate.

A
SSS

B
ASA

C
AAS

D
SAS

Solution

In $$\displaystyle \Delta ABD$$ and $$\displaystyle \Delta CDB$$
$$\displaystyle AD=BC$$
$$\displaystyle \angle BDA=\angle DBC$$ (Alternate angles)
$$\displaystyle BD=BD$$
$$\displaystyle \therefore \quad \Delta ABD\cong \Delta CDB$$ (By SAS postulate)
Question 2
1 / -0

If $$\overline{AC}$$ and $$\overline{BD}$$ intersect at $$O$$ such that $$AO = CO$$ and $$BO = DO$$, then:

A
$$AB = CD$$

B
$$AB \parallel CD$$ and $$AB = CD$$

C
$$AB \parallel CD$$

D
None of these

Solution

In $$\triangle AOB$$ and $$\triangle DOC$$,
side $$OA =$$ side $$OC$$ ... given
side $$OB =$$ side $$OD$$ ... given
$$\angle AOB \cong \angle COD$$ ...Vertically opposite angles
$$\therefore \triangle AOB \cong \triangle COD$$ ...SAS test of congruence.
$$\therefore$$ side $$AB$$ $$\cong$$ side $$CD$$ ...c.s.c.t
$$\therefore \angle BAO \cong \angle DCO$$ ....c.a.c.t
i.e. $$\angle BAC \cong \angle DCA$$ ....$$A-O-C$$ and $$B-O-D$$
$$\therefore AB \parallel CD$$ ....by converse of alternate angles test
So, option B is correct.
Question 3
1 / -0

In $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta DBC,AC=BD$$ and $$\displaystyle \angle ABC=\angle BCD={ 90 }^{ o }$$. Then $$\displaystyle \Delta ABC\cong \Delta DCB$$ by the postulate

A
SSS

B
SAS

C
RHS

D
ASA

Solution

In $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta DBC$$,
$$\displaystyle \angle ABC=\angle BCD$$
$$\displaystyle AC=BD$$ (Givenn)
$$\displaystyle BC=BC$$ (Common Side)
$$\displaystyle \Delta ABC\cong \Delta DCB$$ WCB (By RHS postulate)
Question 4
1 / -0

In the given figure, $$AD=BC, AC=BD.$$ Then $$\triangle PAB$$ is

A
equilateral

B
right angled

C
scalene

D
isosceles

Solution

In $$\triangle ADB$$ and $$\triangle ACB$$
$$AD=BC$$          (Given)
$$AC=BD$$
$$AB=BA$$          (Common)
$$\therefore $$ $$\triangle ADB\cong \triangle ACB$$          (SSS)
$$\Rightarrow $$ $$\angle ABD=\angle CAB$$          (cpct)
$$\Rightarrow $$ $$\angle ABP=\angle PAB$$
$$\Rightarrow $$ $$PA=PB$$          (Sides opp. equal angles are equal)
$$\Rightarrow $$ $$\triangle PAB$$ is isosceles.
Question 5
1 / -0

In the figure, $$\displaystyle PS=QR$$ and $$PQ=SR$$, then choose the correct option.

A
$$\displaystyle \Delta QRP\cong \Delta PSR$$

B
$$\displaystyle \Delta PQR\cong \Delta PRS$$

C
$$\displaystyle \Delta PQR\cong \Delta RSP$$

D
$$\displaystyle \Delta RQP\cong \Delta RSP$$

Solution

In $$\displaystyle \Delta PQR$$ and $$\displaystyle \Delta PSR$$
$$\displaystyle PS=QR,PQ=SR,PR=PR$$
$$\displaystyle \therefore \quad \Delta RQP\cong \Delta RSP$$ (By SSS postulate)
Question 6
1 / -0

In $$ \displaystyle \bigtriangleup ABC,AD\perp BC,\angle B=\angle C $$ and AB = AC State by which property $$ \displaystyle \Delta ADB\cong \Delta ADC$$ ?

A
RHS property

B
SSS property

C
SAS property

D
ASA property

Solution
Question 7
1 / -0

The two quadrilaterals are congruent.
Which angle in quadrilateral WXYZ corresponds to $$\displaystyle \angle BCD$$ in quadrilateral ABCD?

A
$$\displaystyle \angle XYZ$$

B
$$\displaystyle \angle XWZ$$

C
$$\displaystyle \angle WZY$$

D
$$\displaystyle \angle WXY$$

Solution

Clearly, $$\angle WXY$$ corresponds to $$\angle BCD$$ because $$BC$$ faces the angles marked with one arc and three arc and $$XY$$ also faces the angles marked with one arc and three arcs.
So, $$\text{D}$$ is the correct option.
Question 8
1 / -0

In the figure given below, $$\displaystyle \Delta ABC\cong \Delta MBC.$$ The corresponding side to $$BA$$ is ......

A
$$BM$$

B
$$AB$$

C
$$AC$$

D
$$MC$$

Solution

Given $$\triangle ABC \cong \triangle MBC$$.

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $$AB=MB$$, $$BC=BC$$, $$AC=MC$$, $$\angle A=\angle M$$, $$\angle B=\angle B$$ and $$\angle C=\angle C$$.

Thus, $$AB=MB$$ $$\implies$$ $$BA=BM$$.

Hence, option $$A$$ is correct.
Question 9
1 / -0

In the figure given below, $$\displaystyle \Delta ABC\cong \Delta XYZ$$. The corresponding side to $$XZ$$ is .......

A
$$AB$$

B
$$BC$$

C
$$AC$$

D
$$YZ$$

Solution

Given $$\triangle ABC \cong \triangle XYZ$$.

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $$AB=XY$$, $$BC=YZ$$, $$AC=XZ$$, $$\angle A=\angle X$$, $$\angle B=\angle Y$$ and $$\angle C=\angle Z$$.

Thus, $$AC=XZ$$.

Hence, option $$C$$ is correct.
Question 10
1 / -0

In the triangles $$ABC$$ and $$PQR,$$ $$AC=QP, BC=RP$$ and $$AB=QR.$$ The measure of $$\angle A$$ is ___.

A
$$\displaystyle { 30 }^{ \circ }$$

B
$$\displaystyle { 120 }^{ \circ }$$

C
$$\displaystyle { 20 }^{ \circ }$$

D
$$\displaystyle { 40 }^{ \circ }$$

Solution

Given: In $$\triangle ABC$$ and $$\triangle PQR,$$

$$\angle C=120^\circ$$ and $$\angle R=40^\circ$$

Now, $$AC=QP$$ ...(1)

and $$ BC=RP$$ ...(2)

and $$AB=QR$$ ...(3)

Hence $$ABC \cong PQR$$ $$($$by $$SSS$$ congruency rule$$)$$

By CPCT, $$ \angle B=\angle R=40^\circ$$

and $$ \angle C=\angle P={ 120 }^{ \circ }$$

and $$ \angle A=\angle Q$$

Now, $$\displaystyle \angle A+\angle B+\angle C={ 180 }^{ o }$$

$$\displaystyle \Rightarrow \angle A+{ 40 }^{ \circ }+{ 120 }^{ \circ }={ 180 }^{ \circ }$$

$$\displaystyle \Rightarrow \angle A={ 20 }^{ \circ}$$