Congruence of Triangles Test

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Congruence of Triangles Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

Ankita wants to prove $$\Delta ABC\cong \Delta DEF$$ using $$SAS$$. She knows $$AB=DE$$ and $$AC=DF$$. What additional piece of information does she need?

A
$$\;\angle A=\angle D$$

B
$$\;\angle C=\angle F$$

C
$$\;\angle B=\angle E$$

D
$$\;\angle A=\angle B$$

Solution

By SAS, two corresponding sides and their included angles should be congruent.
Since $$ AB=DE $$ and $$ AC=DF $$,
The included angles of $$ AB$$ and $$AC = \angle A $$
and included angle of $$ DE$$ and $$DF = \angle D.$$
$$ \Rightarrow \angle A = \angle D $$ is needed.
Question 2
1 / -0

By which congruency property, the two triangles connected by the following figure are congruent.

A
SAS property

B
SSS property

C
RHS property

D
ASA property

Solution

Separate the given figure into two triangles.

$$\Delta ABC$$ and $$\Delta ABD$$

We can observe that,

$$AC = AD$$ {Both are $$x \,\,cm$$}

$$BC = BD$$ {Both are $$y\,\,cm$$}

$$AB = AB$$ {Common Side}

So $$\Delta ABC\, \cong\, \Delta ABD$$ by SSS property.
Question 3
1 / -0

In the given fig. if AD = BC and AD || BC, then:

A
AB = AD

B
AB = DC

C
BC = CD

D
None

Solution

In the give fig. AD = BC, AC = AC,
$$\angle ADC\, =\, \angle ABC$$ ($$\because $$ AD || BC)
By SAS theorem $$\Delta ABC\, \cong\, \Delta CDA$$
So AB = DC.
Question 4
1 / -0

ABC is an isosceles triangle in which the altitudes $$BE$$ and $$CF$$ are drawn to the equal sides $$AC$$ and $$AB$$ respectively. Then

A
$$BE = CF$$

B
$$BE = AB$$

C
$$AB = BC$$

D
$$AC = BC$$

Solution

In $$\triangle ABE$$ and $$\triangle ACF$$,

we have $$\angle BEA=\angle CFA$$ (Each $${ 90 }^{ 0 }$$)

$$\angle A= \angle A$$ (Common angle)

$$AB=\,AC$$ (Given)

$$\therefore \triangle ABE \cong \triangle ACF$$ (By SAS congruence criteria)

$$\therefore BF=\,CF$$ [C.S.C.T]
Question 5
1 / -0

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

A
$$60^{\circ}$$

B
$$120^{\circ}$$

C
$$90^{\circ}$$

D
None

Solution

Given that, in $$\triangle ABC$$,
$$AB = AC, \ BD = DC$$
To find out: $$\angle ADC$$

In $$\triangle ADB$$ and $$\triangle ADC$$,
$$AB = AC\quad \quad [given]$$
$$BD = DC\quad \quad [given]$$
Also, $$AD = AD\quad \quad [common]$$
$$\therefore \ $$ By $$SSS$$ congruence criterion, $$\Delta ADB\, \cong\, \Delta ADC$$.

Hence, $$\angle ADB=\angle ADB$$
Also, $$\angle ADB\, +\, \angle ADC\, =\, 180^{\circ}\quad [linear\ pair]$$
$$\therefore \ \angle ADB\, =\, \angle ADC\, =\, 90^{\circ}$$

Hence, $$\angle ADC=90^\circ$$
Question 6
1 / -0

In the given figure, $$\triangle RTQ\cong \triangle PSQ$$ by ASA congruency condition. Which of the following pairs does not satisfy the condition?

A
$$PQ=QR$$

B
$$\angle P=\angle R$$

C
$$\angle TQP=\angle SQR$$

D
None of these

Solution

Given, $$\triangle RTQ \cong \triangle PSQ$$
Thus, $$PQ = QR$$ (by cpct)
$$\angle P = \angle R$$ (by cpct)

Also, $$\angle RQT = \angle PQS$$ (By CPCT)
$$\angle RQT - \angle TQS = \angle PQS - \angle TQS$$
Thus, $$\angle TQP = \angle SQR$$
Question 7
1 / -0

It is given that $$AB=BC$$ and $$AD=EC.$$ The $$\triangle ABE\cong \triangle CBD$$ by __________ congruency.

A
$$SSS$$

B
$$ASA$$

C
$$SAS$$

D
$$AAS$$

Solution

Given, $$AD=EC$$
$$\Rightarrow $$ $$AD+DE=DE+EC$$
$$\Rightarrow $$ $$AE=DC$$
Also, $$AB=BC$$
$$\Rightarrow $$ $$\angle BCA=\angle BAC$$          (isos. $$\triangle $$ property)
$$\Rightarrow $$ $$\angle BCD=\angle BAE$$
$$\therefore $$ In $$\triangle $$s $$ABE$$ and $$CBD,$$
$$AB=CB$$          (Given)
$$AE=DC$$          (proved above)
$$\angle BAE=\angle BCD$$ (Proved above)
$$\therefore $$ $$\triangle ABE\cong \triangle CBD$$          (SAS)
Question 8
1 / -0

In the given figure, $$OA=OB, OC=OD, \angle AOB=\angle COD.$$ Which of the following statements is true$$\: ?$$

A
$$AC=CD$$

B
$$OA=OD$$

C
$$AC=BD$$

D
$$\angle OCA=\angle ODC$$

Solution

In $$\triangle $$s $$AOC$$ and $$BOD$$
$$OA=OB$$          (Given)
$$OC=OD$$          (Given)
$$\angle AOB-\angle COB=\angle COD-\angle COB,$$
i.e., $$\angle AOC=\angle BOD$$
$$\therefore $$ $$\triangle AOC\cong \triangle BOD$$          (SAS)
$$\Rightarrow $$ $$AC=BD$$          (cpct)
Question 9
1 / -0

In the figure, $$\displaystyle \Delta ABC\cong \Delta ADC$$ by the congruence postulate

A
SSS

B
RHS

C
SAS

D
ASA

Solution

According to Right angle hypotenuse side rule, two right-angled triangles are congruent if the hypotenuses and one pair of corresponding sides are equal.

Hence, the given two triangles are congruent by the RHS rule.
Question 10
1 / -0

For a $$\displaystyle \Delta XYZ, ZY = 12\ m, YX = 8\ m$$ and $$XZ = 10\ m$$. If $$\displaystyle \Delta ZYX\cong \Delta ABC$$, then $$AC =$$ _____ $$m$$.

A
$$10$$

B
$$12$$

C
$$8$$

D
$$1$$

Solution

Given : $$\displaystyle \Delta XYZ, ZY = 12\ m, YX = 8\ m$$ and $$XZ = 10\ m$$.
Also, $$\displaystyle \Delta ZYX\cong \Delta ABC$$.
We know, congruent triangles have their sides congruent.
$$\implies AB=ZY, BC=YX$$ and $$AC=ZX$$.
$$\implies \displaystyle ZX=AC$$.
$$\displaystyle \therefore AC=10\ m$$.

Therefore, option $$A$$ is correct.