Triangles Test

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Triangles Test - 21

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Weekly Quiz Competition

Question 1
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.
Question 2
1 / -0

In $$\Delta ABC$$, if $$\angle A = 50^{\circ}$$ and $$\angle B = 60^{\circ}$$, then the greatest side is :

A
AB

B
BC

C
AC

D
Cannot say

Solution

Using angle sum property of triangle
$$\angle A+\angle B+\angle C={ 180 }^{ \circ }\\ \Rightarrow { 50 }^{ \circ }+{ 60 }^{ \circ }+\angle C={ 180 }^{ \circ }\\ \Rightarrow \angle C={ 180 }^{ \circ }-{ 110 }^{ \circ }={ 70 }^{ \circ }$$
Side opposite to the largest angle is the longest side.
Here $$\angle C$$ is largest and side opposite to it is $$AB$$
So $$AB$$ is the longest side.
Question 3
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 4
1 / -0

In $$\triangle ABC$$, if $$AB >BC$$ then:

A
$$\angle C < \angle A$$

B
$$\angle C = \angle A$$

C
$$\angle C > \angle A$$

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

Solution

In any triangle angle opposite to the longest side is the largest. Angle opposite to $$AB$$ is $$\angle C$$ and angle opposite to $$BC$$ is $$A$$

Here,
$$AB>AC$$
$$\Rightarrow \angle C>\angle A$$

So, option $$C$$ is correct.
Question 5
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 6
1 / -0

In $$\Delta ABC$$, if $$\angle A = 35^{\circ}$$ and $$\angle B = 65^{\circ}$$, then the longest side of the triangle is :

A
AC

B
AB

C
BC

D
None of these

Solution

Using angle sum property of triangle
$$\angle A+\angle B+\angle C={ 180 }^{ \circ }\\ \Rightarrow { 35 }^{ \circ }+{ 65 }^{ \circ }+\angle C={ 180 }^{ \circ }\\ \Rightarrow \angle C={ 180 }^{ \circ }-{ 100 }^{ \circ }={ 80 }^{ \circ }$$
Side opposite to largest angle is the longest side of any triangle.
Here $$\angle C$$ is the largest angle and side opposite to it is $$AB$$
So $$AB$$ is the longest side.
Question 7
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 8
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 9
1 / -0

In $$\Delta ABC, AB = 76$$ cm, $$BC=69$$ cm and $$CA=61$$ cm
the greatest angle ?

A
$$\angle C$$

B
$$\angle A$$

C
$$\angle B$$

D
All are equal

Solution

In a triangle, the greatest angle is opposite the largest side. Here, $$AB$$ is the largest side, hence, $$\angle C$$ is the greatest.
Question 10
1 / -0

By which congruency are the following pairs of triangles congruent:
In $$\Delta\,ABC$$ and $$\Delta PQR$$, $$AB=PQ$$, $$BC=QR$$ and $$AC=PR$$

A
$$ASA$$

B
$$SSS$$

C
$$AAA$$

D
$$ASS$$

Solution

In triangles $$ABC$$ and $$PQR$$, we have
$$AB = PQ$$ ....(Given)
$$AC = PR$$ ....(Given)
$$BC = QR$$ ....(Given)
Thus, $$\triangle ABC \cong \triangle PQR$$ ..... (SSS Postulate)

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Triangles Test - 21

Result

Triangles Test - 21

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Question 1

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

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

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

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

Solution

Question 2

AB

BC

AC

Cannot say

Solution

Question 3

$$SSS$$

$$ASA$$

$$SAS$$

$$RHS$$

Solution

Question 4

$$\angle C < \angle A$$

$$\angle C = \angle A$$

$$\angle C > \angle A$$

$$\angle A = \angle B$$

Solution

Question 5

SSS

ASA

SAS

RHS

Solution

Question 6

AC

AB

BC

None of these

Solution

Question 7

$$BC=DE$$

$$AC=EF$$

$$BC=EF$$

$$AC=DE$$

Solution

Question 8

SAS

ASA

RHS

SSS

Solution

Question 9

$$\angle C$$

$$\angle A$$

$$\angle B$$

All are equal

Solution

Question 10

$$ASA$$

$$SSS$$

$$AAA$$

$$ASS$$

Solution

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