Triangles Test

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Triangles Test - 20

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Question 1
1 / -0

In $$\Delta ABC$$ and $$\Delta DEF$$, AB = DF and $$\angle A = \angle D$$. The two triangles will be congruent by SAS axiom if :

A
BC = EF

B
AC = DE

C
BC = DE

D
AC = EF

Solution

For triangle to be congruent $$SAS$$(Side angle Side) axiom we need to have two equal sides and the included angle same.
In $$\Delta ABC$$ and $$\Delta DFE$$
We have $$AB=DF$$
and $$\angle A=\angle D$$
But $$\angle A$$ is formed by the two sides AB and AC of $$\Delta ABC$$
But $$\angle D$$ is formed by the two sides DE and DF of $$\Delta DFE$$
So we need $$AC=DE$$ for the two triangles to be congruent.
Question 2
1 / -0

In $$\Delta PQR, \angle P = 60^{\circ}$$ and $$\angle Q = 50^{\circ}$$. Which side of the triangle is the longest ?

A
PQ

B
QR

C
PR

D
None

Solution

Using angle sum property of triangle
$$\angle P+\angle Q+\angle R={ 180 }^{ \circ }\\ \Rightarrow { 60 }^{ \circ }+{ 50 }^{ \circ }+\angle R={ 180 }^{ \circ }\\ \Rightarrow \angle R={ 180 }^{ \circ }-{ 110 }^{ \circ }={ 70 }^{ \circ }$$

So $$\angle R$$ is the largest angle and side opposite to largest angle is the longest side.
$$\therefore PQ$$ is the longest side.
Question 3
1 / -0

In the given figure , which of the following statement is true ?

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

B
$$\angle B$$ is the greatest angle in triangle

C
$$\angle B$$ is the smallest angle in triangle

D
$$\angle A$$ is the smallest angle in triangle

Solution

In any triangle angle opposite to the longest side is largest and smallest side is smallest.
Here $$BC$$ is the longest side and angle opposite to it is $$\angle A$$ so $$\angle A$$ is the largest angle.
Smallest side is $$AC$$ and angle opposite to it is $$\angle B$$ so $$\angle B$$ is the smallest angle.
So option $$C$$ is correct.
Question 4
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 $$SSS$$ congruence rule then their corresponding sides are equal.
Here $$\triangle ABC\cong \triangle DEF$$
$$\therefore AB=DE,BC=EF,AC=DF$$ or $$(CA=FD)$$
So option $$C$$ is correct.
Question 5
1 / -0

In $$\Delta ABC, \angle B = 30^{\circ}, \angle C = 80^{\circ}$$ and $$\angle A = 70^{\circ}$$ then,

A
$$AB > BC < AC$$

B
$$AB < BC > AC$$

C
$$AB > BC > AC$$

D
$$AB < BC < AC$$

Solution

In any triangle side opposite to the largest angle is the longest side.
Here $$\angle C$$ is largest and side opposite to it is $$AB$$
$$\therefore AB$$ is the longest side.
Then comes $$\angle A$$ and side opposite to it is $$BC$$.
$$\therefore BC$$ is second longest.
Then comes $$\angle B$$ and side opposite to it is $$AC$$
So it is the smallest side.
So the decreasing order of sides is
$$AB>BC>AC$$
Question 6
1 / -0

In triangles ABC and DEF, AB $$=$$ FD and $$\angle A = \angle D$$. The two triangles will be congruent by
SAS axiom if :

A
BC $$=$$ EF

B
AC $$=$$ DE

C
AC $$=$$ EF

D
BC $$=$$ DE

Solution

For triangle to be congruent by $$SAS$$ axiom two of the sides and the included angle of the triangles must be equal.
Given $$AB=DF$$ and $$\angle A=\angle D$$

No for $$\triangle ABC\cong \triangle DFE$$ by $$SAS$$ axiom we need $$AC=DE$$
So option $$B$$ is correct.
Question 7
1 / -0

In $$\Delta PQR,$$ if $$\angle R > \angle Q,$$ then:

A
$$QR > PR$$

B
$$PQ > PR$$

C
$$PQ < PR$$

D
$$QR < PR$$

Solution

We know that the side opposite to the greater angle is greater.
Given: $$\angle R> \angle Q$$
The side opposite to $$\angle R=PQ$$ and the side opposite to $$\angle Q=PR$$.
Thus, $$PQ>PR$$.
Hence, option $$B$$ is correct.
Question 8
1 / -0

The construction of a triangle ABC, given that BC = 3 cm is possible when difference of AB and AC is equal to :

A
3.2 cm

B
3.1 cm

C
3 cm

D
2.8 cm

Solution

Let the length of $$AB$$ be $$x$$ and $$AC$$ be $$y$$
A triangle can be formed if the sum of any two sides is greater then the third
$$\Rightarrow BC+AC>AB\\ \Rightarrow 3+AC>AB\\ \Rightarrow 3>AB-AC\\ \Rightarrow AB-AC<3$$
So only option $$D$$is correct.
Question 9
1 / -0

ABC is an isosceles triangle with AB $$= $$AC and D is a point on BC such that $$AD \perp BC$$ (Fig. 7.13). To prove that $$\angle BAD = \angle CAD,$$ a student proceeded as follows:

$$\Delta ABD$$ and $$ \Delta ACD,$$
AB $$=$$ AC (Given)
$$\angle B = \angle C$$ (because AB $$=$$ AC)
and $$\angle ADB = \angle ADC$$
Therefore, $$\Delta ABD \cong \Delta ACD (AAS)$$
So, $$\angle BAD = \angle CAD (CPCT)$$
What is the defect in the above arguments?

A
It is defective to use $$\angle ABD = \angle ACD$$ for proving this result.

B
It is defective to use $$\angle ADB = \angle ADC$$ for proving this result.

C
It is defective to use $$\angle BAD = \angle DCA$$ for proving this result.

D
Cannot be determined

Solution

$$\bigtriangleup ABD$$ and $$\bigtriangleup ACD$$
AB=AC (given )
Then $$\angle ABD=\angle ACD$$ ( because AB=AC )
and $$\angle ADB=\angle ADC=90$$( because AD⊥BC )
$$\therefore \bigtriangleup ABD=\bigtriangleup ACD$$
$$\angle BAD=\angle CAD$$
It is defective to use $$\angle ABD=\angle ACD$$ for proving this result
Question 10
1 / -0

Given $$\Delta OAP \cong \Delta OBP$$ in figure, the criteria by which the triangles are congruent is:

A
$$SAS$$

B
$$SSS$$

C
$$RHS$$

D
$$ASA$$

Solution

In $$\triangle OAP$$ and $$\triangle OBP$$
$$OA=OB$$ (given)
$$\angle AOP=\angle BOP$$ (given)
$$OP=OP$$ (common side)
$$\therefore \triangle OAP\cong \triangle OBP$$ by $$SAS$$ congruent rule as two corresponding sides and the included angles of the triangles are equal.

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Triangles Test - 20

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Triangles Test - 20

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SHARING IS CARING

Question 1

BC = EF

AC = DE

BC = DE

AC = EF

Solution

Question 2

PQ

QR

PR

None

Solution

Question 3

$$\angle B=\angle C$$

$$\angle B$$ is the greatest angle in triangle

$$\angle B$$ is the smallest angle in triangle

$$\angle A$$ is the smallest angle in triangle

Solution

Question 4

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

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

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

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

Solution

Question 5

$$AB > BC < AC$$

$$AB < BC > AC$$

$$AB > BC > AC$$

$$AB < BC < AC$$

Solution

Question 6

BC $$=$$ EF

AC $$=$$ DE

AC $$=$$ EF

BC $$=$$ DE

Solution

Question 7

$$QR > PR$$

$$PQ > PR$$

$$PQ < PR$$

$$QR < PR$$

Solution

Question 8

3.2 cm

3.1 cm

3 cm

2.8 cm

Solution

Question 9

It is defective to use $$\angle ABD = \angle ACD$$ for proving this result.

It is defective to use $$\angle ADB = \angle ADC$$ for proving this result.

It is defective to use $$\angle BAD = \angle DCA$$ for proving this result.

Cannot be determined

Solution

Question 10

$$SAS$$

$$SSS$$

$$RHS$$

$$ASA$$

Solution

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