Triangles Test

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Triangles Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

In $$\triangle ABC$$ and $$\triangle DEF$$, $$AB = FD$$ 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

$$SAS=$$ If two pairs of sides of two triangles are equal in length, and the included angles are equal in measurement, then the triangles are congruent.

Here, $$AB = FD$$,
$$\angle A = \angle D$$
So, $$AC = DE$$
Hence, option B is correct.
Question 2
1 / -0

The construction of a triangle $$ABC$$, given that $$BC =$$ $$6$$ cm, $$B =$$ $$45 ^{\circ}$$ is not possible when difference of $$AB$$ and $$AC$$ is equal to:

A
$$6.9$$ cm

B
$$5.2$$ cm

C
$$5.0$$ cm

D
$$4.0$$ cm

Solution

According to the theorem of inequalities, the sum of any two sides of the triangle is greater than the third side.
Therefore, $$AC+BC>AB$$
$$\Rightarrow BC>AB-AC$$
Therefore, only the first option that is $$6.9$$ cm does not satisfy the above equation. Rest all the options satisfy the equation.
Question 3
1 / -0

Two sides of a triangle are of lengths $$4$$ cm and $$1.5$$ cm. The length of the third side of the triangle cannot be

A
$$3.6$$ cm

B
$$4.1$$ cm

C
$$3.8$$ cm

D
$$5.8$$ cm

Solution

Sum of given sides of the triangle is $$5.5$$ cm.
The length of the third side cannot be greater than the sum of the two given sides.
Here, one option gives length as $$5.8$$ cm is not possible.
Question 4
1 / -0

In $$\Delta ABC, \angle A=100^0, \angle B=30^0$$ and $$\angle C=50^0$$ then

A
$$AB > AC$$

B
$$AB < AC$$

C
$$BC < AC$$

D
none of these

Solution

We know that the side opposite to largest angle is longest side of triangle .
$$\angle A$$ is the largest angle and side opposite to it is $$BC$$
So $$BC$$ is the longest side.
Then comes $$\angle C$$ and side opposite to it is $$AB$$.
So $$AB$$ is second longest side.
Then comes $$\angle B$$ and side opposite to it is $$AC$$.
So it is the smallest side.
So the decreasing order of side is
$$BC>AB>AC$$
So option $$A$$ is correct.
Question 5
1 / -0

In $$\Delta PQR$$, $$\angle P =$$$$ 70^{\circ}$$ and $$\angle R = 30^{\circ}.$$ Which side of this triangle is the longest?

A
$$PR$$

B
$$QR$$

C
$$PQ$$

D
All sides are equal.

Solution

Given, $$\triangle PQR$$, $$\angle P = 70^{\circ}$$, $$\angle R = 30^{\circ}$$
Sum of angles of triangle $$= 180^{\circ}$$
$$\implies \angle P + \angle Q + \angle R = 180^{\circ}$$
$$\implies 70^{\circ} + 30^{\circ} + \angle Q = 180^{\circ}$$
$$\implies \angle Q = 80^{\circ}$$
Since $$\angle Q$$ is the largest $$\implies PR$$ will be the longest side this is because when the two sides of a triangle are unequal, the angle opposite to the longer side is larger.
Question 6
1 / -0

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 7
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.
Question 8
1 / -0

In $$\Delta ABC\, \angle A=50^{\circ} $$and$$ \, \angle B=60^{\circ} $$. By arranging the sides of the triangle in ascending order, we get :

A
$$AB < BC < CA$$

B
$$CA < AB < BC$$

C
$$BC < CA < AB$$

D
$$BC < AB < CA$$

Solution

Using angle sum property in $$\triangle ABC$$
$$\angle A+\angle B+\angle C={ 180 }^{ \circ }\\ \Rightarrow { 50 }^{ \circ }+{ 60 }^{ \circ }+\angle C={ 180 }^{ \circ }\\ \Rightarrow \angle C={ 70 }^{ \circ }$$
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 $$\angle B$$ and side opposite to it is $$AC$$
$$\therefore AC$$ is the second longest side.
Then comes $$\angle A$$ and side opposite to it is $$AC$$
$$\therefore BC$$ is the shortest side.
So the increasing order of sides is
$$BC<AC<AB$$
Question 9
1 / -0

In $$\Delta ABC, \angle A=100^{\circ}, \angle B=30^{\circ}$$ and $$\angle C= 50^{\circ}$$,then

A
$$AB>AC$$

B
$$AB=AC$$

C
$$AB<AC$$

D
None of these

Solution

In any triangle side opposite to the largest angle is the longest side.
Here $$\angle A$$ is largest and side opposite to it is $$BC$$
$$\therefore BC$$ is the longest side.
Then comes $$\angle C$$ and side opposite to it is $$AB$$
$$\therefore AB$$ is the second longest side.
Then comes $$\angle B$$ and side opposite to it is $$AC$$
$$\therefore AC$$ is the shortest side.
So the increasing order of sides is
$$AC<AB<BC$$
$$\Rightarrow AB>AC$$
So option $$A$$ is correct.
Question 10
1 / -0

T is a point on side QR of $$\Delta$$ PQR. "S" is a point such that $$RT = ST.$$ then

A
$$PQ + QR > QS$$

B
$$PQ + PR > QS$$

C
$$QR + PR > QS$$

D
$$PQ + PR < QS$$

Solution

In $$\Delta PQR$$,
$$PQ + PR > RQ...........(1)$$
$$RQ = TR + TQ$$

$$Also, in$$ $$ \Delta STQ $$,
$$= S T + TQ > QS...........(2)$$

From (1) and (2) [since ST+TQ=RQ]
$$PQ + PR > QS$$

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Triangles Test - 29

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Triangles Test - 29

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SHARING IS CARING

Question 1

$$BC = EF$$

$$AC = DE$$

$$AC = EF $$

$$ BC = DE$$

Solution

Question 2

$$6.9$$ cm

$$5.2$$ cm

$$5.0$$ cm

$$4.0$$ cm

Solution

Question 3

$$3.6$$ cm

$$4.1$$ cm

$$3.8$$ cm

$$5.8$$ cm

Solution

Question 4

$$AB > AC$$

$$AB < AC$$

$$BC < AC$$

none of these

Solution

Question 5

$$PR$$

$$QR$$

$$PQ$$

All sides are equal.

Solution

Question 6

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

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

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

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

Solution

Question 7

$$SSS$$

$$ASA$$

$$SAS$$

$$RHS$$

Solution

Question 8

$$AB < BC < CA$$

$$CA < AB < BC$$

$$BC < CA < AB$$

$$BC < AB < CA$$

Solution

Question 9

$$AB>AC$$

$$AB=AC$$

$$AB<AC$$

None of these

Solution

Question 10

$$PQ + QR > QS$$

$$PQ + PR > QS$$

$$QR + PR > QS$$

$$PQ + PR < QS$$

Solution

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