Triangles Test

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Triangles Test - 35

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Question 1
1 / -0

In $$\displaystyle \Delta PQR,\angle P={ 67 }^{ o }$$ and $$\displaystyle \angle Q={ 76 }^{ o }$$. Arrange the sides in descending order according to the lengths we get,

A
$$PR> QR > PQ$$

B
$$PQ> QR> PR$$

C
$$PQ> QR>PR$$

D
$$QR>PQ> PR$$

Solution

In $$\displaystyle \Delta PQR,\angle P={ 67 }^{ o }$$ and $$\displaystyle \angle Q={ 76 }^{ o }$$
$$\displaystyle \therefore \quad \angle R={ 37 }^{ o }$$
$$\displaystyle \left( \because \angle P+\angle Q+\angle R={ 180 }^{ o } \right). $$
Now arranging the angles in descending order, we get
$$\displaystyle \angle Q>\angle P > \angle R$$ .
Sides opposite to greater angle are greater in length and the angles and the opposite sides of a triangle have same relation.
Therefore, the sides arranged in descending order, we get $$PR > QR > PQ.$$
Question 2
1 / -0

In $$\displaystyle \Delta XYZ,\angle X={ 60 }^{ o },\angle Y={ 75 }^{ o }$$. The greatest side is ......... and the smallest side is ..............

A
$$XY, YZ$$

B
$$YZ, XZ$$

C
$$XZ, XY$$

D
$$XZ, YZ$$

Solution

In $$\displaystyle \Delta XYZ,\angle X={ 60 }^{ o },\angle Y={ 75 }^{ o }$$
Also, $$\displaystyle \angle X+\angle Y+\angle Z={ 180 }^{ o }$$
$$\displaystyle \therefore \quad \angle Z={ 45 }^{ o }$$
$$\displaystyle \therefore \quad \angle Y$$ is the greatest angle and $$\displaystyle \angle Z$$ is the smallest angle. Thus $$XZ$$ is the greatest side and $$XY$$ is the smallest side.
Question 3
1 / -0

If two sides of a triangle are unequal, the angle opposite to the greater side is ___.

A
smaller

B
equal

C
greater

D
does not exist

Solution

Let we have triangle $$\Delta ABC$$ and we consider a point $$D$$ on $$AC$$ such that $$AB=AD$$
In $$\Delta ABD$$
$$AB=AD$$
$$\to \angle ABD=\angle ADB~~~~-(1)$$ (Angle opposite to equal sides are equal)
In $$\Delta BDC$$
$$\angle BDA>\angle BCD$$ (Exterior angle is greater than interior angle)
$$\angle ABD>\angle BCA~~~~~~-(2)$$ (from eqs.(1))
Also, as $$\angle ABC>\angle ABD\\\to \angle ABC>\angle BCA$$
Hence, angle opposite to greater side is greater.
Question 4
1 / -0

Which of the following sets of side lengths form a triangle?

A
4 m, 3 m, 11 m

B
7 mm, 4 mm, 4 mm

C
3 cm, 1.1 cm, 5 cm

D
3 m, 4 m, 8 m

Solution

A triangle can be formed only if sum of any two sides is greater than the third side.
In option $$B$$ sum of any two sides taken at a time is greater than the third side.
$$7+4 = 11>4$$
$$4+4= 8>7$$
$$4+7= 11>4$$
In all other options, sum of first two sides is less than the third side:
A) $$4+3<11$$
C) $$3+1.1<5 $$
D) $$3+4<8$$
So option $$B$$ is correct.
Question 5
1 / -0

In $$\displaystyle \Delta PQR,\angle Q={ 67 }^{ o },\angle R={ 48 }^{ o }$$. The smallest side is ........ and the greatest side is ........

A
$$PQ, PR$$

B
$$PQ, QR$$

C
$$QR, PR$$

D
$$PR, PQ$$

Solution

In $$\displaystyle \Delta PQR,\quad \angle Q={ 67 }^{ o },\angle R={ 48 }^{ o }$$.
Also, $$\displaystyle \angle P+\angle Q+\angle R={ 180 }^{ o }$$. Thus $$\displaystyle \angle P={ 65 }^{ o }$$.
The greatest angle is $$\displaystyle \angle Q$$ and the smallest angle is $$\displaystyle \angle R$$.
Side opposite to greater angle is greater in length and side opposite to the smaller angle is smallest.
Thus, the smallest side is $$PQ$$ and the greatest side is $$PR.$$
Question 6
1 / -0

In $$\displaystyle \Delta XYZ, XY=6 \ cm, YZ=8 \ cm$$ and $$XZ=9 \ cm.$$ The greatest angle is ___ and the smallest angle is ____.

A
$$\displaystyle \angle X,\angle Y$$

B
$$\displaystyle \angle Z,\angle Y$$

C
$$\displaystyle \angle Y,\angle X$$

D
$$\displaystyle \angle Y,\angle Z$$

Solution

In $$\displaystyle \Delta XYZ$$, side $$XZ$$ is the greatest and side $$XY$$ is the smallest.
Angle opposite to greater side is greater and smaller side is smaller.
Thus, $$\displaystyle \angle XYZ$$ i.e. $$\angle Y $$ is the greatest and $$\displaystyle \angle YZX$$ i.e. $$\angle Z$$ is the smallest.
Question 7
1 / -0

In $$\displaystyle \Delta XYZ,\angle Y={ 90 }^{ o }$$ and $$\angle Z={ 40 }^{ o }$$. Arranging sides (according to length) in ascending order, we get:

A
$$XY< YZ<YZ$$

B
$$XY< YZ< ZX$$

C
$$YZ< ZX< XY$$

D
$$YZ< XY< ZX$$

Solution

$$\displaystyle \Delta XYZ,\angle Y={ 90 }^{ o }$$ and $$\angle Z={ 40 }^{ o }$$
$$\displaystyle \therefore \quad \angle X={ 50 }^{ o }\quad \left( \because \angle X+\angle Y+\angle Z={ 180 }^{ o } \right) $$
Now, arranging the angles in ascending order, we get $$\displaystyle \angle Z,\angle X$$ and $$ \angle Y$$
Sides opposite to greater angle are greater in length and the angles and the opposite sides of a triangle have same relation.
Thus, arranging the sides in ascending order $$XY< YZ < ZX.$$
Question 8
1 / -0

Which is the greatest side in the following triangle?

A
$$AB$$

B
$$BC$$

C
$$AC$$

D
cannot be determined

Solution

$$\displaystyle \angle A={ 69 }^{ o },\angle C={ 32 }^{ o }$$
Now, $$\displaystyle \angle A+\angle B+\angle C={ 180 }^{ o }$$
$$\displaystyle \therefore \quad \angle B={ 79 }^{ o }$$
Thus, $$\displaystyle \angle B$$ is the greatest angle and hence $$AC$$ is the greatest side.
Question 9
1 / -0

In $$\displaystyle \Delta ABC,\angle B={ 55 }^{ o }$$ and $$\displaystyle \angle A={ 45 }^{ o }$$. Arranging the sides in descending order we get:

A
$$AB > BC> AC$$

B
$$BC>AB> AC$$

C
$$AC>AB> BC$$

D
$$AB> AC> BC$$

Solution

In $$\displaystyle \Delta ABC,\angle B={ 55 }^{ o }$$ and $$\displaystyle \angle A={ 45 }^{ o }$$
$$\displaystyle \therefore \quad \angle C={ 80 }^{ o }\left( \because \angle A+\angle B+\angle C={ 180 }^{ o } \right) $$
$$\displaystyle \therefore $$ Arranging the angles in descending order, we get
$$\displaystyle \angle C,\angle B$$ and $$\angle A$$
Since, the angles and the opposite sides of a triangle have same relation
Therefore, arranging the sides in descending order we get, $$AB, AC$$ and $$BC.$$
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.

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Re-Attempt Weekly Quiz Competition

Triangles Test - 35

Result

Triangles Test - 35

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SHARING IS CARING

Question 1

$$PR> QR > PQ$$

$$PQ> QR> PR$$

$$PQ> QR>PR$$

$$QR>PQ> PR$$

Solution

Question 2

$$XY, YZ$$

$$YZ, XZ$$

$$XZ, XY$$

$$XZ, YZ$$

Solution

Question 3

smaller

equal

greater

does not exist

Solution

Question 4

4 m, 3 m, 11 m

7 mm, 4 mm, 4 mm

3 cm, 1.1 cm, 5 cm

3 m, 4 m, 8 m

Solution

Question 5

$$PQ, PR$$

$$PQ, QR$$

$$QR, PR$$

$$PR, PQ$$

Solution

Question 6

$$\displaystyle \angle X,\angle Y$$

$$\displaystyle \angle Z,\angle Y$$

$$\displaystyle \angle Y,\angle X$$

$$\displaystyle \angle Y,\angle Z$$

Solution

Question 7

$$XY< YZ<YZ$$

$$XY< YZ< ZX$$

$$YZ< ZX< XY$$

$$YZ< XY< ZX$$

Solution

Question 8

$$AB$$

$$BC$$

$$AC$$

cannot be determined

Solution

Question 9

$$AB > BC> AC$$

$$BC>AB> AC$$

$$AC>AB> BC$$

$$AB> AC> BC$$

Solution

Question 10

$$10$$

$$12$$

$$8$$

$$1$$

Solution

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