Lines and Angles Test

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Lines and Angles Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

Two angles are called adjacent if

A
they have a common vertex

B
they have a ray in common
C

their other arms lie on the opposite sides of the common arm
D
all the above

Solution

We know that, adjacent angles are angles that have a common vertex and a common side, but do not overlap. Also, their other arms lie on the opposite side of the common arm.

So, all the given options are correct.

Hence, option D is the answer.
Question 2
1 / -0

If the sum of two adjacent angles is $$100^{\circ}$$ and one of them is $$35^{\circ}$$, then the other is :

A
$$70^{\circ}$$

B
$$65^{\circ}$$

C
$$135^{\circ}$$

D
$$145^{\circ}$$

Solution

Let the other angle be $$x$$
Now their sum $$=100^{\circ}$$
$$\Rightarrow x+{ 35 }^{ \circ }={ 100 }^{ \circ }\\ \Rightarrow x={ 100 }^{ \circ }-{ 35 }^{ \circ }={ 65 }^{ \circ }$$
So the other angle is $$65^{\circ}$$
Question 3
1 / -0

Two supplementary angles are in the ratio 4 : 5. The angles are

A
$$90^{\circ}, 90^{\circ}$$

B
$$80^{\circ}, 100^{\circ}$$

C
$$30^{\circ}, 150^{\circ}$$

D
$$45^{\circ}, 45^{\circ}$$

Solution

Let the angles be $$4x$$ and $$5x$$
Angles are supplementary
$$\therefore 4x+5x={ 180 }^{ \circ }\\ \Rightarrow 9x={ 180 }^{ \circ }\\ \Rightarrow x={ 20 }^{ \circ }$$
So the angles are
$$4\times { 20 }^{ \circ }={ 80 }^{ \circ }\\ 5\times { 20 }^{ \circ }={ 100 }^{ \circ }$$
Question 4
1 / -0

The complement of $$(90^o-a)$$ is :

A
$$-a$$

B
$$90^o+a$$

C
$$90^o-a$$

D
$$a$$

Solution

Let the complement be $$y$$.
We know, two angles are complementary if their sum is $${ 90 }^{ \circ }$$.
$$\therefore { 90 }^{ \circ }-a+y={ 90 }^{ \circ }\\ \Rightarrow y=a$$.
Hence, option $$D$$ is correct.
Question 5
1 / -0

In fig, AB and CD are parallel to each others. The value of $$x$$ is :

A
$$90^{\circ}$$

B
$$100^{\circ}$$

C
$$120^{\circ}$$

D
$$140^{\circ}$$

Solution

Draw a line parallel to $$AB$$ and $$CD$$ and passing through $$E$$ as shown in the figure which divides $$x$$ in $$\angle 1$$ and $$\angle 2$$
Now $$AB||EF$$ and ang $$BE$$ is transversal
$$\therefore \angle 1+120^{\circ}=180^{\circ}$$ (Sum of angles made on side side of transversal is supplementary)
$$\Rightarrow \angle 1={ 180 }^{ \circ }-{ 120 }^{ \circ }\\ \Rightarrow \angle 1={ 60 }^{ \circ }$$
Also $$DC||EF$$ and $$DE$$ is transversal
$$\therefore \angle 2+140^{\circ}=180^{\circ}$$ (Sum of angles made on side side of transversal is supplementary)
$$\Rightarrow \angle 2={ 180 }^{ \circ }-{ 140 }^{ \circ }\\ \Rightarrow \angle 1={ 40 }^{ \circ }$$
$$x=\angle 1+\angle 2\\ \Rightarrow x={ 60 }^{ \circ }+{ 40 }^{ \circ }={ 100 }^{ \circ }$$
Question 6
1 / -0

If two supplementary angles are in the ratio 2 : 7, then the angles are :

A
$$35^{\circ}, 145^{\circ}$$

B
$$70^{\circ}, 110^{\circ}$$

C
$$40^{\circ}, 140^{\circ}$$

D
$$50^{\circ}, 130^{\circ}$$

Solution

$$\textbf{ Hint: Sum of supplementary angles is}$$ $$180^\circ$$

$$\textbf{Step1: Evaluate using addition of ratio}$$
$$\text{Given}$$
$$\text{ratio}$$ $$=2:7$$
$$\text{ Let angles as }$$ $$2x$$ $$\text{and}$$ $$7x$$
$$\text{As we know that sum of supplementary angle is }180$$
$$\Rightarrow 2x+7x=180$$
$$\Rightarrow 9x=180$$
$$\Rightarrow x=20$$
$$2x\Rightarrow 2(20)=40$$
$$7x\Rightarrow 7(20)=140$$
$$\text{ Hence, the angles are}$$ $$40^\circ$$ $$140^\circ.$$
$$\textbf{Option C is correct.}$$
Question 7
1 / -0

Find the complement of the angle :
$$\dfrac{1}{4}$$ of a right angle.

A
$$67 \cdot 5^o$$

B
$$57 \cdot 5^o$$

C
$$37 \cdot 5^o$$

D
none of the above

Solution

We know, two angles are complementary when they add up to $$90^o.$$
Given, measure of one complementary angle is $$=\dfrac{1}{4}\times 90^o=\dfrac{45}{2}^o.$$
$$\Rightarrow$$ Measure of other complementary angle $$=90^o-$$ $$\dfrac{45}{2}^o$$ $$=\dfrac{180-45}{2}^o$$ $$=\dfrac{135}{2}^o$$ $$=67.5^o.$$
$$\therefore$$ Measure of a complementary angle of $$\dfrac{1}{4}$$ of a right angle $$=67.5^o$$.
Question 8
1 / -0

In the given figure $$AB \parallel CD$$. find the value of x and y.

A
$$x = 35^{\circ}, y = 65^{\circ}$$

B
$$x = 95^{\circ}, y = 65^{\circ}$$

C
$$x = 85^{\circ}, y = 65^{\circ}$$

D
$$x = 25^{\circ}, y = 65^{\circ}$$

Solution

$$
Given\quad AB\parallel CD\quad \& \quad RQ\quad is\quad the\quad transversal\quad then\\ \angle QHC=\angle AFH={ 115 }^{ o }\quad \quad \left\{ Corresponding\quad angles \right\} \quad \\ Now\quad y+\angle AFH={ 180 }^{ o }\quad \quad \left\{ Linear\quad pair \right\} \\ \Rightarrow y={ 180 }^{ o }-{ 115 }^{ o }\quad \Rightarrow y={ 65 }^{ o }\\ Now\quad AB\parallel CD\quad \& \quad PQ\quad is\quad the\quad transversal\quad then\\ \angle PBF=x\quad \quad \left\{ Corresponding\quad angles \right\} \\ So\quad x={ 85 }^{ o }
$$
Question 9
1 / -0

If two angles are complementary of each other, then each angle is:

A
an obtuse angle

B
a right angle

C
an acute angle

D
a supplementary angle

Solution

We know, two angles whose sum is equal to $$90^o$$ are known as complementary angles.
Also, acute angle is the angle which is greater than $$\displaystyle { 0 }^{ o }$$ and less than $$\displaystyle { 90 }^{ o }$$.
Hence, iftwo angles are complement of each other, then each of them should necessarily be acute angle.
Therefore, option $$C$$ is correct.
Question 10
1 / -0

If one angle of a linear pair is acute, then its other angle will be ______.

A
Obtuse angle

B
Acute angle

C
Right angle

D
None of these

Solution

Given: An angle of a linear pair is acute i.e., it is less than $$90^\circ$$.

We know that linear pair of angles are supplementary i.e., they add up to form $$180^\circ$$

So the other angle must be greater than $$90^\circ$$.
So the other angle should be obtuse.

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Lines and Angles Test - 14

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Lines and Angles Test - 14

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SHARING IS CARING

Question 1

they have a common vertex

they have a ray in common

their other arms lie on the opposite sides of the common arm

all the above

Solution

Question 2

$$70^{\circ}$$

$$65^{\circ}$$

$$135^{\circ}$$

$$145^{\circ}$$

Solution

Question 3

$$90^{\circ}, 90^{\circ}$$

$$80^{\circ}, 100^{\circ}$$

$$30^{\circ}, 150^{\circ}$$

$$45^{\circ}, 45^{\circ}$$

Solution

Question 4

$$-a$$

$$90^o+a$$

$$90^o-a$$

$$a$$

Solution

Question 5

$$90^{\circ}$$

$$100^{\circ}$$

$$120^{\circ}$$

$$140^{\circ}$$

Solution

Question 6

$$35^{\circ}, 145^{\circ}$$

$$70^{\circ}, 110^{\circ}$$

$$40^{\circ}, 140^{\circ}$$

$$50^{\circ}, 130^{\circ}$$

Solution

Question 7

$$67 \cdot 5^o$$

$$57 \cdot 5^o$$

$$37 \cdot 5^o$$

none of the above

Solution

Question 8

$$x = 35^{\circ}, y = 65^{\circ}$$

$$x = 95^{\circ}, y = 65^{\circ}$$

$$x = 85^{\circ}, y = 65^{\circ}$$

$$x = 25^{\circ}, y = 65^{\circ}$$

Solution

Question 9

an obtuse angle

a right angle

an acute angle

a supplementary angle

Solution

Question 10

Obtuse angle

Acute angle

Right angle

None of these

Solution

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