Lines and Angles Test

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Lines and Angles Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

An angle is $$24^{o}$$ more than its complement. The measure of the angle is:

A
$$47^{o}$$

B
$$57^{o}$$

C
$$53^{o}$$

D
$$66^{o}$$

Solution

Let the measure of the angle be $$x^o$$.
Let the measure of its complement be $$(x-24^{o})$$.
Then,
$$x^{o} + (x-24)^{o}$$ = $$90^{o}$$
$$\implies$$  $$x+x=90^o+24$$
$$\implies$$  $$2x=114$$
$$\implies$$  $$x=57$$
$$\implies$$  $$x=57^o$$.

Therefore, option $$B$$ is correct.
Question 2
1 / -0

The supplement of $$45^{o}$$ is

A
$$45^{o}$$

B
$$75^{o}$$

C
$$135^{o}$$

D
$$155^{o}$$

Solution

Two angles are said to be supplementary if the sum of their measures is $$180^{o}$$.
The given angle is $$45^{o}$$
Let the measure of its supplement be $$X^{o}$$.
Then,
$$\implies X+45=180$$
$$\implies X=180-45$$
$$\implies X=135$$
Hence, the supplement of the given angle measures $$135^{o}$$.
Question 3
1 / -0

Triangle DEF in figure is right triangle with $$\angle E=90^{o}$$.
What type of angles are $$\angle D$$ and $$\angle F$$?

A
They are equal angles

B
They form a pair of adjacent angles

C
They are complementary angles

D
They are supplementary angles

Solution

$$\angle D+\angle E+\angle F=180^{o}$$ ....Angle sum property
$$\because \angle E=90^{o}$$
$$\Rightarrow \angle D+\angle F=180^{o}-90^{o}$$
$$\Rightarrow \angle D+\angle F=90^{o}$$
Therefore, angle $$D$$ and $$F$$ are complementary angles.
Question 4
1 / -0

The complement of $$80^{o}$$ is:

A
$$100^{o}$$

B
$$10^{o}$$

C
$$20^{o}$$

D
$$280^{o}$$

Solution

Two angles are said to be complementary, if the sum of their measures is $$90^{o}$$.
The given angle is $$80^{o}$$.
Let the measure of its supplement be $$x^{o}$$.
Then,
$$\implies x^{o}+80^{o}=90^{o}$$
$$\implies x=90^o-80^o$$
$$\implies x=10^o$$.
Hence, the supplement of the given angle measures $$10^{o}$$.
Therefore, option $$B$$ is correct.
Question 5
1 / -0

In the given figure, two straight lines $$AB$$ and $$CD$$ intersect at a point $$O$$ and $$\angle AOC$$ = $$50^{o}$$. Then $$\angle BOD$$ = ?

A
$$40^{o}$$

B
$$50^{o}$$

C
$$130^{o}$$

D
$$60^{o}$$

Solution

(c) $$50^{o}$$
Because,
$$\angle AOC = \angle BOD$$ ....... $$[\because Vertically \ opposite \ angles \ are\ equal]$$
Question 6
1 / -0

The angles between North & West and South & East are

A
complementary

B
supplementary

C
both are acute

D
both are obtuse

Solution

Option (b) is correct.
The angle between the North and West $$=90^o$$
The angle between the South and East $$=90^o$$
Therefore, the angle between North & West and South & East are supplementary angles as they add up to $$180^o$$.
Question 7
1 / -0

Angles which are both supplementary and vertically opposite are

A
$$95^o , 85^o$$

B
$$90^o, 90^o$$

C
$$100^o, 80^o$$

D
$$45^o, 45^o$$

Solution

As we know, vertically opposite angles are equal.
Let the angle be $$y$$
$$\Rightarrow y+y=180^o$$
$$\Rightarrow 2y=180^o$$
$$\Rightarrow y=90^o$$
Therefore, $$90^o$$ is asked angle for each.
Question 8
1 / -0

Angles between South & West and South & East are

A
vertically opposite angles

B
complementary angles

C
making a linear pair

D
adjacent but not supplementary

Solution

We have,
Angle between South and West $$=90^o$$
Angle between South and East $$=90^o$$
Sum of the angles $$=180^o$$

$$\therefore$$ The angles between South and West and South and East are making a linear pair.

Hence, option (C) is correct.
Question 9
1 / -0

If the complement of an angle is $$79^o$$, then the angle will be of :

A
$$1^o$$

B
$$11^o$$

C
$$79^o$$

D
$$101^o$$

Solution

We know, two angles are complementary when they add up to $$90^o.$$

Given, measure of one complementary angle is $$79^o.$$

$$\Rightarrow$$ Measure of other complementary angle $$=90^o-79^o$$ $$=11^o.$$

$$\therefore$$ Measure of a complementary angle of $$ 79^{o}=11^o$$.
Hence, option $$B$$ is correct.
Question 10
1 / -0

The angles $$x$$ and $$90^o -x$$ are:

A
supplementary

B
complementary

C
vertically opposite

D
making a linear pair

Solution

We know, two angles are complementary when they add up to $$90^o.$$

Sum of given angles = $$(90^o-x^o)+$$ $$x^o=90^o$$.
Hence, these are complementary angles.
Therefore, option $$B$$ is correct.

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Re-Attempt Weekly Quiz Competition

Lines and Angles Test - 28

Result

Lines and Angles Test - 28

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SHARING IS CARING

Question 1

$$47^{o}$$

$$57^{o}$$

$$53^{o}$$

$$66^{o}$$

Solution

Question 2

$$45^{o}$$

$$75^{o}$$

$$135^{o}$$

$$155^{o}$$

Solution

Question 3

They are equal angles

They form a pair of adjacent angles

They are complementary angles

They are supplementary angles

Solution

Question 4

$$100^{o}$$

$$10^{o}$$

$$20^{o}$$

$$280^{o}$$

Solution

Question 5

$$40^{o}$$

$$50^{o}$$

$$130^{o}$$

$$60^{o}$$

Solution

Question 6

complementary

supplementary

both are acute

both are obtuse

Solution

Question 7

$$95^o , 85^o$$

$$90^o, 90^o$$

$$100^o, 80^o$$

$$45^o, 45^o$$

Solution

Question 8

vertically opposite angles

complementary angles

making a linear pair

adjacent but not supplementary

Solution

Question 9

$$1^o$$

$$11^o$$

$$79^o$$

$$101^o$$

Solution

Question 10

supplementary

complementary

vertically opposite

making a linear pair

Solution

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