Lines and Angles Test

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Lines and Angles Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

If an angle is $$60^o$$ less than two times of its supplement, then the greater angle is

A
$$100^o$$

B
$$80^o$$

C
$$60^o$$

D
$$120^o$$

Solution

Option (a) is correct.
Let an angle be $$x$$ and $$180^o-x$$ be supplementary angles.
So, according to question,
$$\Rightarrow x=2(180^o -x)-60^o$$
$$\Rightarrow x=360^o -2x-60^o$$
$$\Rightarrow x+2x=300^o$$
$$\Rightarrow 3x=300^o$$
$$\Rightarrow x=\dfrac{300^o}{3}$$
$$\Rightarrow x=100^o$$
$$\Rightarrow 180^0-x=80^0$$
Greater angle $$=100^0$$
Hence, option $$A$$ is correct
Question 2
1 / -0

In a pair of adjacent angles, (i) vertex is always common, (ii) one arm is always common, and (iii) uncommon arms are always opposite rays
Then

A
All (i), (ii) and (iii) are true

B
(iii) is flase

C
(i) is false but (ii) and (iii) are true

D
(ii) is false

Solution

Option $$(B)$$ is correct.
Adjacent angles have a common vertex and a common arm.
but uncommon arms are only opposite in linear pair.
So, they always do not need to be opposite.
So, statement $$(iii)$$ is false
Question 3
1 / -0

Vertically opposite angles are always

A
supplementary

B
complementary

C
adjacent

D
equal
Question 4
1 / -0

In figure, $$POQ$$ is a line, then $$a$$ is equal to

A
$$35^o$$

B
$$100^o$$

C
$$80^o$$

D
$$135^o$$

Solution

We can see in the given figure, $$POQ$$ is a straight line
$$\therefore \angle ROQ +\angle POR=180^o$$
$$\Rightarrow 100^o +a=180^o$$
$$\Rightarrow a=180^o-100^o$$
$$\Rightarrow a=80^o$$
Hence, Option $$C$$ is correct.
Question 5
1 / -0

In Fig., $$PQ = PR, RS = PQ$$ and $$ST || QR$$. If the exterior angle $$RPU$$ is $$140^o$$, the the measures of angle $$TSR$$ is

A
$$55^o$$

B
$$40^o$$

C
$$50^o$$

D
$$45^o$$

Solution

In triangle $$PQR$$,
$$\angle RPU - \angle PQR + \angle PQR$$ ...Exterior angle property
$$\Longrightarrow 2 \angle PQR = 140^o$$ $$PQ = PR$$
$$\Longrightarrow \angle PQR = \dfrac{140^o}{2}$$
$$\Longrightarrow \angle PQR = 70^o$$
$$ST$$ parallel to $$QR$$ and $$QS$$ is a transversal.
$$\therefore \angle PQR = \angle PST = 70^o$$ ...Corresponding angles
Now, in triangle $$QSR$$
$$\therefore \angle SQR - \angle RSQ = 70^o$$
Here, $$PQ$$ is a straight line.
$$\Longrightarrow \angle PST + \angle TSR + \angle RSQ = 180^o$$
From equation $$1$$ and $$2$$
$$\Longrightarrow \angle TSR = 180^o - 70^o - 70^o$$
$$\Longrightarrow TSR = 40^o$$
Question 6
1 / -0

In the given figure, the value of $$x$$ that will make $$AOB$$ a straight line is

A
$$x=40$$

B
$$x=36$$

C
$$x=30$$

D
$$x=25$$

Solution

$$AOB$$ will be a straight line if,
$$3x+15+2x-15^{0}=180^{0}$$
$$\Rightarrow 5x=180^{0}$$
$$\Rightarrow x=36^{0}$$
Question 7
1 / -0

A pair of supplementary angles is

A
$$55^{\circ},115^{\circ}$$

B
$$65^{\circ},125^{\circ}$$

C
$$47^{\circ},133^{\circ}$$

D
$$40^{\circ},50^{\circ}$$

Solution

A pair of angles that sum up to $$180^\circ$$ are known as supplementary angles.

In option $$C,$$
$$47^\circ+133^\circ=180^\circ$$

Hence, option $$C$$ is correct.
Question 8
1 / -0

If an angle measures $$10^{o}$$ more than its complement, then the measure of the angle is:

A
$$40^{o}$$

B
$$55^{o}$$

C
$$35^{o}$$

D
$$50^{o}$$

Solution

Let $$x$$ and $$y$$ are two angles making a complementary angles, and $$x$$ is $$10^o$$ greater than $$y$$.
Then,
$$\implies$$ $$x-(90-x)=10$$
$$\implies$$ $$x-90+x=10$$
$$\implies$$ $$2x=100$$
$$\implies$$ $$x=50^o$$.
Hence, if one angle measures $$10^{0}$$ more than its complement, then measure of the angle $$=50^{0}$$. $$(\because 50^{0}+40^{0}=90^{0})$$
Therefore, option $$D$$ is correct.
Question 9
1 / -0

A pair of complementary angles is:

A
$$130^{0},50^{0}$$

B
$$35^{0},55^{0}$$

C
$$25^{0},75^{0}$$

D
$$27^{0},53^{0}$$

Solution

We know, two angles whose sum is equal to $$90^o$$ are known as complementary angles.

Consider option $$(A)$$.
The angles are $$130^o$$ and $$50^o$$.
Then their sum $$=130^o+50^o=180^o\ne90^o$$.
Hence, the angles are not complementary.

Consider option $$(B)$$.
The angles are $$35^o$$ and $$55^o$$.
Then their sum $$=35^o+55^o=90^o$$.
Hence, the angles are complementary.

Consider option $$(C)$$.
The angles are $$25^o$$ and $$75^o$$.
Then their sum $$=25^o+75^o=100^o\ne90^o$$.
Hence, the angles are not complementary.

Consider option $$(D)$$.
The angles are $$27^o$$ and $$53^o$$.
Then their sum $$=27^o+53^o=80^o\ne90^o$$.
Hence, the angles are not complementary.

Hence, only option $$B$$ is correct.
Question 10
1 / -0

If the difference of two complementary angles is $$10$$, then the smaller angle is:

A
$$40$$

B
$$50$$

C
$$45$$

D
$$35$$

Solution

Let first angle $$= x$$.
Then its complementary angle $$ = 90 - x$$.
Also, given that the difference of the complementary angles is $$10^o$$.
Then,
$$x - (90 - x) = 10$$
$$\implies\, x - 90 + x =10$$
$$\implies\, 2x = 10 + 90 = 100$$
$$\implies\, x =50$$.
Second angle = $$90- 50 =40$$.
Hence, the smaller angle $$= 40$$.

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Lines and Angles Test - 41

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Lines and Angles Test - 41

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Question 1

$$100^o$$

$$80^o$$

$$60^o$$

$$120^o$$

Solution

Question 2

All (i), (ii) and (iii) are true

(iii) is flase

(i) is false but (ii) and (iii) are true

(ii) is false

Solution

Question 3

supplementary

complementary

adjacent

equal

Question 4

$$35^o$$

$$100^o$$

$$80^o$$

$$135^o$$

Solution

Question 5

$$55^o$$

$$40^o$$

$$50^o$$

$$45^o$$

Solution

Question 6

$$x=40$$

$$x=36$$

$$x=30$$

$$x=25$$

Solution

Question 7

$$55^{\circ},115^{\circ}$$

$$65^{\circ},125^{\circ}$$

$$47^{\circ},133^{\circ}$$

$$40^{\circ},50^{\circ}$$

Solution

Question 8

$$40^{o}$$

$$55^{o}$$

$$35^{o}$$

$$50^{o}$$

Solution

Question 9

$$130^{0},50^{0}$$

$$35^{0},55^{0}$$

$$25^{0},75^{0}$$

$$27^{0},53^{0}$$

Solution

Question 10

$$40$$

$$50$$

$$45$$

$$35$$

Solution

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