Lines and Angles Test

Result

Lines and Angles Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

Mark the correct alternative of the following:
The sum of an angle and half of its complementary angle is $$75^o$$. The measure of the angle is?

A
$$40^o$$

B
$$50^o$$

C
$$60^o$$

D
$$80^o$$

Solution

Given, the sum of angle and half its complementary angle is $$75^o$$.
We know, sum of the complementary angles is $$90^{ \circ }$$.
Then, let the angles be $$x$$ and $$(90^o-x)$$.
Now, $$ x+\dfrac{1}{2}(90^o-x)={ 75 }^{ \circ }$$
$$\implies$$ $$ x-\dfrac{1}{2}x+45^o={ 75 }^{ \circ }$$
$$\implies$$ $$ \dfrac{1}{2}x={ 75 }^{ o }-45^o$$
$$\implies$$ $$ \dfrac{1}{2}x=30^o$$
$$\implies$$ $$x= 2\times30^o=60^o$$.

Hence, the measure of the required angle is $$60^o$$.

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

Mark the correct alternative of the following.
The sum of an angle and one third of its supplementary angle is $$90^o$$. The measure of the angle is?

A
$$135^o$$

B
$$120^o$$

C
$$60^o$$

D
$$45^o$$

Solution

Let, required angle be $$x$$ and its supplementary angle be $$(180-x)$$.
$$\therefore \angle x+\angle (180-x)=180^0$$

According to condition, the sum of an angle and one third of its supplementary angle is $$90^0$$
$$\angle x+\dfrac{1}3\angle (180-x)=90^0\\$$
$$\angle x+\dfrac{1}3(180^0-x)=90^0\\$$
$$\dfrac2{3}\angle x+60^0=90^0\\$$
$$\dfrac2{3}\angle x=30^0\\$$
$$\angle x=45^0\\$$
$$\therefore$$ measure of a required angle is $$45^0$$
Question 3
1 / -0

Mark the correct alternative of the following:
Two complementary angles are in the ratio $$2:3$$. The measure of the larger angle is?

A
$$60^o$$

B
$$54^o$$

C
$$66^o$$

D
$$48^o$$

Solution

Given, the complementary angles are in the ratio $$2:3$$.
Let the angles be $$2x$$ and $$3x$$.
We know, sum of the complementary angles is $$90^{ \circ }$$.
$$\Rightarrow 2x+3x={ 90 }^{ \circ }\\ \Rightarrow 5x=9{ 0 }^{ \circ }\\ \Rightarrow x=18^{ \circ }$$

Hence, the angles are:
$$2x=2\times 18^{ \circ }=36^{ \circ }$$
and $$ 3x=3\times 18^{ \circ }=54^{ \circ }$$.

Here, the larger angle is $$54^o$$.

Therefore, option $$B$$ is correct.
Question 4
1 / -0

Mark the correct alternative of the following.
In figure, if AB$$||$$CD then the value of x is?

A
$$34$$

B
$$124$$

C
$$24$$

D
$$158$$

Solution
Question 5
1 / -0

Mark the correct alternative of the following:
In figure, $$AB||CD||EF$$, $$\angle ABG=110^o$$, $$\angle GCD=100^o$$ and $$\angle BGC=x^o$$. The value of $$x$$ is?

A
$$35$$

B
$$50$$

C
$$30$$

D
$$40$$

Solution

In figure $$AB\ ||\ CD\ ||\ EF$$,
$$\angle ABG=110^0, \angle GCD=100^0$$.
Also, $$\angle ABG$$ and $$\angle BGE$$, $$ \angle GCD$$ and $$\angle CGF$$ are pair of interior angles on same side of the transversal are supplementary.

$$\therefore \angle ABG+\angle BGE=180^0$$
$$\therefore 110^0+\angle BGE=180^0$$
$$\therefore \angle BGE=70^0$$.

Similarly,
$$\therefore \angle GCD+\angle CGF=180^0$$
$$\therefore 100^0+\angle CGF=180^0$$
$$\therefore \angle CGF=80^0$$.

Now,
$$\angle BGE+\angle BGC+\angle CGF=180^0$$ ...{sum of angles on one side of a straight line is $$180^o$$}
$$\therefore 70^0+x+80^0=180^0$$
$$\therefore x+150^0=180^0$$
$$\therefore x=30^0$$.

Hence, option $$C$$ is correct.
Question 6
1 / -0

Mark the correct alternative of the following.
In figure, if AC$$||$$DF and AB$$||$$CE, then?

A
$$x=145$$, $$y=223$$

B
$$x=223$$, $$y=145$$

C
$$x=135$$, $$y=233$$

D
$$x=233$$, $$y=135$$

Solution
Question 7
1 / -0

Mark the correct alternative of the following.
In figure, if AB$$||$$CD then $$x=?$$

A
$$54$$

B
$$39$$

C
$$44$$

D
$$64$$

Solution
Question 8
1 / -0

An angle measuring $$270^{o}$$ is:

A
an obtuse angle

B
an acute angle

C
a straight angle

D
a reflex angle

Solution

The given measure of angle is $$270^o.$$
Reflex angles are angles measuring greater than $$180^o$$ and less than $$360^o.$$
$$\therefore$$ $$270^o$$ is a reflex angle.
Hence, option $$D$$ is correct.
Question 9
1 / -0

An angle is its own complement. The measure of the angle is:

A
$$30^{o}$$

B
$$45^{o}$$

C
$$90^{o}$$

D
$$60^{o}$$

Solution

Let the measure of the required angle be $$x^{o}$$.
Since, the angle is its own complement, both the angle and its complement will be equal.
Then,
$$\implies x^{o}+x^{o}=90^{o}$$
$$\implies 2x=90^o$$
$$\implies x=\dfrac{90}{2}^o$$
$$\implies x=45^o$$.
Hence, the required angle measures $$45^{o}$$.
Therefore, option $$B$$ is correct.
Question 10
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.