MCQ Questions for Class 9 Maths Lines And Angles Quiz 1

$4:15$ , the angle formed between the two hands of a clock is:

0%
acute
0%
obtuse
0%
right angle
0%
none of these

Explanation

$4:15$ , the hour will be between

$4$ and

$5$ .

And minute hand will be on

$3$ .

Thus, the angle formed between these two hands will be less than

$90^{o}$ .

Hence, they form a acute angle.

$\angle ABC$ in the following figure is a / an:

0%
acute angle
0%
obtuse angle
0%
reflex angle
0%
straight angle

Explanation

Angles larger than a straight angle but less than complete angle (i.e. between

$180^o$ and

$360^o$ ) are called reflex angles.

Here, the given

$\angle ABC$ measures greater than

$180^o$ but less than

$360^o$ .

Hence, the given angle is a reflex angle.

Therefore, option

$C$ is correct.

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

0%
$90^{\circ}, 90^{\circ}$
0%
$80^{\circ}, 100^{\circ}$
0%
$30^{\circ}, 150^{\circ}$
0%
$45^{\circ}, 45^{\circ}$

Explanation

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

$4\times { 20 }^{ \circ }={ 80 }^{ \circ }\\ 5\times { 20 }^{ \circ }={ 100 }^{ \circ }$

The complement of

$(90^o-a)$ is :

0%
$-a$
0%
$90^o+a$
0%
$90^o-a$
0%
$a$

Explanation

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.

In Fig, POQ is a line. The value of x is:

0%
$20^0$
0%
$25^0$
0%
$30^0$
0%
$35^0$

Explanation

Sum of the angles forming a linear pair is

$180^\circ$

Here angles measuring

$40^\circ, 4x$ and

$3x$ form a linear pair.

So,

$40^\circ+4x+3x=180^\circ$

$7x+40^\circ=180^\circ$

$x=20^\circ$

Find the complement of the angle :

$\dfrac{1}{4}$ of a right angle.

0%
$67 \cdot 5^o$
0%
$57 \cdot 5^o$
0%
$37 \cdot 5^o$
0%
none of the above

Explanation

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 $=67.5^o$ .

State true or false:

The complement of the angle

$(150-a+b)^o$ is

$(a - b - 60)^o$ .

0%
True
0%
False

Explanation

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

Given, measure of one complementary angle is $(150-a+b)^o.$

$\Rightarrow$ Measure of other complementary angle $=90^o-(150-a+b)^o$

$=(90-150+a-b)^o=(-60+a-b)^o=(a-b-60)^o.$

$\therefore$ Measure of the complementary angle of $(150-a+b)^o$ $=(a-b-60)^o.$

Hence, the statement is true and option $A$ is correct.

Find the angle which is

$20^o$ more than its supplement.

0%
$90$
0%
$95$
0%
$100$
0%
$105$

Explanation

Let the required angle be

$x$ , then its supplement

$=(180-x)\quad$

Given that

$x=(180-x)+20$

$\Rightarrow 2x=200\\ \Rightarrow x={ 100 }^{ o }$

State if the following statement is true or false:

An exterior angle of a triangle is equal to the sum of the two interior opposite angles.

0%
True
0%
False
0%
Aambiguous
0%
Data Insufficient

Explanation

Let's try and prove the exterior angle property for a triangle.

For the given triangle $XZY$ ,

$\angle 1+\angle 2+\angle XZY=180^{o}$

Also, $\angle 3+\angle XZY=180^{o}$ ......... (Linear pair of angles)

$\angle 1+\angle 2+\angle XZY=\angle 3 +\angle XZY$

$\Rightarrow \angle 3=\angle 1+\angle 2$ ...... (which is the Exterior angle property).

Therefore, we can say that an exterior angle of a triangle is equal to the sum of the two interior opposite angles.
That is, the statement is true.

Hence, option $A$ is correct.

1958624_243370_ans_01e6e2265d2f4566afa5621c7dd6be7c.png

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

0%
an obtuse angle
0%
a right angle
0%
an acute angle
0%
a supplementary angle

Explanation

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.

Find the angle which is

$30^o$ more than its complement.

0%
$50$
0%
$55$
0%
$60$
0%
$65$

Explanation

Let the required angle be

$x$ .

Then, its complement

$=(90^o-x)$ .

Given, the angle is

$30^o$ more than its complement.

Then,

$x=(90^o-x)+30^o$

$\Rightarrow x+x=90^o+30^o$

$\Rightarrow 2x=120^o\\ \Rightarrow x=\dfrac { 120 }{ 2 } ^o\\ \Rightarrow x={ 60 }^{ o }$ .

Therefore, option

$C$ is correct.

Two adjacent angles whose sum is

$\displaystyle 180^{0}$ is called

0%
complementary angles
0%
linear pair
0%
vertically opposite angles
0%
none

Explanation

Two adjacent angles whose sum is

$180^o$ is called linear pair.

Ex:

$70^o+110^o = 180^o$ then these two angles is called a linear pair.

Two circles APQC and PBDQ intersect each other at the points P and Q and APB and CQD are two parallel straight lines. Then only one of the following statements is always true. Which one is it?

0%
ABDC is a cyclic quadrilateral.
0%
Ac is parallel to BD.
0%
ABDC is a rectangle.
0%
$\angle ACQ$ is a right angle.

If the difference of two complementary angle in

${ 10 }^{ 0 }$ , then the smaller angle is

0%
${ 40 }^{ 0 }$
0%
${ 50 }^{ 0 }$
0%
${ 45 }^{ 0 }$
0%
${ 85 }^{ 0 }$

If an angle

$x^\circ$ is equal to supplement, then the angle is

0%
$\frac{1}{3}$ right angle
0%
$\frac{1}{4}$ straight angle
0%
$\frac{1}{4}$ Complete angle
0%
$\frac{2}{5}$ Right angle

Two complemntary angles are in the ratio 2:The measure of the larger angle is

0%
$60$
0%
$54$
0%
$66$
0%
$48$

Mark the correct alternative of the following.
In figure, if AB

$||$ CD, then

$x=?$

0%
$62$
0%
$42$
0%
$52$
0%
$31$

Explanation

Constructions:- Extended AB to F and then F to E and FE to X

obtuse

$\angle CDE=249^0$

$\therefore \angle CDE=360^0-249^0=111^0$

So,

$\angle CDX=180^0-\angle CDE=180^0-111^0=69^0=\angle AFE$ (corresponding angles)

$\angle AED=\angle A+\angle AFE$ (exterior angle prop.)

$\Rightarrow 2x+13=28+69$

$\Rightarrow2x=84 \Rightarrow x=42^0$

2009864_1677658_ans_7a74fc241cb34d0e8f5478537691e8c7.png

If the angles made by a line with the positive directions of X and Y-axes are complementary angles then the angle made by the line with Z-axis is

0%
0
0%
$\frac {\pi }{3}$
0%
$\frac {\pi }{4}$
0%
$\frac {\pi }{2}$

Explanation

$\textbf{Step 1 : Write in the terms of l , m , n}$

$\text{Given , }\alpha = x$

$\beta = y$

$\gamma = z$

$\alpha + \beta = 90^\circ$

$\beta = 90^\circ - \alpha$

$\text{We know that , }$

$l = \cos \alpha$

$m = \cos \beta$

$n = \cos \gamma$

$l^2 + m^2 + n^2 = 1$

$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$

$\cos^2\alpha + \cos^2 (90^\circ - \alpha) + \cos^2\gamma = 1$

$\cos^2\alpha + \sin^2\alpha + \cos^2\gamma =1$

$1+ cos^2\gamma = 1$

$\cos^2\gamma = 0$

$\cos\gamma= \cos \dfrac{\pi}{2}$

$\gamma = \dfrac{\pi}{2}$

$\textbf{Hence , angle with Z-axis is }$

$\boldsymbol{\mathbf{\dfrac{\pi}{2}}}$

Mark the correct alternative of the following.
In figure, if AB

$||$ CD then the value of x is?

0%
$87$
0%
$93$
0%
$147$
0%
$141$

In fig. , the value of x is

0%
75
0%
65
0%
45
0%
55

The value of the variable (in the given figure) is equal to:

0%
6.5
0%
7
0%
5.5
0%
9

In the given figure, if

$\angle COA = 62^{\circ}$ , then

$x$ is :

0%
$108^{\circ}$
0%
$128^{\circ}$
0%
$118^{\circ}$
0%
$32^{\circ}$

State true or false:

The supplement of the angle

$(120 + a -2b)^o$ is

$(60-a+2b)^o$

0%
True
0%
False

Explanation

Two angles are supplementary if their sum is

${ 180 }^{ o }$ .

$\begin{aligned}{}120^{ o } + a - 2b + 60^{ o } - a + 2b &= \left( {120 ^{ o }+ 60^{ o }} \right) + \left( {a - a} \right) + \left( { - 2b + 2b} \right)\\& = 180^{ o }\end{aligned}$

As the sum of the given angles is

$180^o$ hence they are supplements of each other.

If the sides of a triangle are produced, then the sum of the exterior angle i.e.

$\angle a+\angle b+\angle c$ is equal to:

0%
${ 180 }^{ 0 }$
0%
${ 360 }^{ 0 }$
0%
$90^{ 0 }$
0%
$270^{ 0 }$

Explanation

$\triangle ABC$ ,

$\angle a = \angle ABC + \angle ACB$ (Exterior angle property)

$\angle b = \angle ACB + \angle BAC$ (Exterior angle property)

$\angle C = \angle ABC + \angle BAC$ (Exterior angle property)

Adding all the equations:

$\angle a + \angle b + \angle c = 2 (\angle ABC + \angle ACB + \angle BAC)$

$\angle a + \angle b + \angle c = 2 (180^o)$

$\angle a + \angle b + \angle c = 360^{\circ}$ .

Therefore, option

$B$ is correct.

Find the angle which is equal to its supplement

0%
$80$
0%
$90$
0%
$110$
0%
$154$

Explanation

Let the required angle be

$x$ , then its supplement

$=(180-x)\quad$

Given that

$x=(180-x)$

$\Rightarrow 2x=180\\ \Rightarrow x={ 90 }^{ o }$

State true or false:

Two distinct intersecting lines cannot be parallel to the same line.

0%
True
0%
False

Find the measure of the supplementary angle of

$54^o$

0%
$26^o$
0%
$126^o$
0%
$34^o$
0%
$134^o$

Explanation

$Two\quad angles\quad are\quad supplementary\quad when\quad they\quad add\quad upto\quad form\quad 180\quad degrees\quad .\\ If\quad one\quad angle\quad =\quad 54\\ Let\quad the\quad other\quad angle\quad be\quad x\\ Hence\quad x\quad =\quad 180-54\\ =126\\ Hence\quad supplementary\quad angle\quad of\quad the\quad following\quad angle\quad is\quad 126$

Find the measure of the complementary angle of

$77^o$ .

0%
$43^o$
0%
$70^o$
0%
$47^o$
0%
$13^o$

Explanation

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

Given, measure of one complementary angle is $77^{\circ}$ .

$\Rightarrow$ Measure of other complementary angle $=90^o-77^o$ $=13^o.$

$\therefore$ Measure of a complementary angle of $77^{o}=13^o$ .

Therefore, option $D$ is correct.

Find the measure of the supplementary angle of

$138^\circ$ .

0%
$48^\circ$
0%
$42^\circ$
0%
$52^\circ$
0%
$38^\circ$

Explanation

Two angles are supplementary when they add upto form

$180^o$ .

Given, one angle

$= 138^o$ .

Let the supplement angle be

$x$

Hence,

$x= 180^o-138^o$

$=42^o$

Hence, supplementary angle of the following angle is

$42^o$ .

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

0%
Obtuse angle
0%
Acute angle
0%
Right angle
0%
None of these

Explanation

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.

CBSE Questions for Class 9 Maths Lines And Angles Quiz 1 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Lines And Angles Quiz 1 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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