Pair of interior angles on same side

MCQ Questions for Class 9 Maths Lines And Angles Quiz 10

An angle is more than

$45^o$ . Its complementary angle must be less than

$45^o$ .

0%
True
0%
False

Explanation

Let

$x$ and

$y$ are two angles making a complementary angles, and

$x$ is greater than

$45$ degrees.

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

$\Rightarrow \ y=90^o -x$ .

$90=2\times45$ .

Therefore,

$y$ will be less than

$45$ degrees.

Hence, given statement is true.

Therefore, option

$A$ is correct.

Two supplementary angles always form a linear pair.

0%
True
0%
False

Explanation

The linear pair are angles who are adjacent and supplementary.

But two angles can add up to

$180^0$ that is they are supplementary even if they are not adjacent. In this case they are not a linear pair.

So, given statement is false.

Two adjacent angles always from a linear pair.

0%
True
0%
False

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

0%
All (i), (ii) and (iii) are true
0%
(iii) is flase
0%
(i) is false but (ii) and (iii) are true
0%
(ii) is false

Explanation

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

In which of the following figures,

$a$ and

$b$ are forming a pair of adjacent angles?

Explanation

For adjacent angles both angles must have one vertex and one side common.

$a$ and

$b$ have a common vertex and a common arm that's why they are an adjacent angles.

Hence, Option

$(D)$ is correct.

Two angles making a linear pair are always supplementary.

0%
True
0%
False

Vertically opposite angles are always

0%
supplementary
0%
complementary
0%
adjacent
0%
equal

In figure,

$POQ$ is a line, then

$a$ is equal to

0%
$35^o$
0%
$100^o$
0%
$80^o$
0%
$135^o$

Explanation

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.

In figure,

$a=40^o$ . The value of

$b$ is

0%
$20^o$
0%
$24^o$
0%
$36^o$
0%
$120^o$

Explanation

We can clearly see that,

$5b$ and

$2a$ are linear angles

And

$a=40^0$

$\Rightarrow 5b+2a=180^o$

$\Rightarrow 5b+2\times 40^o =180^o$

$\Rightarrow 5b+80^o =180^o$

$\Rightarrow 5b=100^o$

$\Rightarrow b=\dfrac{100^o}{5}$

$\Rightarrow b=20^o$

So, option

$A$ is correct

In figure, the reflex

$\angle AOB$ is equal to :

0%
$60^{\circ}$
0%
$120^{\circ}$
0%
$300^{\circ}$
0%
$360^{\circ}$

In figure, which one of the following is not true ?

0%
$\angle 1+\angle 5=180^o$
0%
$\angle 2+\angle 5=180^o$
0%
$\angle 3+\angle 8=180^o$
0%
$\angle 2+\angle 3=180^o$

Explanation

Given,

$PQ$ parallel to

$RS$ and line

$l$ is a transversal.

$\therefore \angle 2+\angle 5=180^0$ (1).....[ Co-interior angles=supplementary angles ]

$\angle 3+\angle 8=180^0$ (2) ..... [ Co- interior angles=supplementary angles ]

$\angle 1=\angle 2$ (3)..... [ Vertically opposite angles ]

$\therefore \angle 1+\angle 5=180^o$ [ By equation (1) and (3)]

$\angle 2=\angle 3$ [ Alternate interior angles ]

However,

$\angle 2 +\angle 3=180^o$ cannot be true.

Hence, option

$D$ is correct.

It is possible to have a triangle in which two of the angles are obtuse.

0%
True
0%
False

Explanation

False
By angle sum property we know,

$\angle A+ \angle B+ \angle C =180^{o}$
According to the question, two angles are obtuse,

$\therefore \angle B+ \angle C > 180^{o}$
Thus, Triangle is not possible

In figure, which of the following is true?

0%
$\angle 1=\angle 5$
0%
$\angle 4=\angle 8$
0%
$\angle 5=\angle 8$
0%
$\angle 3=\angle 7$

Explanation

Given,

$PQ$ parallel to

$RS$ and line

$l$ is a transversal.

$\therefore \angle 5=\angle 8$ [ Alternate interior angles ]

Hence, option

$C$ is correct.

All rest pairs in options are supplementary angles.

If an angle is

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

0%
$100^o$
0%
$80^o$
0%
$60^o$
0%
$120^o$

Explanation

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

In Fig.,

$PQ = PR, RS = PQ$ and

$ST || QR$ . If the exterior angle

$RPU$ is

$140^o$ , the the measures of angle

$TSR$ is

0%
$55^o$
0%
$40^o$
0%
$50^o$
0%
$45^o$

Explanation

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

$2$

$\Longrightarrow \angle TSR = 180^o - 70^o - 70^o$

$\Longrightarrow TSR = 40^o$

In the given figure, the value of

$x$ that will make

$AOB$ a straight line is

0%
$x=40$
0%
$x=36$
0%
$x=30$
0%
$x=25$

Explanation

$AOB$ will be a straight line if,

$3x+15+2x-15^{0}=180^{0}$

$\Rightarrow 5x=180^{0}$

$\Rightarrow x=36^{0}$

A pair of supplementary angles is

0%
$55^{\circ},115^{\circ}$
0%
$65^{\circ},125^{\circ}$
0%
$47^{\circ},133^{\circ}$
0%
$40^{\circ},50^{\circ}$

Explanation

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.

If an angle measures

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

0%
$40^{o}$
0%
$55^{o}$
0%
$35^{o}$
0%
$50^{o}$

Explanation

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.

Pair of vertically opposite angles are always supplementary

0%
True
0%
False

A pair of complementary angles is:

0%
$130^{0},50^{0}$
0%
$35^{0},55^{0}$
0%
$25^{0},75^{0}$
0%
$27^{0},53^{0}$

Explanation

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.

$30^{0}$ is one-half of its complement.

0%
True
0%
False

Explanation

Let

$x$ and

$y$ are two angles making a complementary angles, and

$x$ is

$30^o$ .

Then,

$\implies$

$x+y=90^o$

$\implies$

$30^o+y=90^o$

$\implies$

$y=90^o-30^o=60^o$

$\implies$

$y=60^o$ .

Clearly,

$x$ is half of its complement

$y$ .

That is,

$30^o=\dfrac{1}{2}\times60^o$ and is true.

Hence, the statement is true and option

$A$ is correct.

If the difference of two complementary angles is

$10$ , then the smaller angle is:

0%
$40$
0%
$50$
0%
$45$
0%
$35$

Explanation

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

$= 40$ .

In figure, which of the following pairs of angles are corresponding angles:

0%
$\angle 1, \angle 8$
0%
$\angle 2, \angle 6$
0%
$\angle 4, \angle 8$
0%
$\angle 3, \angle 2$

Explanation

Given is an image having a system of two parallel lines being intersected by a transversal.

In a system of two parallel lines being intersected by a transversal, the angles in corresponding corners are called corresponding angles.

For the given system, the pairs of corresponding angles are:

$(i)\ \angle 1,\ \angle 8$

$(ii)\ \angle 4,\ \angle 7$

$(iii)\ \angle 2,\ \angle 5$

$(iv)\ \angle 3,\ \angle 6$

So, amongst the given options,

$\angle 1,\angle 8$ are corresponding angles.

Hence, option A is correct.

2057050_1880659_ans_c3a30e40072c41eea634129468bd02c7.png

Angles

$t$ and

$h$ are

0%
Alternate angles
0%
Corresponding angles
0%
Interior angles
0%
Exterior angles

Explanation

${\textbf{Hint: Use properties of angles made by two parallel lines to solve}}$

${\textbf{Step - 1: Observe the lines, angles from the given figure}}$

${\text{We can see here two parallel lines}}$

${\text{The lines are intersected with a transversal the angles are}}$

${\text{made by this transversal and two parallel lines}}$

${\textbf{Step - 2: Observe the angles position}}$

${\text{Two angles t and h are in the same side of the transversal}}$

${\text{The pair of angles are same relative position also on different intersecting lines.}}$

${\textbf{So angles t and h are corresponding angles i.e, Option B.}}$

$r$ and

$s$ are

0%
Corresponding angles
0%
Alternate interior angles
0%
Pair of interior angles on same side
0%
Alternate exterior angles

A pair of corresponding angle is

0%
$r$ and $q$
0%
$r$ and $s$
0%
$t$ and $q$
0%
$r$ and $t$

An angle is

$18^{\circ}$ less than its complementary angle. The measure of this angle is

0%
$36^{\circ}$
0%
$48^{\circ}$
0%
$83^{\circ}$
0%
$81^{\circ}$

Explanation

$\textbf{Step 1: Form equations using the concept of complementary angles.}$

$\text{Let, the smaller angle be }x$

$\therefore\text{Bigger angle }(x+18^\circ)$

$\text{We know, the sum of complementary angles is }$

$90^\circ$

$\Rightarrow x+(x+18^\circ)=90^\circ$

$\Rightarrow 2x=90^\circ-18^\circ$

$\Rightarrow 2x=72^\circ$

$\Rightarrow x=36^\circ$

$\textbf{Hence, the correct option is A.}$

In the given figure,

$ABC$ is an isosceles triangle with

$AB=AC$ . If

$AE=AF$ and

$\angle BAE=40^0$ , then the measure of the angle

$FEC$ is

0%
$15^0$
0%
$20^0$
0%
$40^0$
0%
$60^0$

Explanation

Let

$\angle ABC=\angle ACB=\theta$ and

$\angle CEF=X$ then

$\angle AFE=\angle AEF=\theta+X$
Now by the exterior angle property

$\angle AEC=\angle ABE+\angle BAE$

$\Rightarrow \theta+2X=40+\theta\Rightarrow x=20^0$
2030518_1962398_ans_5e915c8faff04ea8951aa2b26fbec3da.png

$M$ and

$N$ are

0%
Alternate exterior angles
0%
Alternate interior angles
0%
Corresponding angles
0%
Vertically opposite angles

A pair of alternate exterior angles is

0%
$a$ and $c$
0%
$a$ and $d$
0%
$c$ and $b$
0%
$d$ and $b$

CBSE Questions for Class 9 Maths Lines And Angles Quiz 10 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Lines And Angles Quiz 10 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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