MCQ Questions for Class 9 Maths Lines And Angles Quiz 6

Find the complement of the following angles:

$63^{\circ}$ .

0%
$27^{\circ}$
0%
$54^{\circ}$
0%
$117^{\circ}$
0%
None of these

Explanation

We know, two angles are complementary when their sum is equal to $90^o.$

Given, measure of one angle is $63^o.$

$\Rightarrow$ Measure of its complementary angle $=90^o-63^o$ $=27^o.$

$\therefore$ Measure of complementary angle of $63^{o}=27^o$ .

Option $A$ is correct.

$\angle PQR$ and

$\angle QOR$ form a linear pair. If

$a-b=80$ , find the values of a and b

0%
$a=130^o$
0%
$a=80^o$
0%
$b=50^o$
0%
$b=130^o$

Explanation

$As\quad \angle POR\quad and\quad \angle QOR\quad form\quad a\quad linear\quad pair\quad (given)\\ \angle POR+\angle QOR=180\quad (Linearpairaxiom)\\ or\quad a\quad +\quad b=180...........(1)\\ Also\quad a\quad -\quad b\quad =\quad 80...........(2)[Given]\\ Adding\quad equations(1)\quad and\quad (2),we\quad get\quad 2\quad a\quad =260\\ \quad a\quad =\quad \frac { 260 }{ 2 } =130\\ Substituting\quad the\quad value\quad of\quad a\quad in\quad (1),\\ we\quad get\quad 130+b=180\\ \quad b\quad =180-130=50\\ Hence\quad a\quad =130\quad b\quad =50$

State True or False.

In figure,

$\angle PQR=\angle PRQ$ , then

$\angle PQS=\angle PRT$ .

0%
True
0%
False

Explanation

Let

$\angle PQR=\angle PRQ=x$ (say).....(1)
Now,

$\angle PQS+\angle PQR=180^o$ [Linear pair of angles]
and

$\angle PRT+\angle PRQ=180^o$ [Linear pair of angles]

$\Rightarrow \angle PQS+\angle PQR=\angle PRT+\angle PRQ [\because each=180^o]$

$\Rightarrow \angle PQS+x=\angle PRT+x$ [By (1)]

$\Rightarrow \angle PQS=\angle PRT$ .

Hence, the statement is true and option

$A$ is correct.

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is:

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

Explanation

$\textbf{Step-1: Apply properties of sum of angles of triangle to find it's one angle}$

$\text{Let the angles of a triabgle be}$

$\alpha ,\beta ,\gamma$

$\text{Given}$

$\alpha +\beta =\gamma$

$\text{We now that in a sum of triangles sum of angles is}$

${ 180 }^{ \circ }$

$\text{So,}$

$\alpha +\beta +\gamma ={ 180 }^{ \circ }$

$\Rightarrow 2\gamma ={ 180 }^{ \circ }$

$\Rightarrow \gamma ={ 90 }^{ \circ }$

$\text{So, it is a right angled triangle.}$

$\textbf{Hence option C is correct}$

If the angles of a triangle are in the ratio

$2:3:4$ , find the three angles.

0%
$80^o, 120^o, 160^o$
0%
$20^o, 30^o, 40^o$
0%
$40^o, 60^o, 80^o$
0%
None of these

Explanation

The angles of a triangle are in the ratio

$2:3:4$ .
Let

$x:y:z=2:3:4$ .
Then,

$x=2t, y=3t$ and

$z=4t$ .

We know, by angle sum property, the sum of all angles of a triangles is

$180^0$ .

$\implies$

$2t + 3t + 4t = 180^o$

$\implies$

$9t = 180^o$

$\implies$

$t = 20^o$ .

Therefore,

$x= 2\times 20^o= 40^o; y= 3\times 20^o= 60^o; z = 4\times 20^o= 80^o$ .

Hence, option

$C$ is correct.

In fig, then

$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=$

0%
$180^o$
0%
$270^o$
0%
$360^o$
0%
$540^o$

Explanation

$\triangle AEC$ ,

$\angle A + \angle E + \angle C = 180^o$ .....[Angle sum property].

Also, in

$\triangle BDF$ ,

$\angle B + \angle D + \angle F = 180^o$ .....[Angle sum property].

Adding both the equations, we get,

$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F= 180^o + 180^o = 360^o$ .

Therefore, option

$C$ is correct.

In the given figure, value of

$x$ is .........

0%
45 $^o$
0%
34 $^o$
0%
64 $^o$
0%
109 $^o$

Explanation

In triangle

$\Delta ABC$ ,

$\angle A+\angle ABC+\angle BCA=180^o$

$34^o+ \angle ABC +30^o=180^o$

$\angle ABC=116^o$
So,

$\angle ABC+\angle CBD=180^o$

$116^o+\angle CBD=180^o$

$\angle CBD=180^o-116^o$

$\angle CBD=64^o$
Now use remote interior angle theorem.
So,

$\angle X = \angle CBD+ \angle D=64^o+45^o=109^o$

The measure of angle

$y$ in the given figure is ......... .

0%
$\angle y = 65^{\circ}$
0%
$\angle y = 85^{\circ}$
0%
$\angle y = 75^{\circ}$
0%
$\angle y = 80^{\circ}$

Explanation

$\textbf{Step 1: Using property of angle of straight line calculate y}$

$\text{We know that a straight line extends an angle of }{180^\circ}$

$\implies \angle \text{ABC}+\angle \text{CBD}=180^\circ$

$\implies \angle \text{CBD }=180^\circ - \angle \text{ABC }$

$\implies y^\circ=180^\circ - 115^\circ$

$\implies y^\circ=65^\circ$

$\mathbf{\text{Thus, the measure of angle y}=65^\circ}$

The measure of an angle is three times the measure of its complement. The angles are:

0%
$20^\circ,\, 60^\circ$
0%
$30^\circ,\, 60^\circ$
0%
$45^\circ,\, 135^\circ$
0%
$22.5^\circ,\, 67.5^\circ$

Explanation

Let the complement be

$x$ .

Then the angle

$=3x$ .

We know, the sum of the complementary angles is

${ 90 }^{ \circ }$

$\Rightarrow x+3x={ 90 }^{ \circ }\\ \Rightarrow 4x={ 90 }^{ \circ }\\ \Rightarrow x=22.5^{ \circ }$ .

Hence, the angles are:

$x=1\times 22.5^{ \circ }=22.5^{ \circ }$

and

$3x=3\times 22.5^{ \circ }={ 67.5 }^{ \circ }$ .

Hence, option

$D$ is correct.

If two angles of a triangle are acute angles, then the third angle:

0%
is less than the sum of the two angles
0%
should be an acute angle
0%
is the largest angle of the triangle
0%
is an angle with any measurement less than $180^o$

When the clock shows time

$5:20$ , the angle formed between the two hands is

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

Explanation

Angles

between

${0}^{o}$ and

${90}^{o}$ are acute angles.

Since at the time

$5:20$ , the hands of the clock form an angle less than

${90}^{o}$ , the angle formed between the two hands is an acute angle.

Supplementary angle of

$108.5^{\circ}$ is

0%
$70.5^{\circ}$
0%
$71.5^{\circ}$
0%
$71^{\circ}$
0%
$72.5^{\circ}$

Explanation

Given angle is

$= 108.5^{\circ}$

Let the angle supplementary with the above angle be

$x$

Now, the sum of two supplementary angles

$= 180^{\circ}$

$\Rightarrow x+108.5^{\circ}=180^{\circ}$

$\Rightarrow x=71.5^{\circ}$

Complementary angle of

$72\displaystyle \frac{1}{2}^{\circ}$ is:

0%
$17^{\circ}$
0%
$18\displaystyle \frac{1}{2}^{\circ}$
0%
$21\displaystyle \frac{1}{2}^{\circ}$
0%
$17\displaystyle \frac{1}{2}^{\circ}$

Explanation

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

Given, measure of one complementary angle is $72\dfrac{1}{2}^o=\dfrac{145}{2}^o.$

$\Rightarrow$ Measure of other complementary angle $=90^o-\dfrac{145}{2}^o$ $=\dfrac{35}{2}^o=17\dfrac{1}{2}^o.$

$\therefore$ Measure of a complementary angle of $72\dfrac{1}{2}^o=$ $17\dfrac{1}{2}^o.$

Therefore, option $D$ is correct.

All linear pairs are

0%
supplementary
0%
vertically opposite
0%
right angles
0%
none of these

Explanation

In the given diagram,

$ABC$ is a straight line.

$\angle ABD$ and

$\angle CBD$ are linear pairs.

$\angle ABD=120^\circ$ and

$\angle CBD=60^\circ$

A linear pair forms a straight angle which contains

$180^\circ$ , hence the sum of

$2$ angles measures

$180^\circ,$ which means they are supplementary.

$\therefore$

$\angle ABD+\angle CBD=120^\circ+60^\circ$

$=180^\circ$

So we can say that,

$\angle ABD$ and

$\angle CBD$ are supplementary.

$\therefore$ All linear pairs are supplementary.

1362553_272154_ans_90ae292c61fd4fb3aa594ab7f4b93506.png

$\angle AEC$ and

$\angle CED$ are supplementary angles. If

$\angle CED$ is equal to

$62^{\circ}$ ,what is the measure of

$\angle AEC$ ?

0%
$28^{\circ}$
0%
$150^{\circ}$
0%
$118^{\circ}$
0%
$124^{\circ}$

Explanation

Two angles are supplementary if they add up to

$180^{\circ}$ .
As given,

$\angle AEC$ and

$\angle CED$ are supplementary angles,

So,

$\angle AEC+\angle CED=180^{\circ}$

$\Rightarrow \angle AEC+62^{\circ}=180^{\circ}$

$\Rightarrow\angle AEC=180^{\circ}-62^{\circ}$

$\Rightarrow\angle AEC=118^{\circ}$

Hence, the required answer is

$118^{\circ}$ .

In given figure which one of the following angle is a straight angle

0%
$\angle AOB$
0%
$\angle AOD$
0%
$\angle BOC$
0%
$\angle BOD$

Explanation

Straight angle is equal to

$180^o$ .

So,

$\angle BOD$ and

$\angle AOC$ are straight angles.

Find the value of

$x$ and

$y$ .

0%
$60^{\circ}, 45^{\circ}$
0%
$45^{\circ}, 60^{\circ}$
0%
$45^{\circ}, 45^{\circ}$
0%
$60^{\circ}, 60^{\circ}$

Explanation

From the figure,

$45^\circ$ is the complement of

$y$

That is,

$y + 45^\circ = 90^{\circ}$

$\Rightarrow$

$y = 90^{\circ}-45^\circ$

$\Rightarrow y = 45^{\circ}$

Also,

$30^\circ$ is the complement of

$x$

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

$\Rightarrow x = 90^\circ-30^\circ$

$\Rightarrow x = 60^{\circ}$

A reflex angle is .........a straight angle.

0%
Smaller than
0%
Larger than
0%
Equal to
0%
One-half of

Which of the following is an obtuse angle ?

0%
$92^\circ$
0%
$181^\circ$
0%
$195^\circ$
0%
$83^\circ$

The sum of two acute angles can be:

0%
Acute angles
0%
Obtuse angles
0%
Right angles
0%
All of the above

Explanation

Let one angle is

$\displaystyle { 30 }^{ o }$ and the other angle is

$\displaystyle { 40 }^{ o }$ then sum is

$\displaystyle { 70 }^{ o }$ . which is an acute angle.
Let one angle is

$\displaystyle { 70 }^{ o }$ and other angle is

$\displaystyle { 80 }^{ o }$ , then sum is

$\displaystyle { 150 }^{ o }$ which is a obtuse angle.
Let one angle is

$\displaystyle { 45 }^{ o }$ and another angle is

$\displaystyle { 45 }^{ o }$ then sum is

$\displaystyle { 90 }^{ o }$ which is a right angle.

So correct answer is

$(D)$

Find the complements of

$\displaystyle { 68 }^{ o }$ and

$\displaystyle { 33 }^{ o }$ respectively.

0%
$22^o, 57^o$
0%
$57^o, 22^o$
0%
$33^o, 67^o$
0%
$67^o, 33^o$

Explanation

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

Given, measure of one complementary angle is $68^o.$

$\Rightarrow$ Measure of other complementary angle $=90^o-68^o$ $=22^o.$

$\therefore$ Measure of a complementary angle of $68^{o}=22^o$ .

Also given, measure of one complementary angle is $33^o.$

$\Rightarrow$ Measure of other complementary angle $=90^o-33^o$ $=57^o.$

$\therefore$ Measure of a complementary angle of $33^{o}=57^o$ .

Hence, option $A$ is correct.

Which of the following indicated figures are examples of adjacent angles.

0%
(i) and (iv)
0%
(i) and (ii)
0%
(i) and (iii)
0%
(ii) and (iv)

$\Delta ABC,$

$\angle A = 43^{o}$ and

$\angle C = 70^{o}.$ What is the measure of

$\angle B?$

0%
$63^{o}$
0%
$65^{o}$
0%
$66^{o}$
0%
$67^{o}$

Explanation

By angle sum property, the sum of angles is

$180^o$ .

$\angle A+ \angle B + \angle C = 180^{o}$

$43^{o} +\angle B + 70^{o} = 180^{o}$

$\angle B = 180^{o} - 113^{o}$

$\angle B = 67^{o}.$

The angles of a triangle are in the ratio of

$2:3:4.$ What is the measure of the smallest interior angle of the triangle?

0%
$10^{o}$
0%
$20^{o}$
0%
$30^{o}$
0%
$40^{o}$

Explanation

By angle sum property, the sum of angles is

$180^o$ .

Let

$2x, 3x$ and

$4x$ be the three angles.

$2x + 3x + 4x = 180^{o}$

$9x = 180^{o}$

$x = 20^{o}$
Smallest interior angle

$= 2 \times 20^{o} = 40^{o}.$

Find the measure of an exterior angle at the base of an isosceles triangle whose vertex angle measures

$50^o$ .

0%
$100^o$
0%
$110^o$
0%
$115^o$
0%
$65^o$

Explanation

The

$2$ base angles of an isosceles triangle are equal, so we'll represent each as

$x$ .

Now, by using the angle sum property,

$x + x + 50^o = 180^o$

$\Rightarrow 2x = 180^o - 50^o$

$\Rightarrow 2x = 130^o$

$\Rightarrow x = 65^o$

So, exterior angle is equal to the sum of the two non-adjacent interior angles.

Therefore,

$y = x + 50^o$

$y = 50^o + 65^o$

$= 115^o$

Hence, option

$C$ is correct.

$\Delta PQR, \angle Q = 23^{o}$ and

$\angle R = 20^{o}.$ What is the measure of

$\angle P?$

0%
$134^{o}$
0%
$135^{o}$
0%
$136^{o}$
0%
$137^{o}$

Explanation

By angle sum property, the sum of angles is

$180^o$ .

$\angle P+ \angle Q + \angle R = 180$

$\angle P + 23^{o} + 20^{o} = 180^{o}$

$\angle P = 180^{o} - 43^{o}$

$\angle P = 137^{o}.$

If m

$\angle$

$PRQ = 45^o$ and m

$\angle$

$QPR = 68^o$ . What is m

$\angle$

$PQR$ in the triangle?

0%
$67^o$
0%
$68^o$
0%
$69^o$
0%
$70^o$

Explanation

by angle sum property, the sum of angles is

$180^o$ .

$m\angle PRQ + m\angle QPR + m\angle PQR$

$= 180^o$

$45^o + 68^o +$ m

$\angle$

$PQR = 180^o$
m

$\angle$

$PQR = 180^o - 113^o$
m

$\angle$

$PQR = 67^o$ .

$\Delta ABC, \angle A = 56^{o}$ and

$\angle B = 60^{o}.$ What is the measure of

$\angle C?$

0%
$64^{o}$
0%
$65^{o}$
0%
$66^{o}$
0%
$67^{o}$

Explanation

By angle sum property, the sum of angles is

$180^o$ .

$\angle A+ \angle B + \angle C = 180^{o}$

$56^{o} + 60^{o} + \angle C = 180^{o}$

$\angle C = 180^{o} - 116^{o}$

$\angle C = 64^{o}$

The angles of a triangle are in the ratio of

$3:5:7.$ What is the measure of the largest interior angle of the triangle?

0%
$18^{o}$
0%
$48^{o}$
0%
$84^{o}$
0%
$96^{o}$

Explanation

By angle sum property, the sum of angles is

$180^o$ .

Let

$3x, 5x$ and

$7x$ be the three angles of the triangle.

So,

$3x + 5x + 7x = 180^{o}$

$15x = 180^{o}$

$x = 12^{o}$

Hence, Largest interior angle

$= 7 \times 12^{o} = 84^{o}$

$\triangle ABC$ , if

$\angle B = \angle C = 45^{\circ}$ , which of the following is correct?

0%
$\angle A= 60°$
0%
$\angle A= 70°$
0%
$\angle A= 80°$
0%
$\angle A= 90°$

Explanation

$\angle B=\angle C=45^{\circ}$ and by angle sum property, the sum of angles is

$180^o$ .

$\therefore \angle A+\angle B+\angle C=180^{\circ}$

$\Rightarrow \boxed{\angle A=90^{\circ}}$

CBSE Questions for Class 9 Maths Lines And Angles Quiz 6 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Lines And Angles Quiz 6 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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