MCQ Questions for Class 9 Maths Lines And Angles Quiz 5

If angle

$P$ and angle

$Q$ are supplementary and the measure of angle

$P$ is

$60^o$ , then the measure of angle

$Q$ is:

0%
$120^o$
0%
$60^o$
0%
$30^o$
0%
$20^o$

Explanation

Option (a) is correct.
Given that

$\angle P +\angle Q=180^o$

$\Rightarrow 60^o +\angle Q=180^o$

$\Rightarrow \angle Q=180^o -60^o$

$\Rightarrow \angle Q=120^o$

Two acute angles are congruent.

0%
True
0%
False

Explanation

False

Two acute angles may vary from

$0^{\mathring{}}$ to

$89^{\mathring{}}$ . Hence, acute angles having different measures are not congruent.

Sum of any two angles of a triangle is always greater than the third angle.

0%
True
0%
False

If the sum of two angles is equal to an obtuse angle, then which of the following is not possible?

0%
One obtuse angle and one acute angle
0%
One right angle and one acute angle
0%
Two acute angles
0%
Two right angles

Explanation

Option [D] is the correct answer.

The sum of two right angles is exactly equal to

${180}^{o}$ , which is a straight angle not an obtuse angle.

In the figure, if OP||RS,

$\angle OPQ=110^0 and \angle QRS =130^0$ , then

$\angle PQR$ is equal to

0%
$40^0$
0%
$50^0$
0%
$60^0$
0%
$70^0$

Explanation

As OP||RS,

$\angle$ PQR

$=180 -(180-110)-(180-130)$ =

$60$

If m

$\angle$

$XYZ = 86^o$ and m

$\angle$

$XZY = 23^o$ . What is m

$\angle$

$YXZ$ in the triangle?

0%
$70^o$
0%
$71^o$
0%
$72^o$
0%
$73^o$

Explanation

By angle sum property, the sum of angles is

$180^o$ .

$86^o + 23^o +$ m

$\angle$

$YXZ = 180^o$

$\angle$

$YXZ = 180^o - 109^o$
m

$\angle$

$YXZ = 71 ^o$ .

Find

$x + y + z$

0%
$x+y+z = 180^{0}$
0%
$x+y+z = 360^{0}$
0%
$x+y+z = 480^{0}$
0%
$x+y+z = 540^{0}$

Explanation

Here, by exterior angle property,

$\angle y$ being an exterior angle, will be the sum of opposite interior angles.
Then,

$\angle y= 90^o + 30^o = 120^{\circ}$ .

We know, by straight angle property, sum of angles on a straight line

$= 180^o$ .
Then,

$\angle z +30^o = 180^o$

$\implies$

$\angle z = 150^{\circ}$ .
Also,

$\angle x + 90^o = 180^o$

$\implies$

$\angle x = 90^{\circ}$ .
Therefore,

$x +y +z = 120^o +150^o +90^o =$

$360^{\circ}$ .

Therefore, option

$B$ is correct.

In the given figure,

$\angle QPB$ is,

0%
$60^{\circ}$
0%
$45^{\circ}$
0%
$30^{\circ}$
0%
$15^{\circ}$

Explanation

In the given figure,

$\angle APR + \angle RPQ + \angle QPB = 180^{\circ}$

$2x + 3x+x = 180^{\circ}$

$6x = 180^{\circ}$

$x = 30^{\circ}$

If two supplementary angles are in the ratio 2 : 7, then the angles are :

0%
$35^{\circ}, 145^{\circ}$
0%
$70^{\circ}, 110^{\circ}$
0%
$40^{\circ}, 140^{\circ}$
0%
$50^{\circ}, 130^{\circ}$

Explanation

$\textbf{ Hint: Sum of supplementary angles is}$

$180^\circ$

$\textbf{Step1: Evaluate using addition of ratio}$

$\text{Given}$

$\text{ratio}$

$=2:7$

$\text{ Let angles as }$

$2x$

$\text{and}$

$7x$

$\text{As we know that sum of supplementary angle is }180$

$\Rightarrow 2x+7x=180$

$\Rightarrow 9x=180$

$\Rightarrow x=20$

$2x\Rightarrow 2(20)=40$

$7x\Rightarrow 7(20)=140$

$\text{ Hence, the angles are}$

$40^\circ$

$140^\circ.$

$\textbf{Option C is correct.}$

In the figure, if

$\angle A + \angle B + \angle C + \angle D + \angle E + \angle F = k\times$ right angle, then

$k$ is :

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

$\Rightarrow$ In

$\triangle ABC$

$\Rightarrow$

$\angle A+\angle B+\angle C=180^\circ$ [Sum of interior angles of triangle is

$180^\circ$ ] --- ( 1 )

$\Rightarrow$ In

$\triangle DEF$

$\Rightarrow$

$\angle D+\angle E+\angle F=180^\circ$ [Sum of interior angles of triangle is

$180^\circ$ ] ---( 2 )

$\Rightarrow$

$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=180^\circ+180^\circ$ [Adding ( 1 ) and ( 2 )]

$\Rightarrow$

$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=360^\circ$

$\Rightarrow$

$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=k\times (right\, angle)$ .

$\therefore$

$k=\dfrac{360^o}{90^o}$

$\therefore$

$k=4$

Two angles are called adjacent if

0%
they have a common vertex
0%
they have a ray in common
0%
- their other arms lie on the opposite sides of the common arm
0%
all the above

If one of the angles of a triangle is

$130^0$ , then the angle between the bisectors of the other two angles can be

0%
$50^0$
0%
$65^0$
0%
$145^0$
0%
$155^0$

Explanation

Let

$x$ and

$y$ be the bisected angles.

So, in the original triangle,

sum of angles

$=180^o$

Therefore,

$2x+2y+130=180$

$\Rightarrow 2(x+y)=50$

$\Rightarrow x+y=25$

In the smallest triangle, consisting of original side opposite

$130^o$ .

Therefore,

$25^o+$ Angle between bisectors

$=180^o$

$\Rightarrow$ Angle between bisectors of other two angles

$=155^o$ .

If the sum of two adjacent angles is

$100^{\circ}$ and one of them is

$35^{\circ}$ , then the other is :

0%
$70^{\circ}$
0%
$65^{\circ}$
0%
$135^{\circ}$
0%
$145^{\circ}$

Explanation

Let the other angle be

$x$

Now their sum

$=100^{\circ}$

$\Rightarrow x+{ 35 }^{ \circ }={ 100 }^{ \circ }\\ \Rightarrow x={ 100 }^{ \circ }-{ 35 }^{ \circ }={ 65 }^{ \circ }$

So the other angle is

$65^{\circ}$

The angle which exceeds its complement by

$20^{\circ}$ is:

0%
$45^{\circ}$
0%
$55^{\circ}$
0%
$70^{\circ}$
0%
$110^{\circ}$

Explanation

Let the required angle be

$x$ .

We knows, complementary angles

$=$ Sum of two angles is

$90^o$ .

$\therefore x = (90^o - x ) + 20^o$

$\therefore$

$x = 90^o - x + 20^o$

$\therefore$

$2x = 110^o$

$\therefore x = 55^o$ .

Hence, option

$B$ is correct.

From the adjoining figure, calculate the values of

$a$ .

0%
$46^o$
0%
$38^o$
0%
$56^o$
0%
none of the above

Explanation

We know, by angle sum property, the sum of angles of triangle

$= 180^o$
Then,

$a + 110^o + 32^o = 180^o$

$\implies$

$a + 142^o = 180^o$

$a =180^o-142^o=38^{\circ}$ .

Thus measure of angle

$a$ is

$38^o$ .

Hence, option

$B$ is correct.

In the given figure, AB // CD, EH // BC,

$\angle BAC = 60^{\circ}$ and

$\ \angle DGH = 40^{\circ}$ . Find the measures of

$\angle AEF$ .

0%
$36^o$
0%
$40^o$
0%
$46^o$
0%
none of the above

Explanation

Given,

$\angle DGH = 40^{\circ}$ .

Since

$AB \parallel CD$ ,

$\implies$

$AE \parallel DG$ .

By corresponding angles axiom,

$\angle AEF = \angle DGH = 40^{\circ}$ .

Hence, option

$B$ is correct.

In the given figure,

$ABC$ is a triangle, side

$CB$ is produced to

$E$ and

$\angle A$ :

$\angle B$ :

$\angle C$

$= 2: 1: 3$ . Find

$\angle DBE$ , if

$DB$ is perpendicular to

$AB$ .

0%
$66^o$
0%
$46^o$
0%
$60^o$
0%
none of the above

Explanation

In triangle

$ABC$ ,

$\angle A : \angle B : \angle C = 2: 1: 3$
Let the angles be

$2x, x, 3x$
By angle sum property, the sum of the angles

$2x + x+ 3x = 180^o$

$\implies$

$6x = 180^o$

$\implies$

$x = 30^o$ .

Hence,

$\angle B = 30^o$ .

Now,

$\angle DBE + \angle DBA + \angle ABC = 180^o$ ...[Straight angle property]

$\angle DBE + 90^o + 30^o = 180^o$

$\angle DBE = 180^o - 120^o$

$\angle DBE = 60^{\circ}$ .

Therefore, option

$C$ is correct.

From the given figure, we can find that

$\angle ABC$

$=47^o$ .

0%
True
0%
False

Explanation

Given,

$\angle BAC = 53^o$ ,

$\angle ACD = 100^o$ .

$\angle ACD = \angle BAC + \angle ABC$ ...(By exterior angle property, sum of interior opposite angles is equal to exterior angle)

$\implies$

$100^o = 53^o + \angle ABC$

$\implies$

$\angle ABC = 100^o-53^o=47^{\circ}$ .

Hence, the statement is true.

Therefore, option

$A$ is correct.

The angles of a triangle are in the ratio 2: 1:Is the triangle right-angled triangle?

0%
True
0%
False

Explanation

The angles of the triangle are in the ratio,

$2:1: 3$ .
Let the angles be

$2x, x$ and

$3x$ .

By angle sum property, sum of the angles

$= 180^o$

$2x + x+ 3x = 180^o$

$\implies$

$6x = 180^o$

$\implies$

$x = 30^o$ .

Hence, the angles will be

$x=30^o, \ 2x=60^o$ and

$3x=90^o$ .

Since, one of the angles is

$90^o$ , the triangle is a right angled triangle.

Hence, the statement is true and option

$A$ is correct.

Two lines that are respectively parallel to two intersecting lines, intersect each other.

0%
True
0%
False
0%
Ambiguous
0%
Data Insufficient

Angles forming a linear pair can both be acute angles.

0%
True
0%
False
0%
Ambiguous
0%
Data Insufficient

Find the angle which is

$80^o$ more than its complement.

0%
$75$
0%
$85$
0%
$95$
0%
$100$

Explanation

Let the required angle be

$x$ .

Then, its complement

$=(90^o-x)$ .

Given, the angle is

$80^o$ more than its complement.

Then,

$x=(90^o-x)+80^o$

$\Rightarrow x+x=90^o+80^o$

$\Rightarrow 2x=170^o\\ \Rightarrow x=\dfrac { 170 }{ 2 } ^o\\ \Rightarrow x={ 85 }^{ o }$ .

Therefore, option

$B$ is correct.

In the given figure, AB // CD, EH // BC,

$\angle BAC = 60^{\circ}$ and

$\ \angle DGH = 40^{\circ}$ . Find the measures of

$\angle AFG$ .

0%
$96^o$
0%
$86^o$
0%
$100^o$
0%
none of the above

Explanation

Given that

$\angle BAC=60^o$ and

$\angle DGH=40^o$ .

$\implies$

$\angle EAF=60^o$ .

Also, given,

$AE \parallel DG$ .

Now,

$\angle AEF = \angle DGH = 40^{\circ}$ (Corresponding angles).

Then,

$\angle AFG = \angle AEF + \angle EAF$ (Exterior angle property)

$\implies$

$\angle AFG = 40^{\circ} + 60^{\circ}$

$\implies$

$\angle AFG = 100^{\circ}$ .

Therefore, option

$C$ is correct.

If two adjacent angles are equal, then each angle measures

$90^o$ .

0%
True
0%
False
0%
Ambiguous
0%
Data Insufficient

Explanation

The answer is

$B$
Two adjacent angles measure

$90^o$ , only when the lines are perpendicular to each other.
hence the above statement is False.

In the figure, given AE // BD and AC // ED: find

$\angle b$ .

0%
$65^o$
0%
$36^o$
0%
$45^o$
0%
none of the above

Explanation

In the given figure,

$\angle EDC = 65^o$ (Vertically opposite angles).
Given,

$AC \parallel ED$ .
Thus,

$\angle b = \angle EDC = 65^o$ (Corresponding angles on parallel lines are equal).
Therefore, option

$A$ is correct.

Find the angle which is

$60^o$ more than its complement.

0%
$55$
0%
$75$
0%
$85$
0%
$60$

Explanation

Let the required angle be

$x$ .

Then, its complement

$=(90^o-x)$ .

Given, the angle is

$60^o$ more than its complement.

Then,

$x=(90^o-x)+60^o$

$\Rightarrow x+x=90^o+60^o$

$\Rightarrow 2x=150^o\\ \Rightarrow x=\dfrac { 150 }{ 2 } ^o\\ \Rightarrow x={ 75 }^{ o }$ .

Therefore, option

$B$ is correct.

In the given figure,

$AB || CD$ ,

$EH || BC$ ,

$\angle BAC = 60^{\circ}$ and

$\ \angle DGH = 40^{\circ}$ . Find the measures of

$\angle CFG$ .

0%
$60^o$
0%
$80^o$
0%
$56^o$
0%
none of the above

Two supplementary angles differ by

$48^o$ . Then find these angles.

0%
$36^o, 84^o$
0%
$46^o, 94^o$
0%
$56^o, 104^o$
0%
$66^o, 114^o$

Explanation

We know that sum of supplementary angles is

$180^o$ .

Let one angle be

$x$ and other be

$180^o - x$

Hence,

$x - (180^o- x) =48^o$ ....(Given)

$\Rightarrow x- 180^o+ x= 48^o$

$\Rightarrow 2x= 48^o+ 180^o= 228^o$

$\Rightarrow x= \dfrac { 228^o }{ 2 } = 114^o$

Hence, other angle

$= 180^o- x= 180^o- 114^o= 66^o$

Two angles are

$114^o$ and

$66^o$ .

The angle formed by the pages of an open book is:

0%
Acute
0%
Obtuse
0%
Right
0%
Straight

Explanation

The angle formed by the pages of an open book is obtuse.

Since the angle will always be more than

$90$ but less than

$180$ .

How many pairs of adjacent angles, in all, can you name in figure?

0%
8
0%
7
0%
12
0%
10

Explanation

The pairs of adjacent angles are:

1.

$\angle AOB$ and

$\angle BOC$

$\angle AOB$ and

$\angle BOD$

$\angle AOB$ and

$\angle BOE$

$\angle BOC$ and

$\angle COD$

$\angle BOC$ and

$\angle COE$

$\angle COD$ and

$\angle AOC$

$\angle COD$ and

$\angle DOE$

$\angle DOE$ and

$\angle BOD$

$\angle DOE$ and

$\angle AOD$

10.

$\angle COE$ and

$\angle AOC$

Hence there are 10 pairs of adjacent angles.

CBSE Questions for Class 9 Maths Lines And Angles Quiz 5 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Lines And Angles Quiz 5 - MCQExams.com

0:0:2

Practice Class 9 Maths Quiz Questions and Answers

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