MCQ Questions for Class 9 Maths Circles Quiz 3

In the figure,

$O$ is the center and

$\angle ACB=20^°$ . Then the

$\angle AOB$ is equal to

0%
$40^°$
0%
$60^°$
0%
$20^°$
0%
$30^°$

Explanation

Theorem: Angle subtended by an arc at the centre is double the angle subtended by the same arc at the circumference of a circle

Thus,

$\angle AOB=2\times 20=40^°$

In the figure,

$O$ is the center, then

$\angle ACD$ equals

0%
$20^°$
0%
$60^°$
0%
$45^°$
0%
$30^°$

Explanation

Since,

$\angle AOD=90^°\\ \angle ACD\quad is\quad substended\quad by\quad the\quad same\quad arc.\quad Therefore\\ \angle ACD=\dfrac{\angle AOD}{2}= \frac { 90 }{ 2 } =45^°$

What is name of following shape ?

0%
Cylinder
0%
Cone
0%
Circle
0%
Cube

From the figure, identify the center of the circle.

0%
$A$
0%
$D$
0%
$B$
0%
$C$

Explanation

Center of a circle is a point inside the circle and is at an equal distance from all of the points on its circumference.

In the given figure,

$C$ is such point.
So,

$C$ is the center of the given circle.

Hence, option

$D$ is correct.

In the given figure,

$AB$ is the diameter of a circle with

$O$ and

$AT$ is a tangent. If

$\angle AOQ = 58^{\circ}$ , then the value of

$\angle ATQ$ is _________.

0%
$52^{\circ}$
0%
$61^{\circ}$
0%
$46^{\circ}$
0%
$75^{\circ}$

Explanation

Angle subtended by an arc(here,

$AQ$ ) at the centre is twice the angle subtended by it on the circumference on same side of the chord(here,

$AQ$ )

Given,

$\angle AOQ = 58^{\circ}$

So,

$\angle ABT = \angle ABQ= 29^{\circ}$

Since, radius of circle is perpendicular to the tangent at the point of contact,

$\angle BAT = 90^{\circ}$

$\triangle ABT$ ,

$\angle BAT + \angle ABT + \angle BTA = 180^{\circ}$

$\Rightarrow 90^{\circ} + 29^{\circ} + \angle BTA = 180^{\circ}$

$\Rightarrow \angle BTA = 61^{\circ}$

$\angle ATQ = \angle BTA = 61^{\circ}$

So, Option (B)

Identify the center of the circle.

0%
$A$
0%
$O$
0%
$D$
0%
$C$

Explanation

Center of a circle is a point inside the circle and is at an equal distance from all of the points on its circumference.

In the given figure,

$O$ is such point.
So,

$O$ is the center of the given circle.

Hence, option

$B$ is correct.

Given circle with center

$C$ shows an inscribed quadrilateral. Find

$y$ .

0%
$100^o$
0%
$80^o$
0%
$60^o$
0%
$40^o$

Explanation

Here, the given quadrilateral is cyclic.

We know, the sum of opposite angles of a cyclic quadrilateral is

$180^o$ .

$\therefore y^o+100^o=180^o$
Thus,

$y^o=180^o-100^o=80^o$ .

Hence, option

$B$ is correct.

Two concentric circles intersect at _____ number of points.

0%
$2$
0%
$0$
0%
$1$
0%
None of these

In the given figure,

$O$ is the centre of the circle. Find the value of

$\angle BDC$ .

0%
$30^{o}$
0%
$45^{o}$
0%
$60^{o}$
0%
$75^{o}$

Explanation

In the given figure,

$\angle BDC=\angle BAC$

$\Rightarrow \angle BDC=45^{\circ}$ (Angle subtended by an arc in the same segment are equal.)

In the given circle ABCD, O is the centre and

$\angle BDC$ =

$42^0$ , The

$\angle ACB$ is equal to

0%
$42^0$
0%
$45^0$
0%
$48^0$
0%
$60^0$

Explanation

In the given figure,

$\angle BDC={ 42 }^{ o }$

$\angle BAC={ 42 }^{ o }$ (Angles subtended by the same arc in same segment of a circle are equal )

Now, since

$\angle ABC={ 90 }^{ o }$ (angle in semicircle is a right angle)

By angle sum property,

$\angle BAC+ \angle ABC + \angle ACB={ 180}^{o}$

$42^{o}+90^{o}+\angle ACB=180^{o}$

$\Rightarrow \angle ACB={ 48 }^{ o }$

In the given figure, chord

$BE=BD$ ,

$\angle CBD=33^{o}$ and

$OB \bot AC$ , then

$x+y-z$ is equal to:

0%
$230^{o}$
0%
$237^{o}$
0%
$337^{o}$
0%
None of these

Explanation

$\Delta BED$ ,

$BE=BD$ (given)

$\angle BED=\angle BDE$ (opposite angles are equals).

Given,

$x=\angle BDE$ .

Also,

$\angle OBC=90^{\circ}$ (given)

$\angle OBD+\angle CBD=90^{\circ}$

$\implies$

$\angle OBD+33^{\circ}=90^{\circ}$ ....[Since

$\angle CBD=33^o$ ]

$\implies$

$\angle OBD=90^o-33^o=57 ^{\circ}$ .

Now,

$x+z=180^{\circ}$ (opposite angles of a cyclic quadrilateral)

$\implies$

$z=180^{\circ}-x$ ----- (i).

$\Delta BED$ ,

$\angle BED + \angle BDE + \angle EBD=180^{\circ}$ (By angle sum property of triangle)

$\implies$

$x+y+57^{\circ}=180^{\circ}$

$\implies$

$x+y+57^{\circ}=180^{\circ}-x$

$\implies$

$x+y+57^{\circ}=z$ .....(from (i))

$\implies$

$x+y-z=-57^{\circ}$ .

Hence, option

$D$ is correct.

"The quadrilateral is cyclic if the sum of opposite angles

${180}^{o}$ ". The statement is ____.

0%
True
0%
False

Explanation

We know, the sum of each pair of opposite angles of a cyclic quadrilateral is

$180^o$ .

Hence, the given statement is true and option

$A$ is correct.

In the above figure,

$\overline {AB}$ and

$\overline {AC}$ are equal chords and

$O$ is the centre. If

$\angle BOC=100^{o}$ , then find

$\angle ACO$ .

0%
$10^{o}$
0%
$20^{o}$
0%
$30^{o}$
0%
$25^{o}$

Explanation

Consider the diagram shown in the question.

Given:

$AB=AC$

$\angle BOC=100{}^\circ$

$\Rightarrow \angle BAC=50{}^\circ$ (angle at the circle is half of angle at the centre by same chord $\left( BC \right)$ )

Consider the quadrilateral $OBAC$ .

$\angle BOC$ (of the quadrilateral) $=360{}^\circ -100{}^\circ =260{}^\circ$

Consider triangle $OBC$ .

$\angle OBC=\angle OCB$ ( $OB$ and $OC$ are radius)

Similarly, in $\Delta ABC$ ,

$\angle ABC=\angle ACB$ (Since, $AB=AC$ )

$\angle ABO+\angle OBC=\angle ACO+\angle OCB$

$\Rightarrow \angle ABO=\angle ACO$

We know that interior angles of a quadrilateral are $360{}^\circ$ . Therefore,

$50{}^\circ +260{}^\circ +\angle ABO+\angle ACO=360{}^\circ$

$2\angle ACO=50{}^\circ$

$\angle ACO=25{}^\circ$

Hence, this is the required result.

In figure, the value of

$x$ is

0%
$60^{\circ}$
0%
$15^{\circ}$
0%
$30^{\circ}$
0%
$None\ of\ these$

Explanation

$\text{Theorem: Angles in the same segments are equal}$

$\therefore x^{\circ}=30^{\circ}$

In the given figure

$\displaystyle \angle AOB=80^{\circ}$ The value of x is

0%
$\displaystyle 10^{\circ}$
0%
$\displaystyle 25^{\circ}$
0%
$\displaystyle 40^{\circ}$
0%
$\displaystyle 160^{\circ}$

Explanation

We have,

$\displaystyle \angle AOB=2\angle ACB$
[ Angle of the centre is twice at th angle of circumference \right ]

$\displaystyle 80^{\circ}=2x$

$\displaystyle \therefore x=\frac{80}{2}=40^{\circ}.$

In figure

$A,B$ and

$C$ are three points on a circle with centre

$O$ such that

$\angle{BOC}={30}^{o}$ and

$\angle {AOB}={60}^{o}$ . If

$D$ is a point on the circle other than the arc

$ABC$ , find

$\angle {ADC}$ .

0%
$30^\circ$
0%
$90^\circ$
0%
$45^\circ$
0%
$60^\circ$

Explanation

According to theorem of circles

$\angle AOC = 2\angle ADC$ .....(i)

$\angle AOC=\angle AOB+\angle BOC = 60^{\circ}+30^{\circ}=90^{\circ}$

from equation(i)

$90^{\circ}=2\angle ADC$

$\angle ADC=45^{\circ}$

1053056_1157801_ans_ee5dd6896d9b401eb6ba1774f43f8438.png

In the figure,

$O$ is the centre of the circle. If

$\angle {ADC}={140}^{o}$ , then what is the value of

$x$ ?

0%
${35}^{o}$
0%
${55}^{o}$
0%
${60}^{o}$
0%
${50}^{o}$

Explanation

$OA = OC \Rightarrow \angle OAC = \angle ACO = X$

$\angle ADC+\angle AMC = 180^{\circ}$ (cyclic quadrilateral )

$\angle AMC = 180^{\circ} -140^{\circ} = 40^{\circ}$

$\angle ADC = 2\angle AMC = 2\times 40^{\circ} = 80^{\circ}$

$\triangle AOC$

$\angle ADC+\angle OAC+\angle OCA = 180^{\circ}$

$x+80^{\circ}+x = 180^{\circ}$

$2x = 100^{\circ}$

$x = 50^{\circ}$

State true or false:

Chords of same length subtends equal angles on the same circle.

0%
True
0%
False

We can draw a circle using given three points if given points are

0%
collinear
0%
non collinear
0%
no condition on points
0%
none of these

The perpendicular distance between the biggest chord and the centre is

0%
Zero
0%
Equal
0%
$9 cm$
0%
$10 cm$

Is it possible to draw a circle with given two points?

0%
Yes
0%
No
0%
Can't say anything
0%
None of these

$AB = 12$ cm,

$BC = 16$ cm and

$AB$ is perpendicular to

$BC$ , then the radius of the circle passing through the points

$A$ ,

$B$ and

$C$ is:

0%
$6\ cm$
0%
$8\ cm$
0%
$10\ cm$
0%
$12\ cm$

Explanation

Given- $A, B$ & $C$ are points such that $AB=12cm$ and $BC=16cm$ and $BC\bot AB$ .

To find out- the radius of the circle passing through $A, B$ & $C=?$

Solution- $BC\bot AB$ i.e $\angle ABC={ 90 }^{ 0 }$ and $\Delta ABC$ is a right one with $AC$ as hypotenuse. $\therefore AC=\sqrt { { AB }^{ 2 }+{ BC }^{ 2 } } =\sqrt { { 12 }^{ 2 }+{ 16 }^{ 2 } } cm=20cm$ . So the circle passing through $A, B$ & $C$ will have its diameter as $AC$ .

$\therefore$ Its radius $=\dfrac { 1 }{ 2 } \times 20cm=10cm.$

Ans- Option C.

$AD$ is a diameter of a circle and

$AB$ is a chord. If

$AD = 34\> cm, AB = 30\ cm$ , the distance of

$AB$ from the centre of the circle is

0%
$17\ cm$
0%
$15\ cm$
0%
$4\ cm$
0%
$8\ cm$

Explanation

It is given that, Diameter, $AD = 34$ cm and chord, $AB = 30$ cm

Draw a perpendicular $ON$ from $O$ to $AB$ . It meets $AB$ at $N$ .

$\therefore$ Radius, $AO=\dfrac { 1 }{ 2 } AD=\dfrac { 1 }{ 2 } \times 34$ cm $=17$ cm.

$\because ON\bot AB$

$\therefore ON$ bisects $AB$ at $N$ and $\angle ANO=90^0$

So, $AN=\dfrac { 1 }{ 2 }AB=\dfrac { 1 }{ 2 } \times 30$ cm $=15$ cm

Now, $AO=17$ cm, $AN=15$ cm and $\angle ANO=90^0$

$\therefore \Delta AON$ is a right one with hypotenuse $AO.$

So, by pythagoras theorem, we have

${ AO }^{ 2 }-{ AN }^{ 2 }={ ON }^{ 2 }\Rightarrow { ON }^{ 2 }={ (17 }^{ 2 }{ -15 }^{ 2 })\ { cm }^{ 2 }$

$\Rightarrow ON=8$ cm

So, $AB$ is at a distance of $8$ cm from the centre $O$ .

If chord

$AOB$ is diameter of the circle and

$AC = BC$ , then

$\angle$

$ACB$ is equal to:

0%
$30$
0%
$60$
0%
$90$
0%
$45$

Explanation

$\angle ACB = 90^o$ as it is inscribed in a semi circle.

So option C is the correct answer.

Two chords

$AB$ and

$AC$ of a circle subtends angles equal to

${ 90 }^{ o }$ and

${ 150 }^{ o }$ , respectively at the centre. Find

$\angle BAC$ , if

$AB$ and

$AC$ lie on the opposite sides of the centre.

0%
${ 120 }^{ o }$
0%
${ 240 }^{ o }$
0%
${ 60 }^{ o }$
0%
${ 30 }^{ o }$

Explanation

Given-

$AB$ and

$AC$ are the chords of a circle with centre

$O$ .

$\angle AOB={ 90 }^{ o }$ and

$\angle AOC={ 150 }^{ o }$ .

To find out-

$\angle BAC=?$

Solution-

$BC$ is joined.

Now

$\angle BOC={ 360 }^{ o }-\left( \angle AOB+\angle AOC \right) = { 360 }^{ o }-({ 90 }^{ o }+{ 150 }^{ o })={ 120 }^{ o }$ .

So,

$\angle BAC=\dfrac { 1 }{ 2 } \angle BOC=\dfrac { 1 }{ 2 } \times { 120 }^{ o }={ 60 }^{ o }$ ....(The angle at the centre, subtended by a chord of a circle is double of that a the circumference.)

State true or false:

A circle of radius

$3$ cm can be drawn through two points

$A, B$ such that

$AB = 6$ cm.

0%
False
0%
True
0%
Cannot be determined
0%
None of the above

Explanation

We know, diameter of a circle

$=2\times$ radius of the circle.

A circle with

$AB=6cm$ as diameter will have its radius equal to

$3cm$ .

So option

$B$ is the right answer.

$ABCD$ is a cyclic quadrilateral such that

$A={ 90 }^{ o }, B={ 70 }^{ o }, C={ 95 }^{ o }$ and

$D={ 105 }^{ o }.$

The statement is:

0%
true.
0%
false.
0%
true in special condition.
0%
Data insufficeint

Explanation

Given,

$ABCD$ is a cyclic quadrilateral where

$\angle A={ 90 }^{ o } , \angle B={ 70 }^{ o }, \angle C={ 95 }^{ o }$ &

$\angle D={ 105 }^{ o }.$

Since

$ABCD$ is a cyclic quadrilateral,

then

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

$\angle B+\angle D={ 180 }^{ o }$ ......

$(i)$ (The sum of the opposite angles of a cyclic quadrilateral is

${ 180 }^{ o })$

But here,

$\angle A+\angle C={ 90 }^{ o }+95^{ o }=185^{ o }$ and

$\angle B+\angle D={ 70 }^{ o }+{ 10 }5^{ o }=175^{ o }.$

This is a contradiction to

$(i)$ .

That is, the statement is false.

Hence, option

$B$ is correct.

$ABCD$ is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and if

$\angle ADC={ 140 }^{ o }$ , then

$\angle BAC$ is equal to:

0%
${ 60 }^{ o\quad }$
0%
${ 50 }^{ o\quad }$
0%
${ 40 }^{ o\quad }$
0%
${ 30 }^{ o\quad }$

Explanation

Given:

$ABCD$ is a cyclic quadrilateral with

$AB$ as the diameter of the circle.

Also,

$\angle ADC={ 140 }^{ o }.$

We have to find:

$\angle BAC$ .

Then,

$\angle ADC+\angle ABC={ 180 }^{ o }$ ....(since the sum of the opposite angles of a quadrilateral is

${ 180 }^{ o }$ ).

$\therefore \angle ABC={ 180 }^{ o }-{ 140 }^{ o }={ 40 }^{ o }.$

Also,

$\angle ACB={ 90 }^{ o }$ ....(since the angle subtended by a diameter at the circumference of the circle, is

${ 90 }^{ o }$ ).

$\therefore$ In

$\Delta ABC$ we have,

$\angle ACB={ 90 }^{ o }$ and

$\angle ABC={ 40 }^{ o }.$

So,

$\angle CAB={ 180 }^{ o }-(\angle ACB+\angle ABC)$

$={ 180 }^{ o }-({ 90 }^{ o }+{ 40 }^{ o })$

$\therefore \angle BAC ={ 50 }^{ o }$ .

Hence, option

$B$ is correct.

If arcs

$AXB$ and

$CYD$ of a circle are congruent, then the ratio of

$AB$ and

$CD$ is

0%
$1:1$
0%
$2:1$
0%
$1:2$
0%
$0$

Explanation

$Given-\quad \\ AXB\quad \& \quad CYD\quad are\quad congruent\quad arcs\quad of\quad a\quad circle.\\ to\quad find\quad out-\\ the\quad ratio\quad AB\quad :\quad CD=?\\ Solution-\\ We\quad join\quad AB\quad \& \quad CD.\quad \\ AXB\quad \& \quad CYD\quad are\quad congruent\quad arcs\quad of\quad the\quad circle.\\ \therefore \quad Their\quad repective\quad chords\quad will\quad be\quad equal.\\ So\quad AB=CD.\\ \therefore AB:CD=1:1.\\ Ans-\quad Option\quad A.\\$

In figure if

$\angle AOB={ 90 }^{ o }\ and\ \angle ABC={ 30 }^{ o }$ then angle CAO is equal to

0%
${ 30 }^{ o }$
0%
${ 60 }^{ o }$
0%
${ 90 }^{ o }$
0%
${ 45 }^{ o }$

Explanation

$AB\quad is\quad the\quad chord\quad of\quad a\quad circle\quad with\quad centre\quad O.\\ AB\quad subtends\quad \angle AOB={ 90 }^{ o }\quad to\quad the\quad centre\quad and\quad \angle ACB\quad to\quad \\ the\quad circumference.\\ \angle ABC={ 30 }^{ o }.\\ To\quad find\quad out\quad -\\ \angle CAO=?\\ Solution-\\ AB\quad subtends\quad \angle AOB={ 90 }^{ o }\quad to\quad the\quad centre\quad and\quad \angle ACB\quad to\quad \\ the\quad circumference.\\ \therefore \quad \angle ACB=\frac { 1 }{ 2 } \angle AOB=\frac { 1 }{ 2 } \times { 90 }^{ o }=45^{ o }.(The\quad angle,\quad subtended\quad \\ by\quad a\quad chord\quad to\quad the\quad circumference\quad of\quad a\quad circle,\quad is\quad half\quad of\quad \\ that\quad subtended\quad to\quad the\quad centre).\\ Now\quad in\quad \Delta ACB\quad \quad we\quad have\\ \angle ACB+\angle ABC\quad =45^{ o }+30^{ o }=75^{ o }.\\ \therefore \quad \angle BAC=\quad 180^{ o }-(\angle ACB+\angle ABC)=\quad 180^{ o }-75^{ o }=105^{ o }.\\ (angle\quad sum\quad property\quad of\quad triangles)\\ Now\quad in\quad \Delta AOB\quad we\quad have\quad AO=BO\quad (radii\quad of\quad the\quad same\quad circle).\\ \therefore \quad \angle OAB=\angle OBA\quad i.e\quad \angle OAB+\angle OBA=2\angle OAB.\\ So\angle AOB+\angle OAB+\angle OBA=\angle AOB+2\angle OAB=180^{ o }.\\ (angle\quad sum\quad property\quad of\quad triangles).\\ i.e\quad { 90 }^{ o }+2\angle OAB=180^{ o }\Longrightarrow \angle OAB={ 45 }^{ o }.\\ \therefore \quad \angle CAO=\angle BAC-\angle OAB=105^{ o }-45^{ o }=60^{ o }.\\ Ans-\quad Option\quad B.\\$

CBSE Questions for Class 9 Maths Circles Quiz 3 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Circles Quiz 3 - MCQExams.com

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

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