MCQ Questions for Class 9 Maths Circles Quiz 7

In given figure, if

$\displaystyle \angle OAB$ =

$\displaystyle 40^{\circ},$ then

$\displaystyle \angle ACB$ is equal to:

0%
$\displaystyle 50^{\circ}$
0%
$\displaystyle 40^{\circ}$
0%
$\displaystyle 60^{\circ}$
0%
$\displaystyle 70^{\circ}$

Explanation

In the circle

$OA=OB =$ Radius

So,

$\angle OAB=\angle OBA=40^°$

Therefore,

$\angle AOB=180^°-40^°-40^°=100^°$ {Angle sum property}

We know in a circle the angle subtended by an arc at the center is twice that of the angle made on the circle.

Therefore,

$\angle ACB=\dfrac{\angle AOB}{2}=50^°$

In the figure above AQ = CT and

$\displaystyle \angle QAC=70^{\circ}$ Find

$\displaystyle \angle ACT$

0%
$\displaystyle 70^{\circ}$
0%
$\displaystyle 80^{\circ}$
0%
$\displaystyle 90^{\circ}$
0%
$\displaystyle 110^{\circ}$

Explanation

Given that:

$AQ=CT, \angle QAC=70^\circ$

To find:

$\angle ACT=?$

Solution:

If two opposite sides of a cyclic quadrilateral are equal then the other two opposite sides are parallel to each other .

So, if

$AQ=CT$ then

$QT\parallel AC$

$\angle QAC+\angle QTC=180^\circ$ (Sum of opposite angles of cyclic quadrilateral is

$180^\circ.$ )

or,

$\angle QTC=180^\circ-70^\circ=110^\circ$

Now,

$\angle ACT+\angle QTC=180^\circ$ (Sum of interior angles on the same side of the transversal line is

$180^\circ.$ )

or,

$\angle ACT=180^\circ-110^\circ$

or,

$\angle ACT=70^\circ$

Hence, A is the correct option.

In a cyclic trapezium ABCD,

$\angle{A}=100^\circ$ . Then,

$\angle{D}$ is equal to:

0%
$80^\circ$
0%
$90^\circ$
0%
$100^\circ$
0%
$110^\circ$

Explanation

Consider given trapezium $ABCD$ .

Given, $\angle A={{100}^{\circ}}$ .

Now,

$\angle A=\angle B$ ....[Since, $DC$ is the diameter of the circle]

$\implies$ $\angle B={{100}^{\circ}}$ .

Since, trapezium $ABCD$ is a cyclic quadrilateral,

Sum of opposite angle of trapezium $={{180}^{\circ}}$ ....[Opposite angles of cyclic quadrilateral are supplementary].

$\implies \angle B+\angle D={{180}^{\circ}}$

${{100}^{\circ}}+\angle D={{180}^{\circ}}$

$\angle D={{80}^{\circ}}$ .

Hence, option

$A$ is correct.

1190521_390664_ans_cb7ffb5717114a538aa370df86266c8e.png

For a

$\triangle PQR$ with points

$P, Q$ and

$R$ lying on a circle, if

$RP$ is the diameter of the circle and

$PQ = 5\ cm$ and

$QR = 12\ cm$ . Find radius of the circle.

0%
$6.5\ cm$
0%
$6\ cm$
0%
$8\ cm$
0%
$5.5\ cm$

Explanation

Because angle in a semicircle is always $90^{\circ},$ so

So $\angle RQP = 90^{\circ}$ and $RP$ is the diameter

Thus by pythagorean theorem we have

$RP^2 = RQ^2 + QP^2$
$= 12^2 + 5^2 = 169$

$RP = 13 = d = 2r$

$r = 6.5\ cm$

For a triangle ABC, with BC as the diameter of circle, if radius is 5 cm and AB = 8 cm. Find AC .

0%
6 cm
0%
3 cm
0%
4 cm
0%
8 cm

Explanation

Given,

$\triangle ABC$ is in a semicircle and BC is the diameter, therefore

$\angle BAC=90^o$

So, it is a right angled triangle, therefore

$BA^2+AC^2=BC^2$

$\implies AC^2=10^2-8^2=36$

$\implies AC=6$

Find the values of the x and y, where O denotes the centre of the circle.

0%
$x = 55^°, y = 155^°$
0%
$x = 50^°, y = 155^°$
0%
$x = 155^°, y = 55^°$
0%
None of these

Find the values of

$x, y$ and

$z$ , where

$O$ denotes the centre of the circle.

0%
$x=30^o, y=90^o$ and $z=135^o$
0%
$x=45^o, y=90^o$ and $z=270^o$
0%
$x=30^o, y=60^o$ and $z=270^o$
0%
$x=45^o, y=90^o$ and $z=345^o$

Explanation

Since the given quadrilateral is cyclic quadrilateral, we have,

$3x + x = 180^\circ$

$\Rightarrow 4x =180^\circ$

$\Rightarrow x=45^\circ$ .

By the inscribed angle theorem, we have,

$y = 2\times x$

$\Rightarrow y=90^\circ$ .

We know that,

$y + z= 360^\circ$

$\Rightarrow z=360^\circ- y$

$\Rightarrow z= 360^\circ-90^\circ$

$\Rightarrow z=270^\circ$

Hence, option

$B$ is correct.

Find the values of

$x$ and

$y$ .

0%
$x-y=180^o$
0%
$x+y=180^o$
0%
$x=121^o$ and $y=68^o$
0%
$x=68^o$ and $y=121^o$

Explanation

Since the given quadrilateral is a cyclic quadrilateral, we have,

$112^o + x^o = 180^o$ ...[Opposite angles of cyclic quadrilateral are supplementary]

$\Rightarrow x = 68^o$ .

Similarly,

$59^o + y^o = 180^o$ ...[Opposite angles of cyclic quadrilateral are supplementary]

$\Rightarrow y = 121^o$ .

Hence, option

$D$ is correct.

In the given figure,

$ST$ is a tangent to the circle with centre

$O$ , at

$S$ .
Find the value of

$x$ .

0%
$10^o$
0%
$40^o$
0%
$45^o$
0%
$55^o$

Given a circle with the center indicated, find

$x.$

0%
$100^°$
0%
$80^°$
0%
$50^°$
0%
$40^°$

Explanation

We know that the angle at the centre is twice is the angle at the circumference subtended by the same arc.

Here angle at the circumference is

$x$

Let the triangle be

$ABC$ .

Thus the angle at the center

$B$ is

$2x$ .

Since

$AB$ and

$BC$ are the radius of the circle, they are of the same length.

So,

$\angle \; A = \angle \; C = 50^{\circ}$

Now,

$2x+50^°+50^°=180^°$

$\Rightarrow 2x=180^°-100^°$

$\Rightarrow 2x=80^°$

$\Rightarrow x=40^°$

$ABCD$ is a cyclic quadrilateral with

$AD$ parallel to

$BC$ .

$\angle DCB = 65^o$ . The magnitude of

$\angle CBE$ is:

0%
$100^o$
0%
$110^o$
0%
$115^o$
0%
$120^o$

Explanation

Given, $AD \parallel BC$ .

Since, $ABCD$ is a cyclic quadrilateral.

Then, $\angle OCB + \angle DAB = 180^\circ$ ...[Opposite angles of cyclic quadrilateral are supplementary]

$\implies \angle DAB = 115^\circ$ .

Since, $AD \parallel BC$

$\implies$ $\angle DAB = \angle CBE = 115^\circ$ ...[Corresponding angles].

Hence, option $C$ is correct.

The portion of a circle between two radii and an arc is called

0%
Sector
0%
Segment
0%
Chord
0%
Secant

Find

$PM$ if

$PQ=8 \ cm$

0%
$3 cm$
0%
$4 cm$
0%
$5 cm$
0%
$8 cm$

Explanation

The perpendicular from centre to the chord bisects the chord.

$PM=\dfrac{1}{2}\times8=4$

So, option B is correct.

Find the value of

$x$ and

$y$ .

0%
$x=150^o, y=120^o$
0%
$x=120^o, y=150^o$
0%
$x=100^o, y=130^o$
0%
$x=130^o, y=100^o$

Explanation

The given quadrilateral is a cyclic quadrilateral.
We know, Opposite angles of cyclic quadrilateral are supplementary.

Therefore,

$\angle x +60^o = 180^o$

$\implies$

$\angle x = 180 - 60 = 120^o$ .

Similarly,

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

$\implies$

$\angle y = 180 - 30 = 150^o$ .

Hence,

$x = 120^o$ and

$y=150^o$ .

Therefore, option

$B$ is correct.

Find the measure of

$\widehat{AOB}$ .

0%
$100^\circ$
0%
$104^\circ$
0%
$102^\circ$
0%
$103^\circ$

Explanation

Given,

$\angle ABC=38^\circ,$

$\angle BCD=12^\circ$ and

$BC$ is diameter

Now,

$\angle BAC=\angle BDC=90^\circ$ (triangle with diameter as a base)

$\triangle ABC$

$\angle ACB+\angle BAC+\angle ABC=180^\circ$

$\angle ACB=180^\circ-90^\circ-38^\circ=52^\circ$

Also,

$\angle AOB=2\angle ACB$ (half angle theorem)

$\Rightarrow \angle AOB=2\times 52^\circ=104^\circ$

Angle

$ANB =$ Angle

$AOB =$ Angle

$AMB = ?$

0%
$45^o$
0%
$55^o$
0%
$180^o$
0%
$90^o$

Explanation

We know that all angles inscribed in a semicircle are right angles.

So,

$\angle ANB =$

$\angle AOB =$

$\angle AMB =$

$90^\circ$

In the figure given above, what is

$\angle BYX$ equal to?

0%
$85^{\circ}$
0%
$50^{\circ}$
0%
$45^{\circ}$
0%
$90^{\circ}$

Explanation

We know that angles subtended by an arc on the circumference of the circle on the same segment are equal.

$\angle XBY = \angle XPY = 45^{\circ}$

$\Rightarrow BYX = 180^{\circ} - (50^{\circ} + 45^{\circ}) = 85^{\circ}$ .

In the given figure,

$O$ is the centre of the circle and

$\angle QPR = x^{\circ}; \angle ORQ = y^{\circ}$ . Which statement is true about

$x^{\circ}$ and

$y^{\circ}$ ?

0%
$x^{\circ} + y^{\circ} = 120^{\circ}$
0%
$x^{\circ} + y^{\circ} = 180^{\circ}$
0%
$x^{\circ} + y^{\circ} = 90^{\circ}$
0%
$x^{\circ} + y^{\circ} = 150^{\circ}$

Explanation

$\triangle OQR \quad OR=OQ=$ radius

$\Rightarrow \angle OQR=\angle ORQ=y^o$

Thus by angle sum property, we have

$\angle OQR+\angle ORQ+\angle QOR=180^o \\ 2\angle ORQ+\angle QOR=180^o \\ \angle QOR=180-2\angle ORQ=180-2x^o$

Also,

$\angle QOR=2\angle QPR$ (half angle property)

$180-2x=2y \\ 180=2x+2y \\ x^o+y^o=90^o$

$A,B,C,D$ are four distinct points on a circle whose centre is at

$O$ .
If

$\angle OBD-\angle CDB=\angle CBD-\angle ODB$ , then what is

$\angle A$ equal to ?

0%
${45}^{o}$
0%
${60}^{o}$
0%
${120}^{o}$
0%
${135}^{o}$

What is the relation between '

$p$ ' and '

$q$ '?

0%
$q = 2p$
0%
$p = 2q$
0%
$p = q$
0%
$q > p$

Explanation

$ABCD$ is a cyclic quadrilateral. Hence,

$\angle AOC=180^{\circ}-p$

At point

$O$ there is a complete angle hence,

$\begin{aligned}{}\angle AOC + q& = {360^\circ }\\q& = {360^\circ } - \angle AOC\\q& = {360^\circ } - ({180^\circ } - p)\\q& = {180^\circ } + p\\q& > p\end{aligned}$

Quadrilateral

$ABCD$ and

$ECBA$ are inscribed in a circle. If

$\angle B={ 125 }^{ \circ },$ find the measure of

$\angle E.$

0%
${ 55 }^{ \circ }$
0%
${ 125 }^{ \circ }$
0%
${ 130 }^{ \circ }$
0%
${ 62.5 }^{ \circ }$

Explanation

In a cyclic quadrilateral, we know, opposite angles are supplementary.

Since $AECB$ is a cyclic quadrilateral,

$\angle E + \angle B = 180^\circ$

$\angle E = 180^\circ - \angle B$

$= 180^\circ – 125^\circ$

$= 55^\circ$

Hence, option $A$ is correct.

Given that

$BC=CD$ , what is the value of

$x$ ?

0%
${ 50 }^{ o }$
0%
${ 55 }^{ o }$
0%
${ 60 }^{ o }$
0%
${ 65 }^{ o }$

Explanation

In $\triangle DBC$

Given that $CD = CB$ ,

$\therefore \angle CDB = \angle CBD$ .

Also, $\angle DCB + \angle DBC + \angle CDB = 180^o$ ...[Angle sum property]

$2\angle CDB = 180^o – 50^o$

$\angle CDB = 65^o$ .

In the first circle, $AEBD$ is a cyclic quadrilateral.

Then, opposite angles are supplementary.

$\angle EAB + \angle EDB = 180^o$

$\angle EAB + 180^o - \angle CDB = 180^o$

$\angle EAB = \angle CDB = 65^o$ .

Hence, option $D$ is correct.

In the given figure, if

$\angle AOB = 125^{\circ}$ find the measure of

$\angle COD$

0%
$62.5^{\circ}$
0%
$45^{\circ}$
0%
$35^{\circ}$
0%
$55^{\circ}$

Explanation

Given

$\angle AOB=125^o \angle AOD=90^o$ and

$\angle BOC=90^o, \angle O=360^o$

$360=\angle AOB+\angle DOA+\angle COD+\angle COB$

$\Rightarrow 360=125+90+\angle COD+90$

$\therefore \angle COD = 55^o$

If one angle of cyclic quadrilateral is

$70^o$ , then the angle opposite to it is:

0%
$20^o$
0%
$110^o$
0%
$140^o$
0%
$160^o$

Explanation

In the cyclic quadrilateral,

angles $A + C = 180^{\circ}$ , and angles $B + D = 180^{\circ}$ ......(opposite angles of acyclic quadrilaterals are supplimentry).

So, if one of the angle is $70^{\circ}$ .

Then, angle opposite to it is $= 180^{\circ} - 70^{\circ}=110^o$ .

Hence, option $B$ is correct.

In a cyclic quadrilateral

$ABCD$ ,

$\angle A=5x, \angle C=4x$ , the value of

$x$ is:

0%
$12^o$
0%
$20^o$
0%
$48^o$
0%
$36^o$

Explanation

Opposite angles of a cyclic quadrilateral are supplementary

Then, $\angle A + \angle C = 5x+4x = 180^{\circ}$

$\implies$ $9x = 180 ^{\circ}$

$x= 20^{\circ}$ .

Hence, option $B$ is correct.

Draw a circle with centre O and radius

$2.5$ cm. Draw a chord

$AB$ passing through its centre. The

$\angle AOB$ is:

0%
$180^o$
0%
$270^o$
0%
$90^o$
0%
$0^o$

Explanation

Chord AB passing through the centre of a circle .

Therefore it is the diameter of a circle and diameter makes

$180^{\circ}$ at the centre of a circle

So ,

$\angle{AOB}=180^{\circ}$

option A is the answer.

In the given figure,

$\Delta$ ABC is an isosceles triangle with AB

$=$ AC and

$\angle$ ABC

$=50^o$ . Find

$\angle$ BDC and

$\angle$ BEC respectively.

0%
$100^o, 60^o$
0%
$50^o, 120^o$
0%
$60^o, 80^o$
0%
$80^o, 100^o$

Explanation

$\triangle ABC$ is an isosceles triangle.

$\therefore \angle B=\angle C=50^o$

$\angle BAC=180-50-50=80^o$

$\angle BAC=\angle BDC=80^o$ (angles on circle by the same chord on the same segment are equal)

$\angle BEC$ is supplementary angle of

$\angle BAC$

$\therefore \angle BEC=180-\angle BAC\\=180-80=100^o$

In the given figure,

$O$ is the centre of the circle, Reflex

$\angle AOB=260^\circ$ .

$C$ is a point on the minor arc

$AB$ of the circle. Find

$\angle ACB$ .

0%
$100^\circ$
0%
$150^\circ$
0%
$80^\circ$
0%
$130^\circ$

Explanation

We know, the angle at the center of a circle is twice the angle at the circumference subtended by the same arc. Hence,

$\angle ACB=\dfrac{1}{2}\times reflex(\angle AOB)$

$=\dfrac{260}{2}$

$=130^\circ$

The sum of pairs of opposite angles of a cyclic quadrilateral is:

0%
${360}^{o}$
0%
${90}^{o}$
0%
${180}^{o}$
0%
${60}^{o}$

Explanation

Sum of opposite angles of a cyclic quadrilateral is

$180 ^{o}$

Let

$ABCD$ is cyclic quadrilateral.
To prove that

$\angle A+\angle C=180°$ and

$\angle B+\angle D=180°$

Construction: Join

$OB$ and

$OD$ .

We know that

$\angle BOD=2\angle BAD$ ...(The angle at the centre is twice the angle at the circumference )

$\implies$

$\angle BAD=\dfrac { 1 }{ 2 } \angle BOD$ .

Similarly, we can write

$\angle BCD=\frac { 1 }{ 2 } \angle BOD$ .

Now, add above two equations,

$\angle BAD+\angle BCD=\dfrac { 1 }{ 2 } \angle BOD+\dfrac { 1 }{ 2 } \angle DOB$

$\implies$

$\angle BAD+\angle BCD=\dfrac { 1 }{ 2 } (\angle BOD+\angle DOB)$ ...(complete angle)

$\implies$

$\angle BAD+\angle BCD=\dfrac { 1 }{ 2 } \times 360°=180°$ .

So we get,

$\angle A+\angle C=180°$

And similarly

$\angle B+\angle D=180°$ .

Hence,option

$C$ is correct.

In the adjacent figure, if

$\angle AOC = 110^0$ , then the value of

$\angle D\ and \angle B$ respectively.

0%
$55^0, 125^0$
0%
$55^0, 110^0$
0%
$110^0, 25^0$
0%
$125^0, 55^0$

Explanation

Given,

$\angle AOC = 110^0$
Also, we know

$\angle AOC = 2 \angle ADC$ (angle subtended by an arc at the center is double the angle subtended by it at any point on the circumference)

$\therefore \angle ADC = 55^0$

Also,

$\angle B + \angle D = 180^0$ (ABCD is a cyclic quadrilateral and opposite angles of it are supplementary)

$\implies \angle B = 125^0$

CBSE Questions for Class 9 Maths Circles Quiz 7 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Circles Quiz 7 - MCQExams.com

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

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