MCQ Questions for Class 9 Maths Circles Quiz 6

In the given figure, if

$O$ is centre, then external

$\angle AOC=$

0%
$190^{\circ}$
0%
$160^{\circ}$
0%
$200^{\circ}$
0%
$185^{\circ}$

Explanation

$In\quad the\quad given\quad figure\\ \angle ADC+\angle CDE=180^° (Linear\quad pair\quad of\quad angles)\\ \angle CDE=80^° (Given)\\ \angle ADC=180^°-\angle CDE\\ \quad \quad \quad \quad \quad =180^°-80^°=100^°\\ Also,\angle AOC=2\angle ADC\\ [\because Angle\quad at\quad centre\quad is\quad double\quad the\quad angle\quad on\quad remaining\quad part\quad of\quad circle]\\ Hence\angle AOC=2\times 100^°=200^°\\ Hence\quad option(C)\quad is\quad right\quad answer$

In the given figure,

$\angle DBC=22^{\circ}\;and\;\angle DCB=78^{\circ}$ then

$\angle BAC$ is equal to:

0%
$90^{\circ}$
0%
$80^{\circ}$
0%
$78^{\circ}$
0%
$22^{\circ}$

Explanation

Let

$\angle BAC = x$

$\therefore \angle BDC = x$ (angle subtended by chord on same segment are equal)

$\Delta BDC$ ,

$x+22^{\circ}+78^{\circ} = 180^{\circ}$

$x+100 = 180^{\circ}$

$\boxed {x = 80^{\circ}} = \angle BAC$

Ans (B)

In the given figure determine the value of x.

0%
$90^{\circ}$
0%
$115^{\circ}$
0%
$130^{\circ}$
0%
$65^{\circ}$

In the given figure,

$CDEF$ is a cyclic quadrilateral,

$DE\;$ and

$\;CF$ are produced to

$A\;$ and

$\;B$ respectively such that

$AB\;\parallel\;CD$ . If

$\angle FED=80^{\circ}$ , find

$\angle FBA$ .

0%
$30^{\circ}$
0%
$60^{\circ}$
0%
$80^{\circ}$
0%
$10^{\circ}$

Explanation

$CDEF\quad is\quad a\quad cyclic\quad quadrilateral.\\ \angle FED=80^o(Given)\\ \angle FED+\angle FCD=180^o ...(sum\quad of opposite\quad angles\quad of\quad cyclic\quad quadrilateral\quad is\quad 180^o)\\ \Rightarrow FCD=180^o-80^o=100^o.\\ Since\quad AB\parallel CD,\\ \Rightarrow ABCD\quad is\quad a\quad parallelogram.\\ \angle ABC+\angle BCD=180^o ...(Adjacent\quad angles\quad of\quad parallelogram.\quad are\quad supplementary)\\ \angle BCD=\angle FCD=100^o\\ \Rightarrow \angle ABC=180^o-100^o=80^o\\ \angle ABC=\angle FBA=80^o.\\ Hence,\quad option\quad C \quad is\quad correct.$

In figure,

$O$ is centre, then

$\angle BXD=$

0%
$65^{\circ}$
0%
$60^{\circ}$
0%
$70^{\circ}$
0%
$55^{\circ}$

Explanation

Given

$\angle AOC={ 95 }^{ o }$

$\angle ABC=\angle ADC=\cfrac { 95^o }{ 2 } \; \; (Half\; Angle)\\ \angle EBX=\angle EOX={ 180 }^{ o }-\cfrac { 95^o }{ 2 } =\cfrac { 265^o }{ 2 }$

In quadrilateral

$BEXD$

$\angle BED+\angle EBX+\angle BXD+\angle XDE=360^o \\ 25^o+\cfrac{265^o}{2}+\angle BXD+\cfrac{265^o}{2}=360^o \\ \angle BXD=360^o-265^o-25^o \\ \angle BXD=70^o$

In the figure,

$O$ is the center. If

$\angle MON=80^o$ , then

$\angle MQN$ equals

0%
$40^o$
0%
$160^o$
0%
$100^o$
0%
$10^o$

Explanation

Angle subtended by an arc at the center is double the angle subtended by the same are at any point on the circumference.

$\angle MON\quad =80^o\\ \angle MQN=\frac { 80^o }{ 2 } =40^o$

$PQ$ is a diameter and

$PQRS$ is a cyclic quadrilateral. If

$\angle PSR=150^o$ , then measure of

$\angle RPQ$ is:

0%
$90^{\circ}$
0%
$60^{\circ}$
0%
$30^{\circ}$
0%
None of these

Explanation

$PQRS$ is a cyclic quadrilateral.

Then,

$\angle PQR +\angle PSR= 180^{\circ}$ ...[Opposite angles of cyclic quadrilateral are supplementary]

$\implies$

$\angle PQR +150^o= 180^{\circ}$

$\implies$

$\angle PQR = 180^{\circ} - 150^{\circ} = 30^{\circ}$ .

$\triangle PQR$ ,

$\angle PRQ = 90^{\circ}$ (Angle of a semicircle)
Then,

$\angle RPQ + 90^{\circ} + 30^{\circ} = 180^{\circ}$ ...[Angle sum property]

$\Rightarrow \angle RPQ + 120^{\circ} = 180^{\circ}$

$\Rightarrow \angle RPQ = 60^{\circ}$ .

Hence, option

$B$ is correct.

$ABCD$ is a cyclic quadrilateral inscribed in a circle with the centre

$O$ . Then

$\angle OAD$ is equal to:

0%
$30^{\circ}$
0%
$40^{\circ}$
0%
$50^{\circ}$
0%
$60^{\circ}$

Explanation

$Given-\quad \\ ABCD\quad is\quad a\quad cyclic\quad quadrilateral\quad inscribed\quad in\quad a\\ circle\quad with\quad centre\quad O.\\ OA,\quad OB,\quad OC\quad \& \quad OD\quad have\quad been\quad joined.\\ \angle OAB={ 40 }^{ o },\quad \angle OBC={ 30 }^{ o }\quad \& \quad \angle OCD={ 50 }^{ o }.\\ To\quad find\quad out-\\ \angle OAD=?\\ Solution-\\ OC=OB\quad (radii\quad of\quad the\quad same\quad circle)\\ \therefore \quad \Delta OBC\quad is\quad an\quad isosceles\quad triangle.\\ \therefore \quad \angle OCB=\angle OBC={ 30 }^{ o }.\\ \therefore \quad \angle BCD=\angle OCD+\angle OCB={ 50 }^{ o }+{ 30 }^{ o }={ 80 }^{ o }.\\ Now,\quad ABCD\quad is\quad a\quad cyclic\quad quadrilateral.\\ So,\quad by\quad angle\quad sum\quad property\quad of\quad a\quad cyclic\quad quadrilateral,\\ we\quad get,\\ \angle BAD={ 180 }^{ o }-\angle BCD={ 180 }^{ o }-{ 80 }^{ o }={ 100 }^{ o }.\\ So,\quad \angle OAD=\angle BAD-\angle OAB={ 100 }^{ o }-{ 40 }^{ o }={ 60 }^{ o }.\\ Hence, \quad option\quad D \quad is \quad correct.$

$ABCD$ is a cyclic quadrilateral. Then, find

$\angle x^{\circ}$ as given in the figure.

0%
$50^{\circ}$
0%
$80^{\circ}$
0%
$90^{\circ}$
0%
$100^{\circ}$

Explanation

$\angle EDC+\angle ADC=180^o ...(Linear\quad pair\quad of\quad angles)\\ \angle EDC=80^o ...(Given)\\ Then, \angle EDC+\angle ADC=180^o\\ \angle ADC=180^o-\angle EDC=180^o-80^o=100^o.\\ ABCD\quad is\quad a\quad cyclic\quad quadrilateral.\\ Hence,\quad \angle ADC+\angle ABC=180^o ...(Sum\quad of\quad opposite\quad angles\quad is\quad 180^o)\\ \Rightarrow \angle ABC=180^o-\angle ADC\\ \quad \quad \quad \quad \quad \quad \quad =180^o-100^o\\ \quad \quad \quad \quad \quad \quad \quad =80^o.\\ Now, \angle ABF+\angle ABC=180^o\\ \angle ABF=180^o-\angle ABC\\ \angle ABF=\angle x ...(Given)\\ \Rightarrow \angle x=180^o-80^o\\ \Rightarrow \angle x=100^o.\\ Hence,\quad option \quad D \quad is\quad correct.$

In the given figure, the value of

$'a'$ is:

0%
$30^{\circ}$
0%
$40^{\circ}$
0%
$60^{\circ}$
0%
$90^{\circ}$

Explanation

Given-

$\overline { AOB }$ is a diameter of a given circle.

$ABCD$ is a cyclic quadrilateral.

$AC$ is joined. Also,

$\angle ADC={ 130 }^{ \circ }$ .

Since,

$\overline { AOB }$ is the diameter of the given circle, it subtends

$\angle ACB$ to the circumference at

$C$ , i.e.

$\angle ACB={ 90 }^{ \circ }$ ...[ since it is an angle in a semicircle].

Again,

$\angle ADC+\angle ABC=180^\circ$ ...[ sum of opposite angles of a cyclic quadrilateral is

${ 180 }^{ \circ }$ ]

$\Rightarrow \angle ABC={ 180 }^{ \circ }-\angle ADC$

$\Rightarrow \angle ABC={ 180 }^{ \circ }-130^{ \circ }$

$\Rightarrow\angle ABC={ 42 }^{ \circ}$ .

$\triangle ABC,$

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

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

$={ 180 }^{ \circ }-({ 90 }^{ \circ }+{ 50 }^{ \circ })$

$={ 40 }^{ \circ }$

Hence, option

$B$ is correct.

In the circle, reflex

$\angle AOC=x^o, \angle ABC=y^o$ .

$OABC$ is a parallelogram, then find

$y$ .

0%
$80^o$
0%
$120^o$
0%
$110^o$
0%
Data insufficient

Explanation

In the circle, clearly,

$x=2y$ .

Then,

$\angle A OC=360-x=360-2y$ .

Here,

$ABCD$ is a parallelogram

Then,

${BCO}+ {AOC}=180^o$

$\Rightarrow {BCO}+360-2y=180^o$

$\Rightarrow 2y=BCO+180^o$ .

Similarly,

$ABC+BCO=180^o$

$\Rightarrow y+2y-180=180$

$\Rightarrow 3y=360^o$

$\Rightarrow y=120^o$ .

Hence, option

$B$ is correct.

Find

$\angle ASR$ .

0%
$52^{\circ}$
0%
$78^{\circ}$
0%
$102^{\circ}$
0%
$10^{\circ}$

Explanation

$PABQ\quad is\quad a\quad cyclic\quad quadilateral\\ \angle QPA=78^o...(Given). \\ Then, \angle QPA+\angle ABQ=180^o\\ \Rightarrow 78^o+\angle ABQ=180^o\\ \Rightarrow \angle ABQ=180^o-78^o=102^o....(1).\\ \angle ABR\quad and\quad \angle ABQ\quad are\quad linear\quad pair\quad of\quad angles\quad on\quad straight\quad line\quad QR\\ Then, \angle ABQ+\angle ABR=180^o\\ \Rightarrow \angle ABR=180^o-\angle ABQ=180^o-102^o=78^o.\\ Hence,\quad \angle ABR\quad =78^o.\\ In\quad cyclic\quad quadilateral\quad ABCD,\quad \\ \angle ASR+\angle ABR=180^o\\ \Rightarrow \angle ASR=180^o-78^o\\ \Rightarrow \angle ASR=102^o.\\ Hence,\quad option\quad C \quad is\quad correct.$

In the given figure,

$PQRS$ is a cyclic quadrilateral. Its diagonals

$PR$ and

$QS$ intersect each other at

$T$ . If

$\angle PRS=80^{\circ}\;and\;\angle RQS=50^{\circ}$ , calculate

$\angle PSR$ .

0%
$30^{\circ}$
0%
$50^{\circ}$
0%
$70^{\circ}$
0%
$90^{\circ}$

Explanation

$Given:\\ PQRS\ is\ a\ cyclic\ quadilateral.\\ \angle PRS=80^o....(1)\\ \angle RQS=50^o\\ \angle RQS=\angle RPS\ (Angles\ in\ the\ same\ segment\ are\ equal)\\ \Rightarrow \angle RPS=50^o...(2)\\ In\quad \triangle PSR\\ \angle PRS+\angle RPS+\angle PSR=180^o\ (Angle\ sum\ property)\\ 80^o+50^o+\angle PSR=180^o\ (From\ (1)\ and\ (2))\\ \Rightarrow \angle PSR=180^o-130^o\\ \ \ \ \ \angle PSR=50^o\\ Hence\ option\ (B)\ is\ right\$

In figure,

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

$\angle BCA = 20^{\circ}$ , find

$\angle ADC$ .

0%
15 $^{\circ}$
0%
50 $^{\circ}$
0%
80 $^{\circ}$
0%
40 $^{\circ}$

Explanation

$\Delta ABC$ , we have

$\angle BAC + \angle BCA + \angle ABC = 180^{\circ}$ ....[Angle sum property]

$\Rightarrow 60^{\circ} + 20^{\circ} + \angle ABC = 180^o$

$\therefore \angle ABC = 180^{\circ} - 80^{\circ} = 100^{\circ}$ .

$ABCD$ is a cyclic quadrilateral,

$\therefore \angle ABC + \angle ADC = 180^{\circ}$ .....[Since, opposite angles of a cyclic quadrilateral are supplementary]

$\implies$

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

$\implies \angle ADC = 180^{\circ} - 100^{\circ} = 80^{\circ}$ .

Hence, option

$C$ is correct.

In the figure,

$\angle$ B is equal to:

0%
$85^0$
0%
$95^0$
0%
$70^0$
0%
$115^0$

Explanation

Since,

$APQD$ is a cyclic quadrilateral,

therefore

$\angle A + \angle DQP = 180^o$ ...(opposite angles of cyclic quadrilateral are supplementary)

$\implies$

$\angle DQP = 180^o -85^o =95^o$ .

Now,

$\angle PQC = 180^o - \angle DQP=180^o-95^o=85^o$ ...[linear pairs].

$\therefore \angle B = 180^o -85^o =95^o$ ...(opposite angles of cyclic quadrilateral are supplementary).

Hence, option

$B$ is correct.

In the figure above, if O is the centre of the circle, find the value of y.

0%
40 $^{\circ}$
0%
70 $^{\circ}$
0%
60 $^{\circ}$
0%
30 $^{\circ}$

Explanation

Given that, the centre of the circle is

$O$ and

$\angle BOD=140^\circ$

To find out: The value of

$y$

We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the circle.

$\therefore \ \angle BOD = 2 \angle BAD$

$\Rightarrow 140^{\circ} = 2 \angle BAD$

$\displaystyle \Rightarrow \angle BAD = \frac{1}{2} \times 140^{\circ} = 70^{\circ}$

Now, in cyclic quadrilateral ABCD,

$\angle BAD + \angle BCD = 180^{\circ}$ [Opposite angles of a cyclic quadrilateral are supplementary]

$\therefore \ 70^{\circ} + \angle BCD = 180^{\circ}$

$\therefore \angle BCD = 180^{\circ} - 70^{\circ} = 110^{\circ}$
Also,

$\angle BCD + \angle DCP = 180^{\circ}$ [Linear pair]

$\Rightarrow 110^{\circ} + y = 180^{\circ}$

$\therefore \ y = 70^{\circ}$

Hence, option B is correct.

In a cyclic quadrilateral

$ABDC,\,\angle CAB=80^{\circ}$ and

$\angle ABC=40^{\circ}$ . The measure of the

$\angle ADB$ will be?

0%
$120^{\circ}$
0%
$80^{\circ}$
0%
$60^{\circ}$
0%
$40^{\circ}$

Explanation

$\Delta ACB$ ,

$\because\;\angle ACB=180^{\circ}-(80^{\circ}+40^{\circ})=60^{\circ}$

$\therefore\;\angle ADB=\angle ACB$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=60^{\circ}$

In fig, a is the centre of the circle. If

$\angle$ BOD 160

$^{\circ}$ find the values of x and y.

0%
80 $^{\circ}$ , 100 $^{\circ}$
0%
135 $^{\circ}$ , 45 $^{\circ}$
0%
140 $^{\circ}$ , 40 $^{\circ}$
0%
120 $^{\circ}$ , 60 $^{\circ}$

Explanation

In the cyclic quadrilateral ABCD

$\displaystyle \angle BCD = \frac{1}{2} \angle BOD$
[Angle made by an arc at the centre is double the angle made by it, at any other point on the remaining part of the circle]

$\displaystyle \therefore \angle x = \frac{1}{2} \times 160^{\circ} = 80^{\circ}$

$\therefore x = 80^{\circ}$

$\angle x + \angle y = 180^{\circ}$
[opp. angles of a cyclic quadrilateral]

$\therefore \angle y = 180^{\circ} - \angle x$

$\angle y = 180^{\circ} - 80^{\circ} = 100^{\circ}$

$y = 100^{\circ}$

One of the base angle a cyclic trapezium is double the other. What is the measure of the larger angle?

0%
$60^0$
0%
$80^0$
0%
$75^0$
0%
$120^0$

Explanation

In cyclic trapezium,

Sum of opposite angle

$=180^o$ ...[opposite angles of a cyclic quadrilateral are supplementary].

Let the smaller angle be

$x$

and larger angle

$=2x$ .

Then,

$2x+x=180^o \implies 3x=180^o \implies x=60^o$ .

Therefore, Larger angle

$=2x=2\times 60^o=120^o$ .

Hence, option

$D$ is correct.

Find

$x.$

0%
$40^{\circ}$
0%
$50^{\circ}$
0%
$30^{\circ}$
0%
$60^{\circ}$

Explanation

ABCD is a quadrilateral inside the circle,

$\angle D + \angle B = 180^o$

$130^o + \angle B = 180^o$

$\angle B = 50^o$

AB is diameter

$\angle C = 90^o$ [angle in a semicircle]

$\Delta ABC$

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

$\angle A + 50^o + 90^o = 180^o$

$\Rightarrow \angle A + 140^o = 180^o$

$\Rightarrow \angle A = 180^o - 140^o$

$=40^o$

$\therefore \angle CAB = 40^o \Rightarrow x = 40^o$

option (A) is correct.

2063980_343716_ans_d6ea9bafda544c6592041cd822b30d20.png

ABCD is cyclic quadrilateral. The tangents to a circle at A and C meet at P. If

$\angle$ APC = 50

$^{\circ}$ , then the value of

$\angle$ ADC is:

0%
$\displaystyle 65^{\circ}$
0%
$\displaystyle 70^{\circ}$
0%
$\displaystyle 80^{\circ}$
0%
none of these

Explanation

Given

$\angle APC=50^0$

As the radius is always perpendicular to tangent at the point of contact,

$\angle OCP=\angle OAP=90^0$

$AOCP$ is a quadrilateral, sum of all its interior angles are equal to

$360^0$

$\therefore\ \angle AOC=130^0\quad \quad [360^0-(90^0+90^0+50^0)]$

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

$\therefore \ \angle ADC=\dfrac{130^0}{2}$

$=65^0$

Hence,

$\angle ADC=65^0$ .

$O$ is the centre of the circle and

$A, B$ and

$C$ are points on its circumference and

$\angle AOC = 130^{\circ}$ , find

$\angle ABC$ .

0%
$105^{\circ}$
0%
$115^{\circ}$
0%
$110^{\circ}$
0%
$120^{\circ}$

Explanation

Take any point

$P$ on the circumference of the circle as shown.

Join

$AP$ and

$CP$ .

$\because ABC$ subtends

$\angle AOC$ at centre

$O$ and

$\angle APC$ at any point

$P$ on the circumference of the circle.

$\therefore \angle AOC = 2 \angle APC$ [

$\because$ Angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the circumference]

$\displaystyle \implies \angle APC = \frac{1}{2} \angle AOC ~~~[ \because AOC = 130^{\circ}]$

$\implies$

$\displaystyle \frac{1}{2} \times 130^{\circ} = 65^{\circ}$ .

$\because$

$ABCP$ is a cycle quadrilateral,

$\implies \angle APC + \angle ABC = 180^{\circ}$ ...[

$\because$ sum of opposite angles of a cyclic quadrilateral is

$180^{\circ}$ ]

$\implies 65^{\circ} + \angle ABC = 180^{\circ}$

$\therefore \angle ABC = 180^{\circ} - 65^{\circ} = 115^{\circ}$ .

Hence, option

$B$ is correct.

$ABCD$ is a cyclic quadrilateral, in which

$BC$ is parallel to

$AD$ ,

$\displaystyle\angle ADC=\displaystyle 110^{\circ}$ and

$\displaystyle \angle BAC=50^{\circ}$ . What is the value of

$\displaystyle \angle DAC$ ?

0%
$\displaystyle 40^{\circ}$
0%
$\displaystyle 50^{\circ}$
0%
$\displaystyle 60^{\circ}$
0%
$\displaystyle 64^{\circ}$

Explanation

$ABCD$ is a cyclic quadrilateral.

Then,

$\angle ABC+\angle ADC=180^o$ ...[Opposite angles of a cyclic quadrilateral are supplementary]

$\Rightarrow$

$\angle ABC =180^{\circ}-110^{\circ} =70^{\circ}$ .

$\triangle ABC$ ,

$\angle ABC+\angle ACB+\angle BAC=180^o$ ...[Angle sum property]

$\Rightarrow$

$70^o+\angle ACB+50^o=180^o$

$\Rightarrow$

$\angle ACB=180^{\circ}-(50^{\circ}+70^{\circ})=60^{\circ}$ .

Given,

$BC$ and

$AD$ are parallel,

$\Rightarrow \angle ACB=\angle DAC =60^{\circ}$ ...[Alternate interior angles].

Hence, option

$C$ is correct.

In the given figure, AB=AC and

$\displaystyle \angle ABC=68^{\circ},$ what is the value of

$\displaystyle \angle BPC$ ?

0%
$\displaystyle 40^{\circ}$
0%
$\displaystyle 42^{\circ}$
0%
$\displaystyle 44^{\circ}$
0%
$\displaystyle 48^{\circ}$

Explanation

Theorem: Chord of a circle subtends equal angles at different points on the circumference of the circle on same side

$\Rightarrow \angle BAP=\angle BPC$

$AB=AC\Rightarrow \angle B=\angle C=68^{°}$ (Opposite angles of equal sides are equal)

$\triangle ABC$

$\angle A+ \angle B + \angle C =180^o$ (Sum of angles in a triangle )

$68^o+68^o+ \angle A = 180^o$

$136^o+\angle A = 180^o$

$\Rightarrow \angle A=44^{°}$

$\Rightarrow \angle A=\angle BAC =\angle BPC =44^{°}$ (Angles in same segment are equal )

In the given figure,

$BD=DC$ and

$\displaystyle \angle DBC=30^{\circ}$ what is the value of

$\displaystyle \angle BAC$ ?

0%
$\displaystyle 40^{\circ}$
0%
$\displaystyle 50^{\circ}$
0%
$\displaystyle 60^{\circ}$
0%
$\displaystyle 65^{\circ}$

Explanation

$BD=DC$ ...... (given)

$\Rightarrow \angle DBC=\angle DCB=30^{\circ}$

$\Rightarrow \angle BDC =180^\circ -30^\circ-30^\circ=120^{\circ}$

$ABCD$ is a cyclic quadrilateral ;

$\angle BAC+\angle BDC =180^{\circ}$

$\Rightarrow \angle BAC =180^{\circ}-120^{\circ} =60^{\circ}$

What is the value of

$\displaystyle \angle ABC$ ?

0%
$\displaystyle 105^{o}$
0%
$\displaystyle 110^{o}$
0%
$\displaystyle 115^{o}$
0%
$\displaystyle 120^{o}$

Explanation

Extending A and C onto the circumference to D as shown in the figure

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

$\Rightarrow \angle AOC=2\times \angle ADC$

$\Rightarrow \angle ADC=\dfrac {130}{2}=65^o$

$ABCD$ is a cyclic quadrilateral, sum of its opposite interior angles are equal to

$180^o$

$\Rightarrow \angle ADC+\angle CBA =180^o$

$\Rightarrow \angle ABC =180^o - 65^o=115^o$

ABCD is a cyclic trapezium and

$\displaystyle AB\parallel CD$ . If

$AB$ is the diameter of the circle and

$\displaystyle \angle CAB=30^{\circ}$ , then the value of

$\displaystyle \angle ADC$ is:

0%
$\displaystyle 110^{\circ}$
0%
$\displaystyle 115^{\circ}$
0%
$\displaystyle 120^{\circ}$
0%
$\displaystyle 125^{\circ}$

Explanation

Here,

$\angle CAB=30^o$ ...[Given].

Also,

$\angle ACB=90^o$ ...[Angle inscribed in a semi-circle].

Now, in

$\triangle ACB$ ,

$\angle ACB+\angle CAB+\angle ABC=180^o$ ...[Angle sum property]

$\Rightarrow$

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

$\Rightarrow$

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

$\Rightarrow$

$\angle ABC=60^o$ .

Here,

$ABCD$ is a cyclic quadrilateral.

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

$180^o$ .

$\therefore$

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

$\Rightarrow$

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

$\Rightarrow$

$\angle ADC=180^o-60^o$

$\Rightarrow$

$\angle ADC=120^o.$

Hence, option

$D$ is correct.

Find the value of x from the given figure in which O is the center of the circle

0%
$\displaystyle x=50^{\circ}$
0%
$\displaystyle x=60^{\circ}$
0%
$\displaystyle x=80^{\circ}$
0%
$\displaystyle x=90^{\circ}$

Explanation

$\textbf{Step - 1 : Find the angle at the center,}$

$\text{Given that, angle subtend by segment BC at point A on circle}=40^{\circ}$

$\text{Then, the angle subtend by segment BC at center O}=\angle{\text{BOC}}=40^{\circ}\times2=80^{\circ}$

$\textbf{Step - 2 : Find the value of x}$

$\text{In triangle OBC, OB = OC, as the radii of the same circle.}$

$\text{Hence, }\angle\text{OBC}=\angle\text{OCB}=x, \text{as the angles opposite to equal sides are also equal}$

$\text{Then by angle sum property of triangle,}$

$\angle\text{OBC}+\angle\text{OCB}+\angle\text{BOC}=180^{\circ}\Rightarrow80^{\circ}+x+x=180^{\circ}\Rightarrow2x=180^\circ-80^\circ=100^\circ$

$\therefore{x}=\dfrac{100^\circ}{2}=50^\circ$

$\textbf{Therefore, option A }\mathbf{x=50^\circ}\textbf{ is correct answer.}$

Find the values of

$y$ and

$z$ , where

$O$ denotes the centre of the circle.

0%
$y=70^o$ and $z=35^o$
0%
$y=290^o$ and $z=145^o$
0%
$y=145^o$ and $z=290^o$
0%
$y=35^o$ and $z=70^o$

Explanation

Since the given quadrilateral is a cyclic quadrilateral,

$\angle y + 35^o =180^o$

$\Rightarrow \angle y=180^o-35^o$

$\Rightarrow \angle y=145^o$ .

By the inscribed angle theorem, we have,

$\angle z=2\times \angle y$

$\Rightarrow \angle z=2\times145^o$

$\Rightarrow \angle z = 290^o$ .

Hence, option

$C$ is correct.

In given figure, BC is a diameter of the circle and

$\displaystyle \angle BAO=60^{\circ}\,$ . Then

$\,\angle ADC$ is equal to

0%
$\displaystyle 30^{\circ}$
0%
$\displaystyle 45^{\circ}$
0%
$\displaystyle 60^{\circ}$
0%
$\displaystyle 120^{\circ}$

Explanation

$\triangle AOB$ ,

$OA=OB$

$=>\angle OAB=\angle OBA=60^{0}$

We know that the exterior angle of a triangle is equal to sum of the two opposite angles of the triangle.

$\angle AOC=\angle OAB+\angle OBA$

$\angle AOC=60^{0}+60^{0}$

$\angle AOC = 120^{0}$

We know that the angle subtended at the centre of the circle by an arch is twice the angle subtended at the circumference by the same arc.

$\therefore \angle ADC=\dfrac{1}{2}\angle AOC$

$=\dfrac{120}{2}$

$=60^{0}$

CBSE Questions for Class 9 Maths Circles Quiz 6 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Circles Quiz 6 - MCQExams.com

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

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