MCQ Questions for Class 9 Maths Circles Quiz 8

The shaded portion of the circle represents

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Semicircle
0%
Diameter
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A sector of the circle
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Segment

In the given figure,

$ABCD$ is a cyclic quadrilateral whose side

$AB$ is a diameter of the circle passing through

$A, B, C$ and

$D$ . If

$\angle$ ADC

$=130^o$ , find

$\angle$ BAC.

0%
$45^o$
0%
$58^o$
0%
$60^o$
0%
$40^o$

Explanation

$\angle ACB=90^o$ (angle subtended by the diameter of circle)

and

$\angle ABC$ is the supplementary angle of

$\angle ADC$ ...(opposite angles of cyclic quadrilateral)

$\therefore \angle ABC=180-\angle ADC=180-130=50^o$ .

Now, by angle sum property,

$\angle BAC=180-(\angle ACB+\angle ABC)\\=180-50-90=40^o$ .

Hence, option

$D$ is correct.

Two chords

$AB$ and

$CD$ of lengths

$5\ cm$ and

$11\ cm$ respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between

$AB$ and

$CD$ is

$6\ cm$ , find the radius of the circle.

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$\dfrac{5\sqrt{5}}{2}$
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$\dfrac{11\sqrt{5}}{2}$
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$\dfrac{5\sqrt{5}}{3}$
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$\dfrac{5\sqrt{3}}{2}$

Explanation

We know

$AM = \dfrac{AB}{2} = \dfrac{5}{2} cm$

and

$CN = \dfrac{CD}{2} = \dfrac{11}{2} cm$

Let

$MOZ \times cm \Rightarrow ON = 6 - x cm$

using pythoqeros in

$\Delta AOM$ , we get

$AD^2 = AM^2 + MO^2 \Rightarrow R^2 = x^2 + \left(\dfrac{5}{2} \right)^2$ .... (1)

Using pythaqeros in

$\Delta CON_0$ we get

$CO^2 = ON^2 + CN^2 + R^2 = (6 - x^2) + \left(\dfrac{11}{2} \right)^2$ ... (2)

On subtracting (2) from (1) , we gert

$x^2 - (6 - x)^2 + \left(\dfrac{5}{2} \right)^2 - \left(\dfrac{11}{2} \right)^2 = 0$

$(x + 6 - x) (x - 6 + x) + \dfrac{25}{4} - \dfrac{121}{4} = 0$

$6 (2x - 6) = \dfrac{96}{4} = 4$

$2x = 10$

$x = 5 cm$

on putting value of x in equation (1) , we get

$R^2 = (5)^2 + \left(\dfrac{5}{2} \right)^2 = 25 + \dfrac{25}{4} = \dfrac{125}{4}$

$\Rightarrow R = \dfrac{5 \sqrt{5}}{2} cm$

1052352_878858_ans_255c6425ee184518b3ce288428ffca86.png

A circle (with centre at O) is touching two intersecting lines AX and BY. The two points of contact A and B subtend an angle of

$65^{\circ}$ at any point C on the circumference of the circle. If P is the point of intersection of the two lines, then the measure of

$\angle$ APB is

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$50^{\circ}$
0%
$65^{\circ}$
0%
$90^{\circ}$
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$40^{\circ}$

Explanation

$\angle AOB = 2\angle ACB$ (Angle subtended by an arc at the centre is double the angle subtended by it at the circumference)

$\implies \angle AOB=2 \times 65= 130^{\circ}$

In quadrilateral

$AOBP$

$\angle OAP + \angle OBP + \angle AOB + \angle APB=360^{\circ}$

$\implies 90^{\circ} + 90^{\circ} + \angle APB + 130^{\circ} = 360^{\circ}$

$\implies \angle APB = 360^{\circ} - 310^{\circ}= 50^{\circ}$
1150775_908551_ans_addbc3c7a5f4498dbb3439d84cb00c19.png

In the figure given below, if

$\angle AOP \, = \, 75^\circ$ and

$\angle AOB \, = \, 120^\circ$ , then what is

$\angle AQP$ ?

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$45^{\circ}$
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$37.5^{\circ}$
0%
$30^{\circ}$
0%
$22.5^{\circ}$

Explanation

$2\angle AQP \, = \angle AOP$ (Angle subtended by an arc at the centre is double the angle subtended by the same arc at any other point on the circle ).

$\implies \angle AQP=\dfrac{1}{2}\angle AOP$

$= \cfrac{75^\circ}{2} \, = \, 37.5^{\circ}$

$BD$ is a chord parallel to the diameter

$AC$ of a circle. A point

$B$ is on the perimeter of the circle such that angle

$CBE={ 63 }^{ o }$ . The angle

$DCE$ is equal to:

0%
${ 63 }^{ o }$
0%
${ 36 }^{ o }$
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${ 31.5 }^{ o }$
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${ 27 }^{ o }$

Explanation

$\angle ACE=\angle ABE=27°\\ (\because \angle ABE=27°\quad as\quad \angle ABC=90°)$

$\angle ACE=\angle CED$ (Alternate angles are equal)

$(\because AC||ED)$

BEDC is a cyclic quadrilateral

$\therefore \angle EBC+\angle EDC=180°\\ \therefore \angle EDC=117°\\ \therefore \angle ECD=180°-(117°+27°)=36°$

In a cyclic quadrilateral

$ABCD$ , twice the measure of

$\angle A$ is thrice the measure of

$\angle C$ , find the measure of

$\angle C$ .

0%
$36^{o}$
0%
$72^{o}$
0%
$90^{o}$
0%
$108^{o}$

Explanation

The sum of the opposite angles of a cyclic quadrilateral is supplementary.

So, in cyclic quadrilateral

$ABCD$

$\angle A+\angle C=180^{\circ}$ .............(1).

It is given that

$2\angle A=3\angle C$

$\implies$

$\angle A=\dfrac{3}{2}\angle C$ ............ (2).

From (1) and (2),

$\dfrac{3}{2}\angle C+\angle C=180^{\circ}$

$\implies$

$\dfrac{5}{2}\angle C=180^{\circ}$

$\implies$

$\angle C=\dfrac{180\times 2}{5}=72^{\circ}$ .

Hence, option

$B$ is correct.

Quadilateral ABCD is cyclic. If

$\angle B = 60^o$ , then

$\angle D =$ ____.

0%
$30^o$
0%
$100^o$
0%
$120^o$
0%
$90^o$

Explanation

In a cyclic quadrilateral, the sum of opposite angles

$={180}^{\circ}$ .

Given,

$ABCD$ is a cyclic quadrilateral.

Then,

$\angle{B}$ and

$\angle{D}$ are opposite angles

Hence

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

$\Rightarrow\,\angle{D}={180}^{\circ}-\angle{B}$

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

Therefore,

$\angle{D}={120}^{\circ}$ .

Thus, option

$C$ is correct.

In the given figure, AD=BC,

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

$\angle CBD={ 70 }^{ o }$ . Find

$\angle BCD$ .

0%
$80^0$
0%
$30^0$
0%
$70^0$
0%
None of these

Explanation

In figure,

$ABCD$ is cyclic quadrilateral.

Here,

$AC$ and

$BD$ are diagonals.

Also,

$\angle BAC={30}^{\circ}$ and

$\angle CBD={70}^{\circ }$ .

Now,

$\angle BAC=\angle BDC={30}^{\circ}$ .

Then,

$\angle BAD=\angle BDC+\angle CAD$

$={30}^{\circ}+{70}^{\circ}={100}^{\circ}$ .

Now,

$\angle BCD+\angle BAD={180}^{\circ}$ ...(opposite angles are cyclic quadrilateral)

$\Rightarrow \angle BCD+{100}^{\circ }={180}^{\circ}$

$\Rightarrow \angle BCD={180}^{\circ }-{100}^{\circ}$

$\Rightarrow \angle BCD={80}^{\circ}$ .

Therefore, option

$A$ is correct.

1111857_1102149_ans_8e8a9e021158432fa543b220e0205a56.png

A chord divides a circle into two

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Diameter
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Semicircle
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Sectors
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Segments

The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.

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True
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False

Explanation

${\textbf{Step-1: Drawing the diagram as per given question and proving the required statement}}$

${\text{Here, ABCD is a cyclic quadrilateral}}$

${\text{AH, BF, CF and DH are the angle bisectors of}}$ $\angle$ ${\text{A,}}$ $\angle$ ${\text{B,}}$ $\angle$ ${\text{C,}}$ $\angle$ ${\text{D.}}$

$\angle$ ${\text{FEH =}}$ $\angle$ ${\text{AEB ……. (1) [Vertically opposite angles]}}$

$\angle$ ${\text{FGH =}}$ $\angle$ ${\text{DGC.…… (2) [Vertically opposite angles]}}$

${\text{Adding (1) and (2),}}$

$\angle$ ${\text{FEH +}}$ $\angle$ ${\text{FGH =}}$ $\angle$ ${\text{AEB + DGC …..(3)}}$

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

$\angle$ ${\text{AEB =}}$ ${\text{18}}{{\text{0}}^{\text{0}}}{\text{ - }}\left( {\dfrac{{\text{1}}}{{\text{2}}}\angle {\text{A + }}\dfrac{{\text{1}}}{{\text{2}}}\angle {\text{B}}} \right)$ ${\text{…….(4)}}$

$\angle {\text{DGC = 18}}{{\text{0}}^{\text{0}}}{\text{ - }}\left( {\dfrac{{\text{1}}}{{\text{2}}}\angle {\text{C + }}\dfrac{{\text{1}}}{{\text{2}}}\angle {\text{D}}} \right)$ ${\text{………..(5)}}$

${\text{Substituting (4) and (5) in equation (3)}}$

$\angle {\text{FEH + }}\angle {\text{FGH = 18}}{{\text{0}}^{\text{0}}}{\text{ - }}\left( {\dfrac{{\text{1}}}{{\text{2}}}\angle {\text{A + }}\dfrac{{\text{1}}}{{\text{2}}}\angle {\text{B}}} \right){\text{ + 18}}{{\text{0}}^{\text{0}}}{\text{ - }}\left( {\dfrac{{\text{1}}}{{\text{2}}}\angle {\text{C + }}\dfrac{{\text{1}}}{{\text{2}}}\angle {\text{D}}} \right)$

$\angle {\text{FEH + }}\angle {\text{FGH = 36}}{{\text{0}}^{\text{0}}}{\text{ - }}\dfrac{1}{2}\left( {\angle {\text{A + }}\angle {\text{B + }}\angle {\text{C + }}\angle {\text{D}}} \right)$

$\angle {\text{FEH + }}\angle {\text{FGH = 36}}{{\text{0}}^{\text{0}}}{\text{ - }}\dfrac{1}{2} \times {360^0}$

$\angle {\text{FEH + }}\angle {\text{FGH = 18}}{{\text{0}}^0}$

${\text{Now, the sum of opposite angles of quadrilateral EFGH is}}$ ${\text{18}}{{\text{0}}^0}$

${\text{EFGH is a cyclic quadrilateral}}$

${\textbf{ Hence, the quadrilateral formed by angle bisectors of a cyclic quadrilateral}}$

${\textbf{is also cyclic.}}$

2167323_1162812_ans_c624d803ed544dd2b871622bb3a8f586.PNG

In the diagram above,

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

$x$ is

0%
$100$
0%
$140$
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$150$
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$160$

Explanation

Given

$O$ is centre so

$OR=OS$ (Radius)

$\angle ORS=\angle OSR$ (

$OR=OS$ ) So angle opposite to side will be equal

Now

$\angle POR+\angle OSR=180^0$ (cyclic quadrilateral)

$140+\angle OSR=180^0$

$\angle OSR=40^0$

$\angle ORS=40^0$

$\Delta OSR$

$x^0+\angle ORS+\angle OSR=180^0$

$x^0+80^0=180^0$

$\boxed{x^0=100^0}$

2124692_1161722_ans_7e4eabb4933a459f975951d9aa769a80.png

In the given figure, O is the centre of the circle .What is the value of

$x?$

0%
$115^o$
0%
$125^o$
0%
$155^o$
0%
$230^o$

Explanation

Given:

$O$ is the centre of the circle.

So,

$OP,OR,OQ$ are the radius of the circle.

So,

$OP=OR=OQ$

Now,

Since

$OP=OQ$

$\therefore \triangle OPQ$ is isosceles.

Hence

$\angle OPQ=\angle OQP={ 25 }^{ \circ }$

From angle sum property,

$\angle POQ={ 180 }^{ \circ }-(\angle OPQ+\angle OQP)$

$\\ ={ 180 }^{ \circ }-({ 25 }^{ \circ }+{ 25 }^{ \circ })\\ ={ 180 }^{ \circ }-50^{ \circ }\\ =130^{ \circ }$

Again,

$OQ=OR$

$\therefore \triangle OQR$ is isosceles.

Hence

$\angle ORQ=\angle OQR={ 30 }^{ \circ }$

By angle sum property

$\angle QOR={ 180 }^{ \circ }-(\angle ORQ+\angle OQR)$

$={ 180 }^{ \circ }-(30^{ \circ }+30^{ \circ })\\ ={ 180 }^{ \circ }-60^{ \circ }\\ =120^{ \circ }$

We know that, in a circle the central angle is twice of any inscribed angle subtended by the same arc.

Here,

$\angle POR$ is the central angle and

$\angle PSR$ is the inscribed angle subtended by the same arc.

$\therefore \angle POR=2\angle PSR$

$\Rightarrow \angle POQ+\angle QOR=2\angle PSR$

$\Rightarrow { 130 }^{ \circ }+{ 120 }^{ \circ }=2x$

$\Rightarrow 2x={ 250 }^{ \circ }$

$\Rightarrow x=125^{ \circ }$

If the perimeter and the area of a circle are numerically equal, then the radius of the circle is

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$2\ units$
0%
$p\ units$
0%
$4\ units$
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$7\ units$

Explanation

Given,
Numerically perimete of the circle is equal to area of circle

$\Rightarrow 2\pi r=\pi r^2$

$\Rightarrow 2=r$

$\therefore r=2$

Given: In a circle with chords

$AP = BP$

$arc\ AP=arc\ PB$

0%
True
0%
False

Explanation

Given,

$AP=BP$

from figure

$\Delta OAP$ and

$\Delta OBP$

$OP=OP$ ( Common)

$OA=OB$ (Radius of circle)

$AP=PB$ (given)

$\Delta OAP$ is congurent to

$\Delta OBP$

$\angle OAP=\angle OBP$

$\angle AOP=\angle BOP$ [By

$C.P.C.T.$ ]

$\angle AOP=\angle BOP$

$arc(AP)=arc(PB)$

The sum of the angles in the four segments exterior to a cyclic quadrilateral is equal to

$8$ right angles.

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True
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False

In the figure ,O is the center of the circle. The value of

$(2a + b + C)$ in

0%
${192^ \circ }$
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${180^ \circ }$
0%
${278^ \circ }$
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${266^ \circ }$

In the figure given,

$O$ is the center of the circle, if

$\angle PAB=108^{o}$ , then find

$\angle BCQ$ .

0%
$152^{\circ}$
0%
$120^{\circ}$
0%
$150^{\circ}$
0%
$162^{\circ}$
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None of these

In the given quadrilateral

$PQRS$ , if

$\angle RSP = {80}^{0}$ , then

$\angle RQT=$ ____.

0%
${100}^{0}$
0%
${80}^{0}$
0%
${70}^{0}$
0%
${110}^{0}$

Explanation

Given is a cyclic quadrilateral

$PQRS$ .

We know that,

In cyclic quadrilaterals sum of opposite angles is

$180^0$ .

So,

$\angle RSP + \angle PQR = 180^0$

$\Rightarrow 80^0 + \angle PQR = 180^0$

$\Rightarrow \angle PQR = 100^0$ .

Also, we know that sum of angles on a straight line is

$180^0$ ,

So,

$\angle PQR + \angle RQT = 180^0$

$\Rightarrow 100^0 + \angle RQT = 180^0$

$\Rightarrow \angle RQT = 80^0$ .

Hence, option

$B$ is the correct.

An arc subtends an angle of

$45^{o}$ in its alternate segment then the angle made by its corresponding major arc is at the centre

0%
$270^{o}$
0%
$90^{o}$
0%
$45^{o}$
0%
$315^{o}$

Explanation

Here, major arc

$AB$ is subtending

$45^{\circ}$ in the alternate (minor) segment

To find : angle

$x$

Soluton: As angle around a point is

$360^{\circ}$

Around

$O$ ,

$x + 45^o =360^o$

$\Rightarrow x = 315^{\circ}$

Look at the given figure having

$O$ as the center of the circle and find the value of

$a$ and

$b$ .

0%
$a=55^\circ$ & $b=40^\circ$
0%
$a=40^\circ$ & $b=55^\circ$
0%
$a=45^\circ$ & $b=55^\circ$
0%
$a=40^\circ$ & $b=50^\circ$

BC is the diameter of a circle. Points A and D are situated on the circumference of the semi circle

$\angle$ ABD

$=35^o$ and

$\angle$ BCD

$=60^o$ ,

$\angle$ ADB equal to?

0%
$20^o$
0%
$25^o$
0%
$30^o$
0%
$115^o$

In the given circle,

$AD=BC,\angle BAC={30}^{o}$ and

$\angle CBD={70}^{o}$
Find

$(i)\,\angle BCD$

$(ii)\,\angle BCA$

$(iii)\,\angle ABC$

$(iv)\,\angle ADB$

0%
$(i)\,50^\circ\,(ii)\,80^\circ\,(iii)\,50^\circ\,(iv)\,100^\circ$
0%
$(i)\,80^\circ\,(ii)\,50^\circ\,(iii)\,100^\circ\,(iv)\,50^\circ$
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$(i)\,80^\circ\,(ii)\,130^\circ\,(iii)\,100^\circ\,(iv)\,50^\circ$
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None of these

In the adjoining figure, in

$\triangle OAM \ and \ \triangle OBM$ ,

$OA = OB$

$AM = BM$

$OM = OM$

$\triangle OAM \ \cong \ \triangle OBM$
What can you conclude?

0%
$\angle OMA = \angle OMB \not= 90$
0%
$\angle OMA = \angle OMB = 90$
0%
$\angle AOB = 90$
0%
None of the above

In figure,

$O$ is the centre of the circle, then

0%
$\angle x= \angle y+ \angle z$
0%
$\angle y= \angle x+ \angle z$
0%
$\angle z= \angle x+ \angle y$
0%
$none\ of\ these$

Explanation

On the basis of given figure, we can conclude

$\angle 3=\angle 4$ and

$\angle x=2\angle 3$

$\Rightarrow \angle x=\angle 3+\angle 4$ .....(i)

$\angle y=\angle 3+\angle 1$ ........(ii)

$(i)-(ii)$ gives,

$\angle x-\angle y=\angle 4-\angle 1$ .......(iii)

$\angle 4=\angle z+\angle 1$

$\Rightarrow \angle 4-\angle 1=\angle z$ .........(iv)

From (iii) and (iv) we get,

$\angle x-\angle y=\angle z$

$\Rightarrow \angle z+\angle y=\angle x$

$\therefore \angle x=\angle y+\angle z$

1191591_1346532_ans_b16ef75f5e314ee8baa50601307568ab.png

$ABCD$ is a cyclic quadrilateral such that

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

$\angle DCA=80^{o}$ , then

$\angle DAB=$ ?

0%
$70^{o}$
0%
$100^{o}$
0%
$150^{o}$
0%
$125^{o}$

Explanation

Given,

$\angle ADB = 30^0$ and

$\angle DCA = 80^0$ .

Now,

$\angle DCA = \angle DBA = 80^0$ (Angle subtended by same chord are equal).

$\triangle ADB$ ,

$\angle ADB + \angle DBA + \angle DAB = 180^0$ (Sum of angles in triangle is

$180^0$ )

$\implies$

$\angle DAB = 180^0 - 80^0 - 30^0 = 70^0$ .

Hence, the measure of

$\angle DAB$ is

$70^0.$

Therefore, option

$A$ is correct.

1826852_1417805_ans_e2e198812eed4430b80c852ceb4dbeec.jpg

The geometric shape that has no corners is _____.

0%
square
0%
circle
0%
rectangle
0%
hexagon

State whether following statement is true or false:

The nearer a chord to the centre of a circle, the longer is its length.

0%
True
0%
False

Write True or False and justify your answer in each of the following :
ABCD is a cyclic quadrilateral such that

$\angle$ A =

$90^{\circ} , \angle$ B =

$70^{\circ} , \angle$ C =

$95^{\circ}$ and

$\angle$ D =

$105^{\circ}$ .

0%
True
0%
False

Explanation

We know that opposite angles of a cyclic quadrilateral are supplementary

But here, sum of opposite angles is not equal to

$180^{\circ}$ .

That is,

$\angle A$

$+\angle C$ =

$90^{\circ} + 95^{\circ} = 185^{\circ} \ne 180^o$ .

And,

$\angle B$

$+\angle D$ =

$70^{\circ} + 105^{\circ} = 175^{\circ} \ne 180^o$ .

Hence ,

$ABCD$ is not a cyclic quadrilateral. That is, given statement is false.

Hence, option

$B$ is correct.

Four alternative answers for the following question is given. Choose the correct alternative.
In a cyclic quadrilateral

$\,ABCD$ , twice the measure of

$\angle A$ is thrice the measure of

$\angle C$ . Find the measure of

$\angle C$ ?

0%
$36$
0%
$72$
0%
$90$
0%
$108$

Explanation

Given,

$ABCD$ is a cyclic quadrilateral.

Also,

$2\angle A=3\angle C \, \, \, \, \,.......(1)$ .

Now,

$\angle A+\angle C=180^{o}$ ...(Opposite angles of cyclic quadrilateral are supplementary)

$\Rightarrow \dfrac{3}{2}\angle C+\angle C=180^{o}$ [From

$(1)$ ]

$\Rightarrow \dfrac{5}{2} \angle C=180^{o}$

$\Rightarrow \angle C=\dfrac{2\times180^{o}}{5}=72^{o}$ .

Thus, the measure of

$\angle C$ is

$72^{o}$

Hence, option

$C$ is correct.

CBSE Questions for Class 9 Maths Circles Quiz 8 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Circles Quiz 8 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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