MCQ Questions for Class 9 Maths Circles Quiz 1

In the above figure,

$\angle POR$ is (there

$O$ is centre of Circle)

0%
$60^\circ$
0%
$80^\circ$
0%
$100^\circ$
0%
$160^\circ$

Explanation

$\angle PQR=50^o+30^o=80^o$

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

$\therefore \angle POR = 2 \times 80 = 160^°$

What is the sum of

$\angle{BAD}+\angle{BPR}+\angle{BCD}+\angle{BQR}$ in the above figure?

0%
$540^\circ$
0%
$360^\circ$
0%
$240^\circ$
0%
None of the above

Explanation

In cyclic quadrilateral

$BADC$ ,

$\angle BAD +\angle BCD = 180^{ 0 }$ .....[opposite angles of a cyclic quadrilateral are supplementary]

In cyclic quadrilateral

$BPRQ$ ,

$\angle BPR+\angle BQR=180^0$ ...[opposite angles of a cyclic quadrilateral are supplementary]

Hence the sum of

$\angle{BAD}+\angle{BPR}+\angle{BCD}+\angle{BQR}$

$=180^o + 180^o = 360^o$ .

Therefore, option

$B$ is correct.

In the given figure, the value of

$x$ and

$y$ is:

0%
$20^{\circ}$ , $30^{\circ}$
0%
$36^{\circ}$ , $60^{\circ}$
0%
$15^{\circ}$ , $30^{\circ}$
0%
$25^{\circ}$ , $30^{\circ}$

Explanation

In a cyclic quadrilateral, the opposite angles are supplementary.

That is,

$2x+3x = 180^o$

$\Rightarrow 5x=180^o$

$\Rightarrow x = 36^o$

Similarly,

$y+2y = 180^o$

$\Rightarrow 3y=180^o$

$\Rightarrow y = 60^o$ .

Hence, option

$(B)$ is correct.

In the given figure,

$O$ is the centre of the circle,

$\angle ACB=54^{\circ}$ and

$BCE$ is a straight line. Find

$x.$

0%
$126^{\circ}$
0%
$54^{\circ}$
0%
$108^{\circ}$
0%
$90^{\circ}$

Explanation

$\angle ACE=180^{\circ}-54^{\circ}=126^{\circ}$ (

$BCE$ is a straight line)

$\Rightarrow$

$\angle ADE=\angle ACE=126^{\circ}$ (Angles in the same segement are equal)

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

$\therefore$ Reflex

$\angle AOE=2\times \angle ADE=2\times 126^{\circ}=252^{\circ}$

$\therefore$

$x=360^{\circ}-252^{\circ}=108^{\circ}$

A set of points equidistant from a fixed point in a plane figure is called

0%
triangle
0%
circle
0%
square
0%
none of these

In the given figure,

$\angle M = 75^{\circ}$ , then

$\angle O$ is equal to:

0%
$105^{\circ}$
0%
$100^{\circ}$
0%
$85^{\circ}$
0%
$110^{\circ}$

Explanation

A cyclic quadrilateral is a quadrilateral whose vertices all lie on the circumference of a circle.

We know, opposite angles of a cyclic quadrilateral add up to

$180^o$ .

In the given cyclic quadrilateral,

$\angle{O }+\angle{ M}=180^o$ and

$\angle{L }+\angle{ N}=180^o$ ...[Opposite angles of cyclic quadrilateral].

Then,

$\angle{O }+\angle{ M}=180^o$

$\implies$

$\angle{O }+75^o=180^o$

$\implies$

$\angle{O}$

$=180^o-75^o={105}^{\circ}$ .

Therefore, option

$A$ is correct.

Which of the following is a cyclic quadrilateral?

0%
Rhombus
0%
Rectangle
0%
Parallelogram
0%
Trapezium

Explanation

A cyclic quadrilateral is a quadrilateral with 4 vertices on the circumference of the circle.

Rectangle and Trapezium are cyclic quadrilaterals because their vertices lies on the circumference of a circle as shown in figure .

Since parallelogram and trapezium does not have that property that its opposite angles should add upto

$180^o$ , therefore they are not cyclic quadrilateral.

Hence, options

$B$ and

$D$ are correct.

If the length of the largest chord of a circle is

$17$ cm, find the radius of a circle.

0%
$34$ cm
0%
$8.5$ cm
0%
$\sqrt { 17 }$ cm
0%
$\sqrt { 34 }$ cm

Explanation

Given, the largest chord of a circle

$=17cm$ .
We know that the largest chord of a circle is its diameter.

$\therefore$ The diameter of the given circle

$=17 cm$ .

Hence, radius of the same circle

$=\dfrac { \text {diameter} }{ 2 } =\dfrac { 17 }{ 2 } cm=8.5cm$ .
Therefore, option

$B$ is correct.

In fig. O is the centre of the circle. Determine

$\angle$ AEC

0%
50 $^{\circ}$
0%
68 $^{\circ}$
0%
22 $^{\circ}$
0%
55 $^{\circ}$

Explanation

ABCD is a cyclic quadrilateral in which side AB is produced to D.

Also we know that exterior

$\angle$ of a cyclic quadrilateral is equal to the interior opposite angle

$\therefore \angle CBD = \angle AEC$
But

$\angle CBD = 68^{\circ}$

$\therefore \angle AEC = 68^{\circ}$

In the figure shown above, the radius

$OA$ is equal to the chord

$AB$ . Then, what is

$\angle APB$ ?

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

Explanation

$OAB$ is an equilateral triangle(all sides are equal). So

$\angle AOB = 60^{\circ}$

$\Rightarrow \angle APB = \dfrac {1}{2} \angle AOB = 60^{\circ} \div 2 = 30^{\circ}$ .

In the adjoining figure,

$18 in$ is the length of

0%
Radius
0%
Diameter
0%
Any chord other than the diameter
0%
None of these

How many circles can be drawn using given three non-collinear points?

0%
0
0%
1
0%
6
0%
Infinitely

Given that

$A(1,0), B(3,6), C(2,1).$ Can we draw a circle using points

$A,B$ and

$C.$

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

How many circles pass through one given point?

0%
1
0%
2
0%
more than 2
0%
Infinitely many.

Consider a circle with center O and radius = 24cm. AB = 20cm and CD = 14cm are two chords of the circle. Which chord is farther away from the centre?

0%
AB
0%
CD
0%
Both are equidistant from the centre
0%
None of these.

How many points are sufficient to determine a circle?

0%
1
0%
2
0%
3
0%
None of the above.

The centres of a set of circle,

each of radius $3$ , lies on the circle $x^2 + y^2 = 25$ . The locus of any

point in the set is

0%
$4\leq x^2+y^2\leq 64$
0%
$x^2+y^2\leq 25$
0%
$x^2+y^2\geq 25$
0%
$3\leq x^2+y^2\leq 9$

$ABCD$ is a cyclic quadrilateral such that

$AB$ is a diameter of the circle circumscribing it and

$\displaystyle \angle ADC=140^{\circ}$ , then

$\displaystyle \angle BAC$ is equal to:

0%
$\displaystyle 80^{\circ}$
0%
$\displaystyle 50^{\circ}$
0%
$\displaystyle 40^{\circ}$
0%
$\displaystyle 30^{\circ}$

Explanation

Since

$ABCD$ is a cyclic quadrilateral,

the sum of the opposite angles will be

$180$ degrees.

Thus,

$∠CBA+∠ADC={ 180 }^{ ∘ }$

$\Rightarrow$

$∠CBA={ 180 }^{ ∘ }-∠ADC$

$\Rightarrow$

$∠CBA={ 180 }^{ ∘ }-{ 140 }^{ ∘ }$

$\Rightarrow$

$∠CBA={ 40 }^{ ∘ }$ .

Also, angle in a semicircle in a right angle.

Therefore,

$∠ACB={ 90 }^{ ∘ }$

Now, in

$\triangle ACB$ ,

$∠CBA+∠ACB+∠BAC={ 180 }^{ ∘ }$ ...[angle property of a triangle]

$\Rightarrow$

${ 40 }^{ ∘ }+{ 90 }^{ ∘ }+∠BAC={ 180 }^{ ∘ }$

$\Rightarrow$

$∠BAC={ 180 }^{ ∘ }-{ 130 }^{ ∘ }$

$\Rightarrow$

$∠BAC=50^{ ∘ }$ .

Hence, option

$B$ is correct.

The perpendicular from the centre of the circle _____ the chord.

0%
bisects
0%
trisects
0%
divides in different ratios
0%
None

Any cyclic parallelogram is a ______.

0%
rectangle
0%
rhombus
0%
trapezium
0%
square

Explanation

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

$180^o$ .

The given cyclic quadrilateral is a parallelogram.

Hence, opposite pairs of angles will be congruent.

$\angle 1 + \angle 3 = 180^o$ and

$\angle 2 + \angle 4 = 180^o$

$\Rightarrow 2 \times \angle 1 = 180^o$ and

$2 \times \angle 2 = 180^o$

$\Rightarrow \angle 1 = \angle 2 = 90^o$

Hence,

$\angle 3= \angle 4 = 90^o$

Therefore, the given parallelogram has all its angles equal to

$90^o$ i.e. a rectangle.

Hence, option

$A$ is correct.

An arc is a part of a circle.

0%
True
0%
False

In the given figure, ABCD is a cyclic quadrilateral and PQ is tangent to the circle with centre 'O'. BD is diameter,

$\angle DCQ = 40^o , \angle ABD = 60^o$ . The

$\angle BCP=$ _______.

0%
$50^o$
0%
$100^o$
0%
$60^o$
0%
$20^o$

In the given figure, O is centre of the circle, then

$\angle XOZ$ is ________.

0%
$2 \angle X$
0%
$2 \angle Y$
0%
$2 \angle Z$
0%
$2[ \angle XYZ + \angle YXZ]$

In the diagram, O is the centre of the circle. Find the value of

$x$ .

0%
$111^o$
0%
$123^o$
0%
$69^o$
0%
$49^o$

Explanation

By using circle theorem which states that the measure of an angle subtended by an arc at the center is twice the measure of an angle subtended by the same arc at the circumference of the circle.

The exterior angle at the center is

$360^{o}-138^{o}=222^{o}$ .

So, the value of

$X$ is half of

$222^{o}$ that is

$\dfrac{222^{o}}{2}=111^{o}$ .

Hence, the answer is

$111^{o}$ .

The circumcirlce of the quadrilateral formed by the lines

$x = a,x = 2a,y = - a,y = a$ is

0%
${x^2} + {y^2} - 3ax - {a^2} = 0$
0%
${x^2} + {y^2} + 3ax + {a^2} = 0$
0%
${x^2} + {y^2}-3ax + {a^2} = 0$
0%
${x^2} + {y^2} + 3ax - {a^2} = 0$

In the given figure, in

$\square$ pors.if

$\angle R S P=80^{\circ}$ then

$\angle R Q T=$ ____________

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

Two tangents

$PQ$ and

$PR$ drawn to the circle

$x^2+y^2-2x-4y-20=0$ from the point

$P(16,7)$ . If is the centre of the circle, area of quadrilateral

$PQCR$ is

0%
75
0%
80
0%
85
0%
95

In a circle of radius

$13 ~cm, PQ$ and

$RS$ are two parallel chords of lengths

$24 ~cm$ and

$10~ cm$ respectively. Calculate the distance between the chords if they are

$(i)$ on the same side of the center and

$(ii)$ on the opposite sides of the center.

0%
$5~cm,12~cm$
0%
$7~cm,13~cm$
0%
$5~cm,17~cm$
0%
$7~cm,17~cm$

In Fig. 2 , the measure of

$\angle B C D$ is:

0%
$100 ^ { \circ }$
0%
$70 ^ { \circ }$
0%
$80 ^ { \circ }$
0%
$30 ^ { \circ }$

In a circle of radius

$2018$ having center at

$O$ ,

$OPQR$ is a rectangle with

$Q$ on the circumference of the circle.

$P,R$ points inside the circle such that

$PQ=504$ . Then

$PR=$

0%
$2018$
0%
$504$
0%
$252$
0%
$\sqrt {1008}$

CBSE Questions for Class 9 Maths Circles Quiz 1 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Circles Quiz 1 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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