MCQ Questions for Class 9 Maths Circles Quiz 2

A square is inscribed in the circle

${ x }^{ 2 }+{ y }^{ 2 }-6x+8y-103=0$ with its sides parallel to the coordinate axes . Then the distance of the vertex of this square which is nearest to the origin is :

0%
6
0%
$\sqrt { 137 }$
0%
$\sqrt { 41 }$
0%
13

In the give figure, ABCD is a cyclic quadrilateral,

$\angle C B Q=48^{\circ}$ and a=2b. Then, b is equal to

0%
$48^{\circ}$
0%
$38^{\circ}$
0%
$28^{\circ}$
0%
none of these

The value of x and y in tyhe fighure are measure of angles,then x+y is equal to

0%
$90^{\circ}$
0%
$85^{\circ}$
0%
$75^{\circ}$
0%
$65^{\circ}$

Distance between chords of contact of the tangents of the circle

$x ^ { 2 } + y ^ { 2 } + 2 y x + 27 y + c = 0$ from the origin and the point (

$g, f$ ) is

0%
$\sqrt { g ^ { 2 } + f ^ { 2 } }$
0%
$\frac { \sqrt { g ^ { 2 } + f ^ { 2 } - c } } { 2 }$
0%
$\frac { g ^ { 2 } + f ^ { 2 } - c } { 2 \sqrt { g ^ { 2 } + f ^ { 2 } } }$
0%
$\frac { \sqrt { g ^ { 2 } + f ^ { 2 } + c } } { 2 \sqrt { g ^ { 2 } + f ^ { 2 } } }$

$AB C D$ is cyclic quadrilateral such that

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

$\angle ADC=130^{ { \circ } },$ then the value of

$\angle A B C$ is:

0%
$50 ^ { \circ }$
0%
$60 ^ { \circ }$
0%
$120 ^ { \circ }$
0%
$40 ^ { \circ }$

What will be the area of the largest square that can be cut out of a circle of radius 10 cm?

0%
100 ${ cm }^{ 2 }$
0%
200 ${ cm }^{ 2 }$
0%
300 ${ cm }^{ 2 }$
0%
400 ${ cm }^{ 2 }$

$\Box ABCD$ is a cyclic quadrilateral,

$\angle BAD={ 75 }^{ \circ },m\angle ADC={ 85 }^{ \circ }$ . The bisectors of

$\angle ABC$ and

$\angle BCD$ meet in the point O. Find m

$\angle BOC.$

0%
${ 40 }^{ \circ }$
0%
${ 70 }^{ \circ }$
0%
${ 80 }^{ \circ }$
0%
${ 60 }^{ \circ }$

A,B,C and D be the angles of cyclic quadrilateral taken order , then

$cos({ 180 }^{ \circ }+A)$ +

$cos({ 180 }^{ \circ }+B)$ +

$cos({ 180 }^{ \circ }+C)$ +

$cos({ 180 }^{ \circ }+D)$ is equal to

0%
2 (cosA+cosB )
0%
2 (cosA+cosD)
0%
0
0%
4 cos A

Consider a circle with center O and radius = 20cm. Find the perpendicular distance of a chord of length 32cm from the center of the circle.

0%
20cm
0%
18cm
0%
14cm
0%
12cm

Write True or False and justify your answer in each of the following :
If

$A , B , C$ and

$D$ are four points such that

$\angle BAC$

$=45^{\circ}$ and

$\angle$ BDC =

$45^{\circ}$ , then

$A , B , C , D$ are concyclic.

0%
True
0%
False

Write True or False and justify your answer in each of the following :
If

$A , B , C , D$ are four points such that

$\angle BAC$

$= 30^{\circ}$ and

$\angle$ BDC =

$60^{\circ}$ , then

$D$ is the center of the circle through

$A , B$ and

$C$ .

0%
True
0%
False

Consider a circle with center O and radius = 25cm. Find the perpendicular distance of a chord of length 14cm from the center of the circle.

0%
$20cm$
0%
$24cm$
0%
$18cm$
0%
$14cm$

Consider a chord

$AB = 10cm$ , of a circle with radius

$r$ . A line segment

$OM = 3cm$ from the center O of the circle to AB, bisects AB into two equal parts.
Then what is the value of

$r$ ?

0%
$\sqrt {40}$
0%
$\sqrt {35}$
0%
$\sqrt {34}$
0%
$\sqrt {50}$

Consider a chord

$AB = 24cm$ , of a circle with radius

$r$ . A line segment

$OM = 5cm$ from the center O of the circle to AB, bisects AB into two equal parts.
Then what is the value of

$r$ ?

0%
$13 cm$
0%
$24 \sqrt 2cm$
0%
$20cm$
0%
$19 cm$

Consider a circle with center O and radius = 13cm. Find the perpendicular distance of a chord of length 10cm from the center of the circle.

0%
8cm
0%
10cm
0%
12cm
0%
15cm

Consider a chord

$AB = 8cm$ , of a circle with radius

$r$ . A line segment

$OM = 3cm$ from the center O of the circle to AB, bisects AB into two equal parts.
Then what is the value of

$r$ ?

0%
$3 \sqrt 2cm$
0%
$\sqrt 71cm$
0%
$5cm$
0%
$4 \sqrt 2cm$

To prove "The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord", we prove congruence of which two triangles?

0%
$\triangle AOM \ and \ \triangle BOM$
0%
$\triangle AOM \ and \ \triangle BOA$
0%
$\triangle AOB \ and \ \triangle BOM$
0%
$\triangle AOM \ and \ \triangle MOB$

How many circles pass through 3 non-colinear given points?

0%
1
0%
2
0%
More than 2
0%
Infinitely many

In given figure, is possible to draw another circle using points

$X, Y$ and

$Z$ which will be different from given circle?

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

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

0%
0
0%
1
0%
4
0%
Infinitely

If three points are collinear, then we can draw a circle using given points.

0%
True
0%
False
0%
May be
0%
None of these

In the given image, in which of figure, we can draw another circle using the vertices of triangle different from already existing circle?

0%
Figure 1
0%
Figure 2
0%
Figure 3
0%
Not possible in any of the figure

Which of the following pair of angles are opposite angles of a cyclic quadrilateral?

0%
$131^{\circ}, 28^{\circ}$
0%
$95^{\circ}, 55^{\circ}$
0%
$123^{\circ}, 57^{\circ}$
0%
$64^{\circ}, 52^{\circ}$

Explanation

We know that, the sum of opposite angles of every cyclic quadrilateral is

$180^{\circ}$ .

$ABCD$ is cyclic quadrilateral,

then

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

The sum of the angles

${123}^{\circ}$ and

${57}^{\circ}$ is equal to

$180^{\circ}$ .

The sum of two angles given in every other option, other than option

$C$ is not

$180^{\circ}$ .

Therefore, option

$C$ is correct.

In Fig 10.9

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

0%
$\displaystyle 30^{\circ}$
0%
$\displaystyle 45^{\circ}$
0%
$\displaystyle 90^{\circ}$
0%
$\displaystyle 60^{\circ}$

Explanation

$\triangle AOB$

$OA=OB$

$\therefore \angle OAB+\angle OBA+\angle AOB=180^{0}$

$=>2\angle OAB+90^{0}=180^{0}$

$=>2\angle OAB=180^{0}-90^{0}$

$=90^{0}$

$=>\angle OAB=\frac{90}{2}$

$=45^{0}$

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.

$\therefore \angle ACB=\frac{1}{2}\angle AOB$

$=\frac{90}{2}$

$=45^{0}$

Again in

$\triangle ABC$

$\angle ACB+\angle CBA+\angle BAC=180^{0}$

$=>45^{0}+30^{0}+\angle BAC=180^{0}$

$=>\angle BAC=180^{0}-75^{0}$

$=105^{0}$

$\therefore \angle CAO=\angle BAC-\angle OAB$

$=>\angle CAO=105^{0}-45^{0}$

$=60^{0}$

In the given figure

$O$ is the centre of the circle and

$\displaystyle \angle AOC=120^{\circ}$ . What is the value of

$\displaystyle \angle APC+\angle ABC$ ?

0%
$\displaystyle 120^{\circ}$
0%
$\displaystyle 140^{\circ}$
0%
$\displaystyle 160^{\circ}$
0%
$\displaystyle 180^{\circ}$

Explanation

$APCB$ is a cyclic quadrilateral, sum of its opposite angles are always equal to

$180^o$ .

Here,

$\angle APC$ and

$\angle ABC$ are the opposite angles of the cyclic quadrilateral.

Then clearly,

$\angle APC+\angle ABC=180^o$ .

Therefore, option

$D$ is correct.

Fixed point in the circle is called _____ of the circle.

0%
radius
0%
centre
0%
diameter
0%
none

Explanation

A circle is the set of all those point in a plane whose distance from a fixed point remains constant. Then, this fixed point is called the centre of the circle.

Hence, option $B$ is correct.

In Fig 10.7 if

$\displaystyle \angle DAB=60^{\circ},\angle ABD= 50^{\circ}$ then

$\displaystyle \angle ACB$ is equal to

0%
$\displaystyle 60^{\circ}$
0%
$\displaystyle 50^{\circ}$
0%
$\displaystyle 70^{\circ}$
0%
$\displaystyle 80^{\circ}$

Explanation

$\triangle ADC,$ we have

$\angle DAC=60^0$ and

$\angle ABD=50^0$

By angle sum property, in

$\triangle ABD$

$\angle DAB+\angle ABD+\angle ADB=180^0$

$=>60^°+50^°+\angle ADB=180^°$

$=>\angle ADB=180^°-110^°$

$=>\angle ADB=70^°$

We know that the angles subtended at the circumference by the same arc are equal.

$\therefore \angle ADB=\angle ACB$

$=>\angle ACB=70^0$

In the given figure,

$\displaystyle \triangle ABC$ is an equilateral triangle. What is the value of

$\displaystyle \angle BEC$ ?

0%
$\displaystyle 80^{\circ}$
0%
$\displaystyle 100^{\circ}$
0%
$\displaystyle 120^{\circ}$
0%
$\displaystyle 130^{\circ}$

Explanation

Here,

$ABEC$ is a cyclic quadrilateral.

We know, sum of its opposite angles are equal to

$180^o$ .

$\Rightarrow \angle BAC+\angle BEC =180^{\circ}$

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

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

Hence, option

$C$ is correct.

If an equilateral triangle inscribed in a circle with center

$O$ , then the measure of

$\angle AOB$ is

0%
$60$
0%
$90$
0%
$120$
0%
$180$

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,

$\angle ACB =60^°$

Hence,

$\angle AOB = 120^°$

In Fig 10.4 if

$\displaystyle \angle ABC = 20^{\circ}$ then

$\displaystyle \angle AOC$ is equal to

0%
$\displaystyle 20^{\circ}$
0%
$\displaystyle 40^{\circ}$
0%
$\displaystyle 60^{\circ}$
0%
$\displaystyle 10^{\circ}$

Explanation

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

Thus,

$\angle AOC=2 \times \angle ABC$

$\implies \angle AOC=2\times 20^0$

$\implies \angle AOC=40^0$

CBSE Questions for Class 9 Maths Circles Quiz 2 - MCQExams.com

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Circles Quiz 2 - MCQExams.com

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.