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# RD Sharma MCQs: Circles Notes | EduRev

## Class 9 : RD Sharma MCQs: Circles Notes | EduRev

``` Page 1

Q u e s t i o n : 7 0
If the length of a chord of a circle is 16 cm and is at a distance of 15 cm from the centre of the circle, then the radius of the circle is
a
15 cm
b
16 cm
c
17 cm
d
34 cm
S o l u t i o n :
c 17 cm
We will represent the given data in the figure

In the diagram AB is the given chord of 16 cm length and OM is the perpendicular distance from the centre to AB.
We know that perpendicular from the centre to any chord divides it into two equal parts.
So,  AM = MB =  = 8 cm.
Now consider right triangle OMA and by using Pythagoras theorem

c.
Q u e s t i o n : 7 1
The radius of a circle is 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm in length, is
a
v 5 cm
Page 2

Q u e s t i o n : 7 0
If the length of a chord of a circle is 16 cm and is at a distance of 15 cm from the centre of the circle, then the radius of the circle is
a
15 cm
b
16 cm
c
17 cm
d
34 cm
S o l u t i o n :
c 17 cm
We will represent the given data in the figure

In the diagram AB is the given chord of 16 cm length and OM is the perpendicular distance from the centre to AB.
We know that perpendicular from the centre to any chord divides it into two equal parts.
So,  AM = MB =  = 8 cm.
Now consider right triangle OMA and by using Pythagoras theorem

c.
Q u e s t i o n : 7 1
The radius of a circle is 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm in length, is
a
v 5 cm
b
2v 5 cm
c
2v 7 cm
d
v 7 cm
S o l u t i o n :
b
2v 5 cm
We will represent the given data in the figure
We know that perpendicular drawn from the centre to the chord divides the chord into two equal parts.
So , AM = MB =
AB
2
=
8
2
= 4 cm.
Using Pythagoras theorem in the ?AMO,

Hence, the correct answer is option
b.
Q u e s t i o n : 7 2
If O is the centre of a circle of radius r and AB is a chord of the circle at a distance r/2 from O, then ?BAO =
a
60°
b
45°
c
30°
d
15°
S o l u t i o n :
We will associate the given information in the following figure.
Since AO = r
AM =
given
Extended OM to D where MD =
Consider the triangles AOM and triangle AMD
So by SSS property
Page 3

Q u e s t i o n : 7 0
If the length of a chord of a circle is 16 cm and is at a distance of 15 cm from the centre of the circle, then the radius of the circle is
a
15 cm
b
16 cm
c
17 cm
d
34 cm
S o l u t i o n :
c 17 cm
We will represent the given data in the figure

In the diagram AB is the given chord of 16 cm length and OM is the perpendicular distance from the centre to AB.
We know that perpendicular from the centre to any chord divides it into two equal parts.
So,  AM = MB =  = 8 cm.
Now consider right triangle OMA and by using Pythagoras theorem

c.
Q u e s t i o n : 7 1
The radius of a circle is 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm in length, is
a
v 5 cm
b
2v 5 cm
c
2v 7 cm
d
v 7 cm
S o l u t i o n :
b
2v 5 cm
We will represent the given data in the figure
We know that perpendicular drawn from the centre to the chord divides the chord into two equal parts.
So , AM = MB =
AB
2
=
8
2
= 4 cm.
Using Pythagoras theorem in the ?AMO,

Hence, the correct answer is option
b.
Q u e s t i o n : 7 2
If O is the centre of a circle of radius r and AB is a chord of the circle at a distance r/2 from O, then ?BAO =
a
60°
b
45°
c
30°
d
15°
S o l u t i o n :
We will associate the given information in the following figure.
Since AO = r
AM =
given
Extended OM to D where MD =
Consider the triangles AOM and triangle AMD
So by SSS property
So AD = AO = r and OD=OM+MD=r
Hence ?AOD is equilateral triangle
So
We know that in equilateral triangle altitudes divide the vertex angles
Hence option (c) is correct.
Q u e s t i o n : 7 3
ABCD is a cyclic quadrilateral such that ?ADB = 30° and ?DCA = 80°, then ?DAB =
a
70°
b
100°
c
125°
d
150°
S o l u t i o n :
a 70°
It is given that ABCD is cyclic quadrilateral ?ADB = 90° and ?DCA = 80°. We have to find ?DAB
We have the following figure regarding the given information
?BDA = ?BCA = 30°
Angleinthesamesegmentareequal
Now, since ABCD is a cyclic quadrilateral
So, ?DAB + ?BCD = 180°
Hence the correct answer is option
a.
Q u e s t i o n : 7 4
A chord of length 14 cm is at  a distance of 6 cm from the centre of a circle. The length of another chord at a distance of 2 cm from the centre of the circle is
a
12 cm
b
14 cm
c
16 cm
d
18 cm
S o l u t i o n :
d 18 cm
We are given the chord of length 14 cm and perpendicular distance from the centre to the chord is 6 cm. We are asked to find the length of another chord at a distance of 2 cm from the
centre.
We have the following figure
Page 4

Q u e s t i o n : 7 0
If the length of a chord of a circle is 16 cm and is at a distance of 15 cm from the centre of the circle, then the radius of the circle is
a
15 cm
b
16 cm
c
17 cm
d
34 cm
S o l u t i o n :
c 17 cm
We will represent the given data in the figure

In the diagram AB is the given chord of 16 cm length and OM is the perpendicular distance from the centre to AB.
We know that perpendicular from the centre to any chord divides it into two equal parts.
So,  AM = MB =  = 8 cm.
Now consider right triangle OMA and by using Pythagoras theorem

c.
Q u e s t i o n : 7 1
The radius of a circle is 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm in length, is
a
v 5 cm
b
2v 5 cm
c
2v 7 cm
d
v 7 cm
S o l u t i o n :
b
2v 5 cm
We will represent the given data in the figure
We know that perpendicular drawn from the centre to the chord divides the chord into two equal parts.
So , AM = MB =
AB
2
=
8
2
= 4 cm.
Using Pythagoras theorem in the ?AMO,

Hence, the correct answer is option
b.
Q u e s t i o n : 7 2
If O is the centre of a circle of radius r and AB is a chord of the circle at a distance r/2 from O, then ?BAO =
a
60°
b
45°
c
30°
d
15°
S o l u t i o n :
We will associate the given information in the following figure.
Since AO = r
AM =
given
Extended OM to D where MD =
Consider the triangles AOM and triangle AMD
So by SSS property
So AD = AO = r and OD=OM+MD=r
Hence ?AOD is equilateral triangle
So
We know that in equilateral triangle altitudes divide the vertex angles
Hence option (c) is correct.
Q u e s t i o n : 7 3
ABCD is a cyclic quadrilateral such that ?ADB = 30° and ?DCA = 80°, then ?DAB =
a
70°
b
100°
c
125°
d
150°
S o l u t i o n :
a 70°
It is given that ABCD is cyclic quadrilateral ?ADB = 90° and ?DCA = 80°. We have to find ?DAB
We have the following figure regarding the given information
?BDA = ?BCA = 30°
Angleinthesamesegmentareequal
Now, since ABCD is a cyclic quadrilateral
So, ?DAB + ?BCD = 180°
Hence the correct answer is option
a.
Q u e s t i o n : 7 4
A chord of length 14 cm is at  a distance of 6 cm from the centre of a circle. The length of another chord at a distance of 2 cm from the centre of the circle is
a
12 cm
b
14 cm
c
16 cm
d
18 cm
S o l u t i o n :
d 18 cm
We are given the chord of length 14 cm and perpendicular distance from the centre to the chord is 6 cm. We are asked to find the length of another chord at a distance of 2 cm from the
centre.
We have the following figure
We are given AB = 14 cm, OD = 6 cm, MO = 2 cm, PQ = ?
Since, perpendicular from centre to the chord divide the chord into two equal parts
Therefore
Now consider the ?OPQ in which OM = 2 cm
So using Pythagoras Theorem in ?OPM
Hence, the correct answer is option
d.
Q u e s t i o n : 7 5
One chord of a circle is known to be 10 cm. The radius of this circle must be
a
5 cm
b
greater than 5 cm
c
greater than or equal to 5 cm
d
less than 5 cm
S o l u t i o n :
b greater than 5 cm
We are given length of a chord to be 10 cm and we have to give information about the radius of the circle.
Since in any circle, diameter of the circle is greater then any chord.
So diameter > 10
? 2r > 10
? r > 5 cm
Hence, the correct answer is option (b)

Q u e s t i o n : 7 6
ABC is a triangle with B as right angle, AC = 5 cm and AB = 4 cm. A circle is drawn with A as centre and AC as radius. The length of the chord of this circle passing through C and B is
a
3 cm
b
4 cm
c
5 cm
d
6 cm
S o l u t i o n :
d 6 cm
We are given a right triangle ABC such that , AC = 5 cm, AB = 4 cm. A circle is drawn with A as centre and AC as radius. We have to find the length of the chord of this circle
passing through C and B. We have the following figure regarding the given information.
In the circle produce CB to P. Here PC is the required chord.
We know that perpendicular drawn from the centre to the chord divide the chord into two equal parts.
So,  PC = 2BC
Page 5

Q u e s t i o n : 7 0
If the length of a chord of a circle is 16 cm and is at a distance of 15 cm from the centre of the circle, then the radius of the circle is
a
15 cm
b
16 cm
c
17 cm
d
34 cm
S o l u t i o n :
c 17 cm
We will represent the given data in the figure

In the diagram AB is the given chord of 16 cm length and OM is the perpendicular distance from the centre to AB.
We know that perpendicular from the centre to any chord divides it into two equal parts.
So,  AM = MB =  = 8 cm.
Now consider right triangle OMA and by using Pythagoras theorem

c.
Q u e s t i o n : 7 1
The radius of a circle is 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm in length, is
a
v 5 cm
b
2v 5 cm
c
2v 7 cm
d
v 7 cm
S o l u t i o n :
b
2v 5 cm
We will represent the given data in the figure
We know that perpendicular drawn from the centre to the chord divides the chord into two equal parts.
So , AM = MB =
AB
2
=
8
2
= 4 cm.
Using Pythagoras theorem in the ?AMO,

Hence, the correct answer is option
b.
Q u e s t i o n : 7 2
If O is the centre of a circle of radius r and AB is a chord of the circle at a distance r/2 from O, then ?BAO =
a
60°
b
45°
c
30°
d
15°
S o l u t i o n :
We will associate the given information in the following figure.
Since AO = r
AM =
given
Extended OM to D where MD =
Consider the triangles AOM and triangle AMD
So by SSS property
So AD = AO = r and OD=OM+MD=r
Hence ?AOD is equilateral triangle
So
We know that in equilateral triangle altitudes divide the vertex angles
Hence option (c) is correct.
Q u e s t i o n : 7 3
ABCD is a cyclic quadrilateral such that ?ADB = 30° and ?DCA = 80°, then ?DAB =
a
70°
b
100°
c
125°
d
150°
S o l u t i o n :
a 70°
It is given that ABCD is cyclic quadrilateral ?ADB = 90° and ?DCA = 80°. We have to find ?DAB
We have the following figure regarding the given information
?BDA = ?BCA = 30°
Angleinthesamesegmentareequal
Now, since ABCD is a cyclic quadrilateral
So, ?DAB + ?BCD = 180°
Hence the correct answer is option
a.
Q u e s t i o n : 7 4
A chord of length 14 cm is at  a distance of 6 cm from the centre of a circle. The length of another chord at a distance of 2 cm from the centre of the circle is
a
12 cm
b
14 cm
c
16 cm
d
18 cm
S o l u t i o n :
d 18 cm
We are given the chord of length 14 cm and perpendicular distance from the centre to the chord is 6 cm. We are asked to find the length of another chord at a distance of 2 cm from the
centre.
We have the following figure
We are given AB = 14 cm, OD = 6 cm, MO = 2 cm, PQ = ?
Since, perpendicular from centre to the chord divide the chord into two equal parts
Therefore
Now consider the ?OPQ in which OM = 2 cm
So using Pythagoras Theorem in ?OPM
Hence, the correct answer is option
d.
Q u e s t i o n : 7 5
One chord of a circle is known to be 10 cm. The radius of this circle must be
a
5 cm
b
greater than 5 cm
c
greater than or equal to 5 cm
d
less than 5 cm
S o l u t i o n :
b greater than 5 cm
We are given length of a chord to be 10 cm and we have to give information about the radius of the circle.
Since in any circle, diameter of the circle is greater then any chord.
So diameter > 10
? 2r > 10
? r > 5 cm
Hence, the correct answer is option (b)

Q u e s t i o n : 7 6
ABC is a triangle with B as right angle, AC = 5 cm and AB = 4 cm. A circle is drawn with A as centre and AC as radius. The length of the chord of this circle passing through C and B is
a
3 cm
b
4 cm
c
5 cm
d
6 cm
S o l u t i o n :
d 6 cm
We are given a right triangle ABC such that , AC = 5 cm, AB = 4 cm. A circle is drawn with A as centre and AC as radius. We have to find the length of the chord of this circle
passing through C and B. We have the following figure regarding the given information.
In the circle produce CB to P. Here PC is the required chord.
We know that perpendicular drawn from the centre to the chord divide the chord into two equal parts.
So,  PC = 2BC
Now in ?ABC apply Pythagoras theorem
So,  PC = 2 × BC
= 2 × 3
= 6 cm
Hence, the correct answer is option
d.
Q u e s t i o n : 7 7
If AB, BC and CD are equal chords of a circle with O as centre and AD diameter, than ?AOB =
a
60°
b
90°
c
120°
d
none of these
S o l u t i o n :
a 60°
As we know that equal chords make equal angle at the centre.
Therefore,
?AOB = ?BOC = ?COD ?AOB + ?BOC + ?COD = 180°     [Linear pair] ? 3 ?AOB = 180° ? ?AOB = 60°
Hence, the correct answer is option a
.
Q u e s t i o n : 7 8
Let C be the mid-point of an arc AB of a circle such that m

AB
= 183°. If the region bounded by the arc ACB and the line segment AB is denoted by S, then the centre O of the circle lies
a
in the interior of S
b
in the exertior of S
c
on the segment AB
d
on AB and bisects AB
S o l u t i o n :
a in the interior of S
Given: m

AB
= 183° and C is mid-point of arc ABO is the centre.
With the given information the corresponding figure will look like the following
So the center of the circle lies inside the shaded region S.
Hence, the correct answer is option
a.
```
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## Mathematics (Maths) Class 9

46 videos|323 docs|108 tests

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