RS Aggarwal MCQs: Circles Notes | Study Mathematics (Maths) Class 9 - Class 9

Class 9: RS Aggarwal MCQs: Circles Notes | Study Mathematics (Maths) Class 9 - Class 9

The document RS Aggarwal MCQs: Circles Notes | Study Mathematics (Maths) Class 9 - Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9
 Page 1


              
              
 
 
  
      
     
     
     
Question:68
The radius of a circle is 13 cm and the length of one of its chords is 10 cm. The distance of the chord from the
centre is
a
11.5 cm
b
12 cm
c
v
69 cm
d
23 cm
Solution:
b
12 cm
Let AB be the chord of the given circle with centre O and a radius of 13 cm.
Then, AB = 10 cm and OB = 13 cm
From O, draw OM perpendicular to AB.
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
? BM = 
10
2
cm = 5cm
From the right ?OMB, we have:
OB
2 
= OM
2
 + MB
2
? 13
2
 = OM
2
 + 5
2
? 169 = OM
2
 + 25
? OM
2
 = 169 -25
= 144
? OM =
v
144cm = 12cm
Hence, the distance of the chord from the centre is 12 cm.
( )
Page 2


              
              
 
 
  
      
     
     
     
Question:68
The radius of a circle is 13 cm and the length of one of its chords is 10 cm. The distance of the chord from the
centre is
a
11.5 cm
b
12 cm
c
v
69 cm
d
23 cm
Solution:
b
12 cm
Let AB be the chord of the given circle with centre O and a radius of 13 cm.
Then, AB = 10 cm and OB = 13 cm
From O, draw OM perpendicular to AB.
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
? BM = 
10
2
cm = 5cm
From the right ?OMB, we have:
OB
2 
= OM
2
 + MB
2
? 13
2
 = OM
2
 + 5
2
? 169 = OM
2
 + 25
? OM
2
 = 169 -25
= 144
? OM =
v
144cm = 12cm
Hence, the distance of the chord from the centre is 12 cm.
( )
Question:69
A chord is at a distance of 8 cm from the centre of a circle of radius 17 cm. The length of the chord is
a
25 cm
b
12.5 cm
c
30 cm
d
9 cm
Solution:
c
30 cm
Let AB be the chord of the given circle with centre O and a radius of 17 cm.
From O, draw OM perpendicular to AB.
Then OM = 8 cm and OB = 17 cm
From the right ?OMB, we have:
OB
2 
= OM
2
 + MB
2
? 17
2
 = 8
2
 + MB
2
? 289 = 64 + MB
2
? MB
2
 = 289 -64
= 225
? MB =
v
225cm = 15cm
The perpendicular from the centre of a circle to a chord bisects the chord.
? AB = 2 × MB = 2x15
cm = 30 cm
Hence, the required length of the chord is 30 cm.
Question:70
In the given figure, BOC is a diameter of a circle and AB = AC. Then, ?ABC = ?
a
30°
b
45°
c
60°
d
90°
Solution:
b
45°
Since an angle in a semicircle is a right angle, ?BAC = 90°
? ?ABC + ?ACB = 90°
Now, AB = AC       Given
Page 3


              
              
 
 
  
      
     
     
     
Question:68
The radius of a circle is 13 cm and the length of one of its chords is 10 cm. The distance of the chord from the
centre is
a
11.5 cm
b
12 cm
c
v
69 cm
d
23 cm
Solution:
b
12 cm
Let AB be the chord of the given circle with centre O and a radius of 13 cm.
Then, AB = 10 cm and OB = 13 cm
From O, draw OM perpendicular to AB.
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
? BM = 
10
2
cm = 5cm
From the right ?OMB, we have:
OB
2 
= OM
2
 + MB
2
? 13
2
 = OM
2
 + 5
2
? 169 = OM
2
 + 25
? OM
2
 = 169 -25
= 144
? OM =
v
144cm = 12cm
Hence, the distance of the chord from the centre is 12 cm.
( )
Question:69
A chord is at a distance of 8 cm from the centre of a circle of radius 17 cm. The length of the chord is
a
25 cm
b
12.5 cm
c
30 cm
d
9 cm
Solution:
c
30 cm
Let AB be the chord of the given circle with centre O and a radius of 17 cm.
From O, draw OM perpendicular to AB.
Then OM = 8 cm and OB = 17 cm
From the right ?OMB, we have:
OB
2 
= OM
2
 + MB
2
? 17
2
 = 8
2
 + MB
2
? 289 = 64 + MB
2
? MB
2
 = 289 -64
= 225
? MB =
v
225cm = 15cm
The perpendicular from the centre of a circle to a chord bisects the chord.
? AB = 2 × MB = 2x15
cm = 30 cm
Hence, the required length of the chord is 30 cm.
Question:70
In the given figure, BOC is a diameter of a circle and AB = AC. Then, ?ABC = ?
a
30°
b
45°
c
60°
d
90°
Solution:
b
45°
Since an angle in a semicircle is a right angle, ?BAC = 90°
? ?ABC + ?ACB = 90°
Now, AB = AC       Given
? ?ABC = ?ACB = 45°
Question:71
In the given figure, O is the centre of a circle and ?ACB = 30°. Then, ?AOB = ?
a
30°
b
15°
c
60°
d
90°
Figure
Solution:
c
60°
We know that the angle at the centre of a circle is twice the angle at any point on the remaining part of the
circumference.
Thus, ?AOB = 2 × ?ACB
= 2 ×30°
= 60°
Question:72
In the given figure, O is the centre of a circle. If ?OAB = 40° and C is a point on the circle, then ?ACB = ?
a
40°
b
50°
c
80°
d
100°
Solution:
b
50°
OA = OB
? ?OBA = ?OAB = 40°
Now, ?AOB = 180° -
40°+40° = 100°
? ?ACB =
1
2
?AOB =
1
2
×100 ° = 50°
Question:73
In the given figure, AOB is a diameter of a circle with centre O such that AB = 34 cm and CD is a chord of length 30
cm. Then the distance of CD from AB is
a
8 cm
b
15 cm
( )
Page 4


              
              
 
 
  
      
     
     
     
Question:68
The radius of a circle is 13 cm and the length of one of its chords is 10 cm. The distance of the chord from the
centre is
a
11.5 cm
b
12 cm
c
v
69 cm
d
23 cm
Solution:
b
12 cm
Let AB be the chord of the given circle with centre O and a radius of 13 cm.
Then, AB = 10 cm and OB = 13 cm
From O, draw OM perpendicular to AB.
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
? BM = 
10
2
cm = 5cm
From the right ?OMB, we have:
OB
2 
= OM
2
 + MB
2
? 13
2
 = OM
2
 + 5
2
? 169 = OM
2
 + 25
? OM
2
 = 169 -25
= 144
? OM =
v
144cm = 12cm
Hence, the distance of the chord from the centre is 12 cm.
( )
Question:69
A chord is at a distance of 8 cm from the centre of a circle of radius 17 cm. The length of the chord is
a
25 cm
b
12.5 cm
c
30 cm
d
9 cm
Solution:
c
30 cm
Let AB be the chord of the given circle with centre O and a radius of 17 cm.
From O, draw OM perpendicular to AB.
Then OM = 8 cm and OB = 17 cm
From the right ?OMB, we have:
OB
2 
= OM
2
 + MB
2
? 17
2
 = 8
2
 + MB
2
? 289 = 64 + MB
2
? MB
2
 = 289 -64
= 225
? MB =
v
225cm = 15cm
The perpendicular from the centre of a circle to a chord bisects the chord.
? AB = 2 × MB = 2x15
cm = 30 cm
Hence, the required length of the chord is 30 cm.
Question:70
In the given figure, BOC is a diameter of a circle and AB = AC. Then, ?ABC = ?
a
30°
b
45°
c
60°
d
90°
Solution:
b
45°
Since an angle in a semicircle is a right angle, ?BAC = 90°
? ?ABC + ?ACB = 90°
Now, AB = AC       Given
? ?ABC = ?ACB = 45°
Question:71
In the given figure, O is the centre of a circle and ?ACB = 30°. Then, ?AOB = ?
a
30°
b
15°
c
60°
d
90°
Figure
Solution:
c
60°
We know that the angle at the centre of a circle is twice the angle at any point on the remaining part of the
circumference.
Thus, ?AOB = 2 × ?ACB
= 2 ×30°
= 60°
Question:72
In the given figure, O is the centre of a circle. If ?OAB = 40° and C is a point on the circle, then ?ACB = ?
a
40°
b
50°
c
80°
d
100°
Solution:
b
50°
OA = OB
? ?OBA = ?OAB = 40°
Now, ?AOB = 180° -
40°+40° = 100°
? ?ACB =
1
2
?AOB =
1
2
×100 ° = 50°
Question:73
In the given figure, AOB is a diameter of a circle with centre O such that AB = 34 cm and CD is a chord of length 30
cm. Then the distance of CD from AB is
a
8 cm
b
15 cm
( )
c
18 cm
d
6 cm
Solution:
a
8 cm
Join OC. Then OC = radius = 17 cm
CL =
1
2
CD =
1
2
×30 cm = 15cm
In right ?OLC, we have:
OL
2
 = OC
2
 - CL
2
 = 17
2
 - 15
2
 = 289 -225
= 64
? OL =
v
64 = 8cm
? Distance of CD from AB = 8 cm
Question:74
AB and CD are two equal chords of a circle with centre O such that ?AOB = 80°, then ?COD = ?
a
100°
b
80°
c
120°
d
40°
Solution:
b
80°
Given: AB = CD
We know that equal chords of a circle subtend equal angles at the centre.
? ?COD = ?AOB = 80°
Question:75
( )
Page 5


              
              
 
 
  
      
     
     
     
Question:68
The radius of a circle is 13 cm and the length of one of its chords is 10 cm. The distance of the chord from the
centre is
a
11.5 cm
b
12 cm
c
v
69 cm
d
23 cm
Solution:
b
12 cm
Let AB be the chord of the given circle with centre O and a radius of 13 cm.
Then, AB = 10 cm and OB = 13 cm
From O, draw OM perpendicular to AB.
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
? BM = 
10
2
cm = 5cm
From the right ?OMB, we have:
OB
2 
= OM
2
 + MB
2
? 13
2
 = OM
2
 + 5
2
? 169 = OM
2
 + 25
? OM
2
 = 169 -25
= 144
? OM =
v
144cm = 12cm
Hence, the distance of the chord from the centre is 12 cm.
( )
Question:69
A chord is at a distance of 8 cm from the centre of a circle of radius 17 cm. The length of the chord is
a
25 cm
b
12.5 cm
c
30 cm
d
9 cm
Solution:
c
30 cm
Let AB be the chord of the given circle with centre O and a radius of 17 cm.
From O, draw OM perpendicular to AB.
Then OM = 8 cm and OB = 17 cm
From the right ?OMB, we have:
OB
2 
= OM
2
 + MB
2
? 17
2
 = 8
2
 + MB
2
? 289 = 64 + MB
2
? MB
2
 = 289 -64
= 225
? MB =
v
225cm = 15cm
The perpendicular from the centre of a circle to a chord bisects the chord.
? AB = 2 × MB = 2x15
cm = 30 cm
Hence, the required length of the chord is 30 cm.
Question:70
In the given figure, BOC is a diameter of a circle and AB = AC. Then, ?ABC = ?
a
30°
b
45°
c
60°
d
90°
Solution:
b
45°
Since an angle in a semicircle is a right angle, ?BAC = 90°
? ?ABC + ?ACB = 90°
Now, AB = AC       Given
? ?ABC = ?ACB = 45°
Question:71
In the given figure, O is the centre of a circle and ?ACB = 30°. Then, ?AOB = ?
a
30°
b
15°
c
60°
d
90°
Figure
Solution:
c
60°
We know that the angle at the centre of a circle is twice the angle at any point on the remaining part of the
circumference.
Thus, ?AOB = 2 × ?ACB
= 2 ×30°
= 60°
Question:72
In the given figure, O is the centre of a circle. If ?OAB = 40° and C is a point on the circle, then ?ACB = ?
a
40°
b
50°
c
80°
d
100°
Solution:
b
50°
OA = OB
? ?OBA = ?OAB = 40°
Now, ?AOB = 180° -
40°+40° = 100°
? ?ACB =
1
2
?AOB =
1
2
×100 ° = 50°
Question:73
In the given figure, AOB is a diameter of a circle with centre O such that AB = 34 cm and CD is a chord of length 30
cm. Then the distance of CD from AB is
a
8 cm
b
15 cm
( )
c
18 cm
d
6 cm
Solution:
a
8 cm
Join OC. Then OC = radius = 17 cm
CL =
1
2
CD =
1
2
×30 cm = 15cm
In right ?OLC, we have:
OL
2
 = OC
2
 - CL
2
 = 17
2
 - 15
2
 = 289 -225
= 64
? OL =
v
64 = 8cm
? Distance of CD from AB = 8 cm
Question:74
AB and CD are two equal chords of a circle with centre O such that ?AOB = 80°, then ?COD = ?
a
100°
b
80°
c
120°
d
40°
Solution:
b
80°
Given: AB = CD
We know that equal chords of a circle subtend equal angles at the centre.
? ?COD = ?AOB = 80°
Question:75
( )
In the given figure, CD is the diameter of a circle with centre O and CD is perpendicular to chord AB. If AB = 12 cm
and CE = 3 cm, then radius of the circles is
a
6 cm
b
9 cm
c
7.5 cm
d
8 cm
Solution:
c
7.5 cm
Let OA = OC = r cm.
Then OE = (r - 3) cm and AE =
1
2
AB = 6cm
Now, in right ?OAE, we have:
OA
2
 = OE
2
 +AE
2
 
? (r)
2
 = (r - 3)
2
 + 6
2
? r
2
 = r
2
 + 9 - 6r + 36
? 6r = 45
? r =
45
6
= 7. 5
cm
Hence, the required radius of the circle is 7.5 cm.
Question:76
In the given figure, O is the centre of a circle and diameter AB bisects the chord CD at a point E such that CE = ED
= 8 cm and EB = 4 cm. The radius of the circle is
a
10 cm
b
12 cm
c
6 cm
d
8 cm
Solution:
a
10 cm
Let the radius of the circle be r cm.
Let OD = OB = r cm.
Then OE = (r - 4) cm and ED = 8 cm
Now, in right ?OED, we have:
Read More
73 videos|351 docs|110 tests

Download free EduRev App

Track your progress, build streaks, highlight & save important lessons and more!

Related Searches

Exam

,

Viva Questions

,

Sample Paper

,

ppt

,

Extra Questions

,

practice quizzes

,

pdf

,

RS Aggarwal MCQs: Circles Notes | Study Mathematics (Maths) Class 9 - Class 9

,

RS Aggarwal MCQs: Circles Notes | Study Mathematics (Maths) Class 9 - Class 9

,

RS Aggarwal MCQs: Circles Notes | Study Mathematics (Maths) Class 9 - Class 9

,

past year papers

,

study material

,

MCQs

,

Semester Notes

,

mock tests for examination

,

Previous Year Questions with Solutions

,

Summary

,

Important questions

,

Objective type Questions

,

shortcuts and tricks

,

video lectures

,

Free

;