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Page 1
Question:1
A chord of length 16 cm is drawn in a circle of radius 10 cm. Find the distance of the chord from the centre of the
circle.
Solution:
Let AB be the chord of the given circle with centre O and a radius of 10 cm.
Then AB =16 cm and OB = 10 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 =
16
2
cm = 8 cm
In the right ?OMB, we have:
OB
2
= OM
2
+ MB
2
Pythagorastheorem
? 10
2
= OM
2
+ 8
2
? 100 = OM
2
+ 64
? OM
2
= 100 -64
= 36
? OM =
v
36 cm = 6 cm
Hence, the distance of the chord from the centre is 6 cm.
Question:2
Find the length of a chord which is at a distance of 3 cm from the centre of a circle of radius 5 cm.
Solution:
Let AB be the chord of the given circle with centre O and a radius of 5 cm.
From O, draw OM perpendicular to AB.
Then OM = 3 cm and OB = 5 cm
From the right ?OMB, we have:
OB
2
= OM
2
+ MB
2
Pythagorastheorem
? 5
2
= 3
2
+ MB
2
? 25 = 9 + MB
2
? MB
2
= (25 - 9) = 16
? MB =
v
16 cm = 4 cm
( )
Page 2
Question:1
A chord of length 16 cm is drawn in a circle of radius 10 cm. Find the distance of the chord from the centre of the
circle.
Solution:
Let AB be the chord of the given circle with centre O and a radius of 10 cm.
Then AB =16 cm and OB = 10 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 =
16
2
cm = 8 cm
In the right ?OMB, we have:
OB
2
= OM
2
+ MB
2
Pythagorastheorem
? 10
2
= OM
2
+ 8
2
? 100 = OM
2
+ 64
? OM
2
= 100 -64
= 36
? OM =
v
36 cm = 6 cm
Hence, the distance of the chord from the centre is 6 cm.
Question:2
Find the length of a chord which is at a distance of 3 cm from the centre of a circle of radius 5 cm.
Solution:
Let AB be the chord of the given circle with centre O and a radius of 5 cm.
From O, draw OM perpendicular to AB.
Then OM = 3 cm and OB = 5 cm
From the right ?OMB, we have:
OB
2
= OM
2
+ MB
2
Pythagorastheorem
? 5
2
= 3
2
+ MB
2
? 25 = 9 + MB
2
? MB
2
= (25 - 9) = 16
? MB =
v
16 cm = 4 cm
( )
Since the perpendicular from the centre of a circle to a chord bisects the chord, we have:
AB = 2 × MB = 2 ×4
cm = 8 cm
Hence, the required length of the chord is 8 cm.
Question:3
A chord of length 30 cm is drawn at a distance of 8 cm from the centre of a circle. Find out the radius of the circle.
Solution:
Let AB be the chord of the given circle with centre O. The perpendicular distance from the centre of the circle to the
chord is 8 cm.
Join OB.
Then OM = 8 cm and AB = 30 cm
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
? MB =
AB
2
=
30
2
cm = 15 cm
From the right ?OMB, we have:
OB
2
= OM
2
+ MB
2
? OB
2
= 8
2
+ 15
2
? OB
2
= 64 + 225
? OB
2
= 289
? OB =
v
289 cm = 17 cm
Hence, the required length of the radius is 17 cm.
Question:4
In a circle of radius 5 cm, AB and CD are two parallel chords of lengths 8 cm and 6 cm respectively. Calculate the
distance between the chords if they are
i
on the same side of the centre
ii
on the opposite sides of the centre.
Solution:
We have:
i
Let AB and CD be two chords of a circle such that AB is parallel to CD on the same side of the circle.
Given: AB = 8 cm, CD = 6 cm and OB = OD = 5 cm
Join OL and OM.
( ) ( )
Page 3
Question:1
A chord of length 16 cm is drawn in a circle of radius 10 cm. Find the distance of the chord from the centre of the
circle.
Solution:
Let AB be the chord of the given circle with centre O and a radius of 10 cm.
Then AB =16 cm and OB = 10 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 =
16
2
cm = 8 cm
In the right ?OMB, we have:
OB
2
= OM
2
+ MB
2
Pythagorastheorem
? 10
2
= OM
2
+ 8
2
? 100 = OM
2
+ 64
? OM
2
= 100 -64
= 36
? OM =
v
36 cm = 6 cm
Hence, the distance of the chord from the centre is 6 cm.
Question:2
Find the length of a chord which is at a distance of 3 cm from the centre of a circle of radius 5 cm.
Solution:
Let AB be the chord of the given circle with centre O and a radius of 5 cm.
From O, draw OM perpendicular to AB.
Then OM = 3 cm and OB = 5 cm
From the right ?OMB, we have:
OB
2
= OM
2
+ MB
2
Pythagorastheorem
? 5
2
= 3
2
+ MB
2
? 25 = 9 + MB
2
? MB
2
= (25 - 9) = 16
? MB =
v
16 cm = 4 cm
( )
Since the perpendicular from the centre of a circle to a chord bisects the chord, we have:
AB = 2 × MB = 2 ×4
cm = 8 cm
Hence, the required length of the chord is 8 cm.
Question:3
A chord of length 30 cm is drawn at a distance of 8 cm from the centre of a circle. Find out the radius of the circle.
Solution:
Let AB be the chord of the given circle with centre O. The perpendicular distance from the centre of the circle to the
chord is 8 cm.
Join OB.
Then OM = 8 cm and AB = 30 cm
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
? MB =
AB
2
=
30
2
cm = 15 cm
From the right ?OMB, we have:
OB
2
= OM
2
+ MB
2
? OB
2
= 8
2
+ 15
2
? OB
2
= 64 + 225
? OB
2
= 289
? OB =
v
289 cm = 17 cm
Hence, the required length of the radius is 17 cm.
Question:4
In a circle of radius 5 cm, AB and CD are two parallel chords of lengths 8 cm and 6 cm respectively. Calculate the
distance between the chords if they are
i
on the same side of the centre
ii
on the opposite sides of the centre.
Solution:
We have:
i
Let AB and CD be two chords of a circle such that AB is parallel to CD on the same side of the circle.
Given: AB = 8 cm, CD = 6 cm and OB = OD = 5 cm
Join OL and OM.
( ) ( )
The perpendicular from the centre of a circle to a chord bisects the chord.
? LB =
AB
2
=
8
2
= 4cm
Now, in right angled ?BLO, we have:
OB
2
= LB
2
+ LO
2
? LO
2
= OB
2
- LB
2
? LO
2
= 5
2
- 4
2
? LO
2
= 25 - 16 = 9
? LO = 3 cm
Similarly, MD =
CD
2
=
6
2
= 3cm
In right angled ?DMO, we have:
OD
2
= MD
2
+ MO
2
? MO
2
= OD
2
- MD
2
? MO
2
= 5
2
- 3
2
? MO
2
= 25 - 9 = 16
?MO = 4 cm
? Distance between the chords = (MO - LO) = 4 -3
cm = 1 cm
ii
Let AB and CD be two chords of a circle such that AB is parallel to CD and they are on the opposite sides of the
centre.
Given: AB = 8 cm and CD = 6 cm
Draw OL ? AB and OM ? CD.
Join OA and OC.
OA = OC = 5 cm Radiiofacircle
The perpendicular from the centre of a circle to a chord bisects the chord.
? AL =
AB
2
=
8
2
= 4cm
Now, in right angled ?OLA, we have:
OA
2
= AL
2
+ LO
2
? LO
2
= OA
2
- AL
2
? LO
2
= 5
2
- 4
2
? LO
2
= 25 - 16 = 9
? LO = 3 cm
Similarly, CM =
CD
2
=
6
2
= 3cm
In right angled ?CMO, we have:
OC
2
= CM
2
+ MO
2
( )
( )
( )
( )
Page 4
Question:1
A chord of length 16 cm is drawn in a circle of radius 10 cm. Find the distance of the chord from the centre of the
circle.
Solution:
Let AB be the chord of the given circle with centre O and a radius of 10 cm.
Then AB =16 cm and OB = 10 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 =
16
2
cm = 8 cm
In the right ?OMB, we have:
OB
2
= OM
2
+ MB
2
Pythagorastheorem
? 10
2
= OM
2
+ 8
2
? 100 = OM
2
+ 64
? OM
2
= 100 -64
= 36
? OM =
v
36 cm = 6 cm
Hence, the distance of the chord from the centre is 6 cm.
Question:2
Find the length of a chord which is at a distance of 3 cm from the centre of a circle of radius 5 cm.
Solution:
Let AB be the chord of the given circle with centre O and a radius of 5 cm.
From O, draw OM perpendicular to AB.
Then OM = 3 cm and OB = 5 cm
From the right ?OMB, we have:
OB
2
= OM
2
+ MB
2
Pythagorastheorem
? 5
2
= 3
2
+ MB
2
? 25 = 9 + MB
2
? MB
2
= (25 - 9) = 16
? MB =
v
16 cm = 4 cm
( )
Since the perpendicular from the centre of a circle to a chord bisects the chord, we have:
AB = 2 × MB = 2 ×4
cm = 8 cm
Hence, the required length of the chord is 8 cm.
Question:3
A chord of length 30 cm is drawn at a distance of 8 cm from the centre of a circle. Find out the radius of the circle.
Solution:
Let AB be the chord of the given circle with centre O. The perpendicular distance from the centre of the circle to the
chord is 8 cm.
Join OB.
Then OM = 8 cm and AB = 30 cm
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
? MB =
AB
2
=
30
2
cm = 15 cm
From the right ?OMB, we have:
OB
2
= OM
2
+ MB
2
? OB
2
= 8
2
+ 15
2
? OB
2
= 64 + 225
? OB
2
= 289
? OB =
v
289 cm = 17 cm
Hence, the required length of the radius is 17 cm.
Question:4
In a circle of radius 5 cm, AB and CD are two parallel chords of lengths 8 cm and 6 cm respectively. Calculate the
distance between the chords if they are
i
on the same side of the centre
ii
on the opposite sides of the centre.
Solution:
We have:
i
Let AB and CD be two chords of a circle such that AB is parallel to CD on the same side of the circle.
Given: AB = 8 cm, CD = 6 cm and OB = OD = 5 cm
Join OL and OM.
( ) ( )
The perpendicular from the centre of a circle to a chord bisects the chord.
? LB =
AB
2
=
8
2
= 4cm
Now, in right angled ?BLO, we have:
OB
2
= LB
2
+ LO
2
? LO
2
= OB
2
- LB
2
? LO
2
= 5
2
- 4
2
? LO
2
= 25 - 16 = 9
? LO = 3 cm
Similarly, MD =
CD
2
=
6
2
= 3cm
In right angled ?DMO, we have:
OD
2
= MD
2
+ MO
2
? MO
2
= OD
2
- MD
2
? MO
2
= 5
2
- 3
2
? MO
2
= 25 - 9 = 16
?MO = 4 cm
? Distance between the chords = (MO - LO) = 4 -3
cm = 1 cm
ii
Let AB and CD be two chords of a circle such that AB is parallel to CD and they are on the opposite sides of the
centre.
Given: AB = 8 cm and CD = 6 cm
Draw OL ? AB and OM ? CD.
Join OA and OC.
OA = OC = 5 cm Radiiofacircle
The perpendicular from the centre of a circle to a chord bisects the chord.
? AL =
AB
2
=
8
2
= 4cm
Now, in right angled ?OLA, we have:
OA
2
= AL
2
+ LO
2
? LO
2
= OA
2
- AL
2
? LO
2
= 5
2
- 4
2
? LO
2
= 25 - 16 = 9
? LO = 3 cm
Similarly, CM =
CD
2
=
6
2
= 3cm
In right angled ?CMO, we have:
OC
2
= CM
2
+ MO
2
( )
( )
( )
( )
? MO
2
= OC
2
- CM
2
? MO
2
= 5
2
- 3
2
? MO
2
= 25 - 9 = 16
? MO = 4 cm
Hence, distance between the chords = (MO + LO) = 4 +3
cm = 7 cm
Question:5
Two parallel chords of lengths 30 cm and 16 cm are drawn on the opposite sides of the centre of a circle of radius
17 cm. Find the distance between the chords.
Solution:
Let AB and CD be two chords of a circle such that AB is parallel to CD and they are on the opposite sides of the
centre.
Given: AB = 30 cm and CD = 16 cm
Draw OL ? AB and OM ? CD.
Join OA and OC.
OA = OC = 17 cm Radiiofacircle
The perpendicular from the centre of a circle to a chord bisects the chord.
?
AL =
(
AB
2
)
=
(
30
2
)
= 15 cm
Now, in right angled ?OLA, we have:
OA
2
= AL
2
+ LO
2
? LO
2
= OA
2
- AL
2
? LO
2
= 17
2
- 15
2
? LO
2
= 289 - 225 = 64
? LO = 8 cm
Similarly, CM =
CD
2
=
16
2
= 8 cm
In right angled ?CMO, we have:
?OC
2
= CM
2
+ MO
2
? MO
2
= OC
2
- CM
2
? MO
2
= 17
2
- 8
2
? MO
2
= 289 - 64 = 225
? MO = 15 cm
( ) ( )
Page 5
Question:1
A chord of length 16 cm is drawn in a circle of radius 10 cm. Find the distance of the chord from the centre of the
circle.
Solution:
Let AB be the chord of the given circle with centre O and a radius of 10 cm.
Then AB =16 cm and OB = 10 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 =
16
2
cm = 8 cm
In the right ?OMB, we have:
OB
2
= OM
2
+ MB
2
Pythagorastheorem
? 10
2
= OM
2
+ 8
2
? 100 = OM
2
+ 64
? OM
2
= 100 -64
= 36
? OM =
v
36 cm = 6 cm
Hence, the distance of the chord from the centre is 6 cm.
Question:2
Find the length of a chord which is at a distance of 3 cm from the centre of a circle of radius 5 cm.
Solution:
Let AB be the chord of the given circle with centre O and a radius of 5 cm.
From O, draw OM perpendicular to AB.
Then OM = 3 cm and OB = 5 cm
From the right ?OMB, we have:
OB
2
= OM
2
+ MB
2
Pythagorastheorem
? 5
2
= 3
2
+ MB
2
? 25 = 9 + MB
2
? MB
2
= (25 - 9) = 16
? MB =
v
16 cm = 4 cm
( )
Since the perpendicular from the centre of a circle to a chord bisects the chord, we have:
AB = 2 × MB = 2 ×4
cm = 8 cm
Hence, the required length of the chord is 8 cm.
Question:3
A chord of length 30 cm is drawn at a distance of 8 cm from the centre of a circle. Find out the radius of the circle.
Solution:
Let AB be the chord of the given circle with centre O. The perpendicular distance from the centre of the circle to the
chord is 8 cm.
Join OB.
Then OM = 8 cm and AB = 30 cm
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
? MB =
AB
2
=
30
2
cm = 15 cm
From the right ?OMB, we have:
OB
2
= OM
2
+ MB
2
? OB
2
= 8
2
+ 15
2
? OB
2
= 64 + 225
? OB
2
= 289
? OB =
v
289 cm = 17 cm
Hence, the required length of the radius is 17 cm.
Question:4
In a circle of radius 5 cm, AB and CD are two parallel chords of lengths 8 cm and 6 cm respectively. Calculate the
distance between the chords if they are
i
on the same side of the centre
ii
on the opposite sides of the centre.
Solution:
We have:
i
Let AB and CD be two chords of a circle such that AB is parallel to CD on the same side of the circle.
Given: AB = 8 cm, CD = 6 cm and OB = OD = 5 cm
Join OL and OM.
( ) ( )
The perpendicular from the centre of a circle to a chord bisects the chord.
? LB =
AB
2
=
8
2
= 4cm
Now, in right angled ?BLO, we have:
OB
2
= LB
2
+ LO
2
? LO
2
= OB
2
- LB
2
? LO
2
= 5
2
- 4
2
? LO
2
= 25 - 16 = 9
? LO = 3 cm
Similarly, MD =
CD
2
=
6
2
= 3cm
In right angled ?DMO, we have:
OD
2
= MD
2
+ MO
2
? MO
2
= OD
2
- MD
2
? MO
2
= 5
2
- 3
2
? MO
2
= 25 - 9 = 16
?MO = 4 cm
? Distance between the chords = (MO - LO) = 4 -3
cm = 1 cm
ii
Let AB and CD be two chords of a circle such that AB is parallel to CD and they are on the opposite sides of the
centre.
Given: AB = 8 cm and CD = 6 cm
Draw OL ? AB and OM ? CD.
Join OA and OC.
OA = OC = 5 cm Radiiofacircle
The perpendicular from the centre of a circle to a chord bisects the chord.
? AL =
AB
2
=
8
2
= 4cm
Now, in right angled ?OLA, we have:
OA
2
= AL
2
+ LO
2
? LO
2
= OA
2
- AL
2
? LO
2
= 5
2
- 4
2
? LO
2
= 25 - 16 = 9
? LO = 3 cm
Similarly, CM =
CD
2
=
6
2
= 3cm
In right angled ?CMO, we have:
OC
2
= CM
2
+ MO
2
( )
( )
( )
( )
? MO
2
= OC
2
- CM
2
? MO
2
= 5
2
- 3
2
? MO
2
= 25 - 9 = 16
? MO = 4 cm
Hence, distance between the chords = (MO + LO) = 4 +3
cm = 7 cm
Question:5
Two parallel chords of lengths 30 cm and 16 cm are drawn on the opposite sides of the centre of a circle of radius
17 cm. Find the distance between the chords.
Solution:
Let AB and CD be two chords of a circle such that AB is parallel to CD and they are on the opposite sides of the
centre.
Given: AB = 30 cm and CD = 16 cm
Draw OL ? AB and OM ? CD.
Join OA and OC.
OA = OC = 17 cm Radiiofacircle
The perpendicular from the centre of a circle to a chord bisects the chord.
?
AL =
(
AB
2
)
=
(
30
2
)
= 15 cm
Now, in right angled ?OLA, we have:
OA
2
= AL
2
+ LO
2
? LO
2
= OA
2
- AL
2
? LO
2
= 17
2
- 15
2
? LO
2
= 289 - 225 = 64
? LO = 8 cm
Similarly, CM =
CD
2
=
16
2
= 8 cm
In right angled ?CMO, we have:
?OC
2
= CM
2
+ MO
2
? MO
2
= OC
2
- CM
2
? MO
2
= 17
2
- 8
2
? MO
2
= 289 - 64 = 225
? MO = 15 cm
( ) ( )
Hence, distance between the chords = (LO + MO) = 8 +15
cm = 23 cm
Question:6
In the given figure, the diameter CD of a circle with centre O is perpendicular to chord AB. If AB = 12 cm and CE = 3
cm, calculate the radius of the circle.
Solution:
CD is the diameter of the circle with centre O and is perpendicular to chord AB.
Join OA.
Given: AB = 12 cm and CE = 3 cm
Let OA = OC = r cm Radiiofacircle
Then OE = (r - 3) cm
Since the perpendicular from the centre of the circle to a chord bisects the chord, we have:
AE =
AB
2
=
12
2
cm = 6 cm
Now, in right angled ?OEA, we have:
? OA
2
= OE
2
+ AE
2
? r
2
= (r - 3)
2
+ 6
2
? r
2
= r
2
- 6r + 9 + 36
? r
2
- r
2
+ 6r = 45
? 6r = 45
? r =
45
6
cm = 7. 5 cm
? r = 7.5 cm
Hence, the required radius of the circle is 7.5 cm.
Question:7
In the given figure, a circle with centre O is given in which a diameter AB bisects the chord CD at a point E such
that CE = ED = 8 cm and EB = 4 cm. Find the radius of the circle.
Solution:
AB is the diameter of the circle with centre O, which bisects the chord CD at point E.
( ) ( )
( )
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FAQs on RS Aggarwal Solutions: Circles- 1 - Mathematics (Maths) Class 9
| 1. What are the different elements of a circle? |  |
Ans. A circle has several elements, including the radius, diameter, circumference, and center. The radius is the distance from the center of the circle to any point on its circumference. The diameter is a line segment that passes through the center and connects two points on the circumference. The circumference is the perimeter or the distance around the circle. The center is the point within the circle that is equidistant from all points on its circumference.
| 2. How are the radius and diameter of a circle related? | |
Ans. The radius and diameter of a circle are related by a simple formula. The radius is half the length of the diameter. In other words, if you multiply the radius by 2, you will get the diameter. Conversely, if you divide the diameter by 2, you will get the radius.
| 3. How do you find the circumference of a circle? | |
Ans. The circumference of a circle can be found using the formula C = 2πr, where C represents the circumference and r represents the radius of the circle. Alternatively, you can also use the formula C = πd, where d represents the diameter of the circle. Both formulas will give you the same result.
| 4. What is the relationship between the circumference and the area of a circle? | |
Ans. The circumference and the area of a circle are related, but they measure different aspects of the circle. The circumference is the distance around the circle, while the area is the space enclosed by the circle. The relationship between the two can be expressed by the formula A = πr^2, where A represents the area and r represents the radius of the circle.
| 5. How can you find the area of a sector of a circle? | |
Ans. To find the area of a sector of a circle, you can use the formula A = (θ/360)πr^2, where A represents the area, θ represents the central angle of the sector, and r represents the radius of the circle. This formula calculates the fraction of the total area of the circle that the sector covers based on the central angle.
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Nov 22, 2024 Last updated |
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Additional Information about RS Aggarwal Solutions: Circles- 1 for Class 9 Preparation
RS Aggarwal Solutions: Circles- 1 Free PDF Download
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The content of RS Aggarwal Solutions: Circles- 1 is prepared as per the latest Class 9 syllabus.
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