Solution:
The perpendicular drawn from the center of the circle to the chords bisects it.
Draw two parallel chords AB and CD of lengths 6 cm and 8 cm. Let the circle's center be O. Join one end of each chord to the center. Draw 2 perpendiculars OM and ON to AB and CD, respectively, which bisects the chords.
AB = 6 cm CD = 8 cm MB = 3 cm ND = 4 cm
Given OM = 4 cm and let ON = x cm Consider ΔOMB
By Pythagoras theorem,
OM² + MB² = OB²
4² + 3² = OB²
OB² = 25
OB = 5 cm
OB and OD are the radii of the circle.
Therefore OD = OB = 5 cm.
Consider ΔOND
By Pythagoras theorem,
ON² + ND² = OD²
x² + 4² = 5²
x² = 25 - 16
x² = 9
x = 3
The distance of the chord CD from the center is 3 cm.
☛ Check: NCERT Solutions for Class 9 Maths Chapter 10
Video Solution:
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the center, what is the distance of the other chord from the centre?
Maths NCERT Solutions Class 9 Chapter 10 Exercise 10.6 Question 3
Summary:
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the center, we have found that the distance of the other chord CD from the center is 3 cm.
☛ Related Questions:
Math worksheets and
visual curriculum
Given:
The length of the parallel chords is 8 cm and the diameter of the circle is 10 cm
Concept:
A line passing through the center perpendicularly bisect the chord into two equal parts.
Equal Chords are equal distance from the center.
Calculation:
Construct a circle with center 'O' and draw parallel chord AB and CD, EF is the perpendicular bisector of the chords AB and CD.
AB = CD = 8 cm, R = 10/2 = 5cm
We have to find the length of EF.
In ΔBEO, ∠E is 90°, OB = 5 cm, BE = 8/2 = 4 cm,
By Pythagoras theorem
EO = (OB2 - BE2)1/2
⇒ (52 - 42)1/2
⇒ (25 - 16)1/2
⇒ (9)1/2
⇒ 3 cm
Now, The length of the EF is
⇒ 2 × OE
⇒ 2 × 3
⇒ 6 cm
∴ The required distance between the chords is 6 cm.
Home » Aptitude » Plane Geometry » Question
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The distance between two parallel chords of length 8 cm each in a circle of diameter 10 cm is
As per the given in question , we draw a figure circle
OP = √OA² - AP²
OP = √5² - 4²
OP = √25 - 16
OP = √9 = 3 cm.
∴ QP = 2 × OP = 6 cm.
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100 Qs. 100 Marks 120 Mins
Given:
The length of the parallel chords is 8 cm and the diameter of the circle is 10 cm
Concept:
A line passing through the center perpendicularly bisect the chord into two equal parts.
Equal Chords are equal distance from the center.
Calculation:
Construct a circle with center 'O' and draw parallel chord AB and CD, EF is the perpendicular bisector of the chords AB and CD.
AB = CD = 8 cm, R = 10/2 = 5cm
We have to find the length of EF.
In ΔBEO, ∠E is 90°, OB = 5 cm, BE = 8/2 = 4 cm,
By Pythagoras theorem
EO = (OB2 - BE2)1/2
⇒ (52 - 42)1/2
⇒ (25 - 16)1/2
⇒ (9)1/2
⇒ 3 cm
Now, The length of the EF is
⇒ 2 × OE
⇒ 2 × 3
⇒ 6 cm
∴ The required distance between the chords is 6 cm.
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