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.
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Home » Aptitude » Plane Geometry » Question
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The length of two chords AB and AC of a circle are 8 cm and 6 cm and ∠BAC = 90°, then the radius of circle is
According to question , we draw a figure of a circle with two chords AB and AC ,
∴ BC = √AB² + AC²
BC = √8² + 6²
BC = √64 + 36
BC = √100 = 10 cm
∴ Radius of the circle = 5 cm
Correct Answer:
Description for Correct answer:
According to question,AB and AC are chordsAB = 8AC = 6
\( \Large \therefore \frac{BC}{2}=radius=\frac{10}{2}=5cm \)
Part of solved Geometry questions and answers : >> Elementary Mathematics >> Geometry
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