The doameter of the circle is 20cm. Two parallel chords are drawn in tje circle. They are AB =16cm and CD=12cm,Then CN=?
Answers
Given :- The diameter of the circle is 20cm. Two parallel chords are drawn in tje circle. They are AB = 16cm and CD = 12cm .
To Find :- MN = ?
Answer :-
Join AO which is radius of circle .
In ∆OMA,
→ ∠OMA = 90° { Line from centre to the chord bisect the chord and perpendicular to the chord.}
so,
→ AM = 16/2 = 8 cm .
then,
→ OM = √(AO² - AM²) { By pythagoras theorem}
→ OM = √(10² - 8²) = √(100 - 64) = √36 = 6 cm .
similarly, Join CO which is radius of circle .
In ∆ONC,
→ ∠ONC = 90° { Line from centre to the chord bisect the chord and perpendicular to the chord.}
so,
→ CN = 12/2 = 6 cm .
then,
→ ON = √(AO² - CN²) { By pythagoras theorem}
→ ON = √(10² - 6²) = √(100 - 36) = √64 = 8 cm .
therefore,
→ MN = MO + ON = 6 + 8 = 14 cm (Ans.)
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Given : The diameter of the circle is 20cm.
Two parallel chords are drawn in the circle.
AB =16cm and CD=12cm
To Find : CN
Solution:
Perpendicular from the center of the circle bisects the chord
Hence CN = CD/ 2
=> CN = 12/2
=> CN = 6 cm
Additional Info :
Diameter = 20 cm => Radius = 10 cm
CN = 6 cm
=> ON = √10² - 6² = 8 cm
M is mid point of AB
=> AM = 8 cm
and OM = √10² - 8² = 6 cm
Distance between chords AB and CD = 8 + 6 = 14 cm
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