Two parallel chords AB and CD, of lengths 6 cm and 8 cm respectively are 1 cm apart. The radius of the circle is: B 6cm 1cm 8 cm D (A) 2.5 cm (C) 5 cm (B) 4.0 cm (D) 5.5 cm
Answers
Answer:
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.
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 center ?
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.