Answer:
Draw two parallel chords AB and CD of lengths 5 cm and 11 cm. Let the center of the circle be O. Join one end of each chord to the center.
Draw two perpendiculars OM and ON to AB and CD, respectively, which bisects the chords.
Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its center. If the distance between AB and CD is 6 cm, find the radius of the circle.
Thus, MB = 2.5 cm and ND = 5.5 cm [The perpendicular drawn from the center of the circle to the chords bisects it.]
Let OM = x and ON = 6 - x
Consider ΔOMB
By Pythagoras theorem,
OM2 + MB2 = OB2
x2 + 2.52 = OB2
x2 + 6.25 = OB2..................(1)
Consider ΔOND
By Pythagoras theorem,
ON2 + ND2 = OD2
(6 - x)² + 5.52 = OD2
36 + x2 - 12x + 30.25 = OD2
x2 - 12x + 66.25 = OD2............... (2)
OB and OD are the radii of the circle. Therefore OB = OD.
Thus, OB2 = OD2
Equating (1) and (2) we get,
x2 + 6.25 = x2 - 12x + 66.25
12x = 60
x = 5
Substituting the value of x in (1),
OB2 = x2 + 6.25
OB2 = 52 + 6.25
OB2 = 31.25
OB = 5.59 (approx.)
Thus, we get the radius of the circle = 5.59 cm.
