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 centre. If the distance between AB and CD is 6, find the radius of the circle.
Here, OM ⊥ AB and ON ⊥ CD. is drawn and OB and OD are joined.
We know that AB bisects BM as the perpendicular from the centre bisects chord.
Since AB = 5
BM = AB/2
Similarly, ND = CD/2 = 11/2
Now, let ON be x.
OM = 6−x.
Consider ΔMOB,
OB2 = OM2 + MB2
Consider ΔNOD,
OD2 = ON2 + ND2
\(OD^2 = x^2 + \frac{121}{4}\) ......(2)
We know, OB = OD (radii)
From equation 1 and equation 2 we get
\(36 + x^2 - 12x + \frac{25}{4} = x^2 + \frac{121}{4}\)
12x = \(36 + \frac{25}{4} - \frac{121}{4}\)
= \(\frac{144+25-121}{4}\)
12x = \(\frac{48}{4}\) = 12
x = 1
Now, from equation (2) we have,
OD2= 12 + \(\frac{121}{4}\)
OD = (5/2) × √5 cm