AB and CD are two chords of a circle , AB = 6 cm, CD = 8 cm and AB $$\parallel$$ CD. If the distance between AB and CD is 7 cm and the centre of the circle lies in the interior of the quadrilateral ABCD, then find the radius of the circle (in cm).​

AB and CD are two chords of a circle

AB = 6 cm

CD = 8 cm

AB || CD

Distance between AB and CD = 7 cm

Centre of the circle lies in the interior of the quadrilateral ABCD

In a circle, the perpendicular distance from the center to a chord bisect the chord

Let O be the center of the circle and radius be r.

Let OP = x, OM = 7-x

In ΔPOD: 

OD² = OP² + PD²

OD² = x² + (CD/2)²

OD² = x² + 16  ... (1)

In ΔOBM

OB² = (7-x)² + 9

OB² = 49 + x² - 14x + 9

OB² = 49 + x² - 14x + 9 ... (2)

Since, OD² = OB² (Both are radius)

So, x² + 16 = 49 + x² - 14x + 9

14x = 42

x = 3 cm

OD² = 9 + 16

OD = 5 cm

The correct answer is (a).