Question

2.
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 cm, find the radius of the circle.​

Answer 1

Let O be the centre of the given circle and let its radius be cm.

Draw ON ⊥ AB and OM⊥ CD since, ON ⊥ AB, OM ⊥ CD and AB || CD, therefore points N, O, M are collinear.

Let ON = a cm

∴ OM = (6 – a) cm

Join OA and OC.

Then, OA = OC = b c m

Since, the perpendicular from the centre to a chord of the circle bisects the chord.

Therefore, AN = NB= 2.5 cm and OM = MD = 5.5 cm

In ∆OAN and ∆OCM, we get

OA² = ON²+ AN²

OC² = OM² + CM²

⇒ b² = a² + (2.5)²

and, b² = (6-a)² + (5.5)² …(i)

So, a² + (2.5)² = (6 – a)² + (5.5)²

⇒ a² + 6.25= 36-12a + a² + 30.25

⇒ 12a = 60

⇒ a = 5

On putting a = 5 in Eq. (i), we get

b² = (5)² + (2.5)²

= 25 + 6.25 = 31.25

So, r = √31.25 = 5.6cm (Approx.)